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Title: electrical hand book
Description: basics of electrical engineering that will help you in your study

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Fundamentals of Electrical Engineering I

By:
Don H
...
org/content/col10040/1
...
1/ >

CONNEXIONS
Rice University, Houston, Texas

©2013 Don Johnson
This selection and arrangement of content is licensed under the Creative Commons Attribution License:
http://creativecommons
...
0

Table of Contents
1 Introduction
1
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1
1
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2
1
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4
1
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6
1
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7
Solutions
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1 Complex Numbers
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2 Elemental Signals
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3 Signal Decomposition
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4 Discrete-Time Signals
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5 Introduction to Systems
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6 Simple Systems
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7 Signals and Systems Problems
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31
3 Analog Signal Processing
3
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33
3
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34
3
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37
3
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3
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61
3
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65
3
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66
3
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67
3
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69
3
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74
3
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76
Solutions
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1 Introduction to the Frequency Domain
...
2 Complex Fourier Series
...
3 Classic Fourier Series
...
4 A Signal’s Spectrum
...
5 Fourier Series Approximation of Signals
...
6 Encoding Information in the Frequency Domain
...
7 Filtering Periodic Signals
...
8 Derivation of the Fourier Transform
...
9 Linear Time Invariant Systems
...
10 Modeling the Speech Signal
...
11 Frequency Domain Problems
...
143
5 Digital Signal Processing
5
...
145
5
...
145
5
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149
5
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152
5
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154
5
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156
5
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160
5
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172
5
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173
5
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176
5
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179
5
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180
Solutions
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1 Information Communication
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2 Types of Communication Channels
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3 Wireline Channels
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4 Wireless Channels
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5 Line-of-Sight Transmission
...
6 The Ionosphere and Communications
...
7 Communication with Satellites
...
8 Noise and Interference
...
9 Channel Models
...
10 Baseband Communication
...
11 Modulated Communication
...
12 Signal-to-Noise Ratio of an Amplitude-Modulated Signal
...
13 Digital Communication
...
14 Binary Phase Shift Keying
...
15 Frequency Shift Keying
...
16 Digital Communication Receivers
...
17 Digital Communication in the Presence of Noise
...
18 Digital Communication System Properties
...
19 Digital Channels
...
20 Entropy
...
21 Source Coding Theorem
...
22 Compression and the Huﬀman Code
...
23 Subtleties of Coding
...
24 Channel Coding
...
25 Repetition Codes
...
26 Block Channel Coding
...
27 Error-Correcting Codes: Hamming Distance
...
28 Error-Correcting Codes: Channel Decoding
...
29 Error-Correcting Codes: Hamming Codes
...
30 Noisy Channel Coding Theorem
...
31 Capacity of a Channel
...
32 Comparison of Analog and Digital Communication
...
33 Communication Networks
...
34 Message Routing
...
35 Network architectures and interconnection
...
36 Ethernet
...
37 Communication Protocols
...
38 Information Communication Problems
...
255
7 Appendix
7
...
261
7
...
262
7
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263
Solutions
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266
Attributions
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1 Themes1
From its beginnings in the late nineteenth century, electrical engineering has blossomed from focusing on
electrical circuits for power, telegraphy and telephony to focusing on a much broader range of disciplines
...
This course concentrates
on the latter theme: the representation, manipulation, transmission, and reception of information
by electrical means
...

Information can take a variety of forms
...
Information arises in your thoughts and is represented by speech, which must
have a well deﬁned, broadly known structure so that someone else can understand what you say
...
There, sound energy is
converted back to neural activity, and, if what you say makes sense, she understands what you say
...
Information can take the form of a text ﬁle you type into your word processor
...
From an information theoretic viewpoint, all of
these scenarios are equivalent, although the forms of the information representation—sound waves, plastic
and computer ﬁles—are very diﬀerent
...
Analog information is continuous valued; examples are audio and video
...

The conversion of information-bearing signals from one energy form into another is known as energy
conversion or transduction
...
Conceptually we could use any
form of energy to represent information, but electric signals are uniquely well-suited for information representation, transmission (signals can be broadcast from antennas or sent through wires), and manipulation
(circuits can be built to reduce noise and computers can be used to modify information)
...

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INTRODUCTION

Telegraphy represents the earliest electrical information system, and it dates from 1837
...
Electrical science came of age when James Clerk Maxwell2 proclaimed in 1864 a set of equations
that he claimed governed all electrical phenomena
...
Because of the complexity of Maxwell’s presentation, the
development of the telephone in 1876 was due largely to empirical work
...
This understanding of fundamentals led to a quick
succession of inventions–the wireless telegraph (1899), the vacuum tube (1905), and radio broadcasting–that
marked the true emergence of the communications age
...
Consequently, circuit theory
served as the foundation and the framework of all of electrical engineering education
...
These were the ﬁrst public demonstration of the ﬁrst electronic
computer (1946), the invention of the transistor (1947), and the publication of A Mathematical Theory
of Communication by Claude Shannon (1948)
...
About twenty years later, the laser was invented, which opened even more design possibilities
...
Only once the intended system is speciﬁed can an
implementation be selected
...

note: Thanks to the translation eﬀorts of Rice University’s Disability Support Services3 , this
collection is now available in a Braille-printable version
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zip ﬁle
containing all the necessary
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1
...
Stated in mathematical terms, a signal is merely a function
...
The independent variable of the signal could be time (speech, for
example), space (images), or the integers (denoting the sequencing of letters and numbers in the football
score)
...
2
...
Speech (Section 4
...
The result is
pressure waves propagating in the air, and the speech signal thus corresponds to a function having independent variables of space and time and a value corresponding to air pressure: s (x, t) (Here we use vector
notation x to denote spatial coordinates)
...
An example of the resulting waveform s (x0 , t) is shown in
Figure 1
...

Photographs are static, and are continuous-valued signals deﬁned over space
...
In Figure 1
...

2 http://www-groups
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1: A speech signal’s amplitude relates to tiny air pressure variations
...

(a)

(b)

Figure 1
...
It
contains straight and curved lines, complicated texture, and a face
...
The colors merely help show what
signal values are about the same size
...
Painters long ago
found that mixing together combinations of the so-called primary colors–red, yellow and blue–can produce
very realistic color images
...
INTRODUCTION
00
08
10
18
20
28
30
38
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70
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...
1: The ASCII translation table shows how standard keyboard characters are represented
by integers
...
The numeric codes
are represented in hexadecimal (base-16) notation
...

in space, but a diﬀerent set of colors is used: How much of red, green and blue is present
...

Interesting cases abound where the analog signal depends not on a continuous variable, such as time, but
on a discrete variable
...

1
...
2 Digital Signals
The word “digital” means discrete-valued and implies the signal depends on the integers rather than a
continuous variable
...
Computers rely on the digital representation of information to manipulate and transform
information
...
The ASCII character code
shown in Table 1
...
For example, the ASCII code represents the letter a as
the number 97, the letter A with 65
...
3 Structure of Communication Systems6
The fundamental model of communications is portrayed in Figure 1
...
In this fundamental model, each message-bearing signal, exempliﬁed by s (t), is analog and is a
function of time
...
4 (Deﬁnition of a system))
...
This
graphical representation is known as a block diagram
...
As typiﬁed by the communications model,
how information ﬂows, how it is corrupted and manipulated, and how it is ultimately received is summarized
by interconnecting block diagrams: The outputs of one or more systems serve as the inputs to others
...
org/content/m0002/2
...

5

s(t)
Source

x(t)
Transmitter

message

r(t)
Channel

modulated
message

s(t)
Receiver

corrupted
modulated
message

Sink

demodulated
message

Figure 1
...

x(t)

System

y(t)

Figure 1
...

In the communications model, the source produces a signal that will be absorbed by the sink
...
Signals
can also be functions of two variables—an image is a signal that depends on two spatial variables—or more—
television pictures (video signals) are functions of two spatial variables and time
...
In physical systems, each signal corresponds to an electrical voltage or current
...
However, we ﬁrst need
to understand the big picture to appreciate the context in which the electrical engineer works
...

The block diagram has the message s (t) passing through a block labeled transmitter that produces the
signal x (t)
...

In the case of a computer network, typed characters are encapsulated in packets, attached with a destination
address, and launched into the Internet
...
In any case, the transmitter should
not operate in such a way that the message s (t) cannot be recovered from x (t)
...
(It is ridiculous
to transmit a signal in such a way that no one can recover the original
...
Such cryptographic systems underlie secret
communications
...
Nothing good happens to a
signal in a channel: It can become corrupted by noise, distorted, and attenuated among many possibilities
...
The channel is another system in our block
diagram, and produces r (t), the signal received by the receiver
...
However, because of the channel, the receiver must do its best to
produce a received message s (t) that resembles s (t) as much as possible
...
It is this result that modern communications systems exploit, and why many
communications systems are going “digital
...
1)
details Shannon’s theory of information, and there we learn of Shannon’s result and how to use it
...

7 http://www-gap
...
st-and
...
uk/∼history/Mathematicians/Shannon
...
INTRODUCTION

In the communications model, the source is a system having no input but producing an output; a sink has
an input and no output
...
This understanding
demands two diﬀerent ﬁelds of knowledge
...
4 The Fundamental Signal8
1
...
1 The Sinusoid
The most ubiquitous and important signal in electrical engineering is the sinusoid
...
1)

A is known as the sinusoid’s amplitude, and determines the sinusoid’s size
...
The frequency f has units of Hz (Hertz) or s−1 , and determines
how rapidly the sinusoid oscillates per unit time
...
AM radio stations have carrier
frequencies of about 1 MHz (one mega-hertz or 106 Hz), while FM stations have carrier frequencies of about
100 MHz
...
Clearly,
ω = 2πf
...
Finally, φ is the phase, and
determines the sine wave’s behavior at the origin (t = 0)
...
Note that if φ = − π ,
2
the sinusoid corresponds to a sine function, having a zero value at the origin
...
2)

Thus, the only diﬀerence between a sine and cosine signal is the phase; we term either a sinusoid
...
Here, the independent
variable is n and represents the integers
...

Exercise 1
...
9
...

note: Notice that we shall call either sinusoid an analog signal
...

Exercise 1
...
9
...

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content is available online at ...
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...
5

1
...
2 Communicating Information with Signals
The basic idea of communication engineering is to use a signal’s parameters to represent either real numbers or
other signals
...
To explore the notion of modulation, we can send a real number (today’s
temperature, for example) by changing a sinusoid’s amplitude accordingly
...
We could relate temperature to amplitude by the formula A = A0 (1 + kT ),
where A0 and k are constants that the transmitter and receiver must both know
...
This modulation scheme assumes we can estimate the sinusoid’s amplitude and
frequency; we shall learn that this is indeed possible
...
We have exploited all of the sinusoid’s two
parameters
...
This simple notion corresponds to how a modem works
...
We’ll learn how
this is done in subsequent modules, and more importantly, we’ll learn what the limits are on such digital
communication schemes
...
5 Introduction Problems9
Problem 1
...

a) What is the period of s (t) = A sin (2πf0 t + φ)?
b) What is the rms value of this signal? How is it related to the peak value?
c) What is the period and rms value of the depicted (Figure 1
...
” What
is the expression for the voltage provided by a wall socket? What is its rms value?
Problem 1
...
” Modems are used not only for connecting computers to telephone lines, but also for connecting digital (discrete-valued) sources to generic channels
...
org/content/m10353/2
...

8

CHAPTER 1
...
Consequently, the modem’s transmitted
signal that represents a single bit has the form
x (t) = A sin (2πf0 t) , 0 ≤ t ≤ T
Within each bit interval T , the amplitude is either A or zero
...
If N amplitude values are used, what is the resulting
datarate?
d) The classic communications block diagram applies to the modem
...

Problem 1
...
A transmission is sent for a period of time T (known as the
transmission or baud interval) and equals the sum of two amplitude-weighted carriers
...

a) What is the smallest transmission interval that makes sense to use with the frequencies given above?
In other words, what should T be so that an integer number of cycles of the carrier occurs?
b) Sketch (using Matlab) the signal that modem produces over several transmission intervals
...

c) Using your signal transmission interval, how many amplitude levels are needed to transmit ASCII
characters at a datarate of 3,200 bits/s? Assume use of the extended (8-bit) ASCII code
...
If we have N1 values for amplitude A1 , and N2
values for A2 , we have N1 N2 possible symbols that can be sent during each T second interval
...

9

Solutions to Exercises in Chapter 1
Solution to Exercise 1
...
6)
As cos (α + β) = cos (α) cos (β) − sin (α) sin (β), cos (2π (f + 1) n) = cos (2πf n) cos (2πn) − sin (2πf n) sin (2πn) =
cos (2πf n)
...
2 (p
...
See the plot in the module Elemental Signals
(Section 2
...
6: Square Wave)
...
INTRODUCTION

Chapter 2

Signals and Systems
2
...
Representing sinusoids in terms of
complex exponentials is not a mathematical oddity
...
Understanding information and power system
designs and developing new systems all hinge on using complex numbers
...

2
...
1 Deﬁnitions
The notion of the square root of −1 originated with the quadratic formula: the solution of certain quadratic
√
equations mathematically exists only if the so-called imaginary quantity −1 could be deﬁned
...
Ampère3
used the symbol i to denote current (intensité de current)
...
By then, using i for current was entrenched
and electrical engineers chose j for writing complex numbers
...
A complex number, z, consists of the ordered pair
(a,b), a is the real component and b is the imaginary component (the j is suppressed because the imaginary
component of the pair is always in the second position)
...
Note that
a and b are real-valued numbers
...
1 (The Complex Plane) shows that we can locate a complex number in what we call the complex
plane
...
From
analytic geometry, we know that locations in the plane can be expressed as the sum of vectors, with the
vectors corresponding to the x and y directions
...
This representation is known as the Cartesian
form of z
...

Some obvious terminology
...
We consider the real part as a function that works by selecting that component of a complex number
not multiplied by j
...
Again, both the real and imaginary parts of a complex number are real-valued
...
org/content/m0081/2
...

2 http://www-groups
...
st-and
...
uk/∼history/Mathematicians/Euler
...
dcs
...
ac
...
html

11

12

CHAPTER 2
...
1: A complex number is an ordered pair (a,b) that can be regarded as coordinates in the
plane
...

opposite sign
...
1)

Using Cartesian notation, the following properties easily follow
...
This property follows from the laws of vector
addition
...

• The product of j and a real number is an imaginary number: ja
...
Consequently, multiplying a complex number
by j rotates the number’s position by 90 degrees
...
1
(Solution on p
...
)
Use the deﬁnition of addition to show that the real and imaginary parts can be expressed as a
∗
∗
sum/diﬀerence of a complex number and its conjugate
...

2
2j
Complex numbers can also be expressed in an alternate form, polar form, which we will ﬁnd quite useful
...
The Cartesian form of a
complex number can be re-written as
a + jb =

a2 + b2

√

a
b
+j√
2
2 + b2
+b
a

a2

By forming a right triangle having sides a and b, we see that the real and imaginary parts correspond to the
cosine and sine of the triangle’s base angle
...

z = a + jb = r∠θ
√
r = |z| = a2 + b2
a = r cos (θ)
b = r sin (θ)
θ = arctan

b
a

The quantity r is known as the magnitude of the complex number z, and is frequently written as |z|
...
In using the arc-tangent formula to ﬁnd the angle, we must take
into account the quadrant in which the complex number lies
...
2
Convert 3 − 2j to polar form
...
31
...
1
...
2)

To show this result, we use Euler’s relations that express exponentials with imaginary arguments in terms
of trigonometric functions
...
3)

cos (θ) =

ejθ + e−jθ
2

sin (θ) =

ejθ − e−jθ
2j

(2
...

ex = 1 +

x
x2
x3
+
+
+
...

1!
2!
3!

because j 2 = −1, j 3 = −j, and j 4 = 1
...

2!
1!
3!
The real-valued terms correspond to the Taylor’s series for cos (θ), the imaginary ones to sin (θ), and Euler’s
ﬁrst √
relation results
...
Because of the relationship
r = a2 + b2 , we see that multiplying the exponential in (2
...

2
...
3 Calculating with Complex Numbers
Adding and subtracting complex numbers expressed in Cartesian form is quite easy: You add (subtract) the
real parts and imaginary parts separately
...
5)

To multiply two complex numbers in Cartesian form is not quite as easy, but follows directly from following
the usual rules of arithmetic
...
6)
= a1 a2 − b1 b2 + j (a1 b2 + a2 b1 )
Note that we are, in a sense, multiplying two vectors to obtain another vector
...

Exercise 2
...
31
...
SIGNALS AND SYSTEMS

Division requires mathematical manipulation
...

a1 + jb1
z1
=
z2
a2 + jb2
a1 + jb1 a2 − jb2
=
a2 + jb2 a2 − jb2
(a1 + jb1 ) (a2 − jb2 )
=
a2 2 + b2 2
a1 a2 + b1 b2 + j (a2 b1 − a1 b2 )
=
a2 2 + b2 2

(2
...

The properties of the exponential make calculating the product and ratio of two complex numbers much
simpler when the numbers are expressed in polar form
...
8)

r1 ejθ1
r1
z1
=
= ej(θ1 −θ2 )
z2
r2 ejθ2
r2

To multiply, the radius equals the product of the radii and the angle the sum of the angles
...
When the original
complex numbers are in Cartesian form, it’s usually worth translating into polar form, then performing the
multiplication or division (especially in the case of the latter)
...

Example 2
...
What we’ll need to understand the
circuit’s eﬀect is the transfer function in polar form
...
9)
s2 + s + 1
s = j2πf

(2
...
Thus,
s+2
j2πf + 2
=
s2 + s + 1
−4π 2 f 2 + j2πf + 1
=

4 + 4π 2 f 2 ejarctan(πf )
2

(1 − 4π 2 f 2 ) + 4π 2 f 2 e
=

4 + 4π 2 f 2
j
e
1 − 4π 2 f 2 + 16π 4 f 4

”
“
2πf
jarctan 1−4π2 f 2

“
“
””
2πf
arctan(πf )−arctan 1−4π2 f 2

(2
...
12)

(2
...
2 Elemental Signals4
Elemental signals are the building blocks with which we build complicated signals
...
Exactly what we mean by the “structure of a signal” will unfold in
this section of the course
...
Very interesting signals are not functions solely of
time; one great example of which is an image
...
Video signals are functions of three variables: two spatial dimensions and time
...

2
...
1 Sinusoids
Perhaps the most common real-valued signal is the sinusoid
...
14)

For this signal, A is its amplitude, f0 its frequency, and φ its phase
...
2
...

s (t) = Aej(2πf0 t+φ)
= Aejφ ej2πf0 t

(2
...
Aejφ is known as the signal’s complex amplitude
...

The complex amplitude is also known as a phasor
...
In fact,
early in the twentieth century, mathematicians thought engineers would not be suﬃciently sophisticated
to handle complex exponentials even though they greatly simpliﬁed solving circuit problems
...
1) for a review of
complex numbers and complex arithmetic
...
The sinusoid consists of two frequency components: one at the frequency +f0 and
the other at −f0
...

cos (2πf t) =

ej2πf t + e−j2πf t
2

(2
...
17)

ej2πf t = cos (2πf t) + j sin (2πf t)
4 This

content is available online at ...
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...
invent
...
html

(2
...
SIGNALS AND SYSTEMS

Figure 2
...

Its real and imaginary parts are sinusoids
...
A fundamental relationship is T = f
...
Thus, sinusoidal signals can be expressed as either the real
or the imaginary part of a complex exponential signal, the choice depending on whether cosine or
sine phase is needed, or as the sum of two complex exponentials
...

A cos (2πf t + φ) = Re Aejφ ej2πf t

(2
...
20)

Using the complex plane, we can envision the complex exponential’s temporal variations as seen in the
above ﬁgure (Figure 2
...
The magnitude of the complex exponential is A, and the initial value of the

17

Exponential
1
e–1
t

τ

Figure 2
...

complex exponential at t = 0 has an angle of φ
...
The number of times per second we go around the
circle equals the frequency f
...
The projections onto the real and imaginary axes of the rotating
vector representing the complex exponential signal are the cosine and sine signal of Euler’s relation ((2
...

2
...
3 Real Exponentials
As opposed to complex exponentials which oscillate, real exponentials (Figure 2
...

t

s (t) = e− τ

(2
...
368
...

t

s (t) = Aejφ e− τ ej2πf t

(2
...
For such signals, we can deﬁne
complex frequency as the quantity multiplying t
...
2
...
4) is denoted by u (t), and is deﬁned to be
u (t) =

0 t<0
1 t>0

(2
...
4: The unit step
...
Its value at the origin need not be
deﬁned, and doesn’t matter in signal theory
...
For example, to mathematically represent turning on an oscillator, we can write it as the product of a sinusoid and a step: s (t) = A sin (2πf t) u (t)
...
SIGNALS AND SYSTEMS

2
...
5 Pulse
The unit pulse (Figure 2
...



0, t < 0
p∆ (t) = 1, 0 < t < ∆
(2
...
5: The pulse
...

2
...
6 Square Wave
The square wave (Figure 2
...
It too has an amplitude and a
period, which must be speciﬁed to characterize the signal
...

Square Wave
A

T

t

Figure 2
...

2
...
Rather, a signal expert looks for ways of decomposing
a given signal into a sum of simpler signals, which we term the signal decomposition
...
In
writing a signal as a sum of component signals, we can change the component signal’s gain by multiplying
it by a constant and by delaying it
...

Example 2
...

p∆ (t) = u (t) − u (t − ∆)
6 This

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...
25)

19
Thus, the pulse is a more complex signal than the step
...

Exercise 2
...
31
...

Because the sinusoid is a superposition of two complex exponentials, the sinusoid is more complex
...
Clearly, the word “complex” is used in two diﬀerent
ways here
...
16)) as a sum of a sine and
a cosine
...
Thus, the complex exponential is more fundamental, and Euler’s
relation does not adequately reveal its complexity
...
4 Discrete-Time Signals7
So far, we have treated what are known as analog signals and systems
...
Discrete-time
signals (Section 5
...
One of the fundamental results
of signal theory (Section 5
...
This result is important because discrete-time signals can
be manipulated by systems instantiated as computer programs
...

As important as such results are, discrete-time signals are more general, encompassing signals derived
from analog ones and signals that aren’t
...
We must deal with such symbolic valued (p
...

As with analog signals, we seek ways of decomposing real-valued discrete-time signals into simpler components
...
For symbolic-valued signals, the approach is diﬀerent:
We develop a common representation of all symbolic-valued signals so that we can embody the information
they contain in a uniﬁed way
...

2
...
1 Real- and Complex-valued Signals
A discrete-time signal is represented symbolically as s (n), where n = {
...
We usually draw
discrete-time signals as stem plots to emphasize the fact they are functions deﬁned only on the integers
...
A delayed unit sample has the
expression δ (n − m), and equals one when n = m
...
7: The discrete-time cosine signal is plotted as a stem plot
...
org/content/m0009/2
...

20

CHAPTER 2
...
4
...

s (n) = ej2πf n

(2
...
4
...
As opposed to analog complex
exponentials and sinusoids that can have their frequencies be any real value, frequencies of their discrete1
time counterparts yield unique waveforms only when f lies in the interval − 2 , 1
...

ej2π(f +m)n = ej2πf n ej2πmn
(2
...

2
...
4 Unit Sample
The second-most important discrete-time signal is the unit sample, which is deﬁned to be
δ (n) =

1 if n = 0
0 otherwise

(2
...
8: The unit sample
...
7 (DiscreteTime Cosine Signal), reveals that all signals consist of a sequence of delayed and scaled unit samples
...

∞

s (m) δ (n − m)

s (n) =

(2
...

Discrete-time systems can act on discrete-time signals in ways similar to those found in analog signals
and systems
...
In fact, a special class of
analog signals can be converted into discrete-time signals, processed with software, and converted back into
an analog signal, all without the incursion of error
...

21

2
...
5 Symbolic-valued Signals
Another interesting aspect of discrete-time signals is that their values do not need to be real numbers
...
Such characters certainly aren’t real numbers, and as a collection of
possible signal values, they have little mathematical structure other than that they are members of a set
...
, aK } which
comprise the alphabet A
...
They could represent keyboard characters, bytes (8-bit quantities),
integers that convey daily temperature
...
Furthermore,
the transmission and reception of discrete-time signals, like e-mail, is accomplished with analog signals and
systems
...

2
...
Mathematically, we represent what a system does by the notation
y (t) = S [x (t)], with x representing the input signal and y the output signal
...
9: The system depicted has input x (t) and output y (t)
...
In many ways, systems are like functions, rules that yield a
value for the dependent variable (our output signal) for each value of its independent variable (its input
signal)
...
We term S [·] the input-output
relation for the system
...
For the mathematically inclined, a system is a
functional: a function of a function (signals are functions)
...
Interconnection topologies can be quite complicated, but usually consist of weaves of
three basic interconnection forms
...
5
...
10: Interconnecting systems so that one system’s output serves as the input to another is
the cascade conﬁguration
...
Mathematically,
w (t) = S1 [x (t)], and y (t) = S2 [w (t)], with the information contained in x (t) processed by the ﬁrst, then
the second system
...
For example, in
8 This

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...
SIGNALS AND SYSTEMS

the fundamental model of communication (Figure 1
...

2
...
2 Parallel Interconnection

x(t)

S1[•]

x(t)

+

y(t)

S2[•]

x(t)

Figure 2
...

A signal x (t) is routed to two (or more) systems, with this signal appearing as the input to all systems
simultaneously and with equal strength
...
Two or more systems operate on x (t) and their
outputs are added together to create the output y (t)
...

2
...
3 Feedback Interconnection

x(t)

e(t)

+

S1[•]

y(t)

–
S2[•]
Figure 2
...

The subtlest interconnection conﬁguration has a system’s output also contributing to its input
...
The mathematical
statement of the feedback interconnection (Figure 2
...
The input e (t) equals the input signal minus the output of some other system’s
output to y (t): e (t) = x (t) − S2 [y (t)]
...

For example, in a car’s cruise control system, x (t) is a constant representing what speed you want, and y (t)
is the car’s speed as measured by a speedometer
...

23

2
...
Why the following are categorized as “simple” will only become evident towards the end of the course
...
6
...
We like to think of these as having controllable parameters,
like amplitude and frequency
...

Simply writing an expression for the signals they produce speciﬁes sources
...

2
...
2 Ampliﬁers
An ampliﬁer (Figure 2
...

y (t) = Gx (t)

(2
...
13: An ampliﬁer
...
If less than one, the ampliﬁer actually attenuates
...
You control the gain by turning the volume control
...
6
...
14: delay) when the output signal equals the input signal at an
earlier time
...
31)

Delay
τ

τ

Figure 2
...

Here, τ is the delay
...
Thus, if the delay is positive, the output emerges later than the input,
and plotting the output amounts to shifting the input plot to the right
...
Such systems are diﬃcult to build (they would have to produce
signal values derived from what the input will be), but we will have occasion to advance signals in time
...
org/content/m0006/2
...

24

CHAPTER 2
...
6
...

y (t) = x (−t)
(2
...
15: A time reversal system
...

Exercise 2
...
31
...
In other words, if
we have two systems in cascade, does the output depend on which comes ﬁrst? Determine if the
ordering matters for the cascade of an ampliﬁer and a delay and for the cascade of a time-reversal
system and a delay
...
6
...
Derivative systems operate in a straightforward way: A ﬁrst-derivative system
d
would have the input-output relationship y (t) = dt x (t)
...
It is a signal theory convention that the elementary integral operation have
a lower limit of −∞, and that the value of all signals at t = −∞ equals zero
...
33)

−∞

2
...
6 Linear Systems
Linear systems are a class of systems rather than having a speciﬁc input-output relation
...
They
have the property that when the input is expressed as a weighted sum of component signals, the output
equals the same weighted sum of the outputs produced by each component
...
34)

for all choices of signals and gains
...

• S [Gx (t)] = GS [x (t)] The colloquialism summarizing this property is “Double the input, you double
the output
...

25
• S [0] = 0 If the input is identically zero for all time, the output of a linear system must be zero
...

Just why linear systems are so important is related not only to their properties, which are divulged throughout
this course, but also because they lend themselves to relatively simple mathematical analysis
...
The equation above (2
...
For example, if
x (t) = e−t + sin (2πf0 t)
the output S (x (t)) of any linear system equals
y (t) = S e−t + S [sin (2πf0 t)]

2
...
7 Time-Invariant Systems
Systems that don’t change their input-output relation with time are said to be time-invariant
...
6
...

y (t) = S [x (t)] =⇒ y (t − τ ) = S [x (t − τ )]
(2
...
Thus, a time-invariant
system responds to an input you may supply tomorrow the same way it responds to the same input applied
today; today’s output is merely delayed to occur tomorrow
...
Much of
the signal processing and system theory discussed here concentrates on such systems
...
Nonlinear ones abound, but characterizing them
so that you can predict their behavior for any input remains an unsolved problem
...
1

26

CHAPTER 2
...
7 Signals and Systems Problems10
Problem 2
...

a) −1√
b) 1+2 3j
π
c) 1 + j + ej 2
π
π
d) ej 3 + ejπ + e−j 3
Problem 2
...
For example, there should be two
square-roots, three cube-roots, etc
...
Find the following roots
...
3: Cool Exponentials
Simplify the following (cool) expressions
...
4: Complex-valued Signals
Complex numbers and phasors play a very important role in electrical engineering
...

a) Find the phasor representation for each, and re-express each as the real and imaginary parts of a
complex exponential
...

i) 3 sin (24t)
√
ii) 2 cos 2π60t + π
4
iii) 2 cos t + π + 4 sin t −
6

π
3

b) Show that for linear systems having real-valued outputs for real inputs, that when the input is the
real part of a complex exponential, the output is the real part of the system’s output to the complex
exponential (see Figure 2
...

S Re Aej2πf t

= Re S Aej2πf t

Problem 2
...

Explicitly indicate the value of the complex amplitude V and the complex frequency s
...

a)
b)
c)
d)

v (t) = cos (5t)
v (t) = sin 8t + π
4
v (t) = e−t
v (t) = e−3t sin 4t +

10 This

3π
4

content is available online at ...
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...
16

e)
f)
g)
h)

v (t) = 5e2t sin (8t + 2π)
v (t) = −2
v (t) = 4 sin (2t) + 3 cos (2t)√
v (t) = 2 cos 100πt + π − 3 sin 100πt +
6

π
2

Problem 2
...
17) as a linear combination of delayed and weighted step
functions and ramps (the integral of a step)
...
7: Linear, Time-Invariant Systems
When the input to a linear, time-invariant system is the signal x (t), the output is the signal y (t) (Figure 2
...

a) Find and sketch this system’s output when the input is the depicted signal (Figure 2
...

b) Find and sketch this system’s output when the input is a unit step
...
8: Linear Systems
The depicted input (Figure 2
...

a) What is the system’s output to a unit step input u (t)?
b) What will the output be when the input is the depicted square wave (Figure 2
...
9: Communication Channel
A particularly interesting communication channel can be modeled as a linear, time-invariant system
...
22)
...
23) x1 (t)?
b) What will be the received signal when the transmitter sends the pulse signal (Figure 2
...
10: Analog Computers
So-called analog computers use circuits to solve mathematical problems, particularly when they involve
diﬀerential equations
...

d
y (t) + ay (t) = x (t)
dt
In this equation, a is a constant
...
What is
the total energy expended by the input?
b) Instead of a unit step, suppose the input is a unit pulse (unit-amplitude, unit-duration) delivered to
the circuit at time t = 10
...

28

CHAPTER 2
...
17

…

t

29

x(t)

y(t)
1

1

1

2

t

3

1

2

3

–1
Figure 2
...
5
1

2

t

3

Figure 2
...
20

x(t)
2
•••
1

2

3

4

t

–2
Figure 2
...
22

2

t

t

30

CHAPTER 2
...
23

t

31

Solutions to Exercises in Chapter 2
Solution to Exercise 2
...
12)
z + z ∗ = a + jb + a − jb = 2a = 2Re [z]
...
2 (p
...
The
√
2
distance from the origin to the complex number is the magnitude r, which equals 13 = 32 + (−2)
...
588 radians (−33
...
The ﬁnal answer is 13∠ (−33
...

Solution to Exercise 2
...
13)
2
zz ∗ = (a + jb) (a − jb) = a2 + b2
...

Solution to Exercise 2
...
19)
∞
n
sq (t) = n=−∞ (−1) ApT /2 t − n T
2
Solution to Exercise 2
...
24)
In the ﬁrst case, order does not matter; in the second it does
...
“Time-reverse”
means t → −t
Case 1 y (t) = Gx (t − τ ), and the way we apply the gain and delay the signal gives the same result
...
Delay then time-reverse: y (t) =
x (−t − τ )
...
SIGNALS AND SYSTEMS

Chapter 3

Analog Signal Processing
3
...
Over the years, electric signals have been found to be the easiest to use
...
Thus, we need to delve into the world of electricity
and electromagnetism
...
In many cases, they
make nice examples of linear systems
...

Voltage is electric potential and represents the “push” that drives electric charge from one place to another
...
A battery
generates, through electrochemical means, excess positive charge at one terminal and negative charge at the
other, creating an electric ﬁeld
...
When a conductor connects the positive and negative potentials,
current ﬂows, with positive current indicating that positive charge ﬂows from the positive terminal to the
negative
...
Because electrons have a negative charge, electrons
move in the opposite direction of positive current ﬂow: Negative charge ﬂowing to the right is equivalent to
positive charge moving to the left
...
Electric charge can arise from many sources, the simplest being the electron
...
“Flow” thus means that electrons hop from atom to atom
driven along by the applied electric potential
...

Electrical engineers call these holes, and in some materials, particularly certain semiconductors, current
ﬂow is actually due to holes
...
Here, neurons
“communicate” using propagating voltage pulses that rely on the ﬂow of positive ions (potassium and sodium
primarily, and to some degree calcium) across the neuron’s outer wall
...

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...
ANALOG SIGNAL PROCESSING

i

+
v
–

Figure 3
...

Current ﬂows through circuit elements, such as that depicted in Figure 3
...
For every circuit element we deﬁne
a voltage and a current
...
In
deﬁning the v-i relation, we have the convention that positive current ﬂows from positive to negative voltage
drop
...
Current has
units of amperes, and is named for the French physicist Ampère3
...
Again using the convention shown in Figure 3
...

p (t) = v (t) i (t)
A positive value for power indicates that at time t the circuit element is consuming power; a negative value
means it is producing power
...
Just as in all areas of physics and chemistry, power is the rate at which energy is
consumed or produced
...

t

E (t) =

p (α) dα
−∞

Again, positive energy corresponds to consumed energy and negative energy corresponds to energy production
...
The units of energy
are joules since a watt equals joules/second
...
1
(Solution on p
...
)
Residential energy bills typically state a home’s energy usage in kilowatt-hours
...
2 Ideal Circuit Elements4
The elementary circuit elements—the resistor, capacitor, and inductor— impose linear relationships between
voltage and current
...
bioanalytical
...
htm
3 http://www-groups
...
st-and
...
uk/∼history/Mathematicians/Ampere
...
org/content/m0012/2
...

35

3
...
1 Resistor
i

R

+
v
–

Figure 3
...
v = Ri

The resistor is far and away the simplest circuit element
...

v (t) = Ri (t)
Resistance has units of ohms, denoted by Ω, named for the German electrical scientist Georg Ohm5
...
Conductance
has units of Siemens (S), and is named for the German electronics industrialist Werner von Siemens6
...
A resistor’s instantaneous
power consumption can be written one of two ways
...
As the resistance becomes zero, the voltage goes to zero
for a non-zero current ﬂow
...
A superconductor physically
realizes a short circuit
...
2
...
3: Capacitor
...
The constant of proportionality, the capacitance, has units of farads (F), and is named for the
English experimental physicist Michael Faraday7
...

i (t) = C

d
1
v (t) or v (t) =
dt
C

t

i (α) dα
−∞

5 http://www-groups
...
st-and
...
uk/∼history/Mathematicians/Ohm
...
siemens
...
html
7 http://www
...
org
...
html

(3
...
ANALOG SIGNAL PROCESSING

If the voltage across a capacitor is constant, then the current ﬂowing into it equals zero
...
The power consumed/produced by a voltage applied to a
capacitor depends on the product of the voltage and its derivative
...

E (t) =

3
...
3 Inductor
i

L

+
v
–

d
Figure 3
...
v = L dt i (t)

The inductor stores magnetic ﬂux, with larger valued inductors capable of storing more ﬂux
...
The diﬀerential and integral
forms of the inductor’s v-i relation are
v (t) = L

1
d
i (t) or i (t) =
dt
L

t

v (α) dα

(3
...
2
...
5: The voltage source on the left and current source on the right are like all circuit elements
in that they have a particular relationship between the voltage and current deﬁned for them
...

8 http://www
...
edu/archives//ihd/jhp/

37
Sources of voltage and current are also circuit elements, but they are not linear in the strict sense of linear
systems
...

As for the current source, i = −is regardless of the voltage
...
Current sources, on the other hand, are much
harder to acquire; we’ll learn why later
...
3 Ideal and Real-World Circuit Elements9
Source and linear circuit elements are ideal circuit elements
...
For example, the 1 kΩ
resistor you can hold in your hand is not exactly an ideal 1 kΩ resistor
...
The fourth band on resistors speciﬁes their tolerance; 10% is common
...
At very high frequencies, the way the resistor is constructed introduces inductance and capacitance
eﬀects
...

On the other hand, physical circuit elements can be readily found that well approximate the ideal, but
they will always deviate from the ideal in some way
...
5 V voltage source
...

3
...
A simple resistive circuit is shown in Figure 3
...

This circuit is the electrical embodiment of a system having its input provided by a source system producing
vin (t)
...
Recasting this problem mathematically, we need to solve some set of equations so
that we relate the output voltage vout to the source voltage
...
Until we have more
knowledge about how circuits work, we must write a set of equations that allow us to ﬁnd all the voltages
and currents that can be deﬁned for every circuit element
...
You can deﬁne the directions
for current ﬂow and positive voltage drop any way you like
...
Do recall in deﬁning your voltage and current variables (Section 3
...
Once you
deﬁne voltages and currents, we need six non-redundant equations to solve for the six unknown voltages and
currents
...
The
v-i relations for the resistors give us two more
...
Said another way, we need the laws that govern the electrical connection of
circuit elements
...

Two nodes are explicitly indicated in Figure 3
...
Electrical engineers tend to draw circuit diagrams—schematics— in a rectilinear
fashion
...
org/content/m0013/2
...

content is available online at ...
27/>
...
ANALOG SIGNAL PROCESSING

i 1 + v1 –

vin

+
–

R1

i

+

R2

vin

vout

+
–

–

(a)

Source

+
v
–

iout

R1

+

R2

vout
–

(b)

vin(t)

System

vout(t)

(c)

Figure 3
...
On the bottom is the block diagram that corresponds to the circuit
...

As shown in the middle, we analyze the circuit—understand what it accomplishes—by deﬁning currents
and voltages for all circuit elements, and then solving the circuit and element equations
...
This line simply means that the two elements are connected
together
...
4
...
4
...
These laws can help us analyze this circuit
...
4
...
What this law means physically is
that charge cannot accumulate in a node; what goes in must come out
...
6, below
we have a three-node circuit and thus have three KCL equations
...

Given any two of these KCL equations, we can ﬁnd the other by adding or subtracting them
...
The convention is to
discard the equation for the (unlabeled) node at the bottom of the circuit
...
2
(Solution on p
...
)
In writing KCL equations, you will ﬁnd that in an n-node circuit, exactly one of them is always
redundant
...

3
...
2 Kirchhoﬀ ’s Voltage Law (KVL)
The voltage law says that the sum of voltages around every closed loop in the circuit must equal zero
...
KVL expresses the fact that electric ﬁelds are conservative: The total work performed
in moving a test charge around a closed path is zero
...
7: The circuit shown is perhaps the simplest circuit that performs a signal processing function
...

In writing KVL equations, we follow the convention that an element’s voltage enters with a plus sign when
traversing the closed path, we go from the positive to the negative of the voltage’s deﬁnition
...
8

For the example circuit (Figure 3
...

v-i:

v = vin
v1 = R1 i1
vout = R2 iout

KCL:

−i − i1 = 0
i1 − iout = 0

KVL:

−v + v1 + vout = 0

We have exactly the right number of equations! Eventually, we will discover shortcuts for solving circuit
problems; for now, we want to eliminate all the variables but vout and determine how it depends on vin and
on resistor values
...
Substituting into it the resistor’s
v-i relation, we have vin = R1 i1 + R2 iout
...
Though
not obvious, it is the simplest way to solve the equations
...
Solving for the current in the output resistor, we have
in
iout = R1v+R2
...
To ﬁnd any other circuit quantities, we can back substitute this answer
into our original equations or ones we developed along the way
...
ANALOG SIGNAL PROCESSING

we obtain the quantity we seek
...
3
(Solution on p
...
)
Referring back to Figure 3
...
What kind of system
does our circuit realize and, in terms of element values, what are the system’s parameter(s)?

3
...
We
should examine whether these circuits variables obey the Conservation of Power principle: since a circuit is
a closed system, it should not dissipate or create energy
...
Later, we will prove that because of KVL and KCL
all circuits conserve power
...
34, the instantaneous power consumed/created by every circuit element equals the
product of its voltage and current
...

P =
vk ik
k

Recall that each element’s current and voltage must obey the convention that positive current is deﬁned to
enter the positive-voltage terminal
...
Because the total power in a circuit must be zero (P = 0), some
circuit elements must create power while others consume it
...
4
...
The voltage across the resistor R2 is the output voltage and we found it to equal vout = R1 +R2 vin
...
This result should not be surprising since
we showed (p
...

v2
or i2 R
R

(3
...
But where does a resistor’s power
go? By Conversation of Power, the dissipated power must be absorbed somewhere
...
Current ﬂowing through a resistor makes it hot; its
power is dissipated by heat
...
In fact, the resistance of a wire of length L and cross-sectional area A is given by
R=

ρL
A

The quantity ρ is known as the resistivity and presents the resistance of a unit-length unit crosssectional area material constituting the wire
...
Most materials
11 This

content is available online at ...
5/>
...
9: The circuit shown is perhaps the simplest circuit that performs a signal processing function
...

have a positive value for ρ, which means the longer the wire, the greater the resistance and thus
the power dissipated
...
Superconductors have zero
resistivity and hence do not dissipate power
...
4
(Solution on p
...
)
Calculate the power consumed/created by the resistor R1 in our simple circuit example
...
The current ﬂowing into the source’s positive terminal is −iout
...

Exercise 3
...
98
...

This result is quite general: sources produce power and the circuit elements, especially resistors, consume it
...

3
...
4), KVL and KCL (Section 3
...
4)) with regard to this circuit (Figure 3
...

Resistors connected in such a way that current from one must ﬂow only into another—currents in all
resistors connected this way have the same magnitude—are said to be connected in series
...

This concept is so pervasive it has a name: voltage divider
...

R2
vout
=
vin
R1 + R2
In this way, we express how the components used to build the system aﬀect the input-output relationship
...
org/content/m10674/2
...

42

CHAPTER 3
...
In any case, this important way of expressing input-output
relationships—as a ratio of output to input—pervades circuit and system theory
...
Because it equals i2 , we have that the
ratio of the source’s voltage to the current ﬂowing out of it equals vin = R1 + R2
...

Resistors in series: The series combination of two resistors acts, as far as the voltage source is
concerned, as a single resistor having a value equal to the sum of the two resistances
...
Thus, the equivalent circuit for a series combination
of resistors is a single resistor having a resistance equal to the sum of its component resistances
...
10: The resistor (on the right) is equivalent to the two resistors (on the left) and has a
resistance equal to the sum of the resistances of the other two resistors
...
Note that in making this equivalent circuit, the output voltage can no longer be deﬁned:
The output resistor labeled R2 no longer appears
...

iout
iin

R1

R2

iin

i1
+
+
v R1 v1 R2

iout
+
v2

Figure 3
...

One interesting simple circuit (Figure 3
...
Here, applying KVL reveals that all the voltages are identical:
v1 = v and v2 = v
...
To write the KCL equation, note that the top
node consists of the entire upper interconnection section
...
Using the
v-i relations, we ﬁnd that
R1
iout =
iin
R1 + R2

43
Exercise 3
...
98
...
11 by a voltage source
...
You can easily show that the parallel
combination of R1 and R2 has the v-i relation of a resistor having resistance

1
R1

+

1
R2

−1

=

R1 R2
R1 +R2
...
As the reciprocal of resistance is conductance (Section 3
...
1:
Resistor), we can say that for a parallel combination of resistors, the equivalent conductance is
the sum of the conductances
...
12

Similar to voltage divider (p
...

The current through a resistor in parallel with another is the ratio of the conductance of the ﬁrst to the sum
of the conductances
...
Expressed in terms of resistances, current
+G
divider takes the form of the resistance of the other resistor divided by the sum of resistances: i2 = R1R1 2 i
...
13

vin

+
–

R1

+

R2

vout

RL

–
source

system

sink

Figure 3
...
9) is attached to an oscilloscope’s input
...

+R

44

CHAPTER 3
...
In system-theory terms, we want to pass our circuit’s output to a sink
...
In circuits, a sink is called a load; thus, we describe a
system-theoretic sink as a load resistance RL
...

We must analyze afresh how this revised circuit, shown in Figure 3
...
Rather than deﬁning eight
variables and solving for the current in the load resistor, let’s take a hint from other analysis (series rules
(p
...
43))
...
Because the voltages are the same, we can ﬁnd the
out
current through each from their v-i relations: i2 = vR2 and iL = vout
...
Said another way, the current
entering the node through R1 must equal the sum of the other two currents leaving the node
...

Let Req denote the equivalent resistance of the parallel combination of R2 and RL
...
The KVL equation written around the leftmost loop has
eq
vin = v1 + vout ; substituting for v1 , we ﬁnd
vin = vout
or

R1
+1
Req

vout
Req
=
vin
R1 + Req

Thus, we have the input-output relationship for our entire system having the form of voltage divider,
but it does not equal the input-output relation of the circuit without the voltage measurement device
...
We should look more carefully to determine if any values for the load resistance would lessen its
impact on the circuit
...
As
1
Req = R2 +
we seek:

1
RL

−1

, the approximation would apply if

1
R2

1
RL

or (R2

RL )
...

Exercise 3
...
98
...
1

R2

R3

R1
R4

Figure 3
...
To apply the series and parallel
combination rules, it is best to ﬁrst determine the circuit’s structure: What is in series with what
and what is in parallel with what at both small- and large-scale views
...
This series combination is in parallel with R1
...

In most cases, this approach works well; try it ﬁrst
...
” A simple check for accuracy is the
units: Each component of the numerator should have the same units (here Ω3 ) as well as in the
denominator (Ω2 )
...
Checking units does not guarantee accuracy,
but can catch many errors
...
In system theory, systems can be cascaded without changing the input-output relation
of intermediate systems
...

Design is in the hands of the engineer; he or she must recognize what have come to be known as loading
eﬀects
...

Because the resistors R1 and R2 can have virtually any value, you can never make the resistance of your
voltage measurement device big enough
...
Electrical engineers deal with this situation through
the notion of speciﬁcations: Under what conditions will the circuit perform as designed? Thus, you will
ﬁnd that oscilloscopes and voltmeters have their internal resistances clearly stated, enabling you to determine
whether the voltage you measure closely equals what was present before they were attached to your circuit
...
The designer of this
circuit must thus specify not only what the attenuation is, but also the resistance values employed so that
integrators—people who put systems together from component systems—can combine systems together and
have a chance of the combination working
...
16 (series and parallel combination rules) summarizes the series and parallel combination results
...
Keep in mind that for series combinations, voltage and
resistance are the key quantities, while for parallel combinations current and conductance are more important
...

Exercise 3
...
98
...
Which variable (voltage or current)
is the same for each and which diﬀers? What are the equivalent resistances? When resistors are
placed in series, is the equivalent resistance bigger, in between, or smaller than the component
resistances? What is this relationship for a parallel combination?

3
...
Those resistors not involved with variables of interest can be collapsed into a single resistance
...
org/content/m0020/2
...

46

CHAPTER 3
...
16:

(b) parallel combination rule

Series and parallel combination rules
...

i

vin

+
–

+

R1
R2

v
–

Figure 3
...
Let’s consider our simple
attenuator circuit (shown in the ﬁgure (Figure 3
...
We want
to ﬁnd the v-i relation for the output terminal pair, and then ﬁnd the equivalent circuit for the boxed circuit
...
We seek the relation between v and i that describes the kind of element that lurks within
the dashed box
...
4)
R1 + R2
If the source were zero, it could be replaced by a short circuit, which would conﬁrm that the circuit does
indeed function as a parallel combination of resistors
...

47

i

+

Req
veq

+
–

v
–

Figure 3
...

If we consider the simple circuit of Figure 3
...
5)

Comparing the two v-i relations, we ﬁnd that they have the same form
...

+R
Thus, from viewpoint of the terminals, you cannot distinguish the two circuits
...

For any circuit containing resistors and sources, the v-i relation will be of the form
v = Req i + veq

(3
...
18
...
In the example (Example 3
...
Because
Thévenin’s theorem applies in general, we should be able to make measurements or calculations only from
the terminals to determine the equivalent circuit
...
18)
...
Because no current ﬂows through the resistor,
the voltage across it is zero (remember, Ohm’s Law says that v = Ri)
...
Now consider
the situation when we set the terminal voltage to zero (short-circuit it) and measure the resulting current
...
From this property, we can determine the equivalent resistance
...
7)
(3
...
9
(Solution on p
...
)
Use the open/short-circuit approach to derive the Thévenin equivalent of the circuit shown in
Figure 3
...

48

CHAPTER 3
...
19

Example 3
...
20

For the circuit depicted in Figure 3
...

Starting with the open/short-circuit approach, let’s ﬁrst ﬁnd the open-circuit voltage voc
...
Thus,
R3 1
voc = R1 +RR+R3 iin
...
In short, R3 does not aﬀect the short-circuit current, and can be
eliminated
...
Thus, the Thévenin
+R
+R2 )
equivalent resistance is R3 (R12 +R3
...
Because the current is now zero, we can replace the current source by an open circuit
...
Thus, Req = (R3 R1 + R2 ), and we obtain the same result
...
21:
All circuits containing sources and resistors can be described by simpler equivalent
circuits
...

As you might expect, equivalent circuits come in two forms: the voltage-source oriented Thévenin equivalent14 and the current-source oriented Mayer-Norton equivalent (Figure 3
...
To derive the latter, the
v-i relation for the Thévenin equivalent can be written as
v = Req i + veq

(3
...
10)

or
i=
v

where ieq = Req is the Mayer-Norton equivalent source
...
21
eq
can be easily shown to have this v-i relation
...

The short-circuit current equals the negative of the Mayer-Norton equivalent source
...
10
Find the Mayer-Norton equivalent circuit for the circuit below
...
22

14 “Finding

Thévenin Equivalent Circuits” ...
98
...
ANALOG SIGNAL PROCESSING

Equivalent circuits can be used in two basic ways
...
Which one is used depends on whether what is attached to the terminals is a series conﬁguration
(making the Thévenin equivalent the best) or a parallel one (making Mayer-Norton the best)
...
When we buy a ﬂashlight battery, either equivalent circuit can accurately describe it
...
Since batteries are labeled
with a voltage speciﬁcation, they should serve as voltage sources and the Thévenin equivalent serves as the
natural choice
...
If we have a load resistance much larger than the battery’s equivalent
+R
resistance, then, to a good approximation, the battery does serve as a voltage source
...
Consider now the Mayer-Norton equivalent; the current through the load resistance is given by
Req
current divider, and equals i = − RL +Req ieq
...
If the load resistance is comparable
to the equivalent resistance, the battery serves neither as a voltage source or a current course
...
On the other hand, if you attach it to a circuit having a
small equivalent resistance, you bought a current source
...
In
1883, he published (twice!) a proof of what is now called the Thévenin equivalent while developing
ways of teaching electrical engineering concepts at the École Polytechnique
...

Hans Ferdinand Mayer: After earning his doctorate in physics in 1920, he turned to communications engineering when he joined Siemens & Halske in 1922
...
During his interesting career, he rose to
lead Siemens’s Central Laboratory in 1936, surreptitiously leaked to the British all he knew of
German warfare capabilities a month after the Nazis invaded Poland, was arrested by the Gestapo
in 1943 for listening to BBC radio broadcasts, spent two years in Nazi concentration camps, and
went to the United States for four years working for the Air Force and Cornell University before
returning to Siemens in 1950
...

Edward L
...
In the same month when Mayer’s paper appeared, Norton wrote in an
internal technical memorandum a paragraph describing the current-source equivalent
...

3
...
23
...
The current through the capacitor is given by i = C dt (vout ),
and this current equals that passing through the resistor
...
dcs
...
ac
...
html
16 http://www
...
rice
...
org/content/m0023/2
...

(3
...
23: A simple RC circuit
...
In contrast
to resistive circuits, where we obtain an explicit input-output relation, we now have an implicit relation
that requires more work to obtain answers
...
Note ﬁrst that even ﬁnding the diﬀerential
equation relating an output variable to a source is often very tedious
...
We would have to slog
our way through the circuit equations, simplifying them until we ﬁnally found the equation that related the
source(s) to the output
...

Although not original with him, Charles Steinmetz18 presented the key paper describing the impedance
approach in 1893
...
To use impedances, we must master complex numbers
...
But more importantly, the impedance concept is central to engineering and physics, having a
reach far beyond just circuits
...
9 The Impedance Concept19
Rather than solving the diﬀerential equation that arises in circuits containing capacitors and inductors, let’s
pretend that all sources in the circuit are complex exponentials having the same frequency
...

For the above example RC circuit (Figure 3
...
The complex
amplitude Vin determines the size of the source and its phase
...
To appreciate
why this should be true, let’s investigate how each circuit element behaves when either the voltage or current
is a complex exponential
...
When v = V ej2πf t ; then i = V ej2πf t
...
Clearly, if the current were assumed to be a
d
complex exponential, so would the voltage
...
Letting the voltage be a complex
j2πf t
exponential, we have i = CV j2πf e

...
Finally,
d
for the inductor, where v = L dt (i), assuming the current to be a complex exponential results in the voltage
having the form v = LIj2πf ej2πf t , making its complex amplitude V = LIj2πf
...

I
This quantity is known as the element’s impedance
...
invent
...
html
19 This

content is available online at ...
23/>
...
ANALOG SIGNAL PROCESSING

i

+
v
–

R

i

+
v
–

C

(a)

i

(b)

Figure 3
...
For example, the magnitude of the capacitor’s impedance is inversely related to frequency, and has a phase of − π
...

Let’s consider Kirchhoﬀ’s circuit laws
...
12)
=0
which means
Vn = 0
(3
...
We can easily imagine that the complex amplitudes
of the currents obey KCL
...
Consequently, the ratio of voltage to
current for each element equals the ratio of their complex amplitudes, which depends only on the source’s
frequency and element values
...
For example, suppose we
had a circuit element where the voltage equaled the square of the current: v (t) = Ki2 (t)
...

Because for linear circuit elements the complex amplitude of voltage is proportional to the complex
amplitude of current— V = ZI — assuming complex exponential sources means circuit elements behave
as if they were resistors, where instead of resistance, we use impedance
...

3
...
What we emphasize here is that it is often easier to ﬁnd
the output if we use impedances
...
A common error in using impedances is keeping the time-dependent part, the complex
exponential, in the fray
...
Only after we ﬁnd the result in the frequency domain do we go back to the time domain and put
things back together again
...
Since you can’t be two places at the same time, you
are faced with solving your circuit problem in one of the two rooms at any point in time
...
Security guards make sure you don’t try
20 This

content is available online at ...
9/>
...
Figure 3
...

R
vin

+
–

C

+
vout
–

frequency-domain
room

time-domain
room

f
Only signals

t
v(t) = Ve j2πft
i(t) = Ie j2πft

Only complex amplitudes

differential equations
KVL, KCL
superposition

impedances
transfer functions
voltage & current divider
KVL, KCL
superposition

vout(t) = …

Vout = Vin•H(f)

Figure 3
...

In the time domain, signals can have any form
...

As we unfold the impedance story, we’ll see that the powerful use of impedances suggested by Steinmetz21
greatly simpliﬁes solving circuits, alleviates us from solving diﬀerential equations, and suggests a general way
of thinking about circuits
...

1
...
We do this because the impedance
approach simpliﬁes ﬁnding how input and output are related
...
We’ll learn how to “get the pulse back” later
...
With a source equaling a complex exponential, all variables in a linear circuit will also be complex
exponentials having the same frequency
...
To ﬁnd these, we consider the source to be a complex number (Vin here)
and the elements to be impedances
...
We can now solve using series and parallel combination rules how the complex amplitude of any variable
relates to the sources complex amplitude
...
3
To illustrate the impedance approach, we refer to the RC circuit (Figure 3
...

21 http://www
...
org/hall_of_fame/139
...
ANALOG SIGNAL PROCESSING

ZR

R
vin

+
vout
–

+
–

C

+

Vin

(a)

ZC

+
Vout

(b)

Figure 3
...
(b) The impedance counterpart for the RC circuit
...

Using impedances, the complex amplitude of the output voltage Vout can be found using voltage
divider:
ZC
Vout =
Vin
ZC + ZR
Vout =
Vout =

1
j2πf C
1
j2πf C +

R

Vin

1
Vin
j2πf RC + 1

If we refer to the diﬀerential equation for this circuit (shown in Circuits with Capacitors and Inductors
d
(Section 3
...
Thus, using impedances is equivalent to
using the diﬀerential equation and solving it when the source is a complex exponential
...
If we cross-multiply the relation
between input and output amplitudes,
Vout (j2πf RC + 1) = Vin
and then put the complex exponentials back in, we have
RCj2πf Vout ej2πf t + Vout ej2πf t = Vin ej2πf t
In the process of deﬁning impedances, note that the factor j2πf arises from the derivative of a complex
exponential
...

RC

d
(vout ) + vout = vin
dt

This is the same equation that was derived much more tediously in Circuits with Capacitors and Inductors
(Section 3
...
Finding the diﬀerential equation relating output to input is far simpler when we use impedances
than with any other technique
...
11
(Solution on p
...
)
Suppose you had an expression where a complex amplitude was divided by j2πf
...
11 Power in the Frequency Domain22
Recalling that the instantaneous power consumed by a circuit element or an equivalent circuit that represents
a collection of elements equals the voltage times the current entering the positive-voltage terminal, p (t) =
v (t) i (t), what is the equivalent expression using impedances? The resulting calculation reveals more about
power consumption in circuits and the introduction of the concept of average power
...

v (t) = |V | cos (2πf t + φ)
i (t) = |I| cos (2πf t + θ)
Here, the complex amplitude of the voltage V equals |V |ejφ and that of the current is |I|ejθ
...
1
...

v (t) =
i (t) =

1
2
1
2

V ej2πf t + V ∗ e−j2πf t
Iej2πf t + I ∗ e−j2πf t

Multiplying these two expressions and simplifying gives
p (t) =
=
=

1
∗
∗
j4πf t
+ V ∗ I ∗ e−j4πf t
4 V I + V I + V Ie
1
1
∗
j4πf t
2 Re [V I ] + 2 Re V Ie
1
1
∗
2 Re [V I ] + 2 |V ||I| cos (4πf t + φ + θ)

We deﬁne 1 V I ∗ to be complex power
...
The second
term varies with time at a frequency twice that of the source
...

From another viewpoint, the real-part of complex power represents long-term energy consumption/production
...
” Consequently, the most convenient deﬁnition of the average power consumed/produced by any circuit is in terms of complex amplitudes
...
14)

Exercise 3
...
98
...
What phase
relationship between voltage and current maximizes the average power? In other words, how are φ
and θ related for maximum power dissipation?
Because the complex amplitudes of the voltage and current are related by the equivalent impedance, average
power can also be written as
1
1
1
2
2
Pave = Re [Z] |I| = Re
|V |
2
2
Z
These expressions generalize the results (3
...
We have derived a fundamental
result: Only the real part of impedance contributes to long-term power dissipation
...
Capacitors and inductors dissipate no power in the long term
...
If you turn on a constant
voltage source in an RC-circuit, charging the capacitor does consume power
...
org/content/m17308/1
...

56

CHAPTER 3
...
13
(Solution on p
...
)
In an earlier problem (Section 1
...
1: RMS Values), we found that the rms value of a sinusoid was
√
its amplitude divided by 2
...
12 Equivalent Circuits: Impedances and Sources23
When we have circuits with capacitors and/or inductors as well as resistors and sources, Thévenin and MayerNorton equivalent circuits can still be deﬁned by using impedances and complex amplitudes for voltage and
currents
...
Thus, we have Thévenin and Mayer-Norton equivalent circuits as shown in Figure 3
...

Example 3
...
29 (Simple RC Circuit)
...
The open-circuit voltage corresponds to the transfer function we have
already found
...
The equivalent impedance can be found by setting
the source to zero, and ﬁnding the impedance using series and parallel combination rules
...
Thus, Zeq = R j2πf C = 1+j2πf RC
...
Note in particular that j2πf RC must be dimensionless
...
13 Transfer Functions24
The ratio of the output and input amplitudes for Figure 3
...
15)

Implicit in using the transfer function is that the input is a complex exponential, and the output is also a
complex exponential having the same frequency
...
org/content/m0030/2
...

content is available online at ...
19/>
...

I

+

Sources,
Resistors,
Capacitors,
Inductors

V
–

I

I

+

+

Zeq
Veq

+

Ieq

V

–

Zeq

–

V
–

Mayer-Norton Equivalent

Thévenin Equivalent

(b) Equivalent circuits with impedances
...
28: Comparing the ﬁrst, simpler, ﬁgure with the slightly more complicated second ﬁgure, we
see two diﬀerences
...
Secondly, the terminal and source variables are now complex amplitudes,
which carries the implicit assumption that the voltages and currents are single complex exponentials, all
having the same frequency
...
29

input amplitude in creating the output amplitude
...
The circuit’s
function is thus summarized by the transfer function
...
Because transfer functions are complex-valued, frequency-dependent quantities, we

58

CHAPTER 3
...
30: A simple RC circuit
...
31: Magnitude and phase of the transfer function of the RC circuit shown in Figure 3
...
(a) |H (f ) | = √
2
(2πf RC) +1

can better appreciate a circuit’s function by examining the magnitude and phase of its transfer function
(Figure 3
...

This transfer function has many important properties and provides all the insights needed to determine
how the circuit functions
...
Recall that sinusoids consist of the sum of two complex exponentials, one having the
negative frequency of the other
...
Do note that the
magnitude has even symmetry: The negative frequency portion is a mirror image of the positive frequency
portion: |H (−f ) | = |H (f ) |
...
These properties of
this speciﬁc example apply for all transfer functions associated with circuits
...

1
The magnitude equals √2 of its maximum gain (one at f = 0) when 2πf RC = 1 (the two terms in
1
the denominator of the magnitude are equal)
...

• For frequencies below this frequency, the circuit does not much alter the amplitude of the complex
exponential source
...
Thus, when the source
frequency is in this range, the circuit’s output has a much smaller amplitude than that of the source
...
In this circuit the cutoﬀ frequency
depends only on the product of the resistance and the capacitance
...
59 × 10−4
...
59 kΩ and
2π
100 nF or 10 Ω and 1
...

The phase shift caused by the circuit at the cutoﬀ frequency precisely equals − π
...
This phase shift corresponds to the diﬀerence between a cosine and a sine
...

First of all, a sinusoid is the sum of two complex exponentials, each having a frequency equal to the negative
of the other
...
If the source is a sine wave, we
know that
vin (t) = A sin (2πf t)
(3
...

vout (t) =

A
A
H (f ) ej2πf t − H (−f ) e−j2πf t
2j
2j

(3
...

Furthermore, |H (−f ) | = |H (f ) | (even symmetry of the magnitude) and ∠H (−f ) = −∠H (f ) (odd symmetry of the phase)
...
18)

The circuit’s output to a sinusoidal input is also a sinusoid, having a gain equal to the magnitude
of the circuit’s transfer function evaluated at the source frequency and a phase equal to the
phase of the transfer function at the source frequency
...

Exercise 3
...
99
...
Show that if the source can
be written as the imaginary part of a complex exponential— vin (t) = Im V ej2πf t — the output
is given by vout (t) = Im V H (f ) ej2πf t
...

The notion of impedance arises when we assume the sources are complex exponentials
...
The diﬀerential equation applies no matter what the source may be
...
In this sense, we have not lost anything by
temporarily pretending the source is a complex exponential
...

3
...
In short, linear circuits are a special case
25 This

content is available online at ...
21/>
...
ANALOG SIGNAL PROCESSING

i

vin

+
–

+

iout

R
L

v

–
Figure 3
...
In particular, suppose these component signals are
complex exponentials, each of which has a frequency diﬀerent from the others
...
Those components having a frequency less than the cutoﬀ frequency pass
through the circuit with little modiﬁcation while those having higher frequencies are suppressed
...
Because low frequencies pass through the ﬁlter, we call it a lowpass ﬁlter to express more
precisely its function
...
Once we ﬁnd the transfer function, we can write the output directly as indicated by the output of
a circuit for a sinusoidal input (3
...

Example 3
...
The source voltage equals Vin = 2 cos (2π60t) + 3
...
Because the input is the sum of two sinusoids–a
constant is a zero-frequency cosine–our approach is
1
...

3
...

ﬁnd the transfer function using impedances;
use it to ﬁnd the output due to each input component;
add the results;
ﬁnd element values that accomplish our design criteria
...

Iout
j2πf L
1
=
·
Vin
R + j2πf L j2πf L
1
=
j2πf L + R
= H (f )
where
voltage divider =

(3
...

61
3
The constant term is easiest to handle
...
Thus, the value we
choose for the resistance will determine the scaling factor of how voltage is converted into current
...
The total
output due to our source is

iout = 2|H (60) | cos (2π60t + ∠H (60)) + 3H (0)

(3
...
Thus, 2πfc L = R, which gives fc = 2πL
...
Suppose we place it at, say, 10 Hz
...
8
...
16
(3
...
3 relative to the constant term’s amplitude of R
...
Having a 100 mH inductor
would require a 6
...
An easily available resistor value is 6
...
To make the resistance bigger would require a proportionally
larger inductor
...
The choice made here represents
only one compromise
...
3
π
0
...
The waveforms for the input and output are shown in Figure 3
...

Note that the sinusoid’s phase has indeed shifted; the lowpass ﬁlter not only reduced the 60 Hz signal’s
amplitude, but also shifted its phase by 90 ◦
...
15 Formal Circuit Methods: Node Method26
In some (complicated) cases, we cannot use the simpliﬁcation techniques—such as parallel or series combination rules—to solve for a circuit’s input-output relation
...
We need a formal method
that produces a small, easy set of equations that lead directly to the input-output relation we seek
...

The node method begins by ﬁnding all nodes–places where circuit elements attach to each other–in the
circuit
...
For the remaining nodes, we deﬁne node
voltages en that represent the voltage between the node and the reference
...
In our example, we
have two node voltages
...
The reason for this simple, but astounding, fact is that a node voltage is
uniquely deﬁned regardless of what path is traced between the node and the reference
...

In some cases, a node voltage corresponds exactly to the voltage across a voltage source
...
For example, in our circuit, e1 = vin ;
thus, we need only to ﬁnd one node voltage
...
In our example, the only circuit equation
is
e2
e2
e2 − vin
+
+
=0
(3
...
org/content/m0032/2
...

62

CHAPTER 3
...
1

Time (s)

Figure 3
...
28Ω and L = 100mH
...
34

A little reﬂection reveals that when writing the KCL equations for the sum of currents leaving a node, that
node’s voltage will always appear with a plus sign, and all other node voltages with a minus sign
...

Also remember to check units at this point: Every term should have units of current
...
23)
e2 =
R1 R2 + R1 R3 + R2 R3
Have we really solved the circuit with the node method? Along the way, we have used KVL, KCL, and
the v-i relations
...
This result guarantees that the node method can be used to “solve” any circuit
...
All circuit variables can be found using the v-i relations and voltage divider
...

3

e1

e2
i

R2
iin

R1

R3

Figure 3
...
The circuit has three nodes, requiring us to deﬁne
two node voltages
...
Rewrite these
equations in the standard set-of-linear-equations form
...

Example 3
...
36), we cannot use the series/parallel combination rules: The vertical
resistor at node 1 keeps the two horizontal 1 Ω resistors from being in series, and the 2 Ω resistor
prevents the two 1 Ω resistors at node 2 from being in series
...
The node
equations are
e1 − vin
e1
e1 − e2
+
+
= 0 (Node 1)
1
1
1
e2 − vin
e2
e2 − e1
+
+
=0
2
1
1

(Node 2)

64

CHAPTER 3
...
36

6
5
5
2
Solving these equations yields e1 = 13 vin and e2 = 13 vin
...

One unfortunate consequence of using the element’s numeric values from the outset is that it
becomes impossible to check units while setting up and solving equations
...
15
What is the equivalent resistance seen by the voltage source?

E

(Solution on p
...
)

+

R1
Vin

+
–

C

R2

Vout
–

Figure 3
...

The node method applies to RLC circuits, without signiﬁcant modiﬁcation from the methods used on
simple resistive circuits, if we use complex amplitudes
...
In the example circuit, we deﬁne complex amplitudes for
the input and output variables and for the node voltages
...

The transfer

65
This circuit diﬀers from the one shown previously (Figure 3
...
What eﬀect has it had on the transfer function, which in the original circuit
1
was a lowpass ﬁlter having cutoﬀ frequency fc = 2πR1 C ? As shown in Figure 3
...

|H(f)|
1
No R2

R1=1, R2=1
0

0

1
R1+R2
1
2πRC 2πR1C•
R2

1

f

Figure 3
...
37 (Node Method and Impedances)
...

When R2 = R1 , as shown on the plot, the passband gain becomes half of the original, and the cutoﬀ
frequency increases by the same factor
...

Exercise 3
...
99
...
Does the addition of the R2 resistor help in circuit design?

3
...

First of all, deﬁne node voltages for all nodes in a given circuit
...
For example, in the portion of a large circuit (Figure 3
...
With these node voltages, we can express the voltage across any element in terms of them
...
The instantaneous power for element 1
becomes
v1 i1 = (eb − ea ) i1 = eb i1 − ea i1
27 This

content is available online at ...
2/>
...
ANALOG SIGNAL PROCESSING

a
1

i1

2

i2

b

3

i3

c

Figure 3
...

For example, for node b, we have eb (i3 − i1 )
...
Consequently, we conclude that the sum of element powers must equal zero in any
circuit regardless of the elements used to construct the circuit
...

In particular, note that the complex amplitudes of voltages and currents obey KVL and KCL, respectively
...
Furthermore, the complex-conjugate of currents also satisﬁes
KCL, which means we also have k Vk Ik ∗ = 0
...
Finding the real-part of this power conservation gives the result that average power
is also conserved in any circuit
...
All we
need is a set of voltages that obey KVL and a set of currents that obey KCL
...

vk (t1 ) ik (t2 ) = 0
k

Even more interesting is the fact that the elements don’t matter
...
We can then make element-for-element replacements and, if the topology has not
changed, we can measure a set of currents
...
17 Electronics28
So far we have analyzed electrical circuits: The source signal has more power than the output variable,
be it a voltage or a current
...
Resistors, inductors,
and capacitors as individual elements certainly provide no power gain, and circuits built of them will not
magically do so either
...
org/content/m0035/2
...

67
gain: electronic circuits
...
The basic idea of the transistor is to let the
weak input signal modulate a strong current provided by a source of electrical power–the power supply–to
produce a more powerful signal
...

The waterpower results from the static pressure of the water in your plumbing created by the water utility
pumping the water up to your local water tower
...
Just as in this analogy, a power supply is a
source of constant voltage as the water tower is supposed to provide a constant water pressure
...
An op-amp is an integrated circuit (a
complicated circuit involving several transistors constructed on a chip) that provides a large voltage gain if
you attach the power supply
...

3
...
Thus, there are four diﬀerent kinds of dependent sources; to describe an op-amp, we
need a voltage-dependent voltage source
...
Dependent sources do not serve as inputs to a circuit like
independent sources
...
The
RLC circuits we have been considering so far are known as passive circuits
...
40: Of the four possible dependent sources, depicted is a voltage-dependent voltage source
in the context of a generic circuit
...
41 (op-amp) shows the circuit symbol for the op-amp and its equivalent circuit in terms of a
voltage-dependent voltage source
...
org/content/m0053/2
...

Signal Model for Bipolar Transistor” ...
ANALOG SIGNAL PROCESSING

Rout

a
a

c
b

c

+
+

Rin

G(ea–eb)

–

–
b

Figure 3
...
Inputs attach to
nodes a and b, and the output is node c
...

Here, the output voltage equals an ampliﬁed version of the diﬀerence of node voltages appearing across
its inputs
...
As in most active circuit
schematics, the power supply is not shown, but must be present for the circuit model to be accurate
...

Because dependent sources cannot be described as impedances, and because the dependent variable
cannot “disappear” when you apply parallel/series combining rules, circuit simpliﬁcations such as current
and voltage divider should not be applied in most cases
...
15)
...
Consider the circuit shown on the top in Figure 3
...

RF

R
–
vin

+

+

+
–

RL

vout
–

RF

R

+

+
Rout
+
–

Rin

v

+
–

–

–Gv

RL

vout
–

Figure 3
...
On the bottom
is the equivalent circuit, and integrates the op-amp circuit model into the circuit
...
As we explore op-amps in more detail in the next section, this conﬁguration will
appear again and again, and its usefulness demonstrated
...
Only two node voltages— v and vout —need be deﬁned; the
remaining nodes are across sources or serve as the reference
...
24)

vout − (−G) v vout − v vout
+
+
=0
(3
...

Solving these to learn how vout relates to vin yields
RF Rout
Rout − GRF

1
1
1
+
+
Rout
Rin
RL

1
1
1
+
+
R Rin
RF

−

1
1
vout = vin
RF
R

(3
...
Once we learn more about op-amps (Section 3
...
Do note that the units check, and that the parameter G of the dependent source is a dimensionless gain
...
19 Operational Ampliﬁers31
Rout

a
a

+
c

b

c

Rin

–

+
–

G(ea–eb)

b

Figure 3
...
Inputs attach to
nodes a and b, and the output is node c
...

Op-amps not only have the circuit model shown in Figure 3
...

• The input resistance, Rin , is typically large, on the order of 1 MΩ
...

• The voltage gain, G, is large, exceeding 105
...

If you were to build such a circuit—attaching a voltage source to node a, attaching node b to the reference,
and looking at the output—you would be disappointed
...

Unmodeled limitations imposed by power supplies: It is impossible for electronic components to yield voltages that exceed those provided by the power supply or for them to yield currents
that exceed the power supply’s rating
...
org/content/m0036/2
...

70

CHAPTER 3
...
Attaching the 1 mv signal not
only would fail to produce a 100 V signal, the resulting waveform would be severely distorted
...
Another
consideration in designing circuits with op-amps is that these element values are typical: Careful control of
the gain can only be obtained by choosing a circuit so that its element values dictate the resulting gain,
which must be smaller than that provided by the op-amp
...
44: The top circuit depicts an op-amp in a feedback ampliﬁer conﬁguration
...

3
...
1 Inverting Ampliﬁer
The feedback conﬁguration shown in Figure 3
...

RF Rout
Rout − GRF

1
1
1
+
+
Rout
Rin
RL

1
1
1
+
+
R Rin
RF

−

1
1
vout = vin
RF
R

(3
...
In choosing element values with respect to op-amp characteristics, we can simplify the expression dramatically
...
This situation drops the term RL from the
second factor of (3
...

• Make the resistor, R, smaller than Rin , which means that the R1 term in the third factor is negligible
...
27) becomes
RF
Rout − GRF

1
1
+
R RF

−

1
1
vout = vout
RF
R

Because the gain is large and the resistance Rout is small, the ﬁrst term becomes −
−

1
G

1
1
+
R RF

−

1
1
vout = vin
RF
R

(3
...
29)

71

• If we select the values of RF and R so that (GR
1
op-amp’s inherent gain, and it will equal − RF
...

RF
vout = −
vin
(3
...
It is always negative, and can be less than one or greater than one in
magnitude
...
Interestingly, note that this relationship does not depend
on the load resistance
...
Thus observation means that, if careful, we can place op-amp circuits in cascade, without
incurring the eﬀect of succeeding circuits changing the behavior (transfer function) of previous ones; see this
problem (Problem 3
...

3
...
2 Active Filters
As long as design requirements are met, the input-output relation for the inverting ampliﬁer also applies
when the feedback and input circuit elements are impedances (resistors, capacitors, and inductors)
...
45:

Vout
Vin

= − ZF
Z

Example 3
...
We want the transfer function
between the output and input voltage to be
H (f ) =

K
1 + jf
fc

where K equals the passband gain and fc is the cutoﬀ frequency
...
With the transfer function of the above op-amp circuit in mind,
let’s consider some choices
...
This choice means the feedback impedance is a resistor and that the
fc
input impedance is a series combination of an inductor and a resistor
...

1
1
• ZF = 1+ jf , Z = K
...

Since this admittance is a sum of admittances, this expression suggests

72

CHAPTER 3
...
46

1
the parallel combination of a resistor (value = 1 Ω) and a capacitor (value = fc F)
...
Consider the general RC parallel
1
combination; its admittance is RF +j2πf C
...

Creating a speciﬁc transfer function with op-amps does not have a unique answer
...
43) and gain (increase in power and amplitude) can result
...
Signals
transmitted over the telephone have an upper frequency limit of about 3 kHz
...
3 × 10−5
...
A 1 µF
capacitor and a 330 Ω resistor, 10 nF and 33 kΩ, and 10 pF and 33 MΩ would all theoretically work
...
Recall that we must have R < Rin
...
We also need to ask for less gain than the op-amp can
provide itself
...
We must have |ZF | < 105 for all frequencies of
R
RF

interest
...
As this impedance decreases with frequency, the design speciﬁcation of
R
RF
= 10 means that this criterion is easily met
...
Additional considerations like parts cost might
enter into the picture
...

For resistors, having values r10d , easily obtained values of r are 1, 1
...
3, 4
...
8, and the decades
span 0–8
...
17
(Solution on p
...
)
What is special about the resistor values; why these rather odd-appearing values for r?

3
...
3 Intuitive Way of Solving Op-Amp Circuits
When we meet op-amp design speciﬁcations, we can simplify our circuit calculations greatly, so much so that
we don’t need the op-amp’s circuit model to determine the transfer function
...

When we take advantage of the op-amp’s characteristics—large input impedance, large gain, and small
output impedance—we note the two following important facts
...
The voltage produced by the dependent source is 105 times the
v
voltage v
...
For example,

73

+ v –
R
e

RF
iin=0
–
+
+

+

RL

–

vout
–

Figure 3
...
Consequently,
we can ignore iin in our calculations and assume it to be zero
...
This means that in op-amp circuits, the voltage across the op-amp’s input is basically
zero
...
47 (op-amp)
...
Thus, the
in
current through the resistor R equals vR
...
Because the current going into the op-amp is zero, all of the current ﬂowing through R ﬂows
R
through the feedback resistor (iF = i)! The voltage across the feedback resistor v equals vinR F
...
Using this approach makes analyzing
new op-amp circuits much easier
...

Example 3
...
48: Two-source, single-output op-amp circuit example
...
48 (Two Source Circuit)
...
By superposition, we know

74

CHAPTER 3
...
31)

When we start from scratch, the node joining the three resistors is at the same potential as the
reference, e ≈ 0, and the sum of currents ﬂowing into that node is zero
...
Because the feedback resistor is essentially in parallel with the
load resistor, the voltages must satisfy v = −vout
...

What utility does this circuit have? Can the basic notion of the circuit be extended without
bound?

3
...
Voltage and current sources are (technically) nonlinear devices: stated simply, doubling
the current through a voltage source does not double the voltage
...
Its input-output relation has an exponential form
...
32)

Here, the quantity q represents the charge of a single electron in coulombs, k is Boltzmann’s constant, and
i ( µA)
50
40
30
20
10

i
+
v

0
...
49: v-i relation and schematic symbol for the diode
...

T is the diode’s temperature in K
...
The constant I0 is the
q
leakage current, and is usually very small
...
49 (Diode), the nonlinearity
becomes obvious
...
This situation is
known as forward biasing
...
A less detailed model for the diode has any positive current
ﬂowing through the diode when it is forward biased, and no current when negative biased
...

Because of the diode’s nonlinear nature, we cannot use impedances nor series/parallel combination rules
to analyze circuits containing them
...
org/content/m0037/2
...

Junction: Part II” ...
50

its application, and KVL is a statement about voltage drops around a closed path regardless of whether
the elements are linear or not
...
33)
R
This equation cannot be solved in closed form
...
As an approximation, when vin is positive, current ﬂows through
the diode so long as the voltage vout is smaller than vin (so the diode is forward biased)
...
Thus, at this level of analysis, positive input voltages result in positive output voltages
with negative ones resulting in vout = − (RI0 )
...
51

We need to detail the exponential nonlinearity to determine how the circuit distorts the input voltage
waveform
...
50 (diode circuit) to determine the output voltage
when the input is a sinusoid
...
We plot each term as a
function of vout for various values of the input voltage vin ; where they intersect gives us the output voltage
...
As for the right side, which expresses the diode’s v-i relation, the point at which the curve
crosses the vout axis gives us the value of vin
...
This reduction
is smaller if the straight line has a shallower slope, which corresponds to using a bigger output resistor
...

What utility might this simple circuit have? The diode’s nonlinearity cannot be escaped here, and the
clearly evident distortion must have some practical application if the circuit were to be useful
...

76

CHAPTER 3
...
52

Here is a circuit involving a diode that is actually simpler to analyze than the previous one
...
Thus, the diode’s current is proportional
to the input voltage
...
34)
q
RI0
Clearly, the name logarithmic ampliﬁer is justiﬁed for this circuit
...
21 Analog Signal Processing Problems34
Problem 3
...
53

For each circuit shown in Figure 3
...

a) What is the voltage across each element and what is the voltage v in each case?
b) For the last circuit, are there element values that make the voltage v equal zero for all time? If so,
what element values work?
c) Again, for the last circuit, if zero voltage were possible, what circuit element could substitute for the
capacitor-inductor series combination that would yield the same voltage?

Problem 3
...
54) behavior
...
org/content/m10349/2
...

77
b) Solve these equations for i1 : In other words, express this current in terms of element and source values
by eliminating non-source voltages and currents
...

d) What is the power dissipated by the load resistor RL in this case?

i1
iin

R2

R1

+
–

vin

15 A

(a) Circuit A

20Ω

RL

(b) Circuit B

Figure 3
...
3: Equivalent Resistance
For each of the following circuits (Figure 3
...

Calculate the conductance seen at the terminals for circuit (c) in terms of each element’s conductance
...
How is the circuit (c) derived from circuit (b)?
Problem 3
...
This Principle has important consequences in simplifying the calculation of circuit
variables in multiple source circuits
...
56), ﬁnd the indicated current using any technique you like (you
should use the simplest)
...
This result means that we can think of the current as a superposition of two components,
each of which is due to a source
...

Thus, to ﬁnd the voltage source component, you can set the current source to zero (an open circuit) and
use the usual tricks
...
Calculate the total current i using the Superposition
Principle
...
5: Current and Voltage Divider
Use current or voltage divider rules to calculate the indicated circuit variables in Figure 3
...

Problem 3
...
58)
...
7: Detective Work
In the depicted circuit (Figure 3
...

78

CHAPTER 3
...
55

1/2
vin

+
–

i 1/2
1

1/2

iin

Figure 3
...

b) With is = 2, determine R such that i1 = −1
...
8: Bridge Circuits
Circuits having the form of Figure 3
...

a) What resistance does the current source see when nothing is connected to the output terminals?
b) What resistor values, if any, will result in a zero voltage for vout ?
c) Assume R1 = 1Ω, R2 = 2Ω, R3 = 2Ω and R4 = 4Ω
...
Express your answer as a sinusoid
...
9: Cartesian to Polar Conversion
Convert the following expressions into polar form
...

√
2
a) 1 + −3
4
b) 3 + j
c)

6
2 − j √3

d) 4 − j 3
35 “The

6
2 + j √3

1+j1
2

Complex Plane” ...
57

π
3

2
1

1
1
...
58

π

e) 3ejπ + 4ej 2
√
√
π
f)
3 + j 2 2e−j 4
3
g) 1+j3π

–

–
1

80

CHAPTER 3
...
59

i

R1
+

iin

R3

vout

R2

R4

Figure 3
...
10: The Complex Plane
The complex variable z is related to the real variable u according to
z = 1 + eju
• Sketch the contour of values z takes on in the complex plane
...

z+1
Problem 3
...
What geometric shapes do the following
trace in the complex plane?
a)
b)
c)
d)

ejx
1 + ejx
e−x ejx
π
ejx + ej (x+ 4 )

Problem 3
...
In many cases, they were derived
using this approach
...
61

Problem 3
...
61
...

Problem 3
...
62
...
15: Measurement Chaos
The following simple circuit (Figure 3
...
When the source was sin (2πf0 t), the current i (t) equaled 32 sin 2πf0 t + π and the voltage
4
1
v2 (t) = 3 sin (2πf0 t)
...

c) Construct these impedances from elementary circuit elements
...
16: Transfer Functions
In the following circuit (Figure 3
...

a) Find the transfer function between the source and the indicated output voltage
...

Problem 3
...
65)
...
ANALOG SIGNAL PROCESSING

R1
iout
vin

+

C

R1

iout

L

C

R2
L

iin

+

vin

–

R2

(a) circuit a

(b) circuit b

+

L1
iin

i

R

L2

iin

v

C

1

1

1
2

1

–
(c) circuit c

(d) circuit d

Figure 3
...
63

+
vin

+
–

1
2

4

vout

–
Figure 3
...
18: Circuit Design
a) Find the transfer function between the input and the output voltages for the circuits shown in Figure 3
...

83

1
2
iin

iout
1
2

1
1
2

Figure 3
...
66

b) At what frequency does the transfer function have a phase shift of zero? What is the circuit’s gain at
this frequency?
c) Speciﬁcations demand that this circuit have an output impedance (its equivalent impedance) less than
8Ω for frequencies above 1 kHz, the frequency at which the transfer function is maximum
...

Problem 3
...
Is the power dissipated by the equivalent resistor equal to the sum of the powers dissipated
by the actual resistors comprising the circuit? Let’s start with simple cases and build up to a complete proof
...
Show that the power dissipated by (R1
equals the sum of the powers dissipated by the component resistors
...
Show the same result for this combination
...

R2 )

Problem 3
...
67a represents a simple power transmission system
...
The transmission line consists of a long length of
copper wire and can be accurately described as a 50Ω resistor
...
Why does the generator need to generate more than 1,000 watts
of average power to meet this requirement?
b) Suppose the load is changed to that shown in Figure 3
...
Now how much power must the generator
produce to meet the same power requirement? Why is it more than it had to produce to meet the
requirement for the resistive load?
c) The load can be compensated to have a unity power factor(see Exercise 3
...
The compensation technique is to place a
circuit in parallel to the load circuit
...
ANALOG SIGNAL PROCESSING

IL
+

Rs

RT
100

Vg

100

power generator

lossy power
transmission line

1

load

(a) Simple power transmission system

(b) Modiﬁed load circuit

Figure 3
...
68

d) With this compensated circuit, how much power must the generator produce to deliver 1,000 watts
average power to the load?
Problem 3
...
68) shows a general model for power transmission
...
In most applications, the source
components are ﬁxed while there is some latitude in choosing the load
...
What choice of load impedance creates the largest load voltage? What is the largest load
voltage?
b) If we wanted the maximum current to pass through the load, what would we choose the load impedance
to be? What is this largest current?
c) What choice for the load impedance maximizes the average power dissipated in the load? What is
most power the generator can deliver?
note: One way to maximize a function of a complex variable is to write the expression in terms
of the variable’s real and imaginary parts, evaluate derivatives with respect to each, set both
derivatives to zero and solve the two equations simultaneously
...
22: Big is Beautiful
Sammy wants to choose speakers that produce very loud music
...
”
a) What does this mean in terms of the ampliﬁer’s equivalent circuit?
b) Any speaker Sammy attaches to the terminals can be well-modeled as a resistor
...
What choice would maximize the voltage across the
speakers?

85

i
vin

+
Resistors

+

v

Figure 3
...
What
values for the speaker resistor should be chosen to maximize the power delivered to the speaker?

Problem 3
...
The transmitter
signals will be added together by the channel
...
Each transmitter signal has the form
xi (t) = A sin (2πfi t) , 0 ≤ t ≤ T
where the amplitude is either zero or A and each transmitter uses its own frequency fi
...

The datarate is 10 Mbps
...

b) Find circuits that the receivers could employ to separate unwanted transmissions
...

c) Find the second transmitter’s frequency so that the receivers can suppress the unwanted transmission
by at least a factor of ten
...
24: Circuit Detective Work
In the lab, the open-circuit voltage measured across an unknown circuit’s terminals equals sin (t)
...

4
a) What is the Thévenin equivalent circuit?
b) What voltage will appear if we place a 1F capacitor across the terminals?

Problem 3
...
69
...

a) Sammy measures v = 10 V when a 1 Ω resistor is attached to the terminals
...
Who is correct and why?
b) When nothing is attached to the right-hand terminals, a voltage of v = 1 V is measured
...
What resistor circuit
would be consistent with this and the previous part?

86

CHAPTER 3
...
70

Problem 3
...
The test source vin (t)
equals sin (t) (Figure 3
...

We make the following measurements
...

π
4

...

Problem 3
...
A system is said to be time-invariant if delaying the input delays the output by the
same amount
...
Note
that both linear and nonlinear systems have this property
...

a) Show that if a circuit has ﬁxed circuit elements (their values don’t change over time), its input-output
relationship is time-invariant
...
What is its general form? Examine the derivative(s) of delayed signals
...
Consequently,
show that linear, time-varying systems do not have a transfer function
...
Find the transfer function of the linear,
time-invariant (LTI) one(s)
...
28: Long and Sleepless Nights
Sammy went to lab after a long, sleepless night, and constructed the circuit shown in Figure 3
...
He cannot
remember what the circuit, represented by the impedance Z, was
...

a) What is the Thévenin equivalent circuit seen by the impedance?
b) In searching his notes, Sammy ﬁnds that the circuit is to realize the transfer function
1
H (f ) =
j10πf + 2
Find the impedance Z as well as values for the other circuit elements
...
71

i(t)

vin

Z

+

+

1

–

vout
–

Figure 3
...
73

Problem 3
...
72) was given on a test
...

a) What is voltage vout (t)?
b) What is the impedance Z at the frequency of the source?

Problem 3
...
73) that has two terminals for attaching circuit elements
...
When no source is attached (open-circuited terminals), the voltage across the
4
terminals has the form A sin (4t + φ)
...
ANALOG SIGNAL PROCESSING

+
1/2
vin

+

ZL vout

4

–

–
Figure 3
...
75

Problem 3
...
When he attaches a 1Ω resistor to the circuit’s terminals, he measures
the voltage across the terminals to be 3 sin (t)
...

4
a) What voltage should he measure when he attaches nothing to the mystery circuit?
b) What voltage should Sammy measure if he doubled the size of the capacitor to 2 F and attached it to
the circuit?

Problem 3
...
74) has a transfer function between the output voltage and the source equal
to
−8π 2 f 2
H (f ) =
2 f 2 + 4 + j6πf
−8π
a) Sketch the magnitude and phase of the transfer function
...
Is your answer unique? If so, show it to be so;
if not, give another example
...
33: Analog “Hum” Rejection
“Hum” refers to corruption from wall socket power that frequently sneaks into circuits
...
We want to ﬁnd a circuit that will remove hum from
any signal
...
75) consisting of two
series impedances
...
The Rice engineer must decide between two circuits (Figure 3
...
Which of these will work?

89

C
C

L

L

Figure 3
...
77

1
iin

1
+ vout –
1

1

1

Figure 3
...

c) Sketch the magnitude of the resulting frequency response
...
34: An Interesting Circuit
a) For the circuit shown in Figure 3
...

b) What is the output voltage when the input has the form iin = 5 sin (2000πt)?
Problem 3
...
78)
...
If the source equals
1 + sin (t), what is the output voltage?

90

CHAPTER 3
...
79

1/3
vin

+
–

2

v

–

1/6
4

Figure 3
...
36: An Interesting and Useful Circuit
The depicted circuit (Figure 3
...

The portion of the circuit labeled “Oscilloscope” represents the scope’s input impedance
...
A probe is a device
to attach an oscilloscope to a circuit, and it has the indicated circuit inside it
...
What would be the eﬀect of the oscilloscope’s
input impedance on measured voltages?
b) Using the node method, ﬁnd the transfer function relating the indicated voltage to the source when
the probe is used
...

d) For a particular relationship among the element values, the transfer function is quite simple
...

e) The arrow through C1 indicates that its value can be varied
...
What is the impedance seen by the circuit being measured for this
special value?
Problem 3
...
80)
...

b) What is the impedance “seen” by the capacitor?

91

+

1
vroad

1

+
–

+

1

1

vcar

vroad

+

1

–

–

1

vcar

–

Figure 3
...
82

Problem 3
...
An ELEC 241 student wants to understand the
suspension system on his car
...
A well-designed suspension system will smooth out bumpy roads, reducing the car’s vertical motion
...
The student wants to ﬁnd a simple circuit that will model the car’s motion
...
81)
...

a) Which circuit would you pick? Why?
b) For the circuit you picked, what will be the amplitude of the car’s motion if the road has a displacement
given by vroad (t) = 1 + sin (2t)?
Problem 3
...
82)
...

b) Sketch the magnitude and phase of your transfer function
...

t
c) Find vout (t) when vin (t) = sin 2 + π
...
40: Fun in the Lab
You are given an un-openable box that has two terminals sticking out
...
You measure the voltage sin t + π across the terminals when nothing is connected to them and
4
√
the current 2 cos (t) when you place a wire across the terminals
...

92

CHAPTER 3
...
83

b) You attach a 1 H inductor across the terminals
...
41: Dependent Sources
Find the voltage vout in each of the depicted circuits (Figure 3
...

Problem 3
...
84
...
43: Op-Amp Circuit
The following circuit (Figure 3
...

a) What is the transfer function relating the complex amplitude of the output signal, the current Iout , to
the complex amplitude of the input, the voltage Vin ?
b) What equivalent circuit does the load resistor RL see?
c) Find the output current when vin = V0 e−t/τ
...
44: Why Op-Amps are Useful
The circuit (Figure 3
...

a) Find the transfer function relating the complex amplitude of the voltage vout (t) to the source
...

b) What is the load impedance appearing across the ﬁrst op-amp’s output?
c) Figure 3
...
Find the transfer function for this
op-amp circuit (Figure 3
...
45: Operational Ampliﬁers
Consider the depicted circuit (Figure 3
...

a) Find the transfer function relating the voltage vout (t) to the source
...
3 kΩ, C2 = 0
...
3 kΩ
...

93

+

+

+

—

vin
R1

v out

R2

–

–
(a) op-amp a

R1

R2
–

+

R3
+

+
V(1)
in –

+
–

V(2)
in

Vout

R4

–
(b) op-amp b

+

5
+

Vin

–

+
–

10

5

Vout
–

(c) op-amp c

1
+
vin

+
–

+

2

–

1
2

1
2

4

4

vout

–
(d) op-amp d

Figure 3
...
ANALOG SIGNAL PROCESSING

R
C

+

C
Vin

Iout

R

RL

Figure 3
...
86

1 µF
1 kΩ 10 nF
–

+
+

+

vin –

4
...
87

Problem 3
...
88

RF
R
–
C
Vin

+

R = 1 kΩ
RF = 1 kΩ
C = 80 nF
+

+
Vout

–

–
Figure 3
...
We want fl = 1 kHz and fh = 10 kHz
...
Label important amplitude and phase values
and the frequencies at which they occur
...
Specify component values
...
47: Pre-emphasis or De-emphasis?
In audio applications, prior to analog-to-digital conversion signals are passed through what is known as a
pre-emphasis circuit that leaves the low frequencies alone but provides increasing gain at increasingly
higher frequencies beyond some frequency f0
...
After pre-emphasis, digitization, conversion back to analog and de-emphasis,
the signal’s spectrum should be what it was
...
89) has been designed for pre-emphasis or de-emphasis (Samantha
can’t recall which)
...

b) What is the circuit’s output when the input voltage is sin (2πf t), with f = 4kHz?
c) What circuit could perform the opposite function to your answer for the ﬁrst part?
Problem 3
...
90)
...
ANALOG SIGNAL PROCESSING

R1
Rf

C1

R1

R

–
+

–
+

R2
Vin

+
–

V out

R2 C2
R

–
+

Figure 3
...
91

Problem 3
...
91)
...

b) If the input signal is the sinusoid sin (2πf0 t), what will the output be when f0 is larger than the ﬁlter’s
“cutoﬀ frequency?”

Problem 3
...
92)
...
The photodiode acts as a current source, producing
a current proportional to the light intensity falling upon it
...
Thus, the op-amp stage serves to boost the signal and to ﬁlter
out-of-band noise
...

b) What should the circuit realizing the feedback impedance Zf be so that the transducer acts as a 5 kHz
lowpass ﬁlter?

97

Zf
–
Vout

+

Figure 3
...
93

R2

R1= 1 kΩ
R2= 1 kΩ
C = 31
...
94

c) A clever engineer suggests an alternative circuit (Figure 3
...
Determine
whether the idea works or not
...
If not, show why it does not work
...
51: Reverse Engineering
The depicted circuit (Figure 3
...
They are
trying to keep its use secret; we, representing RU Electronics, have discovered the schematic and want to
ﬁgure out the intended application
...

a) Assuming the diode is a short-circuit (it has been removed from the circuit), what is the circuit’s
transfer function?
b) With the diode in place, what is the circuit’s output when the input voltage is sin (2πf0 t)?
c) What function might this circuit have?

98

CHAPTER 3
...
1 (p
...

Solution to Exercise 3
...
38)
KCL says that the sum of currents entering or leaving a node must be zero
...
Since no currents enter an entire
circuit, the sum of currents must be zero
...
Consequently, specifying n − 1 KCL
equations always speciﬁes the remaining one
...
3 (p
...

+R
Solution to Exercise 3
...
41)
The power consumed by the resistor R1 can be expressed as
(vin − vout ) iout =

R1

2 vin

2

(R1 + R2 )

Solution to Exercise 3
...
41)
R2
1
R1
2
2
vin 2 =
2 vin + +
2 vin
R1 + R2
(R1 + R2 )
(R1 + R2 )
Solution to Exercise 3
...
43)
Replacing the current source by a voltage source does not change the fact that the voltages are identical
...
This result does not depend on the resistor R1 , which means that
R2
we simply have a resistor (R2 ) across a voltage source
...

Solution to Exercise 3
...
44)
R2
2
Req = RR2
...
1
...
01
...
8 (p
...
For a series combination, the equivalent resistance is the sum of the resistances, which will be
larger than any component resistor’s value; for a parallel combination, the equivalent conductance is the sum
of the component conductances, which is larger than any component conductance
...

Solution to Exercise 3
...
47)
voc = R1R2 2 vin and isc = − vin (resistor R2 is shorted out in this case)
...

+R
Solution to Exercise 3
...
49)
ieq = R1R1 2 iin and Req = (R3 R1 + R2 )
...
11 (p
...
Consequently,

1
V ⇔
j2πf

V ej2πf t dt

Solution to Exercise 3
...
55)
For maximum power dissipation, the imaginary part of complex power should be zero
...

99
Solution to Exercise 3
...
55)
Pave = Vrms Irms cos (φ − θ)
...

Solution to Exercise 3
...
59)
The key notion is writing the imaginary part as the diﬀerence between a complex exponential and its complex
conjugate:
V ej2πf t − V ∗ e−j2πf t
(3
...

∗
As H (−f ) = H (f ) , the Superposition Principle says that the output to the imaginary part is
j2πf t
Im V H (f ) e

...

Solution to Exercise 3
...
64)
To ﬁnd the equivalent resistance, we need to ﬁnd the current ﬂowing through the voltage source
...
This
6
1
current equals e1 = 13 vin , making the total current through the voltage source (ﬂowing out of it) 11 vin
...

Solution to Exercise 3
...
65)
Not necessarily, especially if we desire individual knobs for adjusting the gain and the cutoﬀ frequency
...
17 (p
...

100

CHAPTER 3
...
1 Introduction to the Frequency Domain1
In developing ways of analyzing linear circuits, we invented the impedance method because it made solving
circuits easier
...

This notion, which also applies to all linear, time-invariant systems, describes how the circuit responds to a
sinusoidal input when we express it in terms of a complex exponential
...

The study of the frequency domain combines these two notions–a system’s sinusoidal response is easy to
ﬁnd and a linear system’s output to a sum of inputs is the sum of the individual outputs—to develop the
crucial idea of a signal’s spectrum
...
In fact, all signals can be expressed as a superposition of sinusoids
...
For example,
radio, television, and cellular telephones transmit over diﬀerent portions of the spectrum
...
3))
...

4
...
4), we showed that a square wave could be expressed as a superposition of
pulses
...
You would be right and in good company as well
...
They worked on what is now known as the Fourier series: representing any
periodic signal as a superposition of sinusoids
...
Rather, the Fourier
series begins our journey to appreciate how a signal can be described in either the time-domain or the
frequency-domain with no compromise
...
We want to show
that periodic signals, even those that have constant-valued segments like a square wave, can be expressed
1 This

content is available online at ...
10/>
...
org/content/m0042/2
...

3 http://www-groups
...
st-and
...
uk/∼history/Mathematicians/Euler
...
dcs
...
ac
...
html
5 http://www-groups
...
st-and
...
uk/∼history/Mathematicians/Fourier
...
FREQUENCY DOMAIN

102

as sum of harmonically related sine waves: sinusoids having frequencies that are integer multiples of
1
the fundamental frequency
...
The
k
complex Fourier series expresses the signal as a superposition of complex exponentials having frequencies T ,
k = {
...

∞

ck e j

s (t) =

2πkt
T

(4
...
The real and imaginary parts of the Fourier coeﬃcients ck are written in this
unusual way for convenience in deﬁning the classic Fourier series
...
The family of functions ej T
basis functions and form the foundation of the Fourier series
...
They depend on the
signal period T , and are indexed by k
...
Thus, it makes no diﬀerence if we have a time-domain or a frequency-domain
characterization of the signal
...
1
What is the complex Fourier series for a sinusoid?

(Solution on p
...
)

To ﬁnd the Fourier coeﬃcients, we note the orthogonality property

 T if k = l
T
2πkt
2πlt
ej T e−j T dt =
 0 if k = l
0

(4
...
Simply multiply each side of (4
...

1
T
1
c0 =
T

T

s (t) e−j

ck =

0

2πkt
T

dt
(4
...
1
Finding the Fourier series coeﬃcients for the square wave sqT (t) is very simple
...

(4
...
The ﬁnal expression
becomes
k
2
bk = − j2πk (−1) − 1

 2 if k odd
(4
...
,−3,−1,1,3,
...
6)

Consequently, the square wave equals a sum of complex exponentials, but only those having fre1
quencies equal to odd multiples of the fundamental frequency T
...
This index corresponds to the k-th harmonic of the signal’s period
...

Property 4
...

−k
This result follows from the integral that calculates the ck from the signal
...
Similarly,
Im [ck ] = − (Im [c−k ]): The imaginary parts of the Fourier coeﬃcients have odd symmetry
...
This kind of symmetry, ck = c∗ , is
−k
known as conjugate symmetry
...
2:
If s (−t) = s (t), which says the signal has even symmetry about the origin, c−k = ck
...

A real-valued Fourier expansion amounts to an expansion in terms of only cosines, which is the simplest
example of an even signal
...
3:
If s (−t) = −s (t), which says the signal has odd symmetry, c−k = −ck
...
The square wave is a great example of an
odd-symmetric signal
...
4:
j2πkτ
The spectral coeﬃcients for a periodic signal delayed by τ , s (t − τ ), are ck e− T , where ck denotes
the spectrum of s (t)
...
Note that the spectral
T
magnitude is unaﬀected
...

Proof:
1
T

T

s (t − τ ) e−j

2πkt
T

dt =

0

=

2πk(t+τ )
1 T −τ
s (t) e−j T dt
T −τ
2πkt
1 −j 2πkτ T −τ
T
e
s (t) e−j T dt
T
−τ

(4
...
Consequently, it should
T −τ
T
not matter how we integrate over a period, which means that −τ (·) dt = 0 (·) dt, and we have
our result
...
FREQUENCY DOMAIN

104

The complex Fourier series obeys Parseval’s Theorem, one of the most important results in signal
analysis
...

Theorem 4
...

∞

T

1
T

2

2

|ck |

s (t) dt =
0

(4
...
22)
...
1)
...
1: Periodic pulse signal
...
The complex Fourier spectrum of this signal
is given by
j2πk∆
1 ∆ − j2πkt
A
ck =
Ae T dt = −
e −( T ) − 1
T 0
j2πk
At this point, simplifying this expression requires knowing an interesting property
...

ck = Ae−

jπk∆
T

sin

πk∆
T

(4
...

The periodic pulse signal has neither even nor odd symmetry; consequently, no additional symmetry exists
in the spectrum
...

sin πk∆
T
|ck | = A
(4
...
The somewhat complicated
expression for the phase results because the sine term can be negative; magnitudes must be positive, leaving
the occasional negative values to be accounted for as a phase shift of π
...
2: The magnitude and phase of the periodic pulse sequence’s spectrum is shown for positivefrequency indices
...
2 and A = 1
...

∆
Comparing this term with that predicted from delaying a signal, a delay of 2 is present in our signal
...
Thus, our calculated spectrum is consistent with the properties
of the Fourier spectrum
...
2
(Solution on p
...
)
What is the value of c0 ? Recalling that this spectral coeﬃcient corresponds to the signal’s average
value, does your answer make sense?
The phase plot shown in Figure 4
...
10) suggests
...
We must realize that any integer multiple of 2π can be added to a phase
at each frequency without aﬀecting the value of the complex spectrum
...
The phase at index 5 is undeﬁned because the magnitude is zero in this example
...

In addition, we expect a shift of −π in the phase between indices 4 and 6
...
Because we can add 2π without aﬀecting the value of the spectrum
at index 6, the result is a slightly negative number as shown
...
In
phase calculations like those made in MATLAB, values are usually conﬁned to the range [−π, π) by adding
some (possibly negative) multiple of 2π to each phase value
...
FREQUENCY DOMAIN

106

4
...

∞

s (t) = a0 +

2πkt
T

ak cos
k=1

∞

+

bk sin
k=1

2πkt
T

(4
...
The Fourier coeﬃcients, ak and bk , express the real and imaginary parts respectively of the
spectrum while the coeﬃcients ck of the complex Fourier series express the spectrum as a magnitude and
phase
...
11) to the complex Fourier series (4
...

ck =

1
(ak − jbk )
2

Exercise 4
...

(Solution on p
...
)

Just as with the complex Fourier series, we can ﬁnd the Fourier coeﬃcients using the orthogonality
properties of sinusoids
...

T

sin
0
T

sin
0

T

cos
0

2πkt
T

2πkt
T

2πkt
T

sin

cos

cos

2πlt
T

2πlt
T

2πlt
T

dt =




dt = 0 , k ∈ Z l ∈ Z
T
2

dt =





if k = l and k = 0 and l = 0
0 if k = l or k = 0 = l







(4
...

sin (α) sin (β) =
cos (α) cos (β) =
sin (α) cos (β) =

1
2 (cos (α − β) − cos (α + β))
1
2 (cos (α + β) + cos (α − β))
1
2 (sin (α + β) + sin (α − β))

(4
...
Each term in
the sum can be integrated by noticing one of two important properties of sinusoids
...

• The integral of the square of a unit-amplitude sinusoid over a period T equals

T
2

...
The idea is that, because integration is linear, the integration will sift out all but
T
6 This

content is available online at ...
22/>
...

T

s (t) cos
0

2πlt
T

T

∞

2πlt
T

a0 cos

dt =
0

∞

bk

2πkt
T

sin
0

k=1

ak

cos
0

k=1

T

+

T

dt +

2πkt
T

cos

2πlt
T

dt
(4
...
If k = 0 = l, we obtain a0 T
...

1
T
2
ak =
T
2
bk =
T

T

s (t) dt

a0 =

0

T

s (t) cos
0
T

s (t) sin
0

2πkt
dt , k = 0
T
2πkt
dt
T

(4
...
4
The expression for a0 is referred to as the average value of s (t)
...
143
...
5
What is the Fourier series for a unit-amplitude square wave?

(Solution on p
...
)

Example 4
...


 sin 2πt if 0 ≤ t < T
T
2
s (t) =

0 if T ≤ t < T

(4
...
17)

Using our trigonometric identities turns our integral of a product of sinusoids into a sum of integrals
of individual sinusoids, which are much easier to evaluate
...
18)

CHAPTER 4
...
The average value, which corresponds to a0 , equals
of the cosine coeﬃcients are easy to ﬁnd, but yield the complicated result
ak =

2
− π k21
−1
0

k ∈ {2, 4,
...

The remainder

(4
...

4
...
A plot
of the Fourier coeﬃcients as a function of the frequency index, such as shown in Figure 4
...
The word “spectrum” implies
that the independent variable, here k, corresponds somehow to frequency
...
Thus, if we half-wave rectiﬁed a 1 kHz sinusoid, k = 1 corresponds
to 1 kHz, k = 2 to 2 kHz, etc
...
5

0

k

-0
...
5

0

k
0

2

4

6

8

10

Figure 4
...
The index indicates
the multiple of the fundamental frequency at which the signal has energy
...
15)) and the signal from the spectrum (composition)
...
A signal’s frequency
domain expression is its spectrum
...

A fundamental aspect of solving electrical engineering problems is whether the time or frequency domain
provides the most understanding of a signal’s properties and the simplest way of manipulating it
...
As a simple example, suppose
we want to know the (periodic) signal’s maximum value
...
To use a frequency domain approach would require us to ﬁnd the spectrum, form the signal from
the spectrum and calculate the maximum; we’re back in the time domain!
7 This

content is available online at ...
20/>
...
2

0
...
4: Power spectrum of a half-wave rectiﬁed sinusoid
...
A signal’s instantaneous power is deﬁned to be its
square
...
For a periodic
signal, the natural time interval is clearly its period; for non-periodic signals, a better choice would be entire
time or time from onset
...
We deﬁne the rms value of a periodic signal to be
1
T

rms (s) =

T

s2 (t) dt

(4
...
21)

s2 (t) dt
0

Exercise 4
...
143
...

1
power (s) =
T

∞

T

a0 +
0

ak cos
k=1

2πkt
T

∞

+

bk sin
k=1

2πkt
T

2

dt

The square inside the integral will contain all possible pairwise products
...
12) say that most of these cross-terms integrate to zero
...

∞
1
ak 2 + bk 2
(4
...
Furthermore,
the contribution of each term in the Fourier series toward representing the signal can be measured by its
contribution to the signal’s average power
...
The power spectrum, Ps (k), such as shown in Figure 4
...

Exercise 4
...
143
...
FREQUENCY DOMAIN

110

in the fundamental
...

Is this calculation most easily performed in the time or frequency domain?

4
...
2)
...

K

sK (t) = a0 +

ak cos
k=1

K

2πkt
T

+

bk sin
k=1

2πkt
T

(4
...
5 (Fourier Series spectrum of a half-wave rectiﬁed sine wave) shows how this sequence of signals
portrays the signal more accurately as more terms are added
...
When we use a K + 1-term series, the error—the diﬀerence
between the signal and the K + 1-term series—corresponds to the unused terms from the series
...
24)

To ﬁnd the rms error, we must square this expression and integrate it over a period
...
25)
k
k=K+1

Figure 4
...
In particular, the use of four
terms, as shown in the bottom plot of Figure 4
...
The Fourier series in this case converges
quickly to the signal
...
7 (Power spectrum and approximation error for a square wave) to see the power
spectrum and the rms approximation error for the square wave
...
Said another
way, the square-wave’s spectrum contains more power at higher frequencies than does the half-wave-rectiﬁed
sinusoid
...
If fact, after
99 terms of the square wave’s approximation, the error is bigger than 10 terms of the approximation for the
half-wave rectiﬁed sinusoid
...

Exercise 4
...

(Solution on p
...
)

More than just decaying slowly, Fourier series approximation shown in Figure 4
...
Although the square wave’s Fourier series requires
more terms for a given representation accuracy, when comparing plots it is not clear that the two are equal
...
As more terms are added
8 This

content is available online at ...
9/>
...
5

0

k

-0
...
5

0

k
0

2

4

6

8

10

(a)
1
K=0
0
...
5
0

t

1
K=2
0
...
5
0
0

0
...
5

2

t

(b)

Figure 4
...

The index indicates the multiple of the fundamental frequency at which the signal has energy
...
The dashed line is the actual signal, with the solid line showing the ﬁnite series
approximation to the indicated number of terms, K + 1
...
For
the Fourier series approximation for the half-wave rectiﬁed sinusoid (Figure 4
...
What is happening?
Consider this mathematical question intuitively: Can a discontinuous function, like the square wave, be
expressed as a sum, even an inﬁnite one, of continuous signals? One should at least be suspicious, and in
fact, it can’t be thus expressed
...
dcs
...
ac
...
html

CHAPTER 4
...
8
0
...
4
0
...
6: The rms error calculated according to (4
...
The error has been normalized by the rms value of the
signal
...
5

0

k
0

2

4

6

8

10

0

2

4

6

8

10

Relative rms error

1

0
...
7: The upper plot shows the power spectrum of the square wave, and the lower plot the rms
error of the ﬁnite-length Fourier series approximation to the square wave
...

presentation on 1807
...

The extraneous peaks in the square wave’s Fourier series never disappear; they are termed Gibbs’
phenomenon after the American physicist Josiah Willard Gibbs
...

Let’s return to the question of equality; how can the equal sign in the deﬁnition of the Fourier series be

113

1

K=1

0

t

-1
1

K=5

0

t

-1
1

K=11

0

t

-1
1

K=49

0

t

-1
Figure 4
...
The number of terms in the Fourier sum is indicated
in each plot, and the square wave is shown as a dashed line over two periods
...

However, mathematicians later in the nineteenth century showed that the rms error of the Fourier series was
always zero
...
A new deﬁnition of equality is mean-square equality: Two
signals are said to be equal in the mean square if rms (s1 − s2 ) = 0
...
The error diﬀers from zero only at isolated points—whenever the
periodic signal contains discontinuities—and equals about 9% of the size of the discontinuity
...
This eﬀect underlies the reason why deﬁning the
value of a discontinuous function, like we refrained from doing in deﬁning the step function (Section 2
...
4:
Unit Step), at its discontinuity is meaningless
...
The Fourier series value “at” the
discontinuity is the average of the values on either side of the jump
...
FREQUENCY DOMAIN

114

4
...
Refer to the Fundamental Model of Communication
(Figure 1
...
We have an information source, and want to construct
a transmitter that produces a signal x (t)
...
For example, we want to represent typed letters produced by an extremely good typist (a key is
struck every T seconds)
...

K

ck ej

x (t) =

2πkt
T

(4
...
An important aspect of
the spectrum is that each frequency component ck can be manipulated separately: Instead of ﬁnding the
Fourier spectrum from a time-domain speciﬁcation, let’s construct it in the frequency domain by selecting
the ck according to some rule that relates coeﬃcient values to the alphabet
...
Because of the Fourier spectrum’s properties (Property 4
...
103),
the spectrum must have conjugate symmetry
...

Assume we have N letters to encode: {a1 ,
...
One simple encoding rule could be to make a single
Fourier coeﬃcient be non-zero and all others zero for each letter
...
In this way, the nth harmonic of the frequency T is used to represent a letter
...
Another possibility is
T
to consider the binary representation of the letter’s index
...
We can use the pattern of zeros and ones to represent
directly which Fourier coeﬃcients we “turn on” (set equal to one) and which we “turn oﬀ
...
9
(Solution on p
...
)
Compare the bandwidth required for the direct encoding scheme (one nonzero Fourier coeﬃcient
for each letter) to the binary number scheme
...

Since both schemes represent information without loss – we can determine the typed letter uniquely
from the signal’s spectrum – both are viable
...
10
(Solution on p
...
)
Can you think of an information-encoding scheme that makes even more eﬃcient use of the spectrum? In particular, can we use only one Fourier coeﬃcient to represent N letters uniquely?
We can create an encoding scheme in the frequency domain (p
...
But,
as this information-encoding scheme stands, we can represent one letter for all time
...
We could change
the signal’s spectrum every T as each letter is typed
...
For the receiver (see the Fundamental Model
of Communication (Figure 1
...
Figure 4
...

In this Fourier-series encoding scheme, we have used the fact that spectral coeﬃcients can be independently speciﬁed and that they can be uniquely recovered from the time-domain signal over one “period
...
By understanding the Fourier series’ properties (in particular that coeﬃcients are determined only over a T -second interval, we can construct
10 This

content is available online at ...
17/>
...
org/content/m0065/latest/#complex>

11 “Complex

115

x(t)
2

1

0

0

T

2T

3T

t

-1

-2
Figure 4
...
” In this example, only the third and fourth harmonics are used, as shown by the spectral magnitudes corresponding
to each T -second interval plotted below the waveforms
...
This approach represents a simpliﬁcation of how modern modems represent text
that they transmit over telephone lines
...
7 Filtering Periodic Signals12
The Fourier series representation of a periodic signal makes it easy to determine how a linear, time-invariant
ﬁlter reshapes such signals in general
...
Because the Fourier series
represents a periodic signal as a linear combination of complex exponentials, we can exploit the superposition
property
...
Said mathematically,
2πkt
2πkt
k
k
if x (t) = ej T , then the output y (t) = H T ej T because f = T
...

∞
2πkt
k
y (t) =
ck H
ej T
(4
...
Its Fourier coeﬃcients equal
k
ck H T
...
The circuit modiﬁes the magnitude and phase of each Fourier coeﬃcient
...

Example 4
...
31: Magnitude and phase of the transfer function))
H (f ) =
12 This

1
1 + j2πf RC

content is available online at ...
10/>
...
28)

CHAPTER 4
...
2

0
...
2

fc: 100 Hz

0

0

10
20
Frequency (kHz)

0

0

10
20
Frequency (kHz)

1

fc: 10 kHz

0

0
10
20
Frequency (kHz)

1

Amplitude

1

fc: 1 kHz

0

0

1
Time (ms)

2

0

0

1
Time (ms)

2

0

0

1
Time (ms)

2

(b)

Figure 4
...
2), serves as the input to
T
an RC lowpass ﬁlter
...
The ﬁlter’s cutoﬀ frequency was set to
the various values indicated in the top row, which display the output signal’s spectrum and the ﬁlter’s
transfer function
...
(a) Periodic pulse signal (b) Top plots show the pulse signal’s spectrum for various
cutoﬀ frequencies
...

Figure 4
...
Note how the signal’s spectrum extends well above its fundamental frequency
...
As the cutoﬀ frequency decreases (center, then
left), the rounding becomes more prominent, with the leftmost waveform showing a small ripple
...
11
(Solution on p
...
)
What is the average value of each output waveform? The correct answer may surprise you
...
The simple RC ﬁlter used
here has a rather gradual frequency response, which means that higher harmonics are smoothly suppressed
...

More importantly, we have calculated the output of a circuit to a periodic input without writing,
much less solving, the diﬀerential equation governing the circuit’s behavior
...
Using Fourier series, we can calculate how any linear circuit
will respond to a periodic input
...
8 Derivation of the Fourier Transform13
Fourier series clearly open the frequency domain as an interesting and useful way of determining how circuits
and systems respond to periodic input signals
...
We need a deﬁnition for the Fourier spectrum of a signal,
periodic or not
...

Let sT (t) be a periodic signal having period T
...
We denote the spectrum for any assumed value of
the period by ck (T )
...
29)

−T
2

where we have used a symmetric placement of the integration interval about the origin for subsequent derivak
tional convenience
...
Deﬁne
T
2

sT (t) e−j2πf t dt

ST (f ) ≡ T ck (T ) =

(4
...
31)

As the period increases, the spectral lines become closer together, becoming a continuum
...
32)

−∞

with

∞

s (t) e−j2πf t dt

S (f ) =

(4
...
33)) converges
...
4
Let’s calculate the Fourier transform of the pulse signal (Section 2
...
5: Pulse), p (t)
...
10)
...
org/content/m0046/2
...

CHAPTER 4
...
2
T=1

0
0
...
11: The upper plot shows the magnitude of the Fourier series spectrum for the case of T = 1
with the Fourier transform of p (t) shown as a dashed line
...
2, and computed its Fourier series coeﬃcients
...
11 (Spectrum) shows how increasing the period does indeed lead to a continuum of coeﬃcients,
and that the Fourier transform does correspond to what the continuum becomes
...
Thus, the magnitude of the
pulse’s Fourier transform equals |∆sinc (πf ∆) |
...
The
direct Fourier transform (or simply the Fourier transform) calculates a signal’s frequency domain representation from its time-domain variant (4
...
The inverse Fourier transform (4
...
Rather than explicitly writing the required integral, we often
symbolically express these transform calculations as F (s) and F −1 (S), respectively
...
34)

−∞

F −1 (S) = s (t)
∞

S (f ) e+j2πf t df

=

(4
...

note: Recall that the Fourier series for a square wave gives a value for the signal at the discontinuities equal to the average value of the jump
...

Showing that you “get back to where you started” is diﬃcult from an analytic viewpoint, and we won’t try
here
...

119
Exercise 4
...
144
...

What is F (S (f ))? In other words, use the wrong exponent sign in evaluating the inverse Fourier
transform
...
1: Short Table of Fourier Transform Pairs and Table 4
...
Especially
important among these properties is Parseval’s Theorem, which states that power computed in either
domain equals the power in the other
...
36)

−∞

Of practical importance is the conjugate symmetry property: When s (t) is real-valued, the spectrum at
negative frequencies equals the complex conjugate of the spectrum at the corresponding positive frequencies
...

Exercise 4
...
144
...
We are transforming (in the nontechnical meaning of the word) a signal
from one representation to another
...
A signal’s time
and frequency domain representations are uniquely related to each other
...
We can deﬁne an
information carrying signal in either the time or frequency domains; it behooves the wise engineer to use the
simpler of the two
...

This situation mirrors what happens with complex amplitudes in circuits: As we reveal how communications
systems work and are designed, we will deﬁne signals entirely in the frequency domain without explicitly
ﬁnding their time domain variants
...
6) where we deﬁne
Fourier series coeﬃcients according to letter to be transmitted
...
For example, impedances depend on frequency and the time variable cannot appear
...
9) that ﬁnding a linear, time-invariant system’s output in the time domain can
be most easily calculated by determining the input signal’s spectrum, performing a simple calculation in the
frequency domain, and inverse transforming the result
...

The only diﬃculty in calculating the Fourier transform of any signal occurs when we have periodic signals
(in either domain)
...

CHAPTER 4
...
1

Time-Domain

Frequency Domain

Linearity

a1 s1 (t) + a2 s2 (t)

a1 S1 (f ) + a2 S2 (f )

Conjugate Symmetry

s (t) ∈ R

S (f ) = S (−f )

Even Symmetry

s (t) = s (−t)

S (f ) = S (−f )

Odd Symmetry

s (t) = − (s (−t))

S (f ) = − (S (−f ))

Scale Change

s (at)

1
|a| S

Time Delay

s (t − τ )

e−j2πf τ S (f )

Complex Modulation

ej2πf0 t s (t)

S (f − f0 )

Amplitude Modulation by Cosine

s (t) cos (2πf0 t)

Amplitude Modulation by Sine

s (t) sin (2πf0 t)

S(f −f0 )+S(f +f0 )
2
S(f −f0 )−S(f +f0 )
2j

Diﬀerentiation
Integration

d
dt s (t)
t
s (α) dα
−∞

Multiplication by t

ts (t)
∞
−∞

Area
Value at Origin
Parseval’s Theorem

∗

s (t) dt

s (0)
∞
−∞

2

|s (t) | dt

f
a

j2πf S (f )
1
j2πf S (f ) if S
1
d
−j2π df S (f )

(0) = 0

S (0)
∞
−∞
∞
−∞

S (f ) df
2

|S (f ) | df

Table 4
...
5
In communications, a very important operation on a signal s (t) is to amplitude modulate it
...
Its period is fc , and its only nonzero
Fourier coeﬃcients are c±1 = 1
...
Using Euler’s relation, the spectrum of the second term can be derived as
∞

S (f ) ej2πf t df cos (2πfc t)

s (t) cos (2πfc t) =
−∞

Using Euler’s relation for the cosine,
(s (t) cos (2πfc t)) =

(s (t) cos (2πfc t)) =

1
2
1
2

∞

S (f ) ej2π(f +fc )t df +
−∞
∞

S (f − fc ) ej2πf t df +
−∞
∞

(s (t) cos (2πfc t)) =
−∞

1
2
1
2

∞

S (f ) ej2π(f −fc )t df
−∞
∞

S (f + fc ) ej2πf t df
−∞

S (f − fc ) + S (f + fc ) j2πf t
e
df
2

Exploiting the uniqueness property of the Fourier transform, we have
F (s (t) cos (2πfc t)) =

S (f − fc ) + S (f + fc )
2

(4
...
The spectrum of the amplitude modulated signal is shown in Figure 4
...

S(f)

–W

S(f+fc)
–fc–W –fc –fc+W

f

W
X(f)

S(f–fc)
fc–W

fc fc+W

f

Figure 4
...
Its highest
frequency — the largest frequency containing power — is W Hz
...

2

Note how in this ﬁgure the signal s (t) is deﬁned in the frequency domain
...

Exercise 4
...
144
...
12?
Exercise 4
...
144
...

CHAPTER 4
...

Signals such as speech and the Dow Jones averages are baseband signals
...
Since x (t)’s spectrum is conﬁned to a frequency band
not close to the origin (we assume (fc
W )), we have a bandpass signal
...
Thus,
in this example, the bandwidth is 2W Hz
...

4
...
27))
...

Example 4
...
We have expressions for the input’s
spectrum and the system’s frequency response
...
38)

(4
...
40)

You won’t ﬁnd this Fourier transform in our table, and the required integral is diﬃcult to evaluate
as the expression stands
...
In particular, recall Euler’s relation for the sinusoidal term and note the
fact that multiplication by a complex exponential in the frequency domain amounts to a time delay
...

e−jπf ∆

sin (πf ∆)
πf

= e−jπf ∆
=

ejπf ∆ − e−jπf ∆
j2πf

1
1 − e−j2πf ∆
j2πf

(4
...
42)

The table of Fourier transform properties (Table 4
...

1
• Multiplication by j2πf means integration
...

14 This

content is available online at ...
18/>
...

t
1
• The inverse transform of the frequency response is RC e−( RC ) u (t)
...
Let’s
1
start with the product of j2πf (integration in the time domain) and the transfer function:
t
1
1
↔ 1 − e−( RC ) u (t)
j2πf 1 + j2πf RC

(4
...
Because of the Fourier transform’s linearity, we simply
subtract the results
...
44)

Note that in delaying the signal how we carefully included the unit step
...
Thus, the waveforms shown in the Filtering Periodic Signals
(Figure 4
...
We say that
the time constant of an exponentially decaying signal equals the time it takes to decrease by 1
e
of its original value
...

Exercise 4
...
144
...
41) in the order given
...
You should get the same answer
...
We essentially treated multiplication by these
1
factors as if they were transfer functions of some ﬁctitious circuit
...
We even implicitly interpreted the circuit’s
transfer function as the input’s spectrum! This approach to ﬁnding inverse transforms – breaking down a
complicated expression into products and sums of simple components – is the engineer’s way of breaking
down the problem into several subproblems that are much easier to solve and then gluing the results together
...

4
...
1 Transfer Functions
The notion of a transfer function applies well beyond linear circuits
...
Thus,
linear circuits are a special case of linear, time-invariant systems
...
At this point,
you may be concerned that this approach is glib, and rightly so
...

4
...
2 Commutative Transfer Functions
Another interesting notion arises from the commutative property of multiplication (exploited in an example
above (Example 4
...
Consider a
cascade of two linear, time-invariant systems
...
FREQUENCY DOMAIN

124

X (f ) H1 (f ) and it serves as the second system’s input, the cascade’s output spectrum is X (f ) H1 (f ) H2 (f )
...
Furthermore, the cascade acts like a single linear system,
having transfer function H1 (f ) H2 (f )
...
13)
...
Using the fact that op-amp circuits can be connected in cascade with the transfer
function equaling the product of its component’s transfer function (see this analog signal processing problem
(Problem 3
...
We now understand why op-amp implementations
of transfer functions are so important
...
10 Modeling the Speech Signal15
The information contained in the spoken word is conveyed by the speech signal
...
This
modeling eﬀort consists of ﬁnding a system’s description of how relatively unstructured signals, arising from
simple sources, are given structure by passing them through an interconnection of systems to yield speech
...
Because the fundamental equation of acoustics – the wave equation – applies here and
is linear, we can use linear systems in our model with a fair amount of accuracy
...
In many cases, the underlying mathematics
governed by the physics, biology, and/or chemistry of the problem are nonlinear, leaving linear systems
models as approximations
...

Figure 4
...
14 (Model of the Vocal
Tract) shows the model speech production system
...
We concentrate ﬁrst on the vowel production mechanism
...
To visualize this eﬀect, take a rubber band and hold it in front of your lips
...
” If held tautly and close together, blowing through the opening causes the sides of the
rubber band to vibrate
...
You can imagine what the airﬂow
is like on the opposite side of the rubber band or the vocal cords
...
The vocal cords respond to this
input by vibrating, which means the output of this system is some periodic function
...
17
(Solution on p
...
)
Note that the vocal cord system takes a constant input and produces a periodic airﬂow that corresponds to its output signal
...

Singers modify vocal cord tension to change the pitch to produce the desired musical note
...
Certainly in the case of speech and in many other cases
as well, it is the control input that carries information, impressing it on the system’s output
...
In singing, musicality is largely conveyed by pitch; in western
speech, pitch is much less important
...
However, the diﬀerence between a statement and a
question is frequently expressed by pitch changes
...
org/content/m0049/2
...

125

Nasal Cavity
Lips
Teeth
Tongue

Oral Cavity

Vocal Cords

Air Flow

Lungs

Figure 4
...
Air pressure produced by the lungs forces
air through the vocal cords that, when under tension, produce puﬀs of air that excite resonances in the
vocal and nasal cavities
...

For some consonants, the vocal cords vibrate just as in vowels
...
For others, the vocal cords do not produce a periodic output
...
The resulting output airﬂow is quite erratic, so much so that we describe it
as being noise
...

The vocal cords’ periodic output can be well described by the periodic pulse train pT (t) as shown in the
periodic pulse signal (Figure 4
...
The spectrum of this signal (4
...
The primary diﬀerence between adult male and female/prepubescent speech is pitch
...
After puberty,
the vocal cords of males undergo a physical change, which has the eﬀect of lowering their pitch frequency
to the range 80–160 Hz
...
FREQUENCY DOMAIN

126

neural
control
l(t)
Lungs

Vocal
Cords

neural
control
pT(t) Vocal
Tract

s(t)

Figure 4
...
The signals l (t), pT (t), and s (t) are the air
pressure provided by the lungs, the periodic pulse output provided by the vocal cords, and the speech
output respectively
...

Clearly, these come from the same source, but for modeling purposes we describe them separately since
they control diﬀerent aspects of the speech signal
...
This diﬀerence is also readily apparent in the speech signal itself
...
With this
simpliﬁcation, we collapse the vocal-cord-lung system as a simple source that produces the periodic pulse
signal (Figure 4
...
The sound pressure signal thus produced enters the mouth
behind the tongue, creates acoustic disturbances, and exits primarily through the lips and to some extent
through the nose
...
The physics governing the sound disturbances produced in the vocal tract and those of an organ
pipe are quite similar
...
It is these positions that are controlled by the brain to produce the vowel sounds
...
” Rounding the lips, spreading the teeth, and positioning the tongue toward the back
of the oral cavity produces the sound “oh
...
15 (Speech Spectrum)
...
Thus, speech signal processors would say that the sound “oh” has
a higher ﬁrst formant frequency than the sound “ee,” with F2 being much higher during “ee
...
” Rather than serving as a ﬁlter,
rejecting high or low frequencies, the vocal tract serves to shape the spectrum of the vocal cords
...
We know that
the output—the speech signal we utter and that is heard by others and ourselves—will also be periodic
...
15 (Speech Spectrum), where the periodicity is
quite apparent
...
18
(Solution on p
...
)
From the waveform plots shown in Figure 4
...

Since speech signals are periodic, speech has a Fourier series representation given by a linear circuit’s response
to a periodic signal (4
...
Because the acoustics of the vocal tract are linear, we know that the spectrum of
the output equals the product of the pitch signal’s spectrum and the vocal tract’s frequency response
...

S (f ) = PT (f ) HV (f )

(4
...
The Fourier series for the vocal cords’ output,
derived in this equation (p
...
46)
πk
and is plotted on the top in Figure 4
...
If we had, for example, a male speaker with about
a 110 Hz pitch (T ≈ 9
...
16(b) (voice spectrum)
...
5

Amplitude

“ee”

20

0

F1

F2 F3 F4F5
Frequency (Hz)

0
...
5

-20

5000

“ee”

0

0

0
...
01
0
...
02

-0
...
005

0
...
015
Time (s)

0
...
15: The ideal frequency response of the vocal tract as it produces the sounds “oh” and “ee”
are shown on the top left and top right, respectively
...
The bottom plots show speech waveforms
corresponding to these sounds
...
The
measured spectrum certainly demonstrates what are known as pitch lines, and we realize from our model
that they are due to the vocal cord’s periodic excitation of the vocal tract
...
The
model transfer function for the vocal tract makes the formants much more readily evident
...
19
(Solution on p
...
)
The Fourier series coeﬃcients for speech are related to the vocal tract’s transfer function only at
k
the frequencies T , k ∈ {1, 2,
...
9)
...
16 (voice spectrum) would change if the
pitch were twice as high (≈ 300Hz)
...
Engineers typically display how the speech spectrum changes over time
with what is known as a spectrogram (Section 5
...
17 (spectrogram)
...

The fundamental model for speech indicates how engineers use the physics underlying the signal generation process and exploit its structure to produce a systems model that suppresses the physics while
emphasizing how the signal is “constructed
...
We want to determine how to transmit and receive it
...
FREQUENCY DOMAIN

128

(a) pulse
50

Pitch Lines

Spectral Amplitude (dB)

40
30
20
10
0
-10
-20

0

1000

2000
3000
Frequency (Hz)

4000

5000

(b) voice spectrum

Figure 4
...
” The pitch lines corresponding to
harmonics of the pitch frequency are indicated
...
(b) The
vocal tract’s transfer function, HV (f ) and the speech spectrum
...
We see from Figure 4
...

Eﬀective speech transmission systems must be able to cope with signals having this bandwidth
...
2 kHz
...
In our sample utterance, the “ce” sound
in “Rice” contains most of its energy above 3
...
Try this yourself: Call a friend and determine if they
can distinguish between the words “six” and “ﬁx
...
Radio does
support this bandwidth (see more about AM and FM radio systems (Section 6
...

Eﬃcient speech transmission systems exploit the speech signal’s special structure: What makes speech
speech? You can conjure many signals that span the same frequencies as speech—car engine sounds, violin
music, dog barks—but don’t sound at all like speech
...
Speech signals
can be transmitted using less than 1 kbps because of its special structure
...
If you used a speech
transmission system to send a violin sound, it would arrive horribly distorted; speech transmitted the same

129

5000

Frequency (Hz)

4000

3000

2000

1000

0

0

0
...
4

ce

0
...
8

si

1

1
...
17: Displayed is the spectrogram of the author saying “Rice University
...
Below the spectrogram
is the time-domain speech signal, where the periodicities can be seen
...

Exploiting the special structure of speech requires going beyond the capabilities of analog signal processing
systems
...

Fundamentally, we need to do more than ﬁltering to determine the speech signal’s structure; we need to
manipulate signals in more ways than are possible with analog systems
...

4
...
1: Simple Fourier Series
Find the complex Fourier series representations of the following signals without explicitly calculating Fourier
integrals
...
org/content/m10350/2
...

CHAPTER 4
...
18)
...
18

Problem 4
...
19)
...
Hint:
How is this signal related to one for which you already have the series?
Problem 4
...
20)
...

Find the magnitude and phase of this transfer function
...
What is the Fourier series for the output voltage?
Use Matlab to ﬁnd the output’s waveform for the cases T = 0
...
What value of T delineates
the two kinds of results you found? The software in fourier2
...

e) Instead of the depicted circuit, the square wave is passed through a system that delays its input, which
applies a linear phase shift to the signal’s spectrum
...
Use the transfer function
4
of a delay to compute using Matlab the Fourier series of the output
...

Problem 4
...
To assess the quality of
an approximation, the most frequently used error measure is the mean-squared error
...
One convenient way of ﬁnding approximations
˜
for periodic signals is to truncate their Fourier series
...
e
...

131

s(t)
1

1

2

2

t

3

3

(a)

s(t)
1

1

t

(b)

1

s(t)

1

2

3

t

4

(c)

Figure 4
...
20

a) Find a frequency-domain expression for the approximation error when we use the truncated Fourier
series as the approximation
...
What
selection of terms minimizes the mean-squared error? Find an expression for the mean-squared error
resulting from your choice
...
FREQUENCY DOMAIN

132

c) Find the Fourier series for the depicted signal (Figure 4
...
Use Matlab to ﬁnd the truncated approximation and best approximation involving two terms
...

1

s(t)

1

2

t

Figure 4
...
5: Long, Hot Days
The daily temperature is a consequence of several eﬀects, one of them being the sun’s heating
...
The
plot (Figure 4
...

95
14
Temperature
85
13

80
75

Daylight
12

70
65

Daylight Hours

Average High Temperature

90

11

60
55

10

50
0

50

100

150

200
Day

250

300

350

Figure 4
...
The ﬁle temperature
...

a) Let the length of day serve as the sole input to a system having an output equal to the average daily
temperature
...
, c4 ) of the complex Fourier series for each signal
...

d) Because the harmonic distortion is small, let’s concentrate only on the ﬁrst harmonic
...
Characterize and interpret the structure of this model
...

f) Predict what the output would be if the model had no phase shift
...
6: Fourier Transform Pairs
Find the Fourier or inverse Fourier transform of the following
...
7: Duality in Fourier Transforms
“Duality” means that the Fourier transform and the inverse Fourier transform are very similar
...

a) Calculate the Fourier transform of the signal shown below (Figure 4
...

b) Calculate the inverse Fourier transform of the spectrum shown below (Figure 4
...

c) How are these answers related? What is the general relationship between the Fourier transform of s (t)
and the inverse transform of s (f )?

1

s(t)

1

S(f)

f

t

1

1
(a)

(b)

Figure 4
...
8: Spectra of Pulse Sequences
Pulse sequences occur often in digital communication and in other ﬁelds as well
...
24(a))
...
24(b))
...
24(c))
...

d) Using Matlab, plot the magnitudes of the three spectra
...

CHAPTER 4
...
24

T

t

T

t

Figure 4
...
9: Spectra of Digital Communication Signals
One way to represent bits with signals is shown in Figure 4
...
If the value of a bit is a “1,” it is represented
by a positive pulse of duration T
...

To represent a sequence of bits, the appropriately chosen pulses are placed one after the other
...
01010101
...

What is this signal’s bandwidth?
c) Suppose the bit sequence becomes “
...
” Now what is the bandwidth?
Problem 4
...

We want to derive an expression for the time-domain response of the ﬁlter to this input
...
Derive
2
an expression for the ﬁlter’s output to this pulse
...
26

c) The nature of this response should change as the relation between the square wave’s period and the
ﬁlter’s cutoﬀ frequency change
...
11: Mathematics with Circuits
Simple circuits can implement simple mathematical operations, such as integration and diﬀerentiation
...
For example, you could use an integrator in a car to determine distance traveled from
the speedometer
...

Problem 4
...
Sound travels at a relatively
slow speed and our brain uses the fact that sound will arrive at one ear before the other
...
26), a sound coming from the right arrives at the left ear τ seconds after it arrives at the right ear
...
In an attempt to
model what the brain might do, RU signal processors want to design an optimal system that delays each
ear’s signal by some amount then adds them together
...
The idea is to determine the delay values according to some criterion that is based on
what is measured by the two ears
...
How are
these maximum-power processing delays related to τ ?
Problem 4
...
For each of the following (Figure 4
...

The overall transfer function for the cascade (ﬁrst depicted system) is particularly interesting
...
FREQUENCY DOMAIN

136

x(t)

y(t)

H2(f)

H1(f)
(a) system a

x(t)

H1(f)
y(t)

x(t)
H2(f)

x(t)

(b) system b

x(t)

e(t)

y(t)

H1(f)

–
H2(f)
(c) system c

Figure 4
...
28

Problem 4
...
28)
...

Problem 4
...

c) Find an expression for the ﬁlter’s output
...
16: Reverberation
Reverberation corresponds to adding to a signal its delayed version
...

b) A music group known as the ROwls is having trouble selling its recordings
...
Thus, the ROwls’ audio would be separated into two parts (one less
than the frequency f0 , the other greater than f0 ), these would be delayed by τl and τh respectively,
and the resulting signals added
...

c) How does the magnitude of the system’s transfer function depend on the two delays?

Problem 4
...
Here, because of acoustic coupling between the ear
piece and microphone in the handset, what you hear is also sent to the person talking
...
Furthermore, the same problem
applies to you as well: The acoustic coupling occurs in her handset as well as yours
...

b) Find the transfer function between your voice and what the listener hears
...
What simple system
could null the echoes?

Problem 4
...
Typically, if the drug concentration in the patient’s intravenous line is Cd u (t), the
concentration in the patient’s blood stream is Cp (1 − e−at ) u (t)
...
What would the
patient’s drug concentration be if the delivered concentration were a ramp? More precisely, if it were
Cd tu (t)?
c) A clever doctor wants to have the ﬂexibility to slow down or speed up the patient’s drug concentration
...
How
should the delivered drug concentration signal be changed to achieve this concentration proﬁle?

Problem 4
...
Radar transmitters emit a signal that
bounces oﬀ any conducting object
...
In Doppler systems, the object’s speed along the direction of the radar
beam is the feature the design must extract
...

The measured return signal equals B cos (2π ((fc + ∆f) t + ϕ)), where the Doppler oﬀset frequency ∆f equals
10v, where v is the car’s velocity coming toward the transmitter
...

CHAPTER 4
...
How would you change your design, if at all, so that whether the car is going away or
toward the transmitter could be determined?
c) Suppose two objects traveling diﬀerent speeds provide returns
...
20: Demodulating an AM Signal
Let m (t) denote the signal that has been amplitude modulated
...
The
frequency fc is very large compared to the frequency content of the signal
...

a) The so-called coherent demodulator simply multiplies the signal x (t) by a sinusoid having the same
frequency as the carrier and lowpass ﬁlters the result
...

Assume the lowpass ﬁlter is ideal
...
Assuming that the sinusoid of the receiver has a phase φ, how does the output
depend on φ? What is the worst possible value for this phase?
c) The incoherent receiver is more commonly used because of the phase sensitivity problem inherent in
coherent reception
...
Analyze this receiver
...
21: Unusual Amplitude Modulation
We want to send a band-limited signal having the depicted spectrum (Figure 4
...
I
...
Diﬀerent suggests using the square-wave carrier shown below (Figure 4
...

Well, it is diﬀerent, but his friends wonder if any technique can demodulate it
...

Sketch the magnitude of X (f ), being careful to label important magnitudes and frequencies
...
B
...
One friend suggests modulating
x (t) with cos πt , another wants to try modulating with cos (πt) and the third thinks cos 3πt will
2
2
work
...

Which student comes closest to recovering the original signal? Why?

Problem 4
...

While sitting in ELEC 241 class, he falls asleep during a critical time when an AM receiver is being described
...
The
message signal is m (t); it has a bandwidth of W Hz and a magnitude less than 1 (|m (t) | < 1)
...
The instructor drew a diagram (Figure 4
...

a) What are the signals xc (t) and xs (t)?
b) What would you put in for the unknown systems that would guarantee that the ﬁnal output contained
the message regardless of the phase?
Hint: Think of a trigonometric identity that would prove useful
...
What is it?

139

S(f)
1

/4

f

1/4
(a)

1

1

3

t

(b)

Figure 4
...
30

Problem 4
...
The resident electrical engineer
decides that she can choose any carrier frequency and message bandwidth for the station
...
The jamming signal jam (t) is what is known as a sawtooth wave (depicted in the following
ﬁgure (Figure 4
...

jam(t)
A
…
–T

…
T
Figure 4
...

2T

t

CHAPTER 4
...

Problem 4
...
To transmit these two signals simultaneously, the transmitter
ﬁrst forms the sum signal s+ (t) = l (t) + r (t) and the diﬀerence signal s− (t) = l (t) − r (t)
...
The sum signal and the modulated diﬀerence signal are added, the
sum amplitude-modulated to the radio station’s carrier frequency fc , and transmitted
...
32)
...
32

a) What is the expression for the transmitted signal? Sketch its spectrum
...

c) What signal would be produced by a conventional coherent AM receiver that expects to receive a
standard AM signal conveying a message signal having bandwidth W ?

Problem 4
...
As shown (Figure 4
...
In detail
the two message signals m1 (t) and m2 (t) are bandlimited to W Hz and have maximal amplitudes equal to
1
...
The transmitted signal x (t) is given by

 A (1 + am (t)) sin (2πf t) if sin (2πf t) ≥ 0
1
c
c
x (t) =
 A (1 + am2 (t)) sin (2πfc t) if sin (2πfc t) < 0
In all cases, 0 < a < 1
...
You, as the patent examiner, must determine
whether the scheme meets its claims and is useful
...

What is the receiver for this scheme? It would yield both m1 (t) and m2 (t) from x (t)
...

Determine whether this scheme satisﬁes the design criteria, allowing you to grant the patent
...

141
Example Transmitter Waveform
1
...
5
0
-0
...
5

0

1

2

3

4

5
Time

6

7

8

9

10

Figure 4
...
26: A Radical Radio Idea
An ELEC 241 student has the bright idea of using a square wave instead of a sinusoid as an AM carrier
...
In other words, do some combinations of W and T prevent reception?
b) Assuming reception is possible, can standard radios receive this innovative AM transmission? If so,
show how a coherent receiver could demodulate it; if not, show how the coherent receiver’s output
would be corrupted
...

Problem 4
...

r (t) = A (1 + m (t)) cos (2π (fc + f0 ) t)
The message signal has a bandwidth of W Hz and a magnitude less than 1 (|m (t) | < 1)
...
Thus, “oﬀ-the-shelf” coherent
demodulators would assume the carrier frequency has fc Hz
...

a) Sketch the spectrum of the demodulated signal produced by a coherent demodulator tuned to fc Hz
...
28: Signal Scrambling
An excited inventor announces the discovery of a way of using analog technology to render music unlistenable
without knowing the secret recovery method
...
FREQUENCY DOMAIN

142

special periodic signal s (t) that is zero during half of its period, which renders the message unlistenable and
superﬁcially, at least, unrecoverable (Figure 4
...

1 s(t)

T
4

T
2

T

t

Figure 4
...
How would they recover the original message without having detailed knowledge of the
modulating signal?

143

Solutions to Exercises in Chapter 4
Solution to Exercise 4
...
102)
Because of Euler’s relation,
sin (2πf t) =

1 +j2πf t
1
e
− e−j2πf t
2j
2j

(4
...

Solution to Exercise 4
...
105)
c0 = A∆
...

T
Solution to Exercise 4
...
106)
Write the coeﬃcients of the complex Fourier series in Cartesian form as ck = Ak + jBk and substitute into
the expression for the complex Fourier series
...
Because the signal is realvalued, the coeﬃcients of the complex Fourier series have conjugate symmetry: c−k = ck ∗ or A−k = Ak and
B−k = −Bk
...
To obtain the classic Fourier series (4
...

Solution to Exercise 4
...
107)
The average of a set of numbers is the sum divided by the number of terms
...

Solution to Exercise 4
...
107)
2
We found that the complex Fourier series coeﬃcients are given by ck = jπk
...
The coeﬃcients of the sine terms are given by bk = − (2Im [ck )] so that

 4 if k odd
πk
bk =
 0 if k even
Thus, the Fourier series for the square wave is
sq (t) =
k∈{1,3,
...
48)

Solution to Exercise 4
...
109)
√
The rms value of a sinusoid equals its amplitude divided by 2
...

Solution to Exercise 4
...
109)
P∞
(ak 2 +bk 2 )
Total harmonic distortion equals k=21 2 +b1 2

...
However, the numerator equals the square of the signal’s rms value minus the power in
the average and the power in the ﬁrst harmonic
...
8 (p
...

2

CHAPTER 4
...
9 (p
...
Using a binary representation, we need log2 N
...
05 smaller bandwidth
...

Solution to Exercise 4
...
114)
We can use N diﬀerent amplitude values at only one frequency to represent the various letters
...
11 (p
...

Solution to Exercise 4
...
118)
∞

∞

S (f ) e−j2πf t df =

F (S (f )) =
−∞

S (f ) e+j2πf (−t) df = s (−t)
−∞

Solution to Exercise 4
...
119)
∞
∞
F (F (F (F (s (t))))) = s (t)
...
Therefore, two Fourier transforms applied to s (t) yields s (−t)
...

Solution to Exercise 4
...
121)
The signal is the inverse Fourier transform of the triangularly shaped spectrum, and equals s (t) =
2

W sin(πW t)
πW t
Solution to Exercise 4
...
121)
The result is most easily found in the spectrum’s formula: the power in the signal-related part of x (t) is half
the power of the signal s (t)
...
16 (p
...
Multiplying the frequency response
by 1 − e−j2πf ∆ means subtract from the original signal its time-delayed version
...
Subtracting from the undelayed signal
−t

−(t−∆)

1
1
yields RC e RC u (t)− RC e RC u (t − ∆)
...
Because the integral of a sum equals the
sum of the component integrals (integration is linear), we can consider each separately
...
The
integral is provided in the example (4
...

Solution to Exercise 4
...
124)
If the glottis were linear, a constant input (a zero-frequency sinusoid) should yield a constant output
...

Solution to Exercise 4
...
126)
In the bottom-left panel, the period is about 0
...
The bottom-right
panel has a period of about 0
...

Solution to Exercise 4
...
127)
Because males have a lower pitch frequency, the spacing between spectral lines is smaller
...
Doubling the pitch frequency to 300 Hz for Figure 4
...

Chapter 5

Digital Signal Processing
5
...
Digital signals are sequences, functions
deﬁned only for the integers
...
Sequences are fundamentally diﬀerent than continuous-time signals
...

Despite such fundamental diﬀerences, the theory underlying digital signal processing mirrors that for analog signals: Fourier transforms, linear ﬁltering, and linear systems parallel what previous chapters described
...
” We will discover that digital signal processing is not an
approximation to analog processing
...
The music stored on CDs, the speech sent over digital cellular telephones, and the video
carried by digital television all evidence that analog signals can be accurately converted to digital ones and
back again
...

This ﬂexibility has obvious appeal, and has been widely accepted in the marketplace
...
We will
also discover that digital systems enjoy an algorithmic advantage that contributes to rapid processing
speeds: Computations can be restructured in non-obvious ways to speed the processing
...
How do computers perform signal processing?

5
...
2
...

The modern deﬁnition of a computer is an electronic device that performs calculations on data, presenting
the results to humans or other computers in a variety of (hopefully useful) ways
...
org/content/m10781/2
...

a systems viewpoint for the moment, a system that produces its output as rapidly as the input arises is said to
be a real-time system
...
Clearly, we need real-time signal processing systems
...

3 This content is available online at ...
28/>
...
DIGITAL SIGNAL PROCESSING

CPU
Memory
I/O
Interface

Keyboard

CRT

Disks

Network

Figure 5
...

The generic computer contains input devices (keyboard, mouse, A/D (analog-to-digital) converter, etc
...
The computational unit
is the computer’s heart, and usually consists of a central processing unit (CPU), a memory, and an
input/output (I/O) interface
...

• A simple computer operates fundamentally in discrete time
...
This description belies
clock speed: When you say “I have a 1 GHz computer,” you mean that your computer takes 1 nanosecond to perform each step
...
Computational speed is expressed in units of millions of instructions/second (Mips)
...

• Computers perform integer (discrete-valued) computations
...
4 Each computer instruction that performs an elementary numeric calculation —
an addition, a multiplication, or a division — does so only for integers
...
How does
a computer deal with numbers that have digits to the right of the decimal point? This problem is
addressed by using the so-called ﬂoating-point representation of real numbers
...

5
...
2 Representing Numbers
Focusing on numbers, all numbers can represented by the positional notation system
...
The
quantity b is known as the base of the number system
...
, b − 1}

n=

(5
...
d0
...
This same
number in binary (base-2) equals 11001 (1 × 24 + 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20 ) and 19 in hexadecimal
4 An

example of a symbolic computation is sorting a list of names
...
For example, we could use stick ﬁgure counting or Roman numerals
...
Fractions between zero and one are represented the same way
...
, b − 1}

f=

(5
...
Complex numbers (Section 2
...

Humans use base-10, commonly assumed to be due to us having ten ﬁngers
...

d7 d6 d5 d4 d3 d2 d1 d0
unsigned 8-bit integer
s
s

d6 d5 d4 d3 d2 d1 d0
signed 8-bit integer
s

exponent
mantissa
floating point
Figure 5
...
The number of bytes
for the exponent and mantissa components of ﬂoating point numbers varies
...
To represent signed values, we tack on a special bit—the sign bit—to express the sign
...
A byte can therefore
represent an unsigned number ranging from 0 to 255
...
But a computer cannot represent
all possible real numbers
...
While a gigabyte of memory may seem to be a lot, it takes an inﬁnite number of bits
to represent π
...
Large integers can be represented by an ordered sequence of bytes
...
Thus, an unsigned
32-bit number can represent integers ranging between 0 and 232 − 1 (4,294,967,295), a number almost big
enough to enumerate every human in the world!6
Exercise 5
...
192
...

While this system represents integers well, how about numbers having nonzero digits to the right of the
decimal point? In other words, how are numbers that have fractional parts represented? For such numbers,
the binary representation system is used, but with a little more complexity
...

x = m2e
(5
...
The number zero is an exception to this rule, and it is
6 You

need one more bit to do that
...
This convention is known as the hidden-ones notation
...
DIGITAL SIGNAL PROCESSING

the only ﬂoating point number having a zero fraction
...

A computer’s representation of integers is either perfect or only approximate, the latter situation occurring
when the integer exceeds the range of numbers that a limited set of bytes can represent
...
Otherwise, we can only represent the number approximately,
not catastrophically in error as with integers
...
5 equals 0
...
8 However, the number 2
...
In single precision ﬂoating point
numbers, which require 32 bits (one byte for the exponent and the remaining 24 bits for the mantissa), the
number “2
...
600000079
...
This level of accuracy may not suﬃce in numerical calculations
...
The more bits used in the mantissa,
the greater the accuracy
...
Such inexact numbers have an inﬁnite binary representation
...

Exercise 5
...
192
...

So long as the integers aren’t too large, they can be represented exactly in a computer using the binary
positional notation
...
(This statement isn’t quite true; when does addition cause problems?)

5
...
3 Computer Arithmetic and Logic
The binary addition and multiplication tables are
0+0=0
0+1=1
1 + 1 = 10
1+0=1
(5
...
Computers use high and low voltage values to express a bit, and an array of such voltages express
numbers akin to positional notation
...

8 See

if you can ﬁnd this representation
...
The choice
of base deﬁnes which do and which don’t
...
333333
...

10 A carry means that a computation performed at a given position aﬀects other positions as well
...

9 Note

149
Exercise 5
...
192
...
Note the carries that might occur
...
A B, the AND of A and B, represents a statement
that both A and B must be true for the statement to be true
...
A B, the OR of
A and B, yields a value of truth if either is true
...

XOR, the exclusive or operator, equals the union of A B and A B
...
It laid the foundation for what we now
call Boolean algebra, which expresses as equations logical statements
...
This fact makes an
integer-based computational device much more powerful than might be apparent
...
3 The Sampling Theorem11
5
...
1 Analog-to-Digital Conversion
Because of the way computers are organized, signal must be represented by a ﬁnite number of bytes
...
12 Quite surprisingly, the Sampling Theorem allows us to quantize the time axis
without error for some signals
...
In contrast, no one
has found a way of performing the amplitude quantization step without introducing an unrecoverable error
...
Signals processed by digital computers must
be discrete-valued: their values must be proportional to the integers
...

5
...
2 The Sampling Theorem
Digital transmission of information and digital signal processing all require signals to ﬁrst be “acquired” by
a computer
...
Harold Nyquist, a Bell Laboratories engineer, ﬁrst derived
this result, known as the Sampling Theorem, in the 1920s
...
Claude
Shannon, also at Bell Laboratories, revived the result once computers were made public after World War II
...

Clearly, the value of the original signal at the sampling times is preserved; the issue is how the signal values
between the samples can be reconstructed since they are lost in the sampling process
...

The resulting signal, as shown in Figure 5
...
, −1, 0, 1,
...
org/content/m0050/2
...

assume that we do not use ﬂoating-point A/D converters
...
DIGITAL SIGNAL PROCESSING

s(t)

t

s(t)pTs(t)
∆

Ts
t

Figure 5
...

The waveform of an example signal is shown in the top plot and its sampled version in

For our purposes here, we center the periodic pulse signal about the origin so that its Fourier series
coeﬃcients are real (the signal is even)
...
5)

k=−∞

where
sin
ck =

πk∆
Ts

(5
...

To understand how signal values between the samples can be “ﬁlled” in, we need to calculate the sampled
signal’s spectrum
...
7)

k=−∞

Considering each term in the sum separately, we need to know the spectrum of the product of the complex
exponential and the signal
...

∞

s (t) e

j2πkt
Ts

∞

e−j2πf t dt =

−∞

s (t) e−j2π(f − Ts )t dt = S f −
k

−∞

k
Ts

(5
...
4 (aliasing))
...
9)

In general, the terms in this sum overlap each other in the frequency domain, rendering recovery of the
original signal impossible
...

151

S(f)

–W
Aliasing
c-1

c-2
– 2
Ts
c-2
– 2
Ts

– 1 –W
Ts
c-1

– 1
Ts

–W

f

W
X(f)
c0

1
Ts>
2W
c2

c1

W
X(f)
c0

1
Ts

2
Ts
c1

W

1
Ts

f
1
Ts<
2W
c2
2
Ts

f

Figure 5
...
If the
sampling interval Ts is chosen too large relative to the bandwidth W , aliasing will occur
...
Note that if the signal were not
bandlimited, the component spectra would always overlap
...
In this delightful case, we can recover the original signal by lowpass ﬁltering x (t)
with a ﬁlter having a cutoﬀ frequency equal to W Hz
...

Exercise 5
...
192
...
What is the eﬀect of this
parameter on our ability to recover a signal from its samples (assuming the Sampling Theorem’s
two conditions are met)?
1
The frequency 2Ts , known today as the Nyquist frequency and the Shannon sampling frequency,
corresponds to the highest frequency at which a signal can contain energy and remain compatible with
the Sampling Theorem
...

Such systems therefore vary the anti-aliasing ﬁlter’s cutoﬀ frequency as the sampling rate varies
...
They sample at high frequencies, 44
...
05 kHz in our example)
...

Exercise 5
...
192
...
For
1
1
simplicity consider only the spectral repetitions centered at − Ts , 0, Ts
...
5 and 4
...
DIGITAL SIGNAL PROCESSING
where the spectral lines go as the period decreases; some will move to the left and some to the
right
...
As
we narrow the pulse, making ∆ smaller and smaller, the nonzero values of the signal s (t) pTs (t) will simply
be s (nTs ), the signal’s samples
...
In these ways, the
sampling signal captures the sampled signal’s temporal variations in a way that leaves all the original signal’s
structure intact
...
6
(Solution on p
...
)
What is the simplest bandlimited signal? Using this signal, convince yourself that less than two
1
samples/period will not suﬃce to specify it
...
4 Amplitude Quantization13
The Sampling Theorem says that if we sample a bandlimited signal s (t) fast enough, it can be recovered
without error from its samples s (nTs ), n ∈ {
...
Sampling is only the ﬁrst phase of acquiring
data into a computer: Computational processing further requires that the samples be quantized: analog
values are converted into digital (Section 1
...
2: Digital Signals) form
...

Q[s(nTs)]
7
∆
6
5
4
3
2
1
0
–1

–0
...
5

1

s(nTs)

(a)
signal
1

1

sampled signal
7

0
...
5

5

0
...
25

2

-0
...
75
–1

amplitude-quantized
and sampled signal

0

-1

(b)

Figure 5
...
For example, all inputs having values lying between 0
...
75 are assigned the integer
value six and, upon conversion back to an analog value, they all become 0
...
The width of a single
quantization interval ∆ equals 22
...
First it is sampled, then amplitude-quantized to three bits
...
For example the two signal values between 0
...
75
become 0
...
This distortion is irreversible; it can be reduced (but not eliminated) by using more bits
in the A/D converter
...
org/content/m0051/2
...

153
A phenomenon reminiscent of the errors incurred in representing numbers on a computer prevents signal
amplitudes from being converted with no error into a binary number representation
...
Assuming we can scale the signal without
aﬀecting the information it expresses, we’ll deﬁne this range to be [−1, 1]
...
A B-bit converter produces one of the integers
0, 1,
...
Figure 5
...
We deﬁne a quantization interval to be the range of values assigned to the same
integer
...
25; in general, it is
2

...
7
(Solution on p
...
)
Recalling the plot of average daily highs in this frequency domain problem (Problem 4
...

Because values lying anywhere within a quantization interval are assigned the same value for computer
processing, the original amplitude value cannot be recovered without error
...
The integer 6 would be assigned to the amplitude 0
...

The error introduced by converting a signal from analog to digital form by sampling and amplitude quantization then back again would be half the quantization interval for each amplitude value
...
As we have ﬁxed the input-amplitude range,
B
the more bits available in the A/D converter, the smaller the quantization error
...
Assuming the signal is a
sinusoid, the signal power is the square of the rms amplitude: power (s) =
(Figure 5
...

2

=

1
2
...
6: A single quantization interval is shown, along with a typical signal’s value before amplitude
quantization s (nTs ) and after Q (s (nTs ))
...

Its width is ∆ and the quantization error is denoted by
...
To calculate the rms value, we must square the error and average it over the interval
...
10)

1
2

Since the quantization interval width for a B-bit converter equals

2
2B

= 2−((B−1)) , we ﬁnd that the signal-

154

CHAPTER 5
...
5 dB
2

(5
...
The constant
term 10 log10 1
...
76
...
8
(Solution on p
...
)
This derivation assumed the signal’s amplitude lay in the range [−1, 1]
...
9
(Solution on p
...
)
How many bits would be required in the A/D converter to ensure that the maximum amplitude
quantization error was less than 60 db smaller than the signal’s peak value?
Exercise 5
...
192
...
To what signal-to-noise ratio does this correspond?
Once we have acquired signals with an A/D converter, we can process them using digital hardware or
software
...
Why go to all the bother if the same
function can be accomplished using analog techniques? Knowing when digital processing excels and when it
does not is an important issue
...
5 Discrete-Time Signals and Systems14
Mathematically, analog signals are functions having as their independent variables continuous quantities,
such as space and time
...
As
with analog signals, we seek ways of decomposing discrete-time signals into simpler components
...
For symbolic-valued signals, the approach is diﬀerent: We develop a common
representation of all symbolic-valued signals so that we can embody the information they contain in a
uniﬁed way
...

5
...
1 Real- and Complex-valued Signals
A discrete-time signal is represented symbolically as s (n), where n = {
...

sn
1
…
n
…

Figure 5
...
Can you ﬁnd the formula for this
signal?
14 This

content is available online at ...
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...
We can delay a discrete-time signal by an integer just as with analog ones
...

5
...
2 Complex Exponentials
The most important signal is, of course, the complex exponential sequence
...
12)

Note that the frequency variable f is dimensionless and that adding an integer to the frequency of the
discrete-time complex exponential has no eﬀect on the signal’s value
...
13)

This derivation follows because the complex exponential evaluated at an integer multiple of 2π equals one
...

5
...
3 Sinusoids
Discrete-time sinusoids have the obvious form s (n) = A cos (2πf n + φ)
...
This choice of frequency
2 2
interval is arbitrary; we can also choose the frequency to lie in the interval [0, 1)
...

5
...
4 Unit Sample
The second-most important discrete-time signal is the unit sample, which is deﬁned to be
δ (n) =

1 n=0
0 otherwise

(5
...
8: The unit sample
...
7 (Cosine),
reveals that all signals consist of a sequence of delayed and scaled unit samples
...

∞

s (m) δ (n − m)

s (n) =
m=−∞

This kind of decomposition is unique to discrete-time signals, and will prove useful subsequently
...
15)

156

CHAPTER 5
...
5
...

u (n) =

1 n≥0
0 n<0

(5
...
5
...
We do
have real-valued discrete-time signals like the sinusoid, but we also have signals that denote the sequence of
characters typed on the keyboard
...
More
formally, each element of the symbolic-valued signal s (n) takes on one of the values {a1 ,
...
This technical terminology does not mean we restrict symbols to being members of the English or Greek alphabet
...
Whether controlled by software or not, discrete-time systems are
ultimately constructed from digital circuits, which consist entirely of analog circuit elements
...
Understanding how discrete-time and analog signals and systems intertwine is perhaps the main
goal of this course
...
5
...
Because of the role of software in discrete-time systems, many more diﬀerent systems can be
envisioned and “constructed” with programs than can be with analog signals
...
For such signals, systems can be easily produced in
software, with equivalent analog realizations diﬃcult, if not impossible, to design
...
6 Discrete-Time Fourier Transform (DTFT)15
The Fourier transform of the discrete-time signal s (n) is deﬁned to be
∞

s (n) e−j2πf n

S ej2πf =

(5
...
As should be expected, this deﬁnition is linear, with the transform of a
sum of signals equaling the sum of their transforms
...

Exercise 5
...
193
...
Derive this property from the deﬁnition of the DTFT
...
When the signal
2
1
is real-valued, we can further simplify our plotting chores by showing the spectrum only over 0, 2 ; the
spectrum at negative frequencies can be derived from positive-frequency spectral values
...
151)
corresponds to the discrete-time frequency 1
...
org/content/m10247/2
...

157
j2πn
n
The exponential in the DTFT at frequency 1 equals e−( 2 ) = e−(jπn) = (−1) , meaning that discrete-time
2
frequency equals analog frequency multiplied by the sampling interval

fD = fA Ts

(5
...
The aliasing ﬁgure (Figure 5
...
As the duration of each pulse in the periodic
sampling signal pTs (t) narrows, the amplitudes of the signal’s spectral repetitions, which are governed by
the Fourier series coeﬃcients (4
...
Examination of the periodic pulse
signal (Figure 4
...
Thus, to maintain a mathematically viable Sampling Theorem, the amplitude A must
Ts
1
increase as ∆ , becoming inﬁnitely large as the pulse duration decreases
...
1 · Ts and use ampliﬁers to rescale the signal
...
Thus, the Nyquist frequency 2Ts corresponds to the frequency 1
...
1
Let’s compute the discrete-time Fourier transform of the exponentially decaying sequence s (n) =
an u (n), where u (n) is the unit-step sequence
...
19)
−j2πf n

n=0

This sum is a special case of the geometric series
...
20)

Thus, as long as |a| < 1, we have our Fourier transform
...

S ej2πf =

1

|S ej2πf | =

(5
...
22)
2

(1 − a cos (2πf )) + a2 sin2 (2πf )
∠S ej2πf = −tan−1

a sin (2πf )
1 − a cos (2πf )

(5
...
Figure 5
...
We need only consider the spectrum between − 2 and 1
2
to unambiguously deﬁne it
...
10 (Spectra of exponential signals))
...
DIGITAL SIGNAL PROCESSING

|S(ej2πf)|

2

1

f
-2

-1

0

1

2

∠S(ej2πf)
45

-2

-1

1

2

f

-45

Angle (degrees)

Spectral Magnitude (dB)

Figure 5
...
5) is shown over the frequency range [-2,
2], clearly demonstrating the periodicity of all discrete-time spectra
...

20

a = 0
...
5

0
...
5
-10
90
45

a = –0
...
5
-90 a = 0
...
5

Figure 5
...
What is the apparent relationship
between the spectra for a = 0
...
5?

Example 5
...


 1 if 0 ≤ n ≤ N − 1
s (n) =

0 otherwise

(5
...

N −1

e−j2πf n

S ej2πf =

(5
...
26)

for all values of α
...
12
(Solution on p
...
)
Derive this formula for the ﬁnite geometric series sum
...

Applying this result yields (Figure 5
...
)
S ej2πf

1 − e−j2πf N
1 − e−j2πf
−jπf (N −1) sin (πf N )
= e
sin (πf )
=

The ratio of sine functions has the generic form of

sin(N x)
sin(x) ,

(5
...
Thus, our transform can be concisely expressed as S ej2πf = e−(jπf (N −1)) dsinc (πf )
...

Figure 5
...
Can you explain the rather complicated
appearance of the phase?

The inverse discrete-time Fourier transform is easily derived from the following relationship:

1
 1 if m = n
2
e−j2πf m ej2πf n df =
 0 if m = n
−1
2

= δ (m − n)

(5
...
DIGITAL SIGNAL PROCESSING

Therefore, we ﬁnd that
1
2

S e

j2πf

e

j2πf n

1
2

df

s (m) e−j2πf m ej2πf n df

=

1
−2

1
−2

=

m

1
2

e−j2πf (m−n) df

s (m)

(5
...
30)

j2πf

S e

s (n) =

e

j2πf n

df

−1
2

The properties of the discrete-time Fourier transform mirror those of the analog Fourier transform
...
One important common property is Parseval’s
Theorem
...
31)

−1
2

n=−∞

To show this important property, we simply substitute the Fourier transform expression into the frequencydomain expression for power
...
32)
df

1
−2

Using the orthogonality relation (5
...
8: Unit sample)
...
We term n s2 (n) the energy in the discrete-time
signal s (n) in spite of the fact that discrete-time signals don’t consume (or produce for that matter) energy
...

Exercise 5
...
193
...
How is the discrete-time signal energy related to
the total energy contained in s (t)? Assume the signal is bandlimited and that the sampling rate
was chosen appropriate to the Sampling Theorem’s conditions
...
7 Discrete Fourier Transforms (DFT)17
The discrete-time Fourier transform (and the continuous-time transform as well) can be evaluated when we
have an analytic expression for the signal
...
How then would you compute the spectrum? For
example, how did we compute a spectrogram such as the one shown in the speech signal example (Figure 4
...
org/content/m0506/latest/>
content is available online at ...
27/>
...
While in discrete-time we can exactly calculate spectra, for analog signals no similar exact spectrum
computation exists
...
Certainly discrete-time spectral analysis is
more ﬂexible than continuous-time spectral analysis
...
17) is a sum, which conceptually can be easily computed save for two
issues
...
The sum extends over the signal’s duration, which must be ﬁnite to compute the
signal’s spectrum
...

• Continuous frequency
...
Let’s compute the spectrum
k
at a few frequencies; the most obvious ones are the equally spaced ones f = K , k ∈ {0,
...

We thus deﬁne the discrete Fourier transform(DFT) to be
N −1

s (n) e−

S (k) =

j2πnk
K

, k ∈ {0,
...
33)

n=0
k

Here, S (k) is shorthand for S ej2π K
...
Note that you can think
about this computationally motivated choice as sampling the spectrum; more about this interpretation later
...

One way of answering this question is determining an inverse discrete Fourier transform formula: given S (k),
k = {0,
...
, N − 1}? Presumably, the formula will be of the form
j2πnk
K−1
s (n) = k=0 S (k) e K
...
34)

k=0 m=0

Note that the orthogonality relation we use so often has a diﬀerent character now
...
}
2πkm
2πkn
e−j K ej K =

0 otherwise
k=0

(5
...
We can express this result as
K l (δ (m − n − lK))
...
36)

l=−∞

The integers n and m both range over {0,
...
To have an inverse transform, we need the sum to be a
single unit sample for m, n in this range
...
If we evaluate
the spectrum at fewer frequencies than the signal’s duration, the term corresponding to m = n + K will
also appear for some values of m, n = {0,
...
This situation means that our prototype transform
equals s (n) + s (n + K) for some values of n
...
In this way, we can return from
the frequency domain we entered via the DFT
...
DIGITAL SIGNAL PROCESSING
Exercise 5
...
193
...
Given the sampling interpretation of the spectrum,
characterize this eﬀect a diﬀerent way
...
If we write out the
expression for the DFT as a set of linear equations,
s (0) + s (1) + · · · + s (N − 1) = S (0)
s (0) + s (1) e

−j 2π
K

+ · · · + s (N − 1) e−j

2π(N −1)
K

= S (1)

...

...
37)
= S (K − 1)

we have K equations in N unknowns if we want to ﬁnd the signal from its sampled spectrum
...
Our orthogonality relation essentially says that
if we have a suﬃcient number of equations (frequency samples), the resulting set of equations can indeed be
solved
...
The
discrete Fourier transform pair consists of
Discrete Fourier Transform Pair
N −1

S (k) =
s (n) =

s (n) e−j
n=0
N −1
1
S
N
k=0

2πnk
N

(5
...
8 DFT: Computational Complexity18
We now have a way of computing the spectrum for an arbitrary signal: The Discrete Fourier Transform
(DFT) (5
...
An issue
that never arises in analog “computation,” like that performed by a circuit, is how much work it takes to
perform the signal processing operation such as ﬁltering
...
The number of steps,
known as the complexity, becomes equivalent to how long the computation takes (how long must we wait
for an answer)
...
Thus, a procedure’s stated complexity says that the time taken
will be proportional to some function of the amount of data used in the computation and the amount
demanded
...
For each frequency we choose, we
must multiply each signal value by a complex number and add together the results
...
To add the results together, we must keep the real and imaginary parts separate
...
Consequently, each frequency requires 2N + 2 (N − 1) = 4N − 2 basic
computational steps
...

In complexity calculations, we only worry about what happens as the data lengths increase, and take the
dominant term—here the 4N 2 term—as reﬂecting how much work is involved in making the computation
...
This notation is read “order N -squared
...

18 This

content is available online at ...
11/>
...
15
(Solution on p
...
)
In making the complexity evaluation for the DFT, we assumed the data to be real
...
First of all, the spectra of such signals have conjugate symmetry, meaning that negative
frequency components (k = N + 1,
...
33)) can be computed from the
2
corresponding positive frequency components
...
9 Fast Fourier Transform (FFT)19
One wonders if the DFT can be computed faster: Does another computational procedure – an algorithm–
exist that can compute the same quantity, but more eﬃciently
...
Here, we have something
more dramatic in mind: Can the computations be restructured so that a smaller complexity results?
In 1965, IBM researcher Jim Cooley and Princeton faculty member John Tukey developed what is now
known as the Fast Fourier Transform (FFT)
...
Now when the length of data doubles, the spectral computational
time will not quadruple as with the DFT algorithm; instead, it approximately doubles
...
Surprisingly,
historical work has shown that Gauss20 in the early nineteenth century developed the same algorithm, but
did not publish it! After the FFT’s rediscovery, not only was the computation of a signal’s spectrum greatly
speeded, but also the added feature of algorithm meant that computations had ﬂexibility not available to
analog implementations
...
16
(Solution on p
...
)
Before developing the FFT, let’s try to appreciate the algorithm’s impact
...
We want to calculate a transform of a signal that is 10 times longer
...

To derive the FFT, we assume that the signal’s duration is a power of two: N = 2L
...

S (k) = s (0) + s (2) e−j
+ s (1) e−j


2πk
N

= s (0) + s (2) e

2π2k
N

+ · · · + s (N − 2) e−j

+ s (3) e−j
−j 2πk
N
2

2π(N −2)k
N

2π(2+1)k
N

+ · · · + s (N − 1) e−j

2π ( N −1)k
2
−j
N

2
+ · · · + s (N − 2) e


+ s (1) + s (3) e

−j 2πk
N
2

+ · · · + s (N − 1) e

−j

2π

( N −1)k
2
N
2

2π(N −2+1)k
N

(5
...
The ﬁrst one is a DFT of the even-numbered
2
elements, and the second of the odd-numbered elements
...
The half-length transforms are each evaluated at frequency
indices k = 0,
...
Normally, the number of frequency indices in a DFT calculation range between zero
and the transform length minus one
...
The FFT simply reuses the computations made in the
19 This

content is available online at ...
17/>
...
dcs
...
ac
...
html

164

CHAPTER 5
...
12: The initial decomposition of a length-8 DFT into the terms using even- and odd-indexed
inputs marks the ﬁrst phase of developing the FFT algorithm
...

j2πk

half-length transforms and combines them through additions and the multiplication by e− N , which is not
periodic over N , to rewrite the length-N DFT
...
12 (Length-8 DFT decomposition) illustrates this
2
decomposition
...
At this
point, the total complexity is still dominated by the half-length DFT calculations, but the proportionality
coeﬃcient has been reduced
...
Because N = 2L , each of the half-length transforms can be reduced to two quarter-length
transforms, each of these to two eighth-length ones, etc
...
This transform is quite simple, involving only additions
...
12 (Length-8 DFT decomposition))
...
Each pair
requires 6 additions and 4 multiplications, giving a total number of computations equaling 10 · N = 5N
...
Because the number of stages, the number of
times the length can be divided by two, equals log2 N , the complexity of the FFT is O (N log2 N )
...
Let’s look at the details of a length-8
DFT
...
13 (Butterﬂy), we ﬁrst decompose the DFT into two length-4 DFTs, with the
outputs added and subtracted together in pairs
...
13 (Butterﬂy) as the frequency index
goes from 0 through 7, we recycle values from the length-4 DFTs into the ﬁnal calculation because of the
periodicity of the DFT output
...
13 (Butterﬂy))
...
13: The basic computational element of the fast Fourier transform is the butterﬂy
...
Each butterﬂy requires
one complex multiplication and two complex additions
...
By further decomposing the length-4 DFTs into two length-2 DFTs and combining their
outputs, we arrive at the diagram summarizing the length-8 fast Fourier transform (Figure 5
...
Although
most of the complex multiplies are quite simple (multiplying by e−jπ means negating real and imaginary
parts), let’s count those for purposes of evaluating the complexity as full complex multiplies
...

2
Exercise 5
...
193
...
12 (Length-8 DFT
decomposition) aren’t quite the same
...
In number theory, the number of prime factors a given integer has measures how composite
it is
...
In over thirty years of Fourier transform algorithm development, the
original Cooley-Tukey algorithm is far and away the most frequently used
...

Exercise 5
...
193
...
10 Spectrograms21
We know how to acquire analog signals for digital processing (pre-ﬁltering (Section 5
...
3), and A/D conversion (Section 5
...
9)), let’s put these various components together to learn how the spectrogram shown in
Figure 5
...
10), is calculated
...
025 kHz and passed through a 16-bit A/D converter
...
1 kHz
...
The 11
...

Exercise 5
...
193
...
14 (speech spectrogram) the signal lasted a little over 1
...
How long
was the sampled signal (in terms of samples)? What was the datarate during the sampling process
21 This

content is available online at ...
19/>
...
DIGITAL SIGNAL PROCESSING

5000

Frequency (Hz)

4000

3000

2000

1000

0

0

0
...
4

ce

0
...
8

si

1

1
...
14

in bps (bits per second)? Assuming the computer storage is organized in terms of bytes (8-bit
quantities), how many bytes of computer memory does the speech consume?
The resulting discrete-time signal, shown in the bottom of Figure 5
...
To display these spectral changes, the long signal was sectioned into frames:
comparatively short, contiguous groups of samples
...
Each frame is not so long that signiﬁcant signal variations are retained within a
frame, but not so short that we lose the signal’s spectral character
...

An important detail emerges when we examine each framed signal (Figure 5
...
Rectangular))
...
A transform of such a segment reveals a curious oscillation in the spectrum, an artifact
directly related to this sharp amplitude change
...
This shaping is accomplished by multiplying the framed signal by the sequence w (n)
...
A much more graceful
1
window is the Hanning window; it has the cosine shape w (n) = 2 1 − cos 2πn
...
15
N
(Spectrogram Hanning vs
...
Considering the spectrum of the Hanning windowed frame, we ﬁnd that the oscillations resulting
from applying the rectangular window obscured a formant (the one located at a little more than half the

167

256
n

Hanning
Window

Rectangular
Window

FFT (512)

FFT (512)

f

f

Figure 5
...
Computing Figure 5
...
If a
rectangular window is applied (corresponding to extracting a frame from the signal), oscillations appear
in the spectrum (middle of bottom row)
...

n

n

Figure 5
...
Clearly, spectral information
extracted from the bottom plot could well miss important features present in the original
...

Exercise 5
...
193
...
Compare your answer with the length-2N transform of a length-N Hanning
window
...
DIGITAL SIGNAL PROCESSING

n

FFT

FFT

FFT

FFT

Log Spectral Magnitude

FFT

FFT

FFT

f

Figure 5
...
Frames were 256 samples long and a Hanning window was applied
with a half-frame overlap
...

we see that we have managed to amplitude-modulate the signal with the periodically repeated window
(Figure 5
...
To alleviate this problem, frames are overlapped (typically by half a frame
duration)
...

The speech signal, such as shown in the speech spectrogram (Figure 5
...
The spectra
of each of these is calculated, and displayed in spectrograms with frequency extending vertically, window
time location running horizontally, and spectral magnitude color-coded
...
17 (Hanning windows)
illustrates these computations
...
21
(Solution on p
...
)
Why the speciﬁc values of 256 for N and 512 for K? Another issue is how was the length-512
transform of each length-256 windowed frame computed?

5
...
In discrete-time signal processing, we are not limited by hardware considerations
but by what can be constructed in software
...
22
(Solution on p
...
)
One of the ﬁrst analog systems we described was the ampliﬁer (Section 2
...
2: Ampliﬁers)
...

What is the discrete-time implementation of an ampliﬁer? Is this especially hard or easy?
In fact, we will discover that frequency-domain implementation of systems, wherein we multiply the input
signal’s Fourier transform by a frequency response, is not only a viable alternative, but also a computationally
22 This

content is available online at ...
5/>
...
We begin with discussing the underlying mathematical structure of linear, shift-invariant
systems, and devise how software ﬁlters can be constructed
...
12 Discrete-Time Systems in the Time-Domain23
A discrete-time signal s (n) is delayed by n0 samples when we write s (n − n0 ), with n0 > 0
...
As opposed to analog delays (Section 2
...
3: Delay),
discrete-time delays can only be integer valued
...

Linear discrete-time systems have the superposition property
...
40)

A discrete-time system is called shift-invariant (analogous to time-invariant analog systems (p
...
If S (x (n)) = y (n), then a shift-invariant system has
the property
S (x (n − n0 )) = y (n − n0 )
(5
...

We want to concentrate on systems that are both linear and shift-invariant
...
Because we have no physical constraints
in “constructing” such systems, we need only a mathematical speciﬁcation
...
The corresponding discrete-time
speciﬁcation is the diﬀerence equation
...
42)

Here, the output signal y (n) is related to its past values y (n − l), l = {1,
...
The system’s characteristics are determined by the choices for the number
of coeﬃcients p and q and the coeﬃcients’ values {a1 ,
...
, bq }
...
42)
...
We have thus created the convention that a0 is always one
...
We simply express the diﬀerence equation by a program that calculates each output
from the previous output values, and the current and previous inputs
...
A MATLAB program that would
compute the ﬁrst 1000 values of the output has the form
for n=1:1000
y(n) = sum(a
...
*x(n:-1:n-q));
end
An important detail emerges when we consider making this program work; in fact, as written it has (at
least) two bugs
...
values we have not yet computed
...
To compute these values, we would need even earlier values, ad
inﬁnitum
...
org/content/m10251/2
...

170

CHAPTER 5
...
These values can be arbitrary, but the choice
does impact how the system responds to a given input
...
The reason lies in the deﬁnition of a linear system (Section 2
...
6: Linear Systems):
The only way that the output to a sum of signals can be the sum of the individual outputs occurs when the
initial conditions in each case are zero
...
23
(Solution on p
...
)
The initial condition issue resolves making sense of the diﬀerence equation for inputs that start at
some index
...

What is it? How can it be “ﬁxed?”
Example 5
...

y (n) = ay (n − 1) + bx (n)

(5
...
In more detail, let’s
compute this system’s output to a unit-sample input: x (n) = δ (n)
...

y (0) = ay (−1) + b

(5
...
Certainly, the diﬀerence equation would not
describe a linear system (Section 2
...
6: Linear Systems) if the input that is zero for all time did
not produce a zero output
...
For n > 0, the
input unit-sample is zero, which leaves us with the diﬀerence equation y (n) = ay (n − 1) , n > 0
...

y (n) = ay (n − 1) + bδ (n)
n

x (n)

y (n)

−1

0

0

0

1

b

1

0

ba

2

0

ba2

:

0

:

n

0

(5
...
1

Coeﬃcient values determine how the output behaves
...
The eﬀect of the parameter a is more complicated (Table 5
...
If it equals zero,
the output simply equals the input times the gain b
...
The reason for this
terminology is that the unit sample also known as the impulse (especially in analog situations), and
the system’s response to the “impulse” lasts forever
...
When a = 1, the output is a unit step
...
When a = −1, the output changes
sign forever, alternating between b and −b
...

171
x(n)

n

1

y(n)
a = 0
...
5, b = 1

1

n

4

n

y(n)
a = 1
...
18: The input to the simple example system, a unit sample, is shown at the top, with the
outputs for several system parameter values shown below
...
19: The plot shows the unit-sample response of a length-5 boxcar ﬁlter
...
Here, n might correspond to generation
...
If this multiple is less than one, the
population becomes extinct; if greater than one, the population ﬂourishes
...
Here, n indexes the times at
which compounding occurs (daily, monthly, etc
...
In signal processing applications, we typically require that
the output remain bounded for any input
...

Exercise 5
...
42),

(Solution on p
...
)

y (n) = a1 y (n − 1) + · · · + ap y (n − p) + b0 x (n) + b1 x (n − 1) + · · · + bq x (n − q)
does not involve terms like y (n + 1) or x (n + 1) on the equation’s right side
...
4
A somewhat diﬀerent system has no “a” coeﬃcients
...
46)

Because this system’s output depends only on current and previous input values, we need not
be concerned with initial conditions
...
DIGITAL SIGNAL PROCESSING
n = {0,
...
Such systems are said to be FIR(Finite Impulse
Response) because their unit sample responses have ﬁnite duration
...
19) shows that the unit-sample response is a pulse of width q and height 1
...
We’ll derive its
frequency response and develop its ﬁltering interpretation in the next section
...
Thus, the output equals the running average of input’s previous q values
...

5
...
We used
impedances to derive directly from the circuit’s structure the frequency response
...
We proceed as when we used impedances: let the
input be a complex exponential signal
...
These amplitude and phase
changes comprise the frequency response we seek
...

Note that this input occurs for all values of n
...
Assume the
output has a similar form: y (n) = Y ej2πf n
...
42), we have
Y ej2πf n = a1 Y ej2πf (n−1) + · · · + ap Y ej2πf (n−p) + b0 Xej2πf n + b1 Xej2πf (n−1) + · · · + bq Xej2πf (n−q) (5
...

We ﬁnd that any discrete-time system deﬁned by a diﬀerence equation has a transfer function given by
H ej2πf =

b0 + b1 e−j2πf + · · · + bq e−j2πqf
1 − a1 e−j2πf − · · · − ap e−j2πpf

(5
...

Y ej2πf = X ej2πf H ej2πf

(5
...
5
The frequency response of the simple IIR system (diﬀerence equation given in a previous example
(Example 5
...
50)
1 − ae−j2πf
This Fourier transform occurred in a previous example; the exponential signal spectrum (Figure 5
...

When the ﬁlter coeﬃcient a is positive, we have a lowpass ﬁlter; negative a results in a highpass
ﬁlter
...

24 This

content is available online at ...
14/>
...
6
The length-q boxcar ﬁlter (diﬀerence equation found in a previous example (Example 5
...
51)
q m=0
This expression amounts to the Fourier transform of the boxcar signal (Figure 5
...
There we
found that this frequency response has a magnitude equal to the absolute value of dsinc (πf ); see
the length-10 ﬁlter’s frequency response (Figure 5
...
We see that
boxcar ﬁlters–length-q signal averagers–have a lowpass behavior, having a cutoﬀ frequency of 1
...
25
(Solution on p
...
)
Suppose we multiply the boxcar ﬁlter’s coeﬃcients by a sinusoid: bm = 1 cos (2πf0 m) Use Fourier
q
transform properties to determine the transfer function
...
The ﬁlter’s order is given by the number p of denominator coeﬃcients
in the transfer function (if the system is IIR) or by the number q of numerator coeﬃcients if the ﬁlter is
FIR
...
By selecting the coeﬃcients and ﬁlter type, ﬁlters having virtually any frequency response
desired can be designed
...
In the next section, we
detail how analog signals can be ﬁltered by computers, oﬀering a much greater range of ﬁltering possibilities
than is possible with circuits
...
14 Filtering in the Frequency Domain25
Because we are interested in actual computations rather than analytic calculations, we must consider the
details of the discrete Fourier transform
...
Because frequency responses have an explicit frequency-domain speciﬁcation
(5
...
Finding this signal is quite easy
...
Because the input and output of
linear, shift-invariant systems are related to each other by Y ej2πf = H ej2πf X ej2πf , a unit-sample
input, which has X ej2πf = 1, results in the output’s Fourier transform equaling the system’s
transfer function
...
26
This statement is a very important result
...

(Solution on p
...
)

In the time-domain, the output for a unit-sample input is known as the system’s unit-sample response,
and is denoted by h (n)
...

h (n) ↔ H ej2πf

(5
...

H (k) = H ej2πk/N

, k = {0,
...
53)

Computing the inverse DFT yields a length-N signal no matter what the actual duration of the unitsample response might be
...
org/content/m10257/2
...

174

CHAPTER 5
...

If, however, the duration exceeds N , errors are encountered
...
By sampling in the frequency domain, we have the potential for
aliasing in the time domain (sampling in one domain, be it time or frequency, can result in aliasing in the
other) unless we sample fast enough
...
For FIR systems — they by deﬁnition have ﬁnite-duration unit sample
responses — the number of required DFT samples equals the unit-sample response’s duration: N ≥ q
...
27
(Solution on p
...
)
Derive the minimal DFT length for a length-q unit-sample response using the Sampling Theorem
...

Exercise 5
...
194
...
Note
that the corresponding question for IIR ﬁlters is far more diﬃcult to answer: Consider the example
(Example 5
...

For IIR systems, we cannot use the DFT to ﬁnd the system’s unit-sample response: aliasing of the unitsample response will always occur
...
Frequency-domain implementations are restricted to
FIR ﬁlters
...
Assume we have an input signal having duration Nx that we pass through a FIR
ﬁlter having a length-q + 1 unit-sample response
...
54)
This equation says that the output depends on current and past input values, with the input value q samples
previous deﬁning the extent of the ﬁlter’s memory of past input values
...
Thus, the output returns to zero
only after the last input value passes through the ﬁlter’s memory
...
Thus, the output
signal’s duration equals q + Nx
...
29
(Solution on p
...
)
In words, we express this result as “The output’s duration equals the input’s duration plus the
ﬁlter’s duration minus one
...

The main theme of this result is that a ﬁlter’s output extends longer than either its input or its unit-sample
response
...
Thus, the number of values at which we must evaluate the
frequency response’s DFT must be at least q + Nx and we must compute the same length DFT of the input
...
Frequency-domain ﬁltering, diagrammed in
Figure 5
...

175

x(n)

X(k)

Y(k)

DFT

y(n)
IDFT

H(k)
Figure 5
...
The
DFT’s length must be at least the sum of the input’s and unit-sample response’s duration minus one
...

Before detailing this procedure, let’s clarify why so many new issues arose in trying to develop a frequencydomain implementation of linear ﬁltering
...
The Fourier transforms in this result are discretetime Fourier transforms; for example, X ej2πf = n x (n) e−j2πf n
...
The reason why we had to “invent” the discrete Fourier transform (DFT) has the
same origin: The spectrum resulting from the discrete-time Fourier transform depends on the continuous
frequency variable f
...

note: Did you know that two kinds of inﬁnities can be meaningfully deﬁned? A countably
inﬁnite quantity means that it can be associated with a limiting process associated with integers
...
The number of rational numbers is
countably inﬁnite (the numerator and denominator correspond to locating the rational by row and
column; the total number so-located can be counted, voila!); the number of irrational numbers is
uncountably inﬁnite
...
The sampling interval here
1
is K for a length-K DFT: faster sampling to avoid aliasing thus requires a longer transform calculation
...
We simply extend the other two signals with zeros (zero-pad)
to compute their DFTs
...
7
Suppose we want to average daily stock prices taken over last year to yield a running weekly
average (average over ﬁve trading sessions)
...
19)), and the input’s duration is 253 (365 calendar days minus
weekend days and holidays)
...
Because we want to use the FFT, we are restricted to power-of-two
transform lengths
...
As
it turns out, 256 is a power of two (28 = 256), and this length just undershoots our required length
...

Figure 5
...
The MATLAB programs that compute the
ﬁltered output in the time and frequency domains are
Time Domain
h = [1 1 1 1 1]/5;
y = filter(h,1,[djia zeros(1,4)]);
Frequency Domain
h = [1 1 1 1 1]/5;

176

CHAPTER 5
...
21: The blue line shows the Dow Jones Industrial Average from 1997, and the red one the
length-5 boxcar-ﬁltered result that provides a running weekly of this market index
...

DJIA = fft(djia, 512);
H = fft(h, 512);
Y = H
...
To force it to produce a signal having the proper length, the program zero-pads the input
appropriately
...
The frequency domain result will have a small
imaginary component — largest value is 2
...
Because of the unfortunate misﬁt between signal lengths and
favored FFT lengths, the number of arithmetic operations in the time-domain implementation is
far less than those required by the frequency domain version: 514 versus 62,271
...

An interesting signal processing aspect of this example is demonstrated at the beginning and
end of the output
...
The ﬁlter “sees” these initial and ﬁnal values as the diﬀerence
equation passes over the input
...

5
...
For the time-domain, diﬀerence26 This

content is available online at ...
16/>
...
The frequency-domain approach requires three Fourier transforms, each requiring 5K (log2 K) computations for a length-K FFT, and the multiplication of two spectra
2
(6K computations)
...
Thus, we must
compare
(Nx + q) (2q + 1) ↔ 6 (Nx + q) + 5 (Nx + q) log2 (Nx + q)
Exact analytic evaluation of this comparison is quite diﬃcult (we have a transcendental equation to solve)
...

2q + 1 ↔ 6 + 5log2 (Nx + q)
With this manipulation, we are evaluating the number of computations per sample
...
However, for ﬁlter durations greater than about 10, as long as the
input is at least 10 samples, the frequency-domain approach is faster so long as the FFT’s power-of-two
constraint is advantageous
...
24), but
so far we have required the input to have limited duration (so that we could calculate its Fourier transform)
...
To section a signal means expressing it as a linear combination of length-Nx non-overlapping
“chunks
...

∞

∞

x (n − mNx ) ⇒ y (n) =

x (n) =
m=−∞

y (n − mNx )

(5
...
22, note that each ﬁltered section has a duration longer than the input
...

Computational considerations reveal a substantial advantage for a frequency-domain implementation
over a time-domain one
...
Thus, the number of computations for each output is 2q + 1
...
We need only compute two DFTs and
multiply them to ﬁlter a section
...
In addition, we must add the ﬁltered outputs
together; the number of terms to add corresponds to the excess duration of the output compared with the
input (q)
...
For even modest ﬁlter orders, the frequency-domain approach is much faster
...
30
(Solution on p
...
)
Show that as the section length increases, the frequency domain approach becomes increasingly
more eﬃcient
...
Once the ﬁlter is chosen, we should section so that the
required FFT length is precisely a power of two: Choose Nx so that Nx + q = 2L
...
24) with a frequency-domain
implementation requires some additional signal management not required by time-domain implementations
...
Frequencydomain approaches don’t operate on a sample-by-sample basis; instead, they operate on sections
...
Because they
generally take longer to produce an output section than the sampling interval duration, we must ﬁlter one
section while accepting into memory the next section to be ﬁltered
...
Buﬀering can also be used
in time-domain ﬁlters as well but isn’t required
...
DIGITAL SIGNAL PROCESSING

Sectioned Input

n

Filter

Filter

Filtered, Overlapped Sections

Output (Sum of Filtered Sections)
n

Figure 5
...
Each ﬁltered section is added to other outputs that overlap to create the
signal equivalent to having ﬁltered the entire input
...

Example 5
...
The
example shown in Figure 5
...
If it weren’t for the
overlaid sinusoid, discerning the sine wave in the signal is virtually impossible
...

A smart Rice engineer has selected a FIR ﬁlter having a unit-sample response corresponding a
1
period-17 sinusoid: h (n) = 17 1 − cos 2πn , n = {0,
...
Its frequency
17
response (determined by computing the discrete Fourier transform) is shown in Figure 5
...
To
apply, we can select the length of each section so that the frequency-domain ﬁltering approach
is maximally eﬃcient: Choose the section length Nx so that Nx + q is a power of two
...
Filtering with the diﬀerence equation
would require 33 computations per output while the frequency domain requires a little over 16; this
frequency-domain implementation is over twice as fast! Figure 5
...

We note that the noise has been dramatically reduced, with a sinusoid now clearly visible in the
ﬁltered output
...

Exercise 5
...
194
...
What is the source of this delay? Can it be removed?

179

1

h(n)

|H(ej2πf)|

Spectral Magnitude

0
...
5

Frequency

Figure 5
...
This ﬁlter functions as a lowpass ﬁlter having a cutoﬀ frequency of
about 0
...

5
...
3
...
24
...
Bandpass signals can also be ﬁltered digitally, but require a more complicated
system
...
Note that the input and output ﬁlters must be analog
ﬁlters; trying to operate without them can lead to potentially very inaccurate digitization
...
24: To process an analog signal digitally, the signal x (t) must be ﬁltered with an antialiasing ﬁlter (to ensure a bandlimited signal) before A/D conversion
...
The greater the number
of bits in the amplitude quantization portion Q [·] of the A/D converter, the greater the accuracy of the
entire system
...
The resulting output y (n) then drives a
D/A converter and a second anti-aliasing ﬁlter (having the same bandwidth as the ﬁrst one)
...

The sampling interval, which is determined by the analog signal’s bandwidth, thus determines how long our
program has to compute each output y (n)
...
42) is O (p + q)
...

The idea begins by computing the Fourier transform of a length-N portion of the input x (n), multiplying
it by the ﬁlter’s transfer function, and computing the inverse transform of the result
...
org/content/m0511/2
...

180

CHAPTER 5
...
Detailing the complexity, however, we have O (N logN ) for the two
transforms (computed using the FFT algorithm) and O (N ) for the multiplication by the transfer function,
which makes the total complexity O (N logN ) for N input values
...
The complexities of time-domain
and frequency-domain implementations depend on diﬀerent aspects of the ﬁltering: The time-domain implementation depends on the combined orders of the ﬁlter while the frequency-domain implementation depends
on the logarithm of the Fourier transform’s length
...
In the latter situations, it is
the FFT algorithm for computing the Fourier transforms that enables the superiority of frequency-domain
implementations
...
Filtering with a diﬀerence equation is straightforward, and the number of
computations that must be made for each output value is 2 (p + q)
...
32
(Solution on p
...
)
Derive this value for the number of computations for the general diﬀerence equation (5
...

5
...
1: Sampling and Filtering
The signal s (t) is bandlimited to 4 kHz
...

a) What sampling frequency (if any works) can be used to sample the result of passing s (t) through an
RC highpass ﬁlter with R = 10kΩ and C = 8nF?
b) What sampling frequency (if any works) can be used to sample the derivative of s (t)?
c) The signal s (t) has been modulated by an 8 kHz sinusoid having an unknown phase: the resulting
signal is s (t) sin (2πf0 t + φ), with f0 = 8kHz and φ =? Can the modulated signal be sampled so that
the original signal can be recovered from the modulated signal regardless of the phase value φ? If so,
show how and ﬁnd the smallest sampling rate that can be used; if not, show why not
...
2: Non-Standard Sampling
Using the properties of the Fourier series can ease ﬁnding a signal’s spectrum
...
If ck represents the signal’s Fourier series coeﬃcients,
what are the Fourier series coeﬃcients of s t − T ?
2
b) Find the Fourier series of the signal p (t) shown in Figure 5
...

1
c) Suppose this signal is used to sample a signal bandlimited to T Hz
...

d) Does aliasing occur? If so, can a change in sampling rate prevent aliasing; if not, show how the signal
can be recovered from these samples
...
3: A Diﬀerent Sampling Scheme
A signal processing engineer from Texas A&M claims to have developed an improved sampling scheme
...
26)
...
org/content/m10351/2
...

181

p(t)
A
…

∆

∆
T/2

…

∆
3T/2

t

T

2T

∆

∆

–A

Figure 5
...
27

p(t)
A
…

∆

∆

∆

∆

…

t
Ts
4

Ts 5Ts
4
Figure 5
...

b) Will this scheme work? If so, how should TS be related to the signal’s bandwidth? If not, why not?
Problem 5
...

a) What is the minimum sampling rate for this signal suggested by the Sampling Theorem?

182

CHAPTER 5
...
28

b) Because of the particular structure of this spectrum, one wonders whether a lower sampling rate could
be used
...

Problem 5
...

This statement of the Sampling Theorem can be taken to mean that all information about the original
signal can be extracted from the samples
...
In addition to the rms value of a signal, an important aspect of a signal is its peak value, which equals
max { |s (t) | }
...
If we sample it at precisely the Nyquist rate, how
accurately do the samples convey the sinusoid’s amplitude? In other words, ﬁnd the worst case example
...
Assume
the maximum voltage allowed by the converter is Vmax volts and that it quantizes amplitudes to
b bits
...
Assuming the converter rounds, how large is maximum
quantization error?
d) We can describe the quantization error as noise, with a power proportional to the square of the
maximum error
...

Problem 5
...
28)
...

b) Can this signal be used to sample a bandlimited signal having highest frequency W =

1
2T

?

Problem 5
...

Let’s assume we have a B-bit converter
...
The ﬁrst step taken by our simple converter is to represent
the number by a sequence of B pulses occurring at multiples of a time interval T
...
For a 4-bit
converter, the number 13 has the binary representation 1101 (1310 = 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 ) and
would be represented by the depicted pulse sequence
...
We’ll see why that is
...
29

This signal (Figure 5
...
We want to design the ﬁlter
and the parameters ∆ and T so that the output voltage at time 4T (for a 4-bit converter) is proportional
to the number
...
The
requirements are
• The voltage at time t = 4T should diminish by a factor of two the further the pulse occurs from this
time
...

• The 4-bit D/A converter must support a 10 kHz sampling rate
...
How do the converter’s parameters change with sampling rate and number of
bits in the converter?
Problem 5
...

n

a) (−1) s (n)
b) s (n) cos (2πf0 n)
s n
n even
2
c) x (n) =
0
n odd
d) ns (n)

Problem 5
...

π
4n

a) The discrete-time Fourier transform of s (n) =

cos2
0

b) The discrete-time Fourier transform of s (n) =

n n = {−2, −1, 0, 1, 2}
0 otherwise

c) The discrete-time Fourier transform of s (n) =

sin
0

d) The length-8 DFT of the previous signal
...
, 7}
otherwise

184

CHAPTER 5
...
10: Just Whistlin’
Sammy loves to whistle and decides to record and analyze his whistling in lab
...
To analyze the spectrum, he samples
his recorded whistle with a sampling interval of TS = 2
...
Sammy (wisely)
decides to analyze a few samples at a time, so he grabs 30 consecutive, but arbitrarily chosen, samples
...
, 29}

a) Did Sammy under- or over-sample his whistle?
b) What is the discrete-time Fourier transform of x (n) and how does it depend on θ?
c) How does the 32-point DFT of x (n) depend on θ?

Problem 5
...
The
key idea is that a sequence can be written as a weighted linear combination of unit samples
...

δ (n) =

1 n=0
0 otherwise

b) If h (n) denotes the unit-sample response—the output of a discrete-time linear, shift-invariant ﬁlter
to a unit-sample input—ﬁnd an expression for the output
...
If the
input has duration N , what is the duration of the ﬁlter’s output to this signal?
1
d) Let the ﬁlter be a boxcar averager: h (n) = q+1 for n = {0,
...
Let the input
be a pulse of unit height and duration N
...
12: A Digital Filter
A digital ﬁlter has the depicted (Figure 5
...

h(n)
2

1

–1

0

1

2

3

4

n

Figure 5
...

185
c) What is the ﬁlter’s output when the input is sin

πn
4

?

Problem 5
...

b) Find this ﬁlter’s transfer function
...
e
...

c) Suppose we take a sequence and stretch it out by a factor of three
...
, −1, 0, 1,
...
What is the ﬁlter’s output to this input? In
particular, what is the output at the indices where the input x (n) is intentionally zero? Now how
would you characterize this system?

Problem 5
...
The idea is to replace the derivative with a discrete-time approximation and
solve the resulting diﬀerential equation
...

a) What is the diﬀerence equation that must be solved to approximate the diﬀerential equation?
b) When x (t) = u (t), the unit step, what will be the simulated output?
c) Assuming x (t) is a sinusoid, how should the sampling interval T be chosen so that the approximation
works well?

Problem 5
...
The digital ﬁlter described
by the diﬀerence equation
y (n) = x (n) − x (n − 1)
resembles the derivative formula
...

a) What is this ﬁlter’s transfer function?

186

CHAPTER 5
...
31)?

x(n)
3
2
1
0

1

2

3

4

n
5

6

Figure 5
...
Under what conditions will the
d
ﬁlter act like a diﬀerentiator? In other words, when will y (n) be proportional to dt x (t) |t=nTs ?

Problem 5
...

a) What is the length-K DFT of length-N boxcar sequence, where N < K?
b) Consider the special case where K = 4
...

c) If we could use DFTs to perform linear ﬁltering, it should be true that the product of the input’s DFT
and the unit-sample response’s DFT equals the output’s DFT
...
The result
of part (b) would then be the ﬁlter’s output if we could implement the ﬁlter with length-4 DFTs
...
186)?
d) What would you need to change so that the product of the DFTs of the input and unit-sample response
in this case equaled the DFT of the ﬁltered output?

Problem 5
...
He will, of course, use the FFT algorithm,
but he is behind schedule and needs to get his results as quickly as possible
...
The issue is whether he can retrieve
the individual DFTs from the result or not
...
” Show that this approach is too simplistic
...
What approach will work? Hint: Use
the symmetry properties of the DFT
...
18: Discrete Cosine Transform (DCT)
The discrete cosine transform of a length-N sequence is deﬁned to be
N −1

Sc (k) =

2πnk
2N

s (n) cos
n=0

Note that the number of frequency terms is 2N − 1: k = {0,
...

a) Find the inverse DCT
...
You
could only send one, which one would you send?

Problem 5
...
20: Another Digital Filter
A digital ﬁlter is determined by the following diﬀerence equation
...

b) What is the ﬁlter’s transfer function? How would you characterize this ﬁlter (lowpass, highpass, special
purpose, something else)?
c) Find the ﬁlter’s output when the input is the sinusoid sin πn
...
Sammy measures the
output to be y (n) = δ (n) + δ (n − 1)
...
If not, why
not?

Problem 5
...

y (n) =

1
1
1
x (n) + x (n − 1) + x (n − 2)
4
2
4

a) What is the ﬁlter’s transfer function? How would you characterize it?
b) What is the ﬁlter’s output when the input equals cos πn ?
2
c) What is the ﬁlter’s output when the input is the depicted discrete-time square wave (Figure 5
...
DIGITAL SIGNAL PROCESSING

x(n)
1
…

…
n

–1

Figure 5
...
22: A Digital Filter in the Frequency Domain
We have a ﬁlter with the transfer function
H ej2πf = e−j2πf cos (2πf )
operating on the input signal x (n) = δ (n) − δ (n − 2) that yields the output y (n)
...
23: Digital Filters
A discrete-time system is governed by the diﬀerence equation
y (n) = y (n − 1) +

x (n) + x (n − 1)
2

a) Find the transfer function for this system
...
24: Digital Filtering
A digital ﬁlter has an input-output relationship expressed by the diﬀerence equation
y (n) =

x (n) + x (n − 1) + x (n − 2) + x (n − 3)
4

a) Plot the magnitude and phase of this ﬁlter’s transfer function
...
25: Detective Work
The signal x (n) equals δ (n) − δ (n − 1)
...

189
b) You are told that when x (n) served as the input to a linear FIR (ﬁnite impulse response) ﬁlter, the
output was y (n) = δ (n) − δ (n − 1) + 2δ (n − 2)
...

Problem 5
...
} when its input
x (n) equals a unit sample
...

b) Find the output when x (n) = cos (2πf0 n)
...
27: Time Reversal has Uses
A discrete-time system has transfer function H ej2πf
...
The time-reversed signal w (−n) is then passed through the system to yield the
time-reversed output y (−n)
...
28: Removing “Hum”
The slang word “hum” represents power line waveforms that creep into signals because of poor circuit construction
...
What we seek are
ﬁlters that can remove hum
...

a) Find ﬁlter coeﬃcients for the length-3 FIR ﬁlter that can remove a sinusoid having digital frequency
f0 from its input
...
In this way, not only can the fundamental
but also its ﬁrst few harmonics be removed
...

d) Find the diﬀerence equation that deﬁnes this ﬁlter
...
29: Digital AM Receiver
Thinking that digital implementations are always better, our clever engineer wants to design a digital AM
receiver
...
Assume in this problem that the carrier frequency is always a large even multiple of
the message signal’s bandwidth W
...

c) Assuming the channel adds white noise and that a b-bit A/D converter is used, what is the output’s
signal-to-noise ratio?

Problem 5
...

a) What answer should Samantha obtain?

190

CHAPTER 5
...
Does Sammy’s result mean that Samantha’s answer is wrong?
c) The homework problem says to lowpass-ﬁlter the sequence by multiplying its DFT by
H (k) =

1 k = {0, 1, 7}
0 otherwise

and then computing the inverse DFT
...
31: Stock Market Data Processing
Because a trading week lasts ﬁve days, stock markets frequently compute running averages each day over the
previous ﬁve trading days to smooth price ﬂuctuations
...

a) What is the diﬀerence equation governing the ﬁve-day averager for daily stock prices?
b) Design an eﬃcient FFT-based ﬁltering algorithm for the broker
...
32: Echoes
Echoes not only occur in canyons, but also in auditoriums and telephone circuits
...

a) Find the diﬀerence equation of the system that models the production of echoes
...
Suppose the duration of x (n) is 1,000 and that
1
a1 = 2 , n1 = 10, a2 = 1 , and n2 = 25
...

Because of the undecided vote, you must break the tie
...
In other
words, with the echoed signal as the input, what system’s output is the signal x (n)?

Problem 5
...
The ﬁlter is to operate on signals that have a 10 kHz bandwidth, and will serve as a lowpass ﬁlter
...

b) What sampling rate must be used and how many bits must be used in the A/D converter for the
acquired signal’s signal-to-noise ratio to be at least 60 dB? For this calculation, assume the signal is a
sinusoid
...
34: Signal Compression
Because of the slowness of the Internet, lossy signal compression becomes important if you want signals to
be received quickly
...
First of all, he would section the signal into length-N blocks, and compute its N -point DFT
...
The receiver would assemble the transmitted spectrum and compute the inverse DFT,
thus reconstituting an N -point block
...
How long should a section be in the proposed scheme so that the required number of
bits/sample is smaller than that nominally required?
c) Assuming that eﬀective compression can be achieved, would the proposed scheme yield satisfactory
results?

192

CHAPTER 5
...
1 (p
...
For b = 32, we have 2,147,483,647 and for b = 64,
we have 9,223,372,036,854,775,807 or about 9
...

Solution to Exercise 5
...
148)
In ﬂoating point, the number of bits in the exponent determines the largest and smallest representable
numbers
...
7 × 1038 (5
...
For
64-bit ﬂoating point, the largest number is about 109863
...
3 (p
...
We ﬁnd that 110012 + 1112 = 1000002 = 32
...
4 (p
...
Because we are only
concerned with the repetition centered about the origin, the pulse duration has no signiﬁcant eﬀect on
recovering a signal from its samples
...
5 (p
...
5
f

Figure 5
...
The dashed
lines correspond to the frequencies about which the spectral repetitions (due to sampling with Ts = 1)
occur
...

Solution to Exercise 5
...
152)
The simplest bandlimited signal is the sine wave
...
Reducing the sampling rate would result in fewer samples/period, and these samples would
appear to have arisen from a lower frequency sinusoid
...
7 (p
...
Thus, the high temperature’s amplitude
was quantized as a form of A/D conversion
...
8 (p
...
With an A/D range of [−A, A], the
A
quantization interval ∆ = 2A and the signal’s rms value (again assuming it is a sinusoid) is √2
...
9 (p
...
001 results in B = 10 bits
...
10 (p
...
5 = 97
...

Solution to Exercise 5
...
156)
∞

s (n) e−j2π(f +1)n

S ej2π(f +1) =
n=−∞
∞

e−j2πn s (n) e−j2πf n

=

(5
...
12 (p
...

Solution to Exercise 5
...
160)
If the sampling frequency exceeds the Nyquist frequency, the spectrum of the samples equals the analog
spectrum, but over the normalized analog frequency f T
...

Solution to Exercise 5
...
162)
This situation amounts to aliasing in the time-domain
...
15 (p
...
If the data are complex-valued, which demands retaining all frequency values, the complexity is
again the same
...

Solution to Exercise 5
...
163)
If a DFT required 1 ms to compute, and signal having ten times the duration would require 100ms to
compute
...

Solution to Exercise 5
...
165)
The upper panel has not used the FFT algorithm to compute the length-4 DFTs while the lower one has
...

Solution to Exercise 5
...
165)
The transform can have any greater than or equal to the actual duration of the signal
...
Recall that the
FFT is an algorithm to compute the DFT (Section 5
...
Extending the length of the signal this way merely
means we are sampling the frequency axis more ﬁnely than required
...
samples long
...
19 (p
...
2 × 11025 = 13230
...
4 kbps
...

Solution to Exercise 5
...
167)
The oscillations are due to the boxcar window’s Fourier transform, which equals the sinc function
...
21 (p
...
To compute
a longer transform than the input signal’s duration, we simply zero-pad the signal
...
DIGITAL SIGNAL PROCESSING

Solution to Exercise 5
...
168)
In discrete-time signal processing, an ampliﬁer amounts to a multiplication, a very easy operation to perform
...
23 (p
...
To ﬁx it, we must start the
signals later in the array
...
24 (p
...
Thus, such terms can cause diﬃculties
...
25 (p
...

Solution to Exercise 5
...
173)
The DTFT of the unit sample equals a constant (equaling 1)
...

Solution to Exercise 5
...
174)
In sampling a discrete-time signal’s Fourier transform L times equally over [0, 2π) to form the DFT, the
corresponding signal equals the periodic repetition of the original signal
...
57)

i=−∞

To avoid aliasing (in the time domain), the transform length must equal or exceed the signal’s duration
...
28 (p
...
58)

(bm δ (n − m))

y (n) =

(5
...
6) of a length-q boxcar ﬁlter
...
29 (p
...
Thus the statement is correct
...
30 (p
...
The time-domain implementation requires a total of N (2q + 1)
N
computations, or 2q + 1 computations per input value
...
Because we divide again
by Nx to ﬁnd the number of computations per input value in the entire input, this quantity decreases as
Nx increases
...

Solution to Exercise 5
...
178)
The delay is not computational delay here–the plot shows the ﬁrst output value is aligned with the ﬁlter’s ﬁrst
input–although in real systems this is an important consideration
...

All ﬁlters have phase shifts
...
Such ﬁlters
do not exist in analog form, but digital ones can be programmed, but not in real time
...
32 (p
...
Thus, the total number of arithmetic operations
equals 2 (p + q)
...
1 Information Communication1
As far as a communications engineer is concerned, signals express information
...
Information comes neatly packaged in both analog and
digital forms
...
Communication systems endeavor not to manipulate information, but to transmit it from one
place to another, so-called point-to-point communication, from one place to many others, broadcast
communication, or from many to many, like a telephone conference call or a chat room
...

This chapter develops a common theory that underlies how such systems work
...
The question as to which is
better, analog or digital communication, has been answered, because of Claude Shannon’s fundamental work
on a theory of information published in 1948, the development of cheap, high-performance computers, and
the creation of high-bandwidth communication systems
...
In most cases, you should convert all information-bearing signals into discrete-time, amplitudequantized signals
...
Because of the Sampling Theorem, we know how to convert analog signals
into digital ones
...
This startling result has no counterpart in analog systems; AM radio
will remain noisy
...
The audio compact disc (CD) and the digital videodisk
(DVD) are now considered digital communications systems, with communication design considerations used
throughout
...
3: Fundamental model of communication)
...

1
...
What are the channel’s characteristics and how do they aﬀect the transmitted signal?
In short, what are we going to send and how are we going to send it? Interestingly, digital as well as
analog transmission are accomplished using analog signals, like voltages in Ethernet (an example of wireline
communications) and electromagnetic radiation (wireless) in cellular telephone
...
org/content/m0513/2
...

195

196

CHAPTER 6
...
2 Types of Communication Channels2
Electrical communications channels are either wireline or wireless channels
...

Consequently, wireline channels are more private and much less prone to interference
...

Listening in on a conversation requires that the wire be tapped and the voltage measured
...
One simple
example of this situation is cable television
...
Wireless channels are much more public, with a transmitter’s antenna radiating
a signal that can be received by any antenna suﬃciently close enough
...
This feature has two faces: The smiley face says that a receiver can take
in transmissions from any source, letting receiver electronics select wanted signals and disregarding others,
thereby allowing portable transmission and reception, while the frowny face says that interference and noise
are much more prevalent than in wireline situations
...

note: You will hear the term tether-less networking applied to completely wireless computer
networks
...

×E=−

∂
(µH)
∂t

div ( E) = ρ
× H = σE +

∂
( E)
∂t

(6
...
Kirchhoﬀ’s Laws represent special cases of these equations
for circuits
...
Perhaps
the most important aspect of them is that they are linear with respect to the electrical and magnetic ﬁelds
...

note: Nonlinear electromagnetic media do exist
...
Nonlinear media are becoming increasingly important in optic ﬁber communications, which are also governed by Maxwell’s
equations
...
3 Wireline Channels3
Wireline channels were the ﬁrst used for electrical communications in the mid-nineteenth century for the
telegraph
...
The transmitter
simply creates a voltage related to the message signal and applies it to the wire(s)
...
In the case of single-wire communications, the earth is used as the
current’s return path
...
You can imagine that the earth’s electrical characteristics are highly variable, and they are
...

2 This
3 This

content is available online at ...
13/>
...
org/content/m0100/2
...

197

insulation
σ
σ

rd

dielectric

ri

central
conductor
outer
conductor

σd,εd,µd

Figure 6
...
This
type of cable supports broader bandwidth signals than twisted pair, and ﬁnds use in cable television and
Ethernet
...
2: The so-called distributed parameter model for two-wire cables has the depicted circuit
model structure
...

Consequently, most wireline channels today essentially consist of pairs of conducting wires (see Figure 6
...
How
these pairs of wires are physically conﬁgured greatly aﬀects their transmission characteristics
...
Telephone cables are one example of a
twisted pair channel
...
Coaxial cable, fondly called “co-ax” by engineers, is what Ethernet uses as
its channel
...
As
we shall ﬁnd subsequently, several transmissions can share the circuit by amplitude modulation techniques;
commercial cable TV is an example
...
Thus, by the time signals arrive at the receiver, they are
relatively interference- and noise-free
...
2 (Circuit Model for a Transmission Line) for an inﬁnitesimally small length
...
The series resistance
comes from the conductor used in the wires and from the conductor’s geometry
...
Note that all the circuit elements have values expressed by the product of a constant times a
length; this notation represents that element values here have per-unit-length units
...
For coaxial cable, the element values depend on the inner conductor’s
radius ri , the outer radius of the dielectric rd , the conductivity of the conductors σ, and the conductivity

198
σd , dielectric constant

CHAPTER 6
...
2)

2πσd

G=

ln
L=

rd
ri

rd
ri

µd
ln
2π

rd
ri

For twisted pair, having a separation d between the conductors that have conductivity σ and common radius
r and that are immersed in a medium having dielectric and magnetic properties, the element values are then
R=

1
πrδσ

π
arccosh
πσ
G=
arccosh

C=

L=

µ
π

d
2r

(6
...
We express this dependence as v (x, t) and i (x, t)
...
As is customary in analyzing linear
circuits, we express voltages and currents as the real part of complex exponential signals, and write circuit
variables as a complex amplitude—here dependent on distance—times a complex exponential: v (x, t) =
Re V (x) ej2πf t and i (x, t) = Re I (x) ej2πf t
...

KCL at Center Node
I (x) = I (x − ∆x) − V (x) G + j2πf C ∆x

(6
...
5)

V-I relation for RL series

Rearranging and taking the limit ∆x → 0 yields the so-called transmission line equations
...
6)

By combining these equations, we can obtain a single equation that governs how the voltage’s or the current’s
complex amplitude changes with position along the transmission line
...

d2
V (x) = G + j2πf C
dx2

R + j2πf L V (x)

(6
...
8)

Calculating its second derivative and comparing the result with our equation for the voltage can check this
solution
...
9)
dx2
2
= γ V (x)
Our solution works so long as the quantity γ satisﬁes
γ=±

G + j2πf C

R + j2πf L

(6
...
The
quantities V+ and V− are constants determined by the source and physical considerations
...
2 (Circuit Model for a Transmission
Line)
...
Expressing γ in terms of its real
and imaginary parts in our solution shows that such increases are a (mathematical) possibility
...
The ﬁrst term will increase exponentially for x < 0
unless V+ = 0 in this region; a similar result applies to V− for x > 0
...


 V e−(a+jb)x if x > 0
+
V (x) =
(6
...
The space constant, also known as the attenuation constant, is the distance over which the voltage
decreases by a factor of 1
...

The presence of the imaginary part of γ, b (f ), also provides insight into how transmission lines work
...
The complete solution for the voltage has the form
v (x, t) = Re V+ e−ax ej(2πf t−bx)

(6
...
If we could take a snapshot of the
voltage (take its picture at t = t1 ), we would see a sinusoidally varying waveform along the transmission
line
...
If we were to take a second
b
picture at some later time t = t2 , we would also see a sinusoidal voltage
...
Thus, the voltage appeared to move to the right with a speed equal to 2πf (assuming b > 0)
...
13)
R + j2πf L

200

CHAPTER 6
...
12) depend on the values of a and b, and
how they depend on frequency
...
In this case, γ simpliﬁes to

a = 0 and b = 2πf

−4π 2 f 2 LC, which seemingly makes it pure imaginary with

LC
...
14)

LC

For typical coaxial cable, this propagation speed is a fraction (one-third to two-thirds) of the speed of light
...
Since the real part of γ is the attenuation factor a, a
more detailed analysis is required to determine if a = 0 (no attenuation) or is non-zero
...
10) by
γ 2 = (a + jb)2 = G + j2πf C

R + j2πf L

Expanding the expressions, we ﬁnd that
a2 − b2 + j2ab = GR − 4π 2 f 2 LC + j2πf (RC + GL)
Considering the high-frequency limit, the constant term on the right side can be ignored
...

1 R
a=
+ GZ0
2 Z0
Here, Z0 is deﬁned to be L/C and its importance is demonstrated below
...

Exercise 6
...
255
...

By using the second of the transmission line equation (6
...
Considering the spatial region x > 0, for example, we ﬁnd that
d
V (x) = −γV (x) = − R + j2πf L I (x)
dx
which means that the ratio of voltage and current complex amplitudes does not depend on distance
...
15)

= Z0
The quantity Z0 is known as the transmission line’s characteristic impedance
...

lim Z0 =

f →∞

L
C

Typical values for characteristic impedance are 50 and 75 Ω
...
16)

201
A related transmission line is the optic ﬁber
...
In this situation, we don’t have two conductors—in fact we have none—and the
energy is propagating in what corresponds to the dielectric material of the coaxial cable
...

From the encompassing view of Maxwell’s equations, the only diﬀerence is the electromagnetic signal’s frequency
...

Exercise 6
...
255
...
Compare this frequency with that of a mid-frequency cable television signal
...
In wireline
communication, we have a direct, physical connection—a circuit—between transmitter and receiver
...
Transmitted signal amplitude does decay
exponentially along the transmission line
...

6
...
When a voltage is applied to an antenna, it creates an electromagnetic ﬁeld that
propagates in all directions (although antenna geometry aﬀects how much power ﬂows in any given direction)
that induces electric currents in the receiver’s antenna
...
In general terms, the dominant factor is the relation of the antenna’s
size to the ﬁeld’s wavelength
...

For example, a 1 MHz electromagnetic ﬁeld has a wavelength of 300 m
...
Consequently, the lower the
frequency the bigger the antenna must be
...

For most antenna-based wireless systems, how the signal diminishes as the receiver moves further from
the transmitter derives by considering how radiated power changes with distance from the transmitting
antenna
...
Considering a sphere centered at the transmitter, the total power, which is
found by integrating the radiated power over the surface of the sphere, must be constant regardless of the
sphere’s radius
...
Thus, if p (d) represents the power
integrated with respect to direction at a distance d from the antenna, the total power will be p (d) 4πd2
...
org/content/m0101/2
...

202

CHAPTER 6
...

AR =

kAT
d

(6
...
Thus, the further from the transmitter the receiver is located, the weaker
the received signal
...

Exercise 6
...
255
...

1
c = √
µ0 0
=

0

of free

(6
...
Because signals travel at a ﬁnite speed, a receiver senses a transmitted signal only after a
time delay inversely related to the propagation speed:
∆t =

d
c

At the speed of light, a signal travels across the United States in 16 ms, a reasonably small time delay
...

6
...
Losses in wireline
channels are explored in the Circuit Models module (Section 6
...
In wireless
channels, not only does radiation loss occur (p
...

dLOS
earth

}h

R

Figure 6
...
Line-of-sight transmission means
the transmitting and receiving antennae can “see” each other as shown
...

5 This

content is available online at ...
14/>
...

Line-of-sight communication has the transmitter and receiver antennas in visual contact with each other
...
19)
dLOS = 2 2hR + h2 ≈ 2 2Rh
where R is the earth’s radius ( 6
...

Exercise 6
...
255
...
Generalize it to
the case where the antennas have diﬀerent heights (as is the case with commercial radio and cellular
telephone)
...
5
(Solution on p
...
)
Can you imagine a situation wherein global wireless communication is possible with only one transmitting antenna? In particular, what happens to wavelength when carrier frequency decreases?
Using a 100 m antenna would provide line-of-sight transmission over a distance of 71
...
Using such very
tall antennas would provide wireless communication within a town or between closely spaced population
centers
...
This kind of network is known as a relay network
...
6 The Ionosphere and Communications6
If we were limited to line-of-sight communications, long distance wireless communication, like ship-to-shore
communication, would be impossible
...
When the experiment worked, but only at night, physicists scrambled to determine
why (using Maxwell’s equations, of course)
...

What he meant was that at optical frequencies (and others as it turned out), the mirror was transparent,
but at the frequencies Marconi used, it reﬂected electromagnetic radiation back to earth
...
The maximum distance along the earth’s surface that can
R
be reached by a single ionospheric reﬂection is 2Rarccos R+hi , which ranges between 2,010 and 3,000 km
when we substitute minimum and maximum ionospheric altitudes
...

√
The communication delay encountered with a single reﬂection in this channel is
between 6
...

2

2Rhi +hi 2
,
c

which ranges

6
...
Here, ground stations transmit to orbiting satellites that
amplify the signal and retransmit it back to earth
...
TV satellites would require the homeowner to continually adjust his or her antenna
if the satellite weren’t in geosynchronous orbit
...
org/content/m0539/2
...

content is available online at ...
10/>
...
20)

204

CHAPTER 6
...
Calculations yield R = 42200km, which
corresponds to an altitude of 35700km
...
Of great importance in satellite communications is the
transmission delay
...
24 s, a signiﬁcant delay
...
6
(Solution on p
...
)
In addition to delay, the propagation attenuation encountered in satellite communication far exceeds
what occurs in ionospheric-mirror based communication
...
Note that the attenuation calculation in the ionospheric case, assuming the ionosphere
acts like a perfect mirror, is not a straightforward application of the propagation loss formula
(p
...

6
...
It’s time
to be more precise about what these quantities are and how they diﬀer
...
Telephone lines are subject to power-line interference (in
the United States a distorted 60 Hz sinusoid)
...
The problem with such interference is that it occupies the
same frequency band as the desired communication signal, and has a similar structure
...
7
(Solution on p
...
)
Suppose interference occupied a diﬀerent frequency band; how would the receiver remove it?
We use the notation i (t) to represent interference
...

Noise signals have little structure and arise from both human and natural sources
...
Thermal noise
plagues all electronic circuits that contain resistors
...
All channels are subject to
noise, and we need a way of describing such signals despite the fact we can’t write a formula for the noise
signal like we can for interference
...
It is deﬁned entirely
by its frequency-domain characteristics
...

• At each frequency, the phase of the noise spectrum is totally uncertain: It can be any value in between
0 and 2π, and its value at any frequency is unrelated to the phase at any other frequency
...

Because of the emphasis here on frequency-domain power, we are led to deﬁne the power spectrum
...

Ps (f ) ≡ |S (f ) |

2

(6
...
Because signals must have negative frequency components that mirror positive frequency ones, we
routinely calculate the power in a spectral band as the integral over positive frequencies multiplied by two
...
org/content/m0515/2
...

Theorem,” (1) ...
22)

205
Using the notation n (t) to represent a noise signal’s waveform, we deﬁne noise in terms of its power spectrum
...
With this deﬁnition, the power in a frequency
2
band equals N0 (f2 − f1 )
...
122) of the system’s frequency response and the input’s spectrum
...
23)
This result applies to noise signals as well
...

2

6
...

• The transmitted signal is usually not ﬁltered by the channel
...

• The signal propagates through the channel at a speed equal to or less than the speed of light, which
means that the channel delays the transmission
...

Letting α represent the attenuation introduced by the channel, the receiver’s input signal is related to the
transmitted one by
r (t) = αx (t − τ ) + i (t) + n (t)
(6
...
4
...

x(t)

r(t)
Channel

x(t) Delay
τ

Attenuation
α

+

+

r(t)

Interference Noise
i(t)
n(t)
Figure 6
...
3: Fundamental model of communication) has the depicted form
...

Adding the interference and noise is justiﬁed by the linearity property of Maxwell’s equations
...
8
Is this model for the channel linear?

(Solution on p
...
)

As expected, the signal that emerges from the channel is corrupted, but does contain the transmitted signal
...
We characterize the channel’s quality
by the signal-to-interference ratio (SIR) and the signal-to-noise ratio (SNR)
...
Assuming the
signal x (t)’s spectrum spans the frequency interval [fl , fu ], these ratios can be expressed in terms of power
spectra
...
25)
f
2 flu Pi (f ) df
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content is available online at ...
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...
INFORMATION COMMUNICATION
∞

2α2 0 Px (f ) df
SNR =
N0 (fu − fl )

(6
...
Variations in signal-tointerference and signal-to-noise ratios arise from the attenuation because of transmitter-to-receiver distance
variations
...
10 Baseband Communication11
We use analog communication techniques for analog message signals, like music, speech, and television
...

The simplest form of analog communication is baseband communication
...
Transmission and reception of analog signals using analog results
in an inherently noisy received signal (assuming the channel adds noise, which it almost certainly
does)
...

x (t) = Gm (t)

(6
...
You don’t use baseband
communication in wireless systems simply because low-frequency signals do not radiate well
...
Assuming the signal occupies a bandwidth of W Hz (the signal’s spectrum
extends from zero to W ), the receiver applies a lowpass ﬁlter having the same bandwidth, as shown in
Figure 6
...

r(t)

LPF
W

^
m(t)

Figure 6
...

We use the signal-to-noise ratio of the receiver’s output m (t) to evaluate any analog-message communication system
...
The ﬁlter does not aﬀect the signal component—we assume its gain is unity—but does ﬁlter the
noise, removing frequency components above W Hz
...
28)

The signal term power (m) will be proportional to the bandwidth W ; thus, in baseband communication the
signal-to-noise ratio varies only with transmitter gain and channel attenuation and noise level
...
org/content/m0517/2
...

207

cos 2πfct
r(t)

BPF
fc; 2W

~
r(t)

LPF
W

~
R(f)

R(f)

f
fc W c fc+W

f

^
m(t)
^
M(f)

f
fc W c fc+W

f

W

W

f

Figure 6
...
The dashed line indicates the white noise level
...

6
...

Point of Interest: We use analog communication techniques for analog message signals, like
music, speech, and television
...

The key idea of modulation is to aﬀect the amplitude, frequency or phase of what is known as the carrier
sinusoid
...
The amplitude modulated message signal has the form
x (t) = Ac (1 + m (t)) cos (2πfc t)

(6
...
Also, the signal’s amplitude is assumed
to be less than one: |m (t) | < 1
...
5)), we know that the transmitted signal’s spectrum occupies the frequency
range [fc − W, fc + W ], assuming the signal’s bandwidth is W Hz (see the ﬁgure (Figure 6
...
The carrier
frequency is usually much larger than the signal’s highest frequency: (fc
W ), which means that the
transmitter antenna and carrier frequency are chosen jointly during the design process
...
20)
...
6)
...
30)

Because of our trigonometric identities, we know that
cos2 (2πfc t) =

1
(1 + cos (2π2fc t))
2

(6
...

Multiplication by the constant term returns the message signal to baseband (where we want it to be!) while
12 This

content is available online at ...
26/>
...
INFORMATION COMMUNICATION

multiplication by the double-frequency term yields a very high frequency signal
...
Thus, the received signal is
m (t) =

Ac
(1 + m (t))
2

(6
...
9
(Solution on p
...
)
This derivation relies solely on the time domain; derive the same result in the frequency domain
...

Because it is so easy to remove the constant term by electrical means—we insert a capacitor in series with
the receiver’s output—we typically ignore it and concentrate on the signal portion of the receiver’s output
when calculating signal-to-noise ratio
...
12 Signal-to-Noise Ratio of an Amplitude-Modulated Signal13
When we consider the much more realistic situation when we have a channel that introduces attenuation
and noise, we can make use of the just-described receiver’s linear nature to directly derive the receiver’s
output
...
The white noise, on the other hand, should be ﬁltered from the received signal
before demodulation
...

As shown in the triangular-shaped signal spectrum (Figure 6
...

As we derive the signal-to-noise ratio in the demodulated signal, let’s also calculate the signal-to-noise
ratio of the bandpass ﬁlter’s output r (t)
...
This
˜
˜
signal’s Fourier transform equals
αAc
(M (f + fc ) + M (f − fc ))
(6
...
34)
4
Exercise 6
...
256
...
What is it?
2

2

Thus, the total signal-related power in r (t) is α Ac power (m)
...
The so-called received signal-to-noise
ratio — the signal-to-noise ratio after the de rigeur front-end bandpass ﬁlter and before demodulation —
equals
α2 Ac 2 power (m)
(6
...
Clearly, the signal power equals α Ac power(m)
...
Because we are concerned with noise, we must deal with the power spectrum since we
˜
don’t have the Fourier transform available to us
...
org/content/m0541/2
...

(6
...
Thus, the total noise power in this ﬁlter’s output equals
2 · N0 · W · 2 · 1 = N0 W
...
37)

Let’s break down the components of this signal-to-noise ratio to better appreciate how the channel and
the transmitter parameters aﬀect communications performance
...

• More transmitter power — increasing Ac — increases the signal-to-noise ratio proportionally
...

• The signal bandwidth W enters the signal-to-noise expression in two places: implicitly through the
signal power and explicitly in the expression’s denominator
...
On the other
hand, our transmitter enforced the criterion that signal amplitude was constant (Section 6
...
Signal
amplitude essentially equals the integral of the magnitude of the signal’s spectrum
...

Enforcing the signal amplitude speciﬁcation means that as the signal’s bandwidth increases we must decrease the spectral amplitude, with the result that the signal power remains constant
...

• Increasing channel attenuation — moving the receiver farther from the transmitter — decreases the
signal-to-noise ratio as the square
...

• Noise added by the channel adversely aﬀects the signal-to-noise ratio
...
For wireline channels, using baseband or amplitude modulation makes little diﬀerence in
terms of signal-to-noise ratio
...
The one
AM parameter that does not aﬀect signal-to-noise ratio is the carrier frequency fc : We can choose any value
we want so long as the transmitter and receiver use the same value
...
The two resulting transmissions will add, and both receivers
will produce the sum of the two signals
...
As more and more users wish to use radio, we need a forum for agreeing on
carrier frequencies and on signal bandwidth
...
In the United States,
the Federal Communications Commission (FCC) strictly controls the use of the electromagnetic spectrum
for communications
...

Exercise 6
...
256
...
How closely can the carrier frequencies
be while avoiding communications crosstalk? What is the signal bandwidth for commercial AM?
How does this bandwidth compare to the speech bandwidth?

6
...
}—is the goal here
...
Digital communication schemes are very
14 This

content is available online at ...
10/>
...
INFORMATION COMMUNICATION

s0(t)

s1(t)

A
T

T

t
–A

t

Figure 6
...
Once we decide how to represent bits by analog signals that can be transmitted over wireline
(like a computer network) or wireless (like digital cellular telephone) channels, we will then develop a way
of tacking on communication bits to the message bits that will reduce channel-induced errors greatly
...
Thus, if b (n) = 0,
we transmit the signal s0 (t); if b (n) = 1, send s1 (t)
...
In virtually every case, these
signals have a ﬁnite duration T common to both signals; this duration is known as the bit interval
...
Interestingly, baseband
and modulated signal sets can yield the same performance
...

Exercise 6
...
256
...
, −1, 0, 1,
...
14 Binary Phase Shift Keying15
A commonly used example of a signal set consists of pulses that are negatives of each other (Figure 6
...

s0 (t) =

ApT (t)

s1 (t) = −ApT (t)

(6
...
The entire bit stream b (n) is
represented by a sequence of these signals
...
39)

n

and graphically Figure 6
...

This way of representing a bit stream—changing the bit changes the sign of the transmitted signal—is
known as binary phase shift keying and abbreviated BPSK
...
The word “binary” is clear enough (one binary-valued
quantity is transmitted during a bit interval)
...
The word “keying” reﬂects back to the ﬁrst electrical
communication system, which happened to be digital as well: the telegraph
...

1
In this example it equals the reciprocal of the bit interval: R = T
...

The choice of signals to represent bit values is arbitrary to some degree
...
We could also have
made the negative-amplitude pulse represent a 0 and the positive one a 1
...
As
in all communication systems, we design transmitter and receiver together
...
org/content/m10280/2
...

211
x(t)
A

“0”

“1”
T

“1”

2T

“0”

3T

4T

t

–A
(a)

x(t)
“0”
A

“1”

“1”

“0”
t

T

2T

3T

4T

(b)

Figure 6
...

The lower one shows an amplitude-modulated variant suitable for wireless channels
...

s0 (t) =

ApT (t) sin

s1 (t) = −ApT (t) sin

s0(t)

2πkt
T
2πkt
T

(6
...
9

Exercise 6
...
256
...
We’ll show later that indeed both signal sets provide
identical performance levels when the signal-to-noise ratios are equal
...
14
(Solution on p
...
)
Write a formula, in the style of the baseband signal set, for the transmitted signal as shown in the
plot of the baseband signal set16 that emerges when we use this modulated signal
...
The bandwidth is determined by the bit
sequence
...
The worst-case—bandwidth consuming—bit sequence is the alternating one shown in
Figure 6
...
In this case, the transmitted signal is a square wave having a period of 2T
...
org/content/m0542/latest/#ﬁg1001>

212

CHAPTER 6
...
10: Here we show the transmitted waveform corresponding to an alternating bit sequence
...
Thus, strictly speaking, the signal’s bandwidth is inﬁnite
...

The ﬁrst and third harmonics contain that fraction of the total power, meaning that the eﬀective bandwidth
3
of our baseband signal is 2T or, expressing this quantity in terms of the datarate, 3R
...
5 MHz
...
15
(Solution on p
...
)
Show that indeed the ﬁrst and third harmonics contain 90% of the transmitted power
...
16
What is the 90% transmission bandwidth of the modulated signal set?

(Solution on p
...
)

6
...

s0 (t) = ApT (t) sin (2πf0 t)

(6
...
11

The frequencies f0 , f1 are usually harmonically related to the bit interval
...
As can be seen from the transmitted signal for our example bit stream (Figure 6
...

To determine the bandwidth required by this signal set, we again consider the alternating bit stream
...
, and the second having the same structure but interleaved with the ﬁrst and containing s1 (t)
(Figure 6
...

Each component can be thought of as a ﬁxed-frequency sinusoid multiplied by a square wave of period
2T that alternates between one and zero
...
org/content/m0545/2
...

213

A

x(t)
“0”

“1”

“1”

“0”
t

T
Figure 6
...
8)
...
13:
The depicted decomposition of the FSK-modulated alternating bit stream into its
frequency components simpliﬁes the calculation of its bandwidth
...
This quantity’s presence changes the
number of Fourier series terms required for the 90% bandwidth: Now we need only include the zero and
1
1
1
ﬁrst harmonics to achieve it
...

k0
k1
If the two frequencies are harmonics of the bit-interval duration, f0 = T and f1 = T with k1 > k0 , the
bandwidth equals k1 −k0 +1
...
If the diﬀerence is 2, the bandwidths are equal and larger diﬀerences
produce a transmission bandwidth larger than that resulting from using a BPSK signal set
...
16 Digital Communication Receivers18
The receiver interested in the transmitted bit stream must perform two tasks when received waveform r (t)
begins
...
Because transmitter and receiver are designed in concert, both use the same value for the bit
interval T
...
This reference bit sequence is usually the alternating sequence
as shown in the square wave example19 and in the FSK example (Figure 6
...
The receiver knows
what the preamble bit sequence is and uses it to determine when bit boundaries occur
...
Because the receiver usually
does not determine which bit was sent until synchronization occurs, it does not know when during the
preamble it obtained synchronization
...
org/content/m0520/2
...

Bandwidth,” Figure 1 ...
INFORMATION COMMUNICATION

to a second bit sequence
...

• Once synchronized and data bits are transmitted, the receiver must then determine every T seconds
what bit was transmitted during the previous bit interval
...

The receiver for digital communication is known as a matched ﬁlter
...
14: The optimal receiver structure for digital communication faced with additive white noise
channels is the depicted matched ﬁlter
...
14 (Optimal receiver structure), multiplies the received signal by each of
the possible members of the transmitter signal set, integrates the product over the bit interval, and compares
the results
...
For the next bit interval, the multiplication and
integration begins again, with the next bit decision made at the end of the bit interval
...
42)

nT

You may not have seen the argmax notation before
...
argmax equals the value of the index that yields the maximum
...

Let’s assume a perfect channel for the moment: The received signal equals the transmitted one
...
43)
2

r (t) s1 (t) dt = −A T
nT

If bit 1 were sent,
(n+1)T

r (t) s0 (t) dt = −A2 T
nT
(n+1)T

(6
...
17
(Solution on p
...
)
Can you develop a receiver for BPSK signal sets that requires only one multiplier-integrator combination?

215
Exercise 6
...
256
...
Channel attenuation would not aﬀect this
correctness; it would only make the values smaller, but all that matters is which is largest
...
17 Digital Communication in the Presence of Noise20
When we incorporate additive noise into our channel model, so that r (t) = αsi (t)+n (t), errors can creep in
...
14), the integrators’ outputs in the matched
ﬁlter receiver (Figure 6
...
45)

(n+1)T
2

n (t) s1 (t) dt

r (t) s1 (t) dt = (−α) A T +
nT

nT

The quantities containing noise terms cause errors in the receiver’s decision-making process
...
Because the noise has zero average value and has an
equal amount of power in all frequency bands, the values of the integrals will hover about zero
...
If the noise is such that its integral term is more negative than αA2 T ,
then the receiver will make an error, deciding that the transmitted zero-valued bit was indeed a one
...
45)) deﬁnes how large the noise term must be for an incorrect receiver decision to result
...
The signal-diﬀerence energy equals
T

2

(s1 (t) − s0 (t)) dt
0

For our BPSK baseband signal set, the diﬀerence-signal-energy term is 4α2 A4 T 2
...

• Probability Distribution of the Noise Term — The value of the noise terms relative to the
signal terms and the probability of their occurrence directly aﬀect the likelihood that a receiver error
will occur
...

Deriving the following expression for the probability the receiver makes an error on any bit transmission
is complicated but can be found in other modules: here21 and here22
...
46)


2α2 A2 T 
= Q
for the BPSK case
N0
∞

2

Here Q (·) is the integral Q (x) = √1 x e−α /2 dα
...
As Figure 6
...

20 This

content is available online at ...
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...
org/content/m16253/latest/>
22 "Continuous-Time Detection Theory" ...
INFORMATION COMMUNICATION

10 0

10-2
Q(x)
10-4

10-6

10-8
0

1

2

3

4

5

6

Figure 6
...
Note that it decreases very
rapidly for small increases in its arguments
...

Probability of Bit Error

10 0
FSK

10 -2

BPSK

10 -4

10 -6

10 -8
-5

0
5
Signal-to-Noise Ratio (dB)

10

Figure 6
...
The upper curve shows the performance
of the FSK signal set, the lower (and therefore better) one the BPSK signal set
...
We
arrive at a concise expression for the probability the matched ﬁlter receiver makes a bit-reception error when
the BPSK signal set is used
...
16 shows how the receiver’s error rate varies with the signal-to-noise ratio

(6
...

217
Exercise 6
...
256
...

6
...
17) reveals several properties about digital communication
systems
...
In such situations, the receiver performs only slightly better than the “receiver”
that ignores what was transmitted and merely guesses what bit was transmitted
...

• As the signal-to-noise ratio increases, performance gains–smaller probability of error pe – can be easily
obtained
...

In words, one out of one hundred million bits will, on the average, be in error
...
Adding
1 dB improvement in signal-to-noise ratio can result in a factor of ten smaller pe
...
All BPSK signal sets, baseband or
modulated, yield the same performance for the same bit energy
...

Exercise 6
...
256
...

The matched-ﬁlter receiver provides impressive performance once adequate signal-to-noise ratios occur
...
The answer is that the matched-ﬁlter receiver is
optimal: No other receiver can provide a smaller probability of error than the matched ﬁlter
regardless of the SNR
...
The reason
for this result rests in the dependence of probability of error pe on the diﬀerence between the noise-free
integrator outputs: For a given Eb , no other signal set provides a greater diﬀerence
...
Do note the phrase “on the average” here: Errors occur randomly because of the noise
introduced by the channel, and we can only predict the probability of occurrence
...
Suppose the error probability is an impressively
small number like 10−6
...
This error rate is very high,
requiring a much smaller pe to achieve a more acceptable average occurrence rate for errors occurring
...
We do have some tricks up our sleeves, however, that
can essentially reduce the error rate to zero without resorting to expending a large amount of energy at the
transmitter
...
19) and Shannon’s Noisy Channel Coding
Theorem (Section 6
...

6
...
3: Fundamental model of communication)
...
17 (DigMC), the message is a
single bit
...
org/content/m10282/2
...

content is available online at ...
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...
INFORMATION COMMUNICATION
s(m)
Source

Source
Coder

x(t)
Transmitter

r(t)
Channel

1–pe

b(n) 0

s(m)
Source

b(n)

Source
Coder

b(n)
Receiver

pe

0

1–pe
Digital Channel

b(n)

Sink

s(m)
Source
Decoder

pe
1

s(m)
Source
Decoder

Sink

1

Figure 6
...
The symbolic-valued signal s (m) forms the message, and it is
encoded into a bit sequence b (n)
...
Each bit is represented by an analog signal, transmitted through
the (unfriendly) channel, and received by a matched-ﬁlter receiver
...
The lower block diagram shows an equivalent system
b
wherein the analog portions are combined and modeled by a transition diagram, which shows how each
transmitted bit could be received
...

(Section 6
...
15), Digital Communication Receivers (Section 6
...
17), Digital
Communication System Properties28 , and Error Probability29 , can be lumped into a single system known as
the digital channel
...

The probabilities on transitions coming from the same symbol must sum to one
...
The probability of error pe is the sole parameter of
the digital channel, and it encapsulates signal set choice, channel properties, and the matched-ﬁlter receiver
...

6
...
Claude Shannon published
in 1948 The Mathematical Theory of Communication, which became the cornerstone of digital communication
...
In the simplest signal model, each symbol can occur at index
n with a probability Pr [ak ], k = {1,
...
What this model says is that for each signal value a K-sided
coin is ﬂipped (note that the coin need not be fair)
...

0 ≤ Pr [ak ] ≤ 1

(6
...
org/content/m0542/latest/>
signal set" ...
org/content/m0544/latest/>
28 "Digital Communication System Properties" ...
org/content/m0548/latest/>
30 This content is available online at ...
13/>
...
49)

219
This coin-ﬂipping model assumes that symbols occur without regard to what preceding or succeeding symbols
were, a false assumption for typed text
...
The key quantity that
characterizes a symbolic-valued signal is the entropy of its alphabet
...
50)

k

Because we use the base-2 logarithm, entropy has units of bits
...
A zero-probability symbol never occurs;
thus, we deﬁne 0log2 0 = 0 so that such symbols do not aﬀect the entropy
...
In this case, the
entropy equals log2 K
...

Exercise 6
...
256
...
Derive the value of the minimum entropy
alphabet
...
1
A four-symbol alphabet has the following probabilities
...
As
of this alphabet equals
H (A) = −
=−

1
8
1
2

Pr [a3 ] =

1
8

= 2−1 , log2

1
1
1
1
1
1
log
+ log2
+ log2
2 2 2
4
4
8
8
1
1
1
1
(−1) + (−2) + (−3) + (−3)
2
4
8
8

1
+ log2
8

1
2

= −1
...
51)

= 1
...
21 Source Coding Theorem31
The signiﬁcance of an alphabet’s entropy rests in how we can represent it with a sequence of bits
...
We convert back and forth between symbols to bit-sequences with what is known
as a codebook: a table that associates symbols to bit sequences
...

note: You may be conjuring the notion of hiding information from others when we use the
name codebook for the symbol-to-bit-sequence table
...
The codebook terminology
was developed during the beginnings of information theory just after World War II
...
13) is the transmission of
symbolic-valued signals from one place to another
...
Because we want to send
the ﬁle quickly, we want to use as few bits as possible
...
org/content/m0091/2
...

220

CHAPTER 6
...
For example, we could use one
bit for every character: File transmission would be fast but useless because the codebook creates errors
...
Let B (ak )
denote the number of bits used to represent the symbol ak
...
The Source Coding Theorem states that the
average number of bits needed to accurately represent the alphabet need only to satisfy
H (A) ≤ B (A) < H (A) + 1

(6
...
The smaller an alphabet’s entropy, the fewer bits required for digital transmission of ﬁles
expressed in that alphabet
...
2
A four-symbol alphabet has the following probabilities
...
75 bits (Example 6
...
Let’s see if we can ﬁnd a codebook for this four-letter
alphabet that satisﬁes the Source Coding Theorem
...

(a0 ↔ 00) (a1 ↔ 01) (a2 ↔ 10) (a3 ↔ 11)

(6
...
Because the entropy equals 1
...
If we chose a codebook with
diﬀering number of bits for the symbols, a smaller average number of bits could indeed be obtained
...
One codebook like
this is
(a0 ↔ 0) (a1 ↔ 10) (a2 ↔ 110) (a3 ↔ 111)
(6
...
75
...
Using the eﬃcient code, we can
transmit the symbolic-valued signal having this alphabet 12
...
Furthermore, we know that
no more eﬃcient codebook can be found because of Shannon’s Theorem
...
22 Compression and the Huﬀman Code32
Shannon’s Source Coding Theorem (6
...
Here, we have
a symbolic-valued signal source, like a computer ﬁle or an image, that we want to represent with as few
bits as possible
...
Using
a lossy compression scheme means that you cannot recover a symbolic-valued signal from its compressed
version without incurring some error
...

Shannon’s Source Coding Theorem states that symbolic-valued signals require on the average at least
H (A) number of bits to represent each of its values, which are symbols drawn from the alphabet A
...
org/content/m0092/2
...

221
module on the Source Coding Theorem (Section 6
...
What is not discussed there is a procedure for designing an eﬃcient source coder: one guaranteed
to produce the fewest bits/symbol on the average
...

Point of Interest: In the early years of information theory, the race was on to be the ﬁrst to ﬁnd
a provably maximally eﬃcient source coding algorithm
...
We’re pretty sure he received an “A
...

• Form a binary tree to the right of the table
...

Build the tree by merging the two lowest probability symbols at each level, making the probability of
the node equal to the sum of the merged nodes’ probabilities
...

• At each node, label each of the emanating branches with a binary number
...

Example 6
...
1) and Source Coding (Example 6
...
75 bits (Example 6
...
This alphabet has the Huﬀman coding tree shown in
Figure 6
...

Symbol Probability
1
a1
a2
a3
a4

2
1
4
1
8
1
8

Source Code
0
0
10

0
1
2

0
1

1
4

1

110

1
111

Figure 6
...
The binary tree created by the algorithm extends to the right, with the root node (the one
at which the tree begins) deﬁning the codewords
...

The code thus obtained is not unique as we could have labeled the branches coming out of
each node diﬀerently
...
75 bits, which is the Shannon entropy limit for this source alphabet
...
}, our Huﬀman code would produce the bitstream
b (n) = 101100111010
...
Furthermore, we may not be able to achieve the entropy limit
...
75 bits
...
INFORMATION COMMUNICATION
entropy limit is 1
...
The Huﬀman code does satisfy the Source Coding Theorem—its average
length is within one bit of the alphabet’s entropy—but you might wonder if a better code existed
...
We can’t do better
...
22
(Solution on p
...
)
Derive the Huﬀman code for this second set of probabilities, and verify the claimed average code
length and alphabet entropy
...
23 Subtleties of Coding33
In the Huﬀman code, the bit sequences that represent individual symbols can have diﬀering lengths so the
bitstream index m does not increase in lock step with the symbol-valued signal’s index n
...
If our source code averages B (A) bits/symbol and symbols are produced at a rate R, the average
bit rate equals B (A) R, and this quantity determines the bit interval duration T
...
23
Calculate the relation between T and the average bit rate B (A) R
...
257
...
When we use an unequal number
of bits to represent symbols, how does the receiver determine when symbols begin and end? If you created a
source code that required a separation marker in the bitstream between symbols, it would be very ineﬃcient
since you are essentially requiring an extra symbol in the transmission stream
...

As shown in this example (Example 6
...
Huﬀman showed that his (maximally eﬃcient) code had
the preﬁx property: No code for a symbol began another symbol’s code
...

Exercise 6
...
257
...

However, having a preﬁx code does not guarantee total synchronization: After hopping into the middle of a
bitstream, can we always ﬁnd the correct symbol boundaries? The self-synchronization issue does mitigate
the use of eﬃcient source coding algorithms
...
25
(Solution on p
...
)
Show by example that a bitstream produced by a Huﬀman code is not necessarily self-synchronizing
...
Despite the small probabilities of
error oﬀered by good signal set design and the matched ﬁlter, an infrequent error can devastate the ability to
translate a bitstream into a symbolic signal
...

Example 6
...
When ﬁrst deployed in
1844, it communicated text over wireline connections using a binary code—the Morse code—to
represent individual letters
...
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...

223

A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
T
U
V
W
X
Y
Z

%
6
...
32
3
...
97
10
...
68
1
...
63
6
...
06
0
...
07
2
...
73
6
...
87
0
...
87
5
...
68
2
...
70
1
...
25
1
...
06

Morse Code

...

-
...

...

...

...

–
-
...

–
...

...
–
–
...
19: Morse and Huﬀman Codes for American-Roman Alphabet
...
The entropy H (A) of the
this source is 4
...
The average Morse codeword length is 2
...
Adding one more symbol for
the letter separator and converting to bits yields an average codeword length of 5
...
The average
Huﬀman codeword length is 4
...

would tap the message using a telegraph key to another operator, who would relay the message
on to the next operator, presumably getting the message closer to its destination
...
To say it
presaged modern communications would be an understatement
...
The Morse code, shown in Figure 6
...
To separate codes for each letter, Morse code required that a space—a
pause—be inserted between each letter
...
The resulting source code is not within a bit of entropy, and is grossly ineﬃcient (about
25%)
...
19 shows a Huﬀman code for English text, which as we know is eﬃcient
...
INFORMATION COMMUNICATION
s(m)
Source

b(n)
Source
Coder

c(l)
Channel
Coder

c(l)
Digital
Channel

b(n)
Channel
Decoder

s(m)
Source
Decoder

Sink

Figure 6
...
Properly designed channel coding can greatly reduce the probability (from
the uncoded value of pe ) that a data bit b (n) is received incorrectly even when the probability of c (l)
be received in error remains pe or becomes larger
...

6
...
The idea is for the transmitter to send not only the symbolderived bits emerging from the source coder but also additional bits derived from the coder’s bit stream
...
Instead of the communication model
(Figure 6
...
20)
...

Shannon’s Noisy Channel Coding Theorem (Section 6
...

Unfortunately, Shannon did not demonstrate an error correcting code that would achieve this remarkable
feat; in fact, no one has found such a code
...
In any case, that should not prevent us from studying commonly used error correcting codes
that not only ﬁnd their way into all digital communication systems, but also into CDs and bar codes used
on merchandise
...
25 Repetition Codes35
Perhaps the simplest error correcting code is the repetition code
...
Because the error probability pe is always less than 1 , we
2
know that more of the bits should be correct rather than in error
...
For example, let’s consider the three-fold repetition code: for every bit b (n) emerging from the
source coder, the channel coder produces three
...
The coding table (Table 1
...

Thus, if one bit of the three bits is received in error, the receiver can correct the error; if more than
one error occurs, the channel decoder announces the bit is 1 instead of transmitted value of 0
...
This probability of a decoding error
1
is always less than pe , the uncoded value, so long as pe < 2
...
26
(Solution on p
...
)
Demonstrate mathematically that this claim is indeed true
...
26 Block Channel Coding36
Because of the higher datarate imposed by the channel coder, the probability of bit error occurring in the
digital channel increases relative to the value obtained when no channel coding is used
...
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...

content is available online at ...
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...
org/content/m0094/2
...

35 This

225

1

bn

x(t)

0

n
1/R’

2/R’

Digital
Transmitter

t
T

2T

Channel
Coder
1

x(t)

cl

0

l
1/R’

2/R’

Digital
Transmitter

t
T 3T

6T

Figure 6
...
R’ is the datarate produced by the source
coder
...
This reduction
in the bit interval means that the transmitted energy/bit decreases by a factor of three, which results in
an increased error probability in the receiver
...
1: In this example, the transmitter encodes 0 as 000
...
The ﬁrst column lists all possible received data words
and the second the probability of each data word being received
...
When the decoder produces 0, it successfully corrected the
errors introduced by the channel (if there were any; the top row corresponds to the case in which
no errors occurred)
...

duration must be reduced by K in comparison to the no-channel-coding situation, which means the energy
N
per bit Eb goes down by the same amount
...
21: Repetition Code)
...

Such is the sometimes-unfriendly nature of the real world
...
The question thus becomes
does channel coding really help: Is the eﬀective error probability lower with channel coding even though the
error probability for each transmitted bit is larger? The answer is no: Using a repetition code for channel

226

CHAPTER 6
...
The ultimate reason is
the repetition code’s ineﬃciency: transmitting one data bit for every three transmitted is too ineﬃcient for
the amount of error correction provided
...
27
(Solution on p
...
)
Using MATLAB, calculate the probability a bit is received incorrectly with a three-fold repetition
code
...

The repetition code (p
...
For every
K bits that enter the block channel coder, it inserts an additional N − K error-correction bits to produce a
block of N bits for transmission
...

In the three-fold repetition code (p
...
A block code’s coding eﬃciency E equals
the ratio K , and quantiﬁes the overhead introduced by channel coding
...
We represent the fact that the bits sent through the digital channel
operate at a diﬀerent rate by using the index l for the channel-coded bit stream c (l)
...

Does any error-correcting code reduce communication errors when real-world constraints are taken into
account? The answer now is yes
...

N

6
...
The phrase “linear
combination” means here single-bit binary arithmetic
...

c (1) = b (1)
c (2) = b (1)
c (3) = b (1)
or
c = Gb
where


1





G= 1 


1



c (1)





c =  c (2) 


c (3)

b=

b (1)

The length-K (in this simple example K = 1) block of data bits is represented by the vector b, and the
length-N output block of the channel coder, known as a codeword, by c
...

As we consider other block codes, the simple idea of the decoder taking a majority vote of the received bits
won’t generalize easily
...

37 This

content is available online at ...
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...
22:
In a (3,1) repetition code, only two of the possible eight three-bit data blocks are
codewords
...
In the left plot, the ﬁlled circles represent the codewords [0 0 0] and [1 1 1], the only
possible codewords
...
The center plot shows that the
distance between codewords is three
...
” The
right plot shows the data words that result when one error occurs as the codeword goes through the
channel
...
Note that the received data
word groups do not overlap, which means the code can correct all single-bit errors
...
As shown in Figure 6
...
We deﬁne the Hamming distance between binary data words c1 and c2 , denoted by
d (c1 , c2 ) to be the minimum number of bits that must be “ﬂipped” to go from one word to the other
...
In our table of binary arithmetic, we see that adding
a 1 corresponds to ﬂipping a bit
...
We can express the
Hamming distance as
d (c1 , c2 ) = sum ((c1 ⊕ c2 ))
(6
...
28
(Solution on p
...
)
Show that adding the error vector col[1, 0,
...

N −1

The probability of one bit being ﬂipped anywhere in a codeword is N pe (1 − pe )

...

To perform decoding when errors occur, we want to ﬁnd the codeword (one of the ﬁlled circles in Figure 6
...
Note that if a data
word lies a distance of one from two codewords, it is impossible to determine which codeword was actually
sent
...
Thus, to have a code that can correct all single-bit errors, codewords must
have a minimum separation of three
...

Introducing code bits increases the probability that any bit arrives in error (because bit interval durations
decrease)
...
Do we win or
lose by using an error-correcting code? The answer is that we can win if the code is well-designed
...
28)
...
) We also need a systematic way of ﬁnding the codeword closest to any
N

228

CHAPTER 6
...
A much better code than our (3,1) repetition code is the following (7,4) code
...

Error correction amounts to searching for the codeword c closest to the received block c in terms of the
Hamming distance between the two
...
Bad codes would produce blocks close together, which would result
in ambiguity when assigning a block of data bits to a received block
...

dmin = min (d (ci , cj )) , ci = cj

(6
...

Exercise 6
...
258
...
What must the
minimum Hamming distance between codewords dmin be?
How do we calculate the minimum distance between codewords? Because we have 2K codewords, the number
of possible unique pairs equals 2K−1 2K − 1 , which can be a large number
...
Therefore (ci ⊕ cj ) = G ((bi ⊕ bj ))
...
Finding these
codewords is easy once we examine the coder’s generator matrix
...
To ﬁnd
dmin , we need only count the number of bits in each column and sums of columns
...
Considering sums of column pairs
next, note that because the upper portion of G is an identity matrix, the corresponding upper portion of all
column sums must have exactly two bits
...
Triple
sums will have at least three bits because the upper portion of G is an identity matrix
...

229

6
...
One way of checking for errors is to try recreating the error
correction bits from the data portion of the received block c
...
It is formed from
the generator matrix G by taking the bottom, error-correction portion of G and attaching to it an identity
matrix
...
57)

Identity

The parity check matrix thus has size (N − K) × N , and the result of multiplying this matrix with a received
word is a length- (N − K) binary vector
...
For example, the ﬁrst column of G, (1, 0, 0, 0, 1, 0, 1) , is a codeword
...

Exercise 6
...
258
...
In other words, show that HG = 0 an (N − K) × K
matrix of zeroes
...
Because the presence of an error can be mathematically written
as c = (c ⊕ e), with e a vector of binary values having a 1 in those positions where a bit error occurred
...
31
(Solution on p
...
)
T
Show that adding the error vector (1, 0,
...

Consequently, Hc = H (c ⊕ e) = He
...
To perform our
channel decoding,
1
...
if this result is zero, no detectable or correctable error occurred;
3
...
add the error vector thus obtained to the received vector c to correct the error (because (c ⊕ e ⊕ e) = c)
...
Select the data bits from the corrected word to produce the received bit sequence b (n)
...
) error
occurring during the transmission/reception of one codeword can create the same received word as a singleT
T
bit error or no error in another codeword
...
The second results when the ﬁrst one experiences three bit errors
(ﬁrst, second, and sixth bits)
...
Our receiver uses the principle of maximum probability:
An error-free transmission is much more likely than one with three errors if the bit-error probability pe is
small enough
...
32
(Solution on p
...
)
How small must pe be so that a single-bit error is more likely to occur than a triple-bit error?
38 This

content is available online at ...
20/>
...
INFORMATION COMMUNICATION

6
...
We start with single-bit
error patterns, and multiply them by the parity check matrix
...
In our case, single-bit
error patterns give a unique result
...
3

This corresponds to our decoding table: We associate the parity check matrix multiplication result with
the error pattern and add this to the received word
...

As with the repetition code, we must question whether our (7,4) code’s error correction capability compensates for the increased error probability due to the necessitated reduction in bit energy
...
23
(Probability of error occurring) shows that if the signal-to-noise ratio is large enough channel coding yields
a smaller error probability
...
Clearly, our (7,4) channel code does yield smaller
error rates, and is worth the additional systems required to make it work
...
This pleasant property arises because the number of error patterns that can be corrected, 2N −K − 1,
equals the codeword length N
...
4: Hamming Codes) provides the parameters of these codes
...

Unfortunately, for such large blocks, the probability of multiple-bit errors can exceed the number of
single-bit errors unless the channel single-bit error probability pe is very small
...

Exercise 6
...
259
...
30 Noisy Channel Coding Theorem40
As the block length becomes larger, more error correction will be needed
...
org/content/m0097/2
...

content is available online at ...
12/>
...
23: The probability of an error occurring in transmitted K = 4 data bits equals 1 − (1 − pe )4
as (1 − pe )4 equals the probability that the four bits are received without error
...
When
a (7,4) single-bit error correcting code is used, the transmitter reduced the energy it expends during a
single-bit transmission by 4/7, appending three extra bits for error correction
...
Here
6
(7pe ) (1 − pe ) equals the probability of exactly on in seven bits emerging from the channel in error; The
channel decoder corrects this type of error, and all data bits in the block are received correctly
...
33

7

4

0
...
73

31

26

0
...
90

127

120

0
...
4

this question
...

6
...
1 Noisy Channel Coding Theorem
Let E denote the eﬃciency of an error-correcting code: the ratio of the number of data bits to the total
number of bits used to represent them
...

lim Pr [block error] = 0 , E < C

N →∞

(6
...
INFORMATION COMMUNICATION

1
0
...
1

0
...
3
Error Probability (Pe)

0
...
5

1
0
...
24:
The capacity per transmission through a binary symmetric channel is plotted as a
function of the digital channel’s error probability (upper) and as a function of the signal-to-noise ratio
for a BPSK signal set (lower)
...
30
...

lim Pr [block error] = 1
(6
...
Generally, a channel’s capacity changes with the signal-to-noise ratio: As
one increases or decreases, so does the other
...

This result astounded communication engineers when Shannon published it in 1948
...
The key for this capability to exist is
that the code’s eﬃciency be less than the channel’s capacity
...
60)
Figure 6
...
For example, our (7,4)
Hamming code has an eﬃciency of 0
...

6
...
30), Shannon also derived
the capacity for a bandlimited (to W Hz) additive white noise channel
...
” Instead of
constraining channel code eﬃciency, the revised Noisy Channel Coding Theorem states that some errorcorrecting code exists such that as the block length increases, error-free transmission is possible if the source
41 This

content is available online at ...
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...

C = W log2 (1 + SNR) bits/s

(6
...
42 Shannon’s proof of his theorem was very clever, and did not indicate
what this code might be; it has never been found
...
Until the “magic” code is found, more
important in communication system design is the converse
...
For this reason, capacity calculations are
made to understand the fundamental limits on transmission rates
...
34
(Solution on p
...
)
The ﬁrst deﬁnition of capacity applies only for binary symmetric channels, and represents the
number of bits/transmission
...
How would you convert the ﬁrst deﬁnition’s result into units of bits/second?
Example 6
...
The maximum data rate a modem can produce for this wireline
channel and hope that errors will not become rampant is the capacity
...
901 kbps

(6
...

Note that the data rate allowed by the capacity can exceed the bandwidth when the signal-to-noise ratio
1
exceeds 0 dB
...
What kind
of signal sets might be used to achieve capacity? Modem signal sets send more than one bit/transmission
using a number, one of the most popular of which is multi-level signaling
...
For example, two bits
can be sent with a signal set comprised of a sinusoid with amplitudes of ±A and ± A
...
32 Comparison of Analog and Digital Communication43
Analog communication systems, amplitude modulation (AM) radio being a typifying example, can inexpensively communicate a bandlimited analog signal from one location to another (point-to-point communication)
or from one point to many (broadcast)
...
6)
provides the largest possible signal-to-noise ratio for the demodulated message
...
12)
of this receiver thus indicates that some residual error will always be present in an analog system’s output
...

• Eﬃciency: The Source Coding Theorem allows quantiﬁcation of just how complex a given message
source is and allows us to exploit that complexity by source coding (compression)
...
We cannot exploit
signal structure to achieve a more eﬃcient communication system
...
Even though we may send information by way of a noisy channel, digital schemes are capable
of error-free transmission while analog ones cannot overcome channel disturbances; see this problem
(Problem 6
...

42 The bandwidth restriction arises not so much from channel properties, but from spectral regulation, especially for wireless
channels
...
org/content/m0074/2
...

234

CHAPTER 6
...
Any signal that can be transmitted by analog means can be sent by digital means,
with the only issue being the number of bits used in A/D conversion (how accurately do we need
to represent signal amplitude)
...
In addition to digital
communication’s ability to transmit a wider variety of signals than analog systems, point-to-point
digital systems can be organized into global (and beyond as well) systems that provide eﬃcient and
ﬂexible information transmission
...
Even analog-based networks, such as the telephone system, employ modern
computer networking ideas rather than the purely analog systems of the past
...

6
...
3: Fundamental
model of communication)
...
25 describes point-to-point communications
well, wherein the link between transmitter and receiver is straightforward, and they have the channel to
themselves
...
The key aspect, some would say ﬂaw, of this
model is that the channel is dedicated: Only one communications link through the channel is allowed for
all time
...

Communication Network
Source
Sink

Figure 6
...
Messages formed by the source are
transmitted within the network by dynamic routing
...
The longer one would be
used if the direct link were disabled or congested
...
Most communication networks, even modern ones, share many of its aspects
...

• This message is sent to the network by delivery to one of the network’s public entry points
...

• The communications network delivers the message in the most eﬃcient (timely) way possible, trying
not to corrupt the message while doing so
...
org/content/m0075/2
...

235
• The message arrives at one of the network’s exit points, and is delivered to the recipient (what we have
termed the message sink)
...
35
(Solution on p
...
)
Develop the network model for the telephone system, making it as analogous as possible with the
postal service-communications network metaphor
...
What they do care about is message integrity and communications
eﬃciency
...
Communication paths that form the
Internet use wireline, optical ﬁber, and satellite communication links
...
Here the network consisted of telegraph
operators who transmitted the message eﬃciently using Morse code and routed the message so that it took
the shortest possible path to its destination while taking into account internal network failures (downed lines,
drunken operators)
...
Morse code, which assigned a sequence of dots and dashes to each letter of
the alphabet, served as the source coding algorithm
...

Rather than a matched ﬁlter, the receiver was the operator’s ear, and he wrote the message (translating
from received bits to symbols)
...
4),
exceeded the Source Coding Theorem’s upper bound
...
However, many messages share the communications channel between nodes using what we call time-domain multiplexing: Rather than the
continuous communications mode implied in the Model as presented, message sequences are sent, sharing
in time the channel’s capacity
...
Routing in networks is necessarily dynamic: The complete route taken by messages is
formed as the network handles the message, with nodes relaying the message having some notion of the best
possible path at the time of transmission
...
Certainly in the case of the postal system dynamic routing occurs,
and can consider issues like inoperative and overly busy links
...
Modern communication networks
strive to achieve the most eﬃcient (timely) and most reliable information delivery system possible
...
34 Message Routing45
Focusing on electrical networks, most analog ones make ineﬃcient use of communication links because truly
dynamic routing is diﬃcult, if not impossible, to obtain
...
The telephone network is more dynamic,
but once it establishes a call the path through the network is ﬁxed
...
Telephone
network customers would be quite upset if the telephone company momentarily disconnected the path so
that someone else could use it
...

During the 1960s, it was becoming clear that not only was digital communication technically superior,
but also that the wide variety of communication modes—computer login, ﬁle transfer, and electronic mail—
needed a diﬀerent approach than point-to-point
...
Computer networks elaborate the basic
45 This

content is available online at ...
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...
INFORMATION COMMUNICATION

Receiver Address
Transmitter Address
Data Length (bytes)
Data

Error Check
Figure 6
...
A packet, like a letter, contains the destination address, the return address (transmitter address), and the data
...

network model by subdividing messages into smaller chunks called packets(Figure 6
...
The rationale for
the network enforcing smaller transmissions was that large ﬁle transfers would consume network resources
all along the route, and, because of the long transmission time, a communication failure might require
retransmission of the entire ﬁle
...
The analogy is that the postal service,
rather than sending a long letter in the envelope you provide, opens the envelope, places each page in a
separate envelope, and using the address on your envelope, addresses each page’s envelope accordingly, and
mails them separately
...

Communications networks are now categorized according to whether they use packets or not
...
Circuit switching has the advantage that once the route is determined,
the users can use the capacity provided them however they like
...
Packet-switched
networks continuously monitor network utilization, and route messages accordingly
...

6
...
25)—typiﬁes what are known as wide area networks(WANs)
...
“Long”
has no precise deﬁnition, and is intended to suggest that the communication links vary widely
...
Local area networks, LANs, employ
a single communication link and special routing
...
LANs connect
to other LANs and to wide area networks through special nodes known as gateways(Figure 6
...
In the
Internet, a computer’s address consists of a four byte sequence, which is known as its IP address (Internet Protocol address)
...
42
...
32: each byte is separated by a period
...
Computers are also addressed by a more
human-readable form: a sequence of alphabetic abbreviations representing institution, type of institution,
and computer name
...
42
...
32 is the same as soma
...
edu)
...
So-called name servers translate between
alphabetic and numerical forms, and the transmitting computer requests this translation before the message
is sent to the network
...
org/content/m0077/2
...

...
27: The gateway serves as an interface between local area networks and the Internet
...

L
Z0

Z0

Terminator

Terminator
Transceiver

Transceiver

Computer A

Computer B

Figure 6
...
Computers attach to the Ethernet
through an interface known as a transceiver because it sends as well as receives bit streams represented
as analog voltages
...
36 Ethernet48
Ethernet uses as its communication medium a single length of coaxial cable (Figure 6
...
This cable serves
as the “ether,” through which all digital data travel
...
28) through a device known as a transceiver
...
Conceptually it
consists of two op-amps, one applying a voltage corresponding to a bit stream (transmitting data) and another
serving as an ampliﬁer of Ethernet voltage signals (receiving data)
...
Computers are attached in
parallel, resulting in the circuit model for Ethernet shown in Figure 6
...

Exercise 6
...
259
...
Without
scheduling authority, you might well wonder how one computer sends to another without the (large) interference that the other computers would produce if they transmitted at the same time
...
This method relies on the
fact that all packets transmitted over the coaxial cable can be received by all transceivers, regardless of
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INFORMATION COMMUNICATION

xA(t)

Rout

Z0

rA(t)

Transceiver
Rout
xA(t)

+
–

xB(t)

Coax

Rout
+
–

… x (t)
Z

Rout
+

Z0

–

Figure 6
...
The output
resistance Rout must be much larger than Z0 so that the sum of the various transmitter voltages add to
create the Ethernet conductor-to-shield voltage that serves as the received signal r (t) for all transceivers
...

which computer might actually be the intended recipient
...
Each computer goes through the following steps to send a packet
...
The computer senses the voltage across the cable to determine if some other computer is transmitting
...
If another computer is transmitting, wait until the transmissions ﬁnish and go back to the ﬁrst step
...

3
...

4
...
The condition wherein two (or more)
computers’ transmissions interfere with others is known as a collision
...
The longest time any
computer must wait to determine if its transmissions do not encounter interference is 2L , where L is the
c
coaxial cable’s length
...
Assuming a propagation
speed of 2/3 the speed of light, this time interval is more than 10 µs
...
31, the
number of these time intervals required to resolve the collision is, on the average, less than two!
Exercise 6
...
259
...

Thus, despite not having separate communication paths among the computers to coordinate their transmissions, the Ethernet random access protocol allows computers to communicate without only a slight degradation in eﬃciency, as measured by the time taken to resolve collisions relative to the time the Ethernet is
used to transmit information
...
The time required to transmit such
packets equals Pmin , where C is the Ethernet’s capacity in bps
...
If
the minimum transmission time is such that the beginning of the packet has not propagated the full length
of the Ethernet before the end-of-transmission, it is possible that two computers will begin transmission at
the same time and, by the time their transmissions cease, the other’s packet will not have propagated to
the other
...
63)
Pmin >
c
Thus, for the 10 Mbps Ethernet having a 1 km maximum length speciﬁcation, the minimum packet size is
200 bits
...
38
(Solution on p
...
)
The 100 Mbps Ethernet was designed more recently than the 10 Mbps alternative
...
37 Communication Protocols49
The complexity of information transmission in a computer network—reliable transmission of bits across
a channel, routing, and directing information to the correct destination within the destination computers
operating system—demands an overarching concept of how to organize information delivery
...
For example, random access issues in Ethernet are not present in wide-area networks such as
the Internet
...
For example, to use
the telephone network, the protocol is to pick up the phone, listen for a dial tone, dial a number having a
speciﬁc number of digits, wait for the phone to ring, and say hello
...
In technical terms, no one protocol or set of protocols can be used for
any communication situation
...
This grand design of information transmission
organization runs through all modern networks today
...
As
shown in Figure 6
...
Segregation
of information transmission, manipulation, and interpretation into these categories directly aﬀects how communication systems are organized, and what role(s) software systems fulﬁll
...

Exercise 6
...
259
...
IP abbreviates Internet protocol, and governs
gateways (how information is transmitted between networks having diﬀerent internal organizations)
...
Telnet is a protocol that concerns how a person at one computer logs on to another computer
across a network
...
Rather, it establishes
connections between computers and directs each byte (presumed to represent a typed character) to the
49 This

content is available online at ...
19/>
...
INFORMATION COMMUNICATION

Application
http

Presentation

telnet

Session

tcp

Transport

ip

Network

ecc

Data Link

signal set

detail

Physical

ISO Network Protocol Standard
Figure 6
...
Protocols at the lower levels (shown toward the bottom) concern reliable bit transmission
...

Bodies such as the IEEE (Institute for Electronics and Electrical Engineers) and the ISO (International
Standards Organization) deﬁne standards such as this
...

appropriate operation system component at each end
...
That aspect of information transmission is left to protocols at
higher layers
...
These protocols exist independently
of the Internet
...
HTTP (hypertext transfer protocol) frame what messages
contain and what should be done with the data
...

6
...
1: Signals on Transmission Lines
A modulated signal needs to be sent over a transmission line having a characteristic impedance of Z0 = 50Ω
...
5 MHz
...
An
op-amp ﬁlter (Figure 6
...

a) What is the transfer function between the input voltage and the voltage across the transmission line?
b) Find values for the resistors and capacitors so that design goals are met
...
org/content/m10352/2
...

241

R2

R1

C1
–
+

+

Vin

C2

Z0

–

Figure 6
...
2: Noise in AM Systems
The signal s (t) emerging from an AM communication system consists of two parts: the message signal, s (t),
and additive noise
...
32) shows the message spectrum S (f ) and noise power spectrum
PN (f )
...

2

S(f)

PN(f)

A
N0/2

A/2
–W

W

f

–W

W

f

Figure 6
...
What is the signal-to-noise ratio in the upper half of the frequency band?
c) A clever ELEC 241 student suggests ﬁltering the message before the transmitter modulates it so that
the signal spectrum is balanced (constant) across frequency
...
Draw a block diagram of this communication system
...
3: Complementary Filters
Complementary ﬁlters usually have “opposite” ﬁltering characteristics (like a lowpass and a highpass)
and have transfer functions that add to one
...
Each
output can then be transmitted separately and the original signal reconstructed at the receiver
...

242

CHAPTER 6
...

c) What is the receiver’s signal-to-noise ratio? How does it compare to the standard system that sends
the signal by simple amplitude modulation?

Problem 6
...
In this problem, the phase deviation is small
...

a) What is the transmission bandwidth?
b) Find a receiver for this modulation scheme
...

Problem 6
...
The issue concerns the discrete-time signal
s (n) cos (2πf0 n), where the signal s (n) has no special characteristics and the modulation frequency f0 is
known
...
Samantha says that approach won’t work
...
He tells them that if s (n) cos (2πf0 n) and
s (n) sin (2πf0 n) were both available, s (n) can be recovered
...
6: Anti-Jamming
One way for someone to keep people from receiving an AM transmission is to transmit noise at the same
carrier frequency
...
The noise n (t) has a constant power density spectrum
over the bandwidth of the message m (t)
...

2
a) What would be the output of a traditional AM receiver tuned to the carrier frequency fc ?
b) RU Electronics proposes to counteract jamming by using a diﬀerent modulation scheme
...
33)
...

c) The jammer, unaware of the change, is transmitting with a carrier frequency of fc , while the receiver
tunes a standard AM receiver to a harmonic of the carrier frequency
...
33

Problem 6
...
“Random switching” means that one carrier frequency is used for some period of time,
switches to the other for some other period of time, back to the ﬁrst, etc
...
Consequently, the receiver must be
designed to receive the transmissions regardless of which carrier frequency is used
...
The channel adds white noise of spectral height N0
...
8: AM Stereo
Stereophonic radio transmits two signals simultaneously that correspond to what comes out of the left and
right speakers of the receiving radio
...
An amazing aspect of AM stereo is that both signals are transmitted within the
same bandwidth as used to transmit just one
...

x (t) = A (1 + ml (t)) cos (2πfc t) + Amr (t) sin (2πfc t)
a) Find the Fourier transform of x (t)
...
34
...

c) Assume the channel adds white noise to the transmitted signal
...

cos 2πfct

×

LPF
W Hz

×

x(t)

LPF
W Hz

BPF

sin 2πfct
Figure 6
...
INFORMATION COMMUNICATION

Problem 6
...
35) has, despite its complexity,
advantages over the usual amplitude modulation system
...
The channel attenuates the transmitted signal x (t) and adds white
noise of spectral height N0
...
35


 j if f < 0
The transfer function H (f ) is given by H (f ) =
 −j if f > 0
a)
b)
c)
d)

Find an expression for the spectrum of x (t)
...

Show that the usual coherent receiver demodulates this signal
...

Find a superior receiver (one that yields a better signal-to-noise ratio), and analyze its performance
...
10: Multi-Tone Digital Communication
In a so-called multi-tone system, several bits are gathered together and transmitted simultaneously on
diﬀerent carrier frequencies during a T second interval
...
64)

k=0

Here, f0 is the frequency oﬀset for each bit and it is harmonically related to the bit interval T
...

a) Find a receiver for this transmission scheme
...

T
He samples the received signal (sampling interval Ts = N )
...
How would you recommend he implement the receiver?

Problem 6
...
As shown in Figure 6
...

245

Reflected
Path
Direct Path

Transmitter

Figure 6
...
5 times as long
...
Find and sketch the magnitude of the transfer function for the multipath component
of the channel
...
1 MHz for radio), be aﬀected or not?
Analog cellular telephone uses amplitude modulation to transmit voice
...

Problem 6
...
Rather than send signals at diﬀerent frequencies, a clever Rice engineer
suggests using a diﬀerent signal set for each data stream
...
37)
...
37

Thus, bits are represented in data stream 1 by s1 (t) and − (s1 (t)) and in data stream 2 by s2 (t) and
− (s2 (t)), each of which are modulated by 900 MHz carrier
...
Each receiver uses a matched ﬁlter for its receiver
...

a) What is the block diagram describing the proposed system?
b) What is the transmission bandwidth required by the proposed system?

246

CHAPTER 6
...
13: Mixed Analog and Digital Transmission
A signal m (t) is transmitted using amplitude modulation in the usual way
...
In addition to sending this analog signal, the transmitter also wants to
send ASCII text in an auxiliary band that lies slightly above the analog transmission band
...
The transmission signal spectrum is as shown (Figure 6
...

X(f)
B
2W
analog

digital

fc

f

Figure 6
...

b) What is the maximum datarate the scheme can provide in terms of the available bandwidth?
c) Find a receiver that yields both the analog signal and the bit stream
...
14: Digital Stereo
Just as with analog communication, it should be possible to send two signals simultaneously over a digital
channel
...
1 kHz with 16 bits/sample)
...
If b(1) (n) and b(2) (n) each represent a
bit stream, the transmitted signal has the form
b(1) (n) sin (2πfc (t − nT )) p (t − nT ) + b(2) (n) cos (2πfc (t − nT )) p (t − nT )

x (t) = A
n

where p (t) is a unit-amplitude pulse having duration T and b(1) (n), b(2) (n) equal either +1 or -1 according
to the bit being transmitted for each signal
...

a) What value would you choose for the carrier frequency fc ?
b) What is the transmission bandwidth?
c) What receiver would you design that would yield both bit streams?

Problem 6
...
We want to compare the
resulting quality of the received signals
...
Assume the speech signal has a 4 kHz bandwidth
and, in the digital case, is sampled at an 8 kHz rate with eight-bit A/D conversion
...

a) What is the transmission bandwidth of the analog (AM) and digital schemes?
b) Assume the speech signal’s amplitude has a magnitude less than one
...
However, errors in each bit have a diﬀerent impact on the error in
0
the reconstructed speech sample
...

d) In the digital case, the recovered speech signal can be considered to have two noise sources added
to each sample’s true value: One is the A/D amplitude quantization noise and the second is due to
channel errors
...
What
is the signal-to-noise ratio of the received speech signal as a function of pe ?
e) Compute and plot the received signal’s signal-to-noise ratio for the two transmission schemes for a few
values of channel signal-to-noise ratios
...

Problem 6
...

Letter
a
b
c
d
e

Probability
0
...
25
0
...
0625
0
...
5

a) Find this source’s entropy
...

c) Find an unequal-length codebook for this sequence that satisﬁes the Source Coding Theorem
...
17: Source Compression
Consider the following 5-letter source
...
4
0
...
15
0
...
1

Table 6
...
INFORMATION COMMUNICATION

a) Find this source’s entropy
...

c) Find the Huﬀman code for this source
...
18: Speech Compression
When we sample a signal, such as speech, we quantize the signal’s amplitude to a set of integers
...
Although these integers could be represented by
a binary code for digital transmission, we should consider whether a Huﬀman coding would be more eﬃcient
...
mat
...
To simulate a 3-bit converter, we use Matlab’s round function to create quantized amplitudes
corresponding to the integers [0 1 2 3 4 5 6 7]
...
5*y + 3
...
The following Matlab program
computes the number of times each quantized value occurs
...

b) Find the Huﬀman code for this source
...
19: Digital Communication
In a digital cellular system, a signal bandlimited to 5 kHz is sampled with a two-bit A/D converter at its
Nyquist frequency
...

Sample Value

Probability

0

0
...
35

2

0
...
2
Table 6
...
39)
...
39

t
-A/2

t

249
a) What is the datarate of the compressed source?
b) Which choice of signal set maximizes the communication system’s performance?
c) With no error-correcting coding, what signal-to-noise ratio would be needed for your chosen signal set
to guarantee that the bit error probability will not exceed 10−3 ? If the receiver moves twice as far
from the transmitter (relative to the distance at which the 10−3 error rate was obtained), how does
the performance change?

Problem 6
...

Letter

Probability

a

1/3

b

1/3

c

1/4

d

1/12
Table 6
...
Find an
error correcting code for two-bit data blocks that corrects all single-bit errors
...

Problem 6
...
An example
(Figure 6
...

…

…

d
Figure 6
...
In retail stores, laser scanners read this code, and after
accessing a database of prices, enter the price into the cash register
...
Now how many bars are needed to represent each digit?
c) What is the probability that the 11-digit code is read correctly if the probability of reading a single
bit incorrectly is pe ?

250

CHAPTER 6
...
22: Error Correcting Codes
A code maps pairs of information bits into codewords of length-5 as follows
...
9

a) What is this code’s eﬃciency?
b) Find the generator matrix G and parity-check matrix H for this code
...
How many patterns of one, two, and three errors are correctly
decoded?
d) What is the block error probability (the probability of any number of errors occurring in the decoded
codeword)?

Problem 6
...

letter

a

b

c

d

e

f

g

h

i

probability

1
4

1
8

1
8

1
8

1
8

1
16

1
16

1
16

1
16

Table 6
...
How does the resulting code compare with the best
possible code?
b) A clever engineer proposes the following (6, 3) code to correct errors after transmission through a digital
channel
...
What kind of code should be used to transmit data over
this channel?

Problem 6
...
He decides to represent 3-bit data with 6-bit codewords in which
none of the data bits appear explicitly
...

b) Find a 3 × 6 matrix that recovers the data bits from the codeword
...
25: Error Correction?
It is important to realize that when more transmission errors than can be corrected, error correction algorithms believe that a smaller number of errors have occurred and correct accordingly
...

a) How many double-bit errors can occur in a codeword?
b) For each double-bit error pattern, what is the result of channel decoding? Express your result as a
binary error sequence for the data bits
...
26: Selective Error Correction
We have found that digital transmission errors occur with a probability that remains constant no matter
how “important” the bit may be
...
Yet, the former errors have a much larger
impact on the overall signal-to-noise ratio than the latter
...
We use single-bit error correction on the most signiﬁcant four
bits and none on the least signiﬁcant four
...

a) How many error correction bits must be added to provide single-bit error correction on the most
signiﬁcant bits?
b) How large must the signal-to-noise ratio of the received signal be to insure reliable communication?
c) Assume that once error correction is applied, only the least signiﬁcant 4 bits can be received in error
...
INFORMATION COMMUNICATION

Problem 6
...
Very few errors are due to noise in the compact disk player;
most occur because of dust and scratches on the disk surface
...
Assume that scratch and
dust-induced errors are four or fewer consecutive bits long
...
1 kHz
analog-to-digital conversion of each channel of the stereo analog signal
...
As the cartoon (Figure 6
...
The CD player de-interleaves
the coded data, then performs error-correction
...

sample n
sample n+1
sample n+2
sample n+3

1111
1010
0000

4-way
interleaver

1100100111001001

0101
Figure 6
...
28: Communication System Design
RU Communication Systems has been asked to design a communication system that meets the following
requirements
...

• The RUCS engineers ﬁnd that the entropy H of the sampled message signal depends on how many
bits b are used in the A/D converter (see table below)
...

• Once received, the message signal must have a signal-to-noise ratio of at least 20 dB
...
19

4

3
...
28

6

5
...
13

Can these speciﬁcations be met? Justify your answer
...
29: HDTV
As HDTV (high-deﬁnition television) was being developed, the FCC restricted this digital system to use in
the same bandwidth (6 MHz) as its analog (AM) counterpart
...
The least-acceptable picture received by television
sets located at an analog station’s broadcast perimeter has a signal-to-noise ratio of about 10 dB
...
30: Digital Cellular Telephones
In designing a digital version of a wireless telephone, you must ﬁrst consider certain fundamentals
...
The signal-to-noise ratio of the allocated wireless channel, which has a 5 kHz bandwidth, measured
100 meters from the tower is 70 dB
...
Can a digital cellphone system be
designed according to these criteria?
Problem 6
...
The access
algorithm works as follows
...

• If none or more than one head comes up, the N computers will either remain silent (no heads) or a
collision will occur (more than one head)
...

a) What is the optimal probability to use for ﬂipping the coin? In other words, what should p be to
maximize the probability that exactly one computer transmits?
b) What is the probability of one computer transmitting when this optimal value of p is used as the
number of computers grows to inﬁnity?
c) Using this optimal probability, what is the average number of coin ﬂips that will be necessary to resolve
the access so that one computer successfully transmits?
d) Evaluate this algorithm
...
32: Repeaters
Because signals attenuate with distance from the transmitter, repeaters are frequently employed for both
analog and digital communication
...
42)
...
However, the signal the repeater receives contains white noise as well as the
transmitted signal
...

transmitter

repeater

D/2

receiver

D/2
D
Figure 6
...
INFORMATION COMMUNICATION

a) What is the block diagram for this system?
b) For an amplitude-modulation communication system, what is the signal-to-noise ratio of the demodulated signal at the receiver? Is this better or worse than the signal-to-noise ratio when no repeater is
present?
c) For digital communication, we must consider the system’s capacity
...
33: Designing a Speech Communication System
We want to examine both analog and digital communication alternatives for a dedicated speech transmission
system
...
The wireless link between transmitter and receiver
is such that 200 watts of power can be received at a pre-assigned carrier frequency
...
1 watt/kHz
...
What is the received signal-to-noise
ratio under these design constraints?
b) How many bits must be used in the A/D converter to achieve the same signal-to-noise ratio?
c) Is the bandwidth required by the digital channel to send the samples without error greater or smaller
than the analog bandwidth?

Problem 6
...
Analog
You are the Chairman/Chairwoman of the FCC
...
5 MHz has been allocated
for a new “high-quality” AM band
...

a) How many stations can be allocated to this band and with what carrier frequencies?
b) Looking ahead, conversion to digital transmission is not far in the future
...

c) Without employing compression, how many digital radio stations could be allocated to the band if
each station used BPSK modulation? Evaluate this design approach
...
1 (p
...
For coaxial cable, c =
For twisted pair, c =

√1
µ

d
arccosh( 2r )
d
( 2r )

δ
2r +arccosh

√ 1
µd

d

...

Solution to Exercise 6
...
201)
You can ﬁnd these frequencies from the spectrum allocation chart (Section 7
...
Light in the middle of the
visible band has a wavelength of about 600 nm, which corresponds to a frequency of 5 × 1014 Hz
...

Thus, the visible electromagnetic frequencies are over six orders of magnitude higher!
Solution to Exercise 6
...
202)
As shown previously (6
...
The inverse-square law
governs free-space propagation because such propagation is lossless, with the inverse-square law a consequence
of the conservation of power
...

Solution to Exercise 6
...
203)

d1

d2

h1

h2
R

R

R

Figure 6
...
The line-of-sight
distance between two earth-based antennae equals
dLOS =

2h1 R + h1 2 +

2h2 R + h2 2

(6
...
If one antenna is at ground elevation, say h2 = 0, the other antenna’s range is 2h1 R
...
5 (p
...
Assuming a distance between the two of 80 km, the relation λf = c gives a corresponding
frequency of 3
...
Such low carrier frequencies would be limited to low bandwidth analog communication
and to low datarate digital communications
...

Solution to Exercise 6
...
204)
Transmission to the satellite, known as the uplink, encounters inverse-square law power losses
...
Reﬂection is the same as transmitting
exactly what arrives, which means that the total loss is the product of the uplink and downlink losses
...
The ionosphere begins at an altitude of about 50 km
...
8 × 10−8 ; for Marconi, it was proportional to
4
...
Marconi was very lucky
...
7 (p
...

256

CHAPTER 6
...
8 (p
...

Solution to Exercise 6
...
208)
The signal-related portion of the transmitted spectrum is given by X (f ) = 1 M (f − fc ) + 1 M (f + fc )
...

1
1
1
1
X (f − fc ) + X (f + fc ) = (M (f − 2fc ) + M (f )) + (M (f + 2fc ) + M (f ))
2
2
4
4
(6
...

Solution to Exercise 6
...
208)
The key here is that the two spectra M (f − fc ), M (f + fc ) do not overlap because we have assumed
that the carrier frequency fc is much greater than the signal’s highest frequency
...

Solution to Exercise 6
...
209)
Separation is 2W
...
Speech is well contained in this bandwidth,
much better than in the telephone!
Solution to Exercise 6
...
210)
∞
x (t) = n=−∞ sb(n) (t − nT )
...
13 (p
...

Solution to Exercise 6
...
211)
x (t) =

(−1)
n

b(n)

ApT (t − nT ) sin

2πkt
T

Solution to Exercise 6
...
212)
The harmonic distortion is 10%
...
16 (p
...

Solution to Exercise 6
...
214)
In BPSK, the signals are negatives of each other: s1 (t) = − (s0 (t))
...
Choosing the largest therefore amounts to
choosing which one is positive
...
If it is positive, we are done
...

Solution to Exercise 6
...
215)
2

The matched ﬁlter outputs are ± A2T because the sinusoid has less power than a pulse having the same
amplitude
...
19 (p
...
Stated in terms of Eb , the
diﬀerence equals 2αEb just as in the baseband case
...
20 (p
...
The noise power remains the same as
2
in the BPSK case, which from the probability of error equation (6
...

257
Solution to Exercise 6
...
219)
1
1
1
Equally likely symbols each have a probability of K
...
To
k K log2 K
prove that this is the maximum-entropy probability assignment, we must explicitly take into account that
probabilities sum to one
...
Pr [a0 ] appears twice in the entropy formula: the terms Pr [a0 ] log2 (Pr [a0 ]) and (1 − Pr [a0 ] + · · · + Pr [aK−2 ]) log2 (1 − Pr [a0 ] + · · · + Pr [aK−2 ])
...
The derivative equals
log2 (Pr [a0 ]) − log2 (1 − Pr [a0 ] + · · · + Pr [aK−2 ]), and all other derivatives have the same form (just substitute your letter’s index)
...
For the minimum
entropy answer, one term is 1log2 1 = 0, and the others are 0log2 0, which we deﬁne to be zero also
...

Solution to Exercise 6
...
222)
The Huﬀman coding tree for the second set of probabilities is identical to that for the ﬁrst (Figure 6
...
The average code length is 2 1 + 1 2 + 5 3 + 20 3 = 1
...
The entropy calculation
4
1
1
1
1
1
1
1
1
is straightforward: H (A) = − 2 log 2 + 4 log 4 + 5 log 5 + 20 log 20 , which equals 1
...

Solution to Exercise 6
...
222)
1
T = B(A)R
...
24 (p
...
Note that we must start at the beginning of the bit stream; jumping into the middle does
not guarantee perfect decoding
...

Solution to Exercise 6
...
222)
Consider the bitstream
...
taken from the bitstream 0|10|110|110|111|
...
If we had a ﬁxed-length code (say 00,01,10,11), the situation
is much worse
...
26 (p
...
Because this is an upward-going
parabola, we need only check where its roots are
...
Consequently in the range 0 ≤ pe ≤ 1 the error rate produced by coding is smaller
...
27 (p
...
47):
pe = Q

2α2 Eb
N0

2

3

...
Plotting this reveals that the increase in bit-error probability out of the channel
3N0
because of the energy reduction is not compensated by the repetition coding
...
INFORMATION COMMUNICATION

10 0

Error Probability with and without (3,1) Repetition Coding

10 -1

Error Probability

Coded
Uncoded
10 -2

10 -3

10 -4 0
10

10 1
Signal-to-Noise Ratio
Figure 6
...
28 (p
...
2), adding 0 to a binary value results in that binary value while adding 1
results in the opposite binary value
...
29 (p
...
30 (p
...

Because the code is linear—sum of any two codewords is a codeword—we can generate all codewords as
sums of columns of G
...

Solution to Exercise 6
...
229)
In binary arithmetic see this table51 , adding 0 to a binary value results in that binary value while adding 1
results in the opposite binary value
...
32 (p
...
org/content/m0095/latest/#table1>

259
ability

N
N −3
pe 3 (1 − pe )

...
31
...
33 (p
...
The number of non-zero vectors
2
resulting from Hc must equal or exceed the sum of these two numbers
...
68)

The ﬁrst two solutions that attain equality are (5,1) and (90,78) codes
...
(Perfect codes satisfy relations like (6
...
)
Solution to Exercise 6
...
233)
To convert to bits/second, we divide the capacity stated in bits/transmission by the bit interval duration T
...
35 (p
...
Dialing the
telephone number informs the network of who will be the message recipient
...
Your friend receives the message via the
same device—the handset—that served as the network entry point
...
36 (p
...
The transfer function to some other transceiver’s receiver circuit
is Rout divided by this load
...
37 (p
...

Transmitters must sense a collision before packet transmission ends
...

Solution to Exercise 6
...
239)
The cable must be a factor of ten shorter: It cannot exceed 100 m
...

Solution to Exercise 6
...
239)
When you pick up the telephone, you initiate a dialog with your network interface by dialing the number
...
The route remains ﬁxed as long as the call persists
...

260

CHAPTER 6
...
1 Decibels1
The decibel scale expresses amplitudes and power values logarithmically
...

power (s, in decibels) = 10 log10
amplitude (s, in decibels) = 20 log10

power (s)
power (s0 )
amplitude (s)
amplitude (s0 )

(7
...
Quantifying
power or amplitude in decibels essentially means that we are comparing quantities to a standard or that we
want to express how they changed
...

Exercise 7
...
265
...
Who is this measure named for?
The consistency of these two deﬁnitions arises because power is proportional to the square of amplitude:
power (s) ∝ amplitude2 (s)

(7
...
3)

Because of this consistency, stating relative change in terms of decibels is unambiguous
...
Converting decibel values back and forth is fun,
and tests your ability to think of decibel values as sums and/or diﬀerences of the well-known values and of
ratios as products and/or quotients
...

√
For example, to ﬁnd the decibel value for 2, we halve the decibel value for 2; 26 dB equals 10 + 10 + 6 dB
that corresponds to a ratio of 10 × 10 × 4 = 400
...

One reason decibels are used so much is the frequency-domain input-output relation for linear systems:
Y (f ) = X (f ) H (f )
...
org/content/m0082/2
...

261

262

CHAPTER 7
...
5
3

10

5

4

6

5

7

8

9

10

10

0
...
1: Common values for the decibel
...

amplitude at a given frequency we simply add the ﬁlter’s gain in decibels (relative to a reference of one) to the
input amplitude at that frequency
...

7
...
2
...
For example, you select 6 numbers out
of 60
...
The chances of winning equal the number of diﬀerent length-k sequences that can be chosen
...
Now the order matters,
and many more choices are possible than when order does not matter
...
In digital communications, for
example, you might ask how many possible double-bit errors can occur in a codeword
...
Solving these kind of
problems amounts to understanding permutations- the number of ways of choosing things when order
matters as in baseball lineups - and combinations- the number of ways of choosing things when order does
not matter as in lotteries and bit errors
...
If we are to pick k numbers from a pool of n, we have n choices
for the ﬁrst one
...
The number of length-two ordered sequences is
therefore be n (n − 1)
...
(n − k + 1)
...
1
...

When order does not matter, the number of combinations equals the number of permutations divided by
the number of orderings
...
Thus, once we
choose the nine starters for our baseball game, we have 9! = 362, 880 diﬀerent lineups! The symbol for the
n!
combination of k things drawn from a pool of n is n and equals (n−k)!k!
...
2
(Solution on p
...
)
What are the chances of winning the lottery? Assume you pick 6 numbers from the numbers 1-60
...
For example, Newton derived that the n-th power of a sum
n
obeyed the formula (x + y) = n xn + n xn−1 y + n xn−2 y 2 + · · · + n y n
...
org/content/m10262/2
...

263
Exercise 7
...
265
...
The probability of any particular two-bit error sequence is
n−2
p2 (1 − p)

...
Note that the probability that zero or one or two, etc
...
Can you prove this?
0 (1 − p) + 1 p(1 − p)
2
n

7
...
With increased use of the radio spectrum for both public and private use, this regulation has become
increasingly important
...
Detailed radio carrier frequency assignments are much
too detailed to present here
...
org/content/m0083/2
...

264

CHAPTER 7
...
5
2190
...
AND TIME SIGNAL

2501

AERONAUTICAL
RADIONAVIGATION

STANDARD FREQ
...
7
29
...
89
29
...
0

2107

STANDARD FREQ
...
01
25
...
21
25
...
55
25
...
1
26
...
48
26
...
96
27
...
41
27
...
0

3 MHz

AMATEUR

24
...
99
25
...
855
23
...
2
23
...
2

300

ISM – 27
...
163 MHz

3000

30 MHz

2900

Radiolocation

Space Research
STANDARD FREQ
...
AND TIME SIGNAL(25,000 kHz)

110

130
2000

MOBILE**

AMATEUR SATELLITE

235
...
45
21
...
924
22
...
0
222
...
0

1900

Mobile

Amateur

2483
...
0

Radiolocation

2417
2450

FIXED

EARTH
EXPLORATION
SAT
...
68
19
...
990
19
...
005
20
...
0

Radiolocation
Radiolocation

1625

1705

AMATEUR SATELLITE

BROADCASTING
FIXED
AERONAUTICAL MOBILE (R)

2655
BROADCASTING
SATELLITE

SPACE RESEARCH
(Passive)

MARITIME
RADIONAVIGATION

3

FIXED

MOBILE

30
...
Astronomy (Passive)
(Passive)

Space Research

STANDARD FREQ
...
5
FIXED SATELLITE (E-S)

Amateur

MOBILE

BROADCASTING
SATELLITE

RADIO
ASTRONOMY

27
...
5

Radiolocation

FIXED

RADIODETERMINATION SAT
...
25
25
...
0
24
...
6

MARITIME Aeronautical
Mobile
MOBILE

2300

Fixed

AERONAUTICAL
RADIONAVIGATION

Earth
Exploration
Satellite (S-S)

MOBILE

MOBILE

FIXED

300
...
0

Amateur

2290

2310
BROADCASTING
SATELLITE

RADIOLOCATION

METEOROLOGICAL
AIDS

265
...
55
EARTH EXPL
...
(Passive)

AMATEUR SATELLITE

RADIOLOCATION

MOBILE**

Radio- Fixed
location

AMATEUR
Amateur
AMATEUR

MOBILE
SPACE RES
...
(s-E)(s-s)

BROADCASTING

BROADCASTING

AMATEUR

MARITIME MOBILE
FIXED
Space Research
STAND
...
& TIME SIG
...
0

MOBILE

FIXED

MOBILE

FIXED

23
...

Satellite (Active)

Land Mobile
MOBILE

1850

17
...
97
18
...
068
18
...
78
18
...
0

FIXED
FIXED

17
...
55

AERONAUTICAL MOBILE (R)
AERONAUTICAL MOBILE (OR)
FIXED
AMATEUR
AMATEUR SATELLITE
Mobile
FIXED
MARITIME MOBILE

173
...
4
174
...
36

FIXED
BROADCASTING

59

70

FIXED

248
...
6

RADIOLOCATION

241
...
5
22
...
990
15
...
010
15
...
0
238
...
0
22
...
0
149
...
05
150
...
2475
157
...
1875
157
...
575
161
...
775
162
...
4

ISM – 2450
...
2

METEOROLOGICAL
SATELLITE (s-E)

1646
...
5
1668
...
AND TIME SIGNAL(15,000 kHz)
Space Research
STANDARD FREQ
...
EARTH EXPL
...
(Passive)

MOBILE

AMATEUR SATELLITE
SPACE RES
...

RADIO
MOBILE** ASTRONOMY RESEARCH SAT
...
125 ± 0
...
0 ± 1GHz

MOBILE

FIXED
RADIOLOCATION

Radiolocation

EARTH EXPL
...
(Passive)

MOBILE

FIXED

EARTH
EXPLORATION
SATELLITE
(Passive)

SPACE
RESEARCH
(Passive)

RADIO
ASTRONOMY

FIXED

FIXED
SATELLITE (S-E)

20
...

(Passive)

FIXED

231
...
7

METEOROLOGICAL
SATELLITE (s-E) ††

METEOROLOGICAL
AIDS (RADIOSONDE)
METEOROLOGICAL
AIDS (Radiosonde) ††

FIXED

Standard
Frequency and
Time Signal
Satellite (S-E)

202
...
0

FIXED

MOBILE
SATELLITE (S-E)

AERO
...

SPACE RESEARCH (Passive)

MOBILE

Fixed

200
...
8

MOBILE

FIXED SATELLITE (S-E)

FIXED
SATELLITE
(E-S)

MOBILE

FIXED

PLEASE NOTE: THE SPACING ALLOTED THE SERVICES IN THE
SPECTRUM SEGMENTS SHOWN IS NOT PROPORTIONAL TO THE
ACTUAL AMOUNT OF SPECTRUM OCCUPIED
...
EXPLORATION SAT
...
6

SPACE
EARTH EXPL
...
(Passive) (Passive) SAT
...
7
17
...
0

17
...
0

190
...
1
17
...
6

MET SAT
...

(s-E)

176
...
AND TIME SIGNAL (60 kHz)

13
...
0
14
...
35

AMATEUR

146
...
2
13
...
36
13
...
6

AMATEUR SATELLITE

144
...
(E-S) (S-E)
FIXED
SATELLITE
(S-E)

EARTH
SPACE RESEARCH EXPLORATION
(Passive)
SATELLITE (Passive)

FIXED

Radiolocation

Radiolocation
RADIOLOCATION

FIXED SATELLITE (E-S)

174
...

Space Res
...
0

Aeronautical
Radionavigation

EARTH

INTERMOBILE RESEARCH SATELLITE EXPLORATION
SAT
...
0

1558
...
6
1613
...
5
1645
...
0

MOBILE SATELLITE (E-s)

136
...
0
137
...
175
137
...
0

AMATEUR SATELLITE

AMATEUR

FIXED

INTERSATELLITE

MARITIME MOBILE SATELLITE (E-s)

1549
...
0125

AERONAUTICAL
MOBILE (R)
MET
...
(S-E)
MET
...
(S-E)
MET
...
(S-E)
MET
...
(S-E)

MARITIME
MOBILE

E
RC

M
NA
O
TIONAL TELEC

10 25Hz

3 x 10-7Å

SPACE RES
...
35
15
...
7

1850-1910 AND 1930-1990 MHzARE ALLOCA TED TO PCS; 1910-1930 MHz
IS DESIGNATED FOR UNLICENSED PCS DEVICES

RADIO
ASTRONOMY

FIXED
FIXED

AERONAUTICAL RADIONAVIGATION RADIONAV
...
RADIONAVIGATION RADIO DET
...
(E-S) MOBILE SAT
...
RADIONAV
...
SAT
...
(E-S) RADIO ASTRONOMY
AERO
...
RADIO DET
...
(E-S) MOBILE SAT
...
(S-E)

15
...
Satellite
Space Operations
(S-E)
(S-E)
SPACE RES
...
(S-E)
SPACE RES
...
(S-E)
SPACE RES
...
(S-E)
SPACE RES
...
(S-E)

Space Research
(S-E)
MOB
...
(S-E)
Mob
...
(S-E)
MOB
...
(S-E)
Mob
...
(S-E)

12
...
7145

Space Research
EARTH EXPL
...

(Passive)

1530

1545

Mobile Satellite (S- E)

128
...
0

Aeronautical
Radionavigation
(Radio Beacons)

150
...
SAT
...

(Passive)
(Passive)

MOBILE

FIXED
Land Mobile 14
...
(E-S) Satellite (E-S) 14
...
0

Aeronautical
Mobile

-4
Å3

FIXED
SATELLITE (S-E)

MOBILE SATELLITE (S-E)
AERONAUTICAL MOBILE SATELLITE (R)
(space to Earth)
AERONAUTICAL MOBILE SATELLITE (R)
(space to Earth)

12
...
2

1429
1432
1435

FIXED

1022Hz

Land Mobile Satellite (E-S)

11
...
9375
123
...
5875

AERONAUTICAL
MOBILE (R)
AERONAUTICAL MOBILE (R)
137-139 SPACE RESEARCH (SPACE TO EARTH)

Radiolocation

MOBILE
SATELLITE

RADIONAVIGATION

x

RADIONAVIGATION
SATELLITE

MOBILE

-3
Å3

1021Hz

MOBILE

Fixed (TLM) Land Mobile (TLM & TLC)
Fixed (TLM) Land Mobile (TLM & TLC)

MOBILE

MOBILE
(AERONAUTICAL TELEMETERING)

11
...
275
11
...

MARITIME MOBILE SAT

...
TLM)
(Space to Earth)
(Space to Earth)
MARITIME MOBILE SATELLITE
MOBILE SATELLITE (S-E)
(space to Earth)

AERONAUTICAL MOBILE (OR)
AERONAUTICAL MOBILE (R)

1400
1427

30

MARITIME
MOBILE

x

FIXED

Fixed

1390

9
...
995
10
...
005
10
...
15

FIXED

1

Amateur Satellite

FIXED

14
...
(E-S) Satellite (E-S)

FIXED SATELLITE (E-S)

AMATEUR SATELLITE 144
...
25
13
...
0

AMATEUR
RADIOLOCATION

FIXED

FIXED

Space Research (E-S)

AERONAUTICAL RADIONAV
...
and
Space
Research
Time Signal
Satellite (E-S)

SPACE
OPERATION
(E-S) (1999)

12
...
0

FIXED

MOBILE

kHz

BROADCASTING
FIXED
STANDARD FREQ
...

AERONAUTICAL MOBILE (R)
AMATEUR

ISM – 13
...
007 MHz

INTERSATELLITE

MOBILE

RADIOLOCATION

FIXED

1 0 -2Å3

1020Hz

FIXED
SATELLITE (E-S)

SPACE
RESEARCH
(Passive)

30

9
...
975

RADIOLOCATION
RADIOLOCATION † (1999)

8
...
965
9
...
7

126
...
1
8
...
05

FIXED

MARITIME MOBILE

AERONAUTICAL
MOBILE (R)
AERONAUTICAL MOBILE
AERONAUTICAL MOBILE

1350

BROADCASTING
SATELLITE

MARITIME
RADIONAVIGATION
(RADIO BEACONS)

19
...
5 ±
...
2

7
...
AND TIME SIGNAL (20 kHz)

AERONAUTICAL
RADIONAVIGATION

FIXED
SATELLITE
(S-E)

FIXED

EARTH
EXPLORATION
SATELLITE
(Passive)

SPACE
RESEARCH
(Passive)

RADIO
ASTRONOMY

1017Hz

1 x 10Å 0
3

RADIOLOCATION

RADIOLOCATION

7
...
3

74
...
8
75
...
4
76
...
0
1215

RADIONAVIGATION
SATELLITE (S-E)

AMATEUR SATELLITE

AMATEUR

AERONAUTICAL MOBILE (OR)

11
...
0

300

ISM – 6
...
015 MHz

10
...
685
6
...
0

88
...
45

10
...
68
5
...
0

BROADCASTING
(FM RADIO)

EARTH EXPL
...
525

BROADCASTING
(TV CHANNELS 5-6)

SPACE
RESEARCH (Passive)

FIXED

BROADCASTING
(TV CHANNELS 2-4)

0

RADIO
ASTRONOMY

105
...
2

MOBILE

890
902

10
...
55
10
...
95

Amateur

Amateur

FIXED

MOBILE**

AMATEUR

Radiolocation

MOBILE
SATELLITE

RADIONAVIGATION

RADIONAVIGATION
SATELLITE

MOBILE

1015Hz

3 x 10 3Å

Radiolocation

RADIOLOCATION

495

MOBILE (DISTRESS AND CALLING)
4
...
003
5
...
060

AERONAUTICAL MOBILE (R)

RADIOLOCATION

9
...

SAT
...
2

Radiolocation

102
...
0

RADIOLOCATION
RADIOLOCATION

100
...
0 ± 13 MHz

1013Hz

Infrared

3 x 10 5Å

INFRARED

1 THz

0
...
0

9
...
85

50
...
3 cm

300 GHz

100 GHz

Radar
Radar
Bands

EHF

8
...
0

MOBILE

STANDARD FREQ
...

608
...
0

806

8
...
45

MOBILE
SATELLITE (S-E)

SPACE RESEARCH (S-E)

Radiolocation

LAND
MOBILE

MOBILE

FIXED

MOBILE
SATELLITE
(E-S)

30 cm

1 GHz

FIXED
SATELLITE
(E-S)

X

UHF

84
...
65
4
...
75

46
...
0

54
...
0

3 MHz

LAND
MOBILE

512
...
125

415

435

AERONAUTICAL MOBILE (R)

6
...
19
SPACE RESEARCH (E-S)
7
...
25
Fixed
7
...
45
FIXED
Mobile
MET
...
55
Mobile
FIXED
FIXED
Satellite (S-E)
SATELLITE (S-E)
7
...
90
FIXED
MOBILE
Fixed
SATELLITE (E-S) SATELLITE (S-E)
8
...

Mobile
SATELLITE (E-S) SATELLITE(S-E) FIXED
Satellite (E-S)
8
...

EARTH EXPL
...
(S-E)
(E-S)
(E-S)
(no airborne) 8
...

SATELLITE
FIXED (E-S)(no airborne)
SATELLITE (S-E)
(E-S)
8
...
65

7
...
0
4
...
875

FIXED SATELLITE (S-E)

FIXED

MARITIME MOBILE

4
...
0

UNITED

RADIO ASTRONOMY
39
...
0

460
...
0
38
...
6

5
...
46
5
...
0
37
...
0

FIXED

FIXED
SATELLITE (E-S)

MOBILE

FREQUENCY

INTER-SATELLITE

MOBILE
LAND MOBILE

MOBILE

FIXED
SATELLITE (E-S)
MOBILE

ALLOCATIONS

LAND MOBILE

RADIO SERVICES COLOR LEGEND

AERONAUTICAL
MOBILE

MOBILE

FIXED

6
...
0

THE RADIO SPECTRUM

AERONAUTICAL
MOBILE SATELLITE

FIXED

FIXED

5
...
0

ISM – 40
...
02 MHz

Amateur

5
...
35

AERONAUTICAL
RADIONAVIGATION
(RADIO BEACONS)

Aeronautical
Mobile

AERONAUTICAL
MOBILE (R)

MOBILE

FIXED

LAND MOBILE
Radio Astronomy
LAND MOBILE
RADIO ASTRONOMY

FIXED

95
...
25 ±
...
8 ±
...
0

86
...

SATELLITE (Passive)

MOBILE

Radiolocation

RADIORadioLOCATION
location
Radiolocation
RADIONAVIGATION
MARITIME
Radiolocation
RADIONAVIGATION
MARITIME
METEOROLOGICAL Radiolocation
RADIONAVIGATION
AIDS

BROAD- BROADCASTING CASTING
SATELLITE

FIXED
RADIOLOCATION SATELLITE
(E-S)

30 MHz

RADIOLOCATION
AERONAUTICAL
RADIONAV
...
5

402
...
0
406
...
1

BROADCASTING
(TV CHANNELS 21-36)

51
...
230

300

3
...
0

401
...
0

3
...
0

400
...
15

420
...
6

AERONAUTICAL
RADIONAVIGATION

81
...
0

3
...
025

AERONAUTICAL MOBILE (R)

MOBILE*

MOBILE

399
...
0

50
...
5

MOBILE

MOBILE

FIXED

4
...
77

MOBILE

LAND MOBILE

4
...
0

FIXED

RADIOLOCATION

RADIOLOCATION SATELLITE

RADIOLOCATION

4
...
56

32
...
66

FIXED

RADIO ASTRONOMY

EARTH
EXPLORATION
SATELLITE
(Passive)

MOBILE

INTERSATELLITE

SPACE
RESEARCH
(Passive)

FIXED

3,000 m

100 kHz

LF

AM Broadcast
Ultra-sonics

EARTH
EXPLORATION
SAT
...
2

59
...
0

MOBILE

450
...
6

FIXED

LAND MOBILE

FIXED

58
...
Satellite
(E-S)

LAND MOBILE

FIXED

MOBILE

AERONAUTICAL
RADIONAVIGATION

AMATEUR

FIXED

MOBILE
SATELLITE

47
...
(Passive)

MOBILE

FIXED
SATELLITE (S-E)
FIXED
SATELLITE (S-E)

47
...
MOBILE
(E-S) SAT
...
91
FIXED
MOBILE
SATELLITE SATELLITE
FIXED
MOBILE
(E-S)
(E-S)
74
...
5

66
...
SAT
...

(S-E)

SPACE RES
...
4

43
...
25

AMATEUR

RADIONAVIGATION
SATELLITE

FI XED
MOBILE
SATELLITE (E-S) SATELLITE (E-S)

Fixed

MARITIME MOBILE
SATELLITE

MOBILE

RADIO
MOBILE
SATELLITE NAVIGATION

MARITIME
RADIONAVIGATION

FIXED
SATELLITE
(E-S)

MOBILE

FIXED

3 x 10 5m

3 kHz

FIXED

SPACE
RESEARCH

MOBILE
...
(S-E)

MET
...
Met
...

Satellite (E-S)
(Radiosonde) Space to Earth (E-S)

FIXED
SATELLITE
(S-E)

MOBILE
SATELLITE
(E-S)

FIXED
SATELLITE
(E-S)

3 x 10 6m

1 kHz

Audible Range
Sonics

100 Hz

EARTH
SPACE
EXPLORATION
RESEARCH SATELLITE

SPACE
RESEARCH
(Passive)

STD
...
& TIME SIGNAL SAT
...
1 MHz)
MET
...
7

AERONAUTICAL
RADIONAVIGATION

42
...
65

4
...
6

MET
...
5

AMATEUR SATELLITE

RADIO
ASTRONOMY

BROADCASTING

Radiolocation
FIXED SAT
...

(S-E)

FIXED

FIXED

MOBILE

MOBILE
335
...
6

MOBILE

RADIO
SATELLITE
ASTRONOMY
(E-S)

RADIONAVIGATION
SATELLITE

300 MHz

AERO
...
(Ground)

FIXED

AERONAUTICAL RADIONAVIGATION

37
...
5

AMATEUR

SPACE OPERATION

SPACE RESEARCH

RADIOLOCATION
RADIOLOCATION

AERO
...
(Ground)

322
...
0

MOBILE* *

METEOROLOGICAL
AIDS

METEOROLOGICAL
SATELLITE

AERONAUTICAL
RADIONAVIGATION
(Ground)

36
...
0

3
...
3

39
...
1

Radiolocation

EARTH
EXPLORATION
SATELLITE
(Passive)

MOBILE

(Passive)

SPACE
RESEARCH

FIXED

FIXED
SATELLITE (S-E)

MOBILE

Radiolocation

RADIOLOCATION

33
...
4

MOBILE

FIXED
FIXED

3
...
0

Radiolocation

RADIOLOCATION

RADIONAVIGATION

31
...
3

BROADCASTING
SATELLITE

STANDARD FREQUENCY
AND TIME SIGNAL SATELLITE

MOBILE
EARTH
EXPLORATION
SAT
...
0
FIXED

EARTH EXPLORATION
SATELLITE

MOBILE SATELLITE

MOBILE
SATELLITE
(E-S)

FIXED
SATELLITE
(E-S)

SPACE
RESEARCH
(Passive)

RADIO
ASTRONOMY

3 GHz

30 GHz

* EXCEPT AERO MOBILE (R)

** EXCEPT AERO MOBILE

‡‡ BAND TO BE DESIGNATED FOR MIXED USE

# BAND ALLOCATED TO PERSONAL COMMUNICATIONS SERVICES (PCS)

30
...
Frequency
and Time Signal
Satellite (S-E)

FIXED

GOVERNMENT/ NON-GOVERNMENT SHARED

FIXED SATELLITE

GOVERNMENT EXCLUSIVE

ACTIVITY CODE

DESCRIPTION

Capital Letters between oblique strokes

NON-GOVERNMENT EXCLUSIVE

EXAMPLE

1st Capital with lower case letters

Capital Letters

/BROADCASTING/

ALLOCATION USAGE DESIGNATION

SERVICE

Mobile

U
...
DEPARTMENT OF COMMERCE

National Telecommunications and Information Administration
Office of Spectrum Management

M

March 1996

FIXED

MINISTRATIO N

Primary

T

AD

E

T OF CO
TMEN
MM

A
IC A
M
TIO NS & INF OR

R
PA
DE

N

Secondary

Permitted

U
...

UN

IO

Aeronautical
Mobile

Maritime
Radionavigation
(Radio Beacons)

Aeronautical
Radionavigation
(Radio Beacons)

275
285
300

265

Solutions to Exercises in Chapter 7
Solution to Exercise 7
...
261)
Alexander Graham Bell
...
In other words, percentage, not absolute diﬀerences, matter to us
...
If we used Bels, they would be decimal fractions,
which aren’t as elegant
...
2 (p
...

=
54!6!
6
Solution to Exercise 7
...
263)
n
Because of Newton’s binomial theorem, the sum equals (1 + 1) = 2n
...
Keywords
do not necessarily appear in the text of the page
...
Ex
...
1 (1) Terms are referenced by the page they appear on
...
apples, 1

A active circuits, 67
address, 235
algorithm, 163, 163
aliasing, 150
alphabet, § 2
...
11(207)
Ampere, § 2
...
1(33)
ampliﬁer, § 2
...
4(6), 6, § 2
...
1(261)
amplitude modulate, 120
amplitude modulation, § 6
...
12(208)
analog, 1, 19, § 3
...
6(156),
§ 5
...
16(179), § 6
...
10(206),
§ 6
...
12(208), § 6
...
21(76)
analog signal, § 1
...
10(206)
analog signals, § 5
...
16(179)
analog-to-digital (A/D) conversion, § 5
...
1(11)
ARPANET, § 6
...
21(76)
attenuation, § 2
...
9(205), § 6
...
31(232)
bandpass ﬁlter, 128
bandpass signal, 122
bandwidth, § 4
...
9(205),
§ 6
...
14(210), § 6
...
10(206), 206
baseband signal, 122, § 6
...
14(210), 210
binary symmetric channel, § 6
...
14(210)
bit stream, § 6
...
17(215)
bits, § 6
...
26(224), 226
block diagram, § 1
...
5(21)
boolean arithmetic, § 5
...
13(172)
BPSK, § 6
...
19(217)
bridge circuits, 78
broadcast, § 6
...
36(237)
broadcast communication, 195
broadcast mode, § 6
...
15(176), 177
butterﬂy, § 5
...
2(34), § 3
...
30(230), 231, § 6
...
36(237)
carrier, § 1
...
11(207), 207
Cartesian form, § 2
...
5(21)
channel, § 1
...
9(205), § 6
...
31(232)
channel coder, 224, § 6
...
25(224), § 6
...
27(226)
channel decoding, § 6
...
1(33)
circuit, § 3
...
2(34), § 3
...
8(50), § 3
...
20(74)
circuit model, § 6
...
34(235), 235
circuits, 33
clock speed, § 5
...
2(34)
coaxial cable, § 6
...
21(219), 219

INDEX
codeword, § 6
...
2(262)
coding, § 6
...
26(224), 226
coherent, 207
coherent receiver, § 6
...
36(237), 238
combination, § 7
...
2(262)
communication, § 1
...
32(233)
communication channel, § 6
...
5(202),
§ 6
...
7(203), § 6
...
9(205)
communication network, § 6
...
35(236), § 6
...
34(235),
§ 6
...
37(239)
communication systems, § 6
...
20(218)
Complementary ﬁlters, 241
complex, § 2
...
4(19)
complex amplitude, 15
complex amplitudes, 56
complex conjugate, 11
complex exponential, 11, § 2
...
2(101)
complex frequency, 17
complex number, § 2
...
4(19)
complexity, § 2
...
2(15)
compression, § 6
...
21(219),
§ 6
...
23(222)
computational advantage, 163
computational complexity, § 5
...
9(163), § 5
...
37(239)
Computer networks, 234
computer organization, § 5
...
2(34), 35
conductor, § 3
...
2(101), 103
Cooley-Tukey algorithm, § 5
...
4(6)
countably inﬁnite, 175

267
current, § 3
...
2(34)
current divider, § 3
...
22(220), 220
datarate, § 6
...
31(232)
De-emphasis circuits, 95
decibel, § 7
...
29(230)
decoding, § 6
...
4(19)
decomposition, § 2
...
1(33)
DFT, § 5
...
9(163), § 5
...
14(173)
diﬀerence equation, § 5
...
14(173)
digital, 1, § 5
...
32(233), § 6
...
1(195), § 6
...
14(210), § 6
...
16(213),
§ 6
...
18(217), § 6
...
20(218), § 6
...
22(220),
§ 6
...
25(224), § 6
...
27(226), § 6
...
29(230),
§ 6
...
31(232), § 6
...
19(217)
digital communication receivers, § 6
...
19(217)
digital ﬁlter, § 5
...
16(179)
digital signal, § 1
...
6(156),
§ 5
...
11(168), § 5
...
14(173), § 5
...
16(179)
digital sources, § 6
...
21(219),
§ 6
...
23(222)
diode, § 3
...
7(160), 161,
§ 5
...
9(163), § 5
...
14(173)
discrete-time, § 2
...
6(156),
§ 5
...
15(176), § 5
...
14(173), § 5
...
16(179)
discrete-time Fourier transform, § 5
...
11(168),
§ 5
...
13(172)
discrete-valued, 149
domain, 236
Doppler, 137
double precision ﬂoating point, § 5
...
29(230)
DSP, § 5
...
6(156), § 5
...
11(168), § 5
...
14(173),
§ 5
...
16(179)

E eﬃciency, § 6
...
17(180)
electrical, 235
electrical engineering, § 1
...
1(33)
electronic circuits, 67
electronics, § 3
...
19(69)
element, § 3
...
2(34)
elemental signals, § 2
...
1(33), 34, § 3
...
20(218), 219, § 6
...
12(56)
Equivalent Circuits, § 3
...
29(230), § 6
...
29(230), § 6
...
25(224),
§ 6
...
27(226)
error correction, § 6
...
26(224),
§ 6
...
28(229), § 6
...
19(217)
error-correcting code, § 6
...
28(229)
ethernet, § 6
...
2(15), § 4
...
2(101)
Euler’s relation, § 2
...
4(19), § 3
...
2(34)
Faraday, § 3
...
9(163), § 5
...
5(21)
FFT, § 5
...
15(176)
ﬁlter, 60
ﬁltering, § 4
...
14(173), § 5
...
16(179)
FIR, 172
ﬁxed, 236
ﬁxed rate, 221
Flexibility, 234
ﬂoating point, § 5
...
2(34)
form, 163
formal circuit method, § 3
...
20(74)

forward biasing, 74
Fourier coeﬃcients, § 4
...
3(106),
106
Fourier series, § 4
...
3(106),
§ 4
...
9(122)
fourier spectrum, § 4
...
1(101), 101, § 4
...
6(156), § 5
...
10(165),
§ 5
...
4(6), 6, § 2
...
3(106),
§ 6
...
3(263), 263
frequency domain, § 3
...
1(101),
§ 5
...
15(176)
frequency response, 56, § 5
...
15(212)
frequency-shift keying, 212
FSK, § 6
...
19(217)
functional, 21
fundamental assumption, 36
fundamental frequency, 102, 125
fundamental model of communication,
§ 6
...
33(234)
Fundamental Model of Digital
Communication, 224, § 6
...
6(23), 23
gateway, § 6
...
3(106)
generator matrix, § 6
...
29(230)
geometric series, 157
geosynchronous orbits, § 6
...
4(108)
half-wave rectiﬁer, 75
Hamming, § 6
...
29(230)
Hamming codes, 230
Hamming distance, § 6
...
10(165), 166
harmonically, 102
Heaviside, § 6
...
1(1)
Henry, § 3
...
1(1)
hole, § 3
...
22(220), § 6
...
22(220), § 6
...
22(220),
221

I

i, § 2
...
2(15)
imaginary number, § 2
...
1(11), 11
impedance, 51, § 3
...
10(52),
§ 3
...
2(34), § 3
...
1(1), 1, § 1
...
1(195),
§ 6
...
6(203), § 6
...
8(204),
§ 6
...
10(206), § 6
...
12(208), § 6
...
14(210),
§ 6
...
16(213), § 6
...
18(217), § 6
...
21(219),
§ 6
...
23(222), § 6
...
26(224), § 6
...
28(229),
§ 6
...
30(230), § 6
...
32(233), § 6
...
34(235),
§ 6
...
36(237)
information theory, § 1
...
5(21)
input resistance, 69
input-output relationship, § 3
...
17(66)
integrator, § 2
...
8(204), 204
internet, § 6
...
37(239)
internet protocol address, § 6
...
8(117)
inverting ampliﬁer, 70
ionosphere, § 6
...
35(236), 236

J j, § 2
...
1(1)
joules, 34

K KCL, 40, § 3
...
16(65)
Kirchhoﬀ, § 3
...
15(61)
Kirchhoﬀ’s Laws, 38
KVL, 40, § 3
...
16(65)

L LAN, § 6
...
20(74)
line-of-sight, § 6
...
2(34), 34, § 5
...
7(115)
linear codes, § 6
...
6(23)
load, § 3
...
35(236)
Local area networks, 236
logarithmic ampliﬁer, 76
logarithmically, 261
long-distance, § 6
...
7(203)
long-distance transmission, § 6
...
2(262)
lowpass ﬁlter, 60, § 4
...
1(11)
Marconi, § 6
...
16(213), 214
Maxwell’s equations, § 6
...
7(45)
Mayer-Norton equivalent, 49
mean-square equality, 113
message, § 6
...
34(235)
model of communication, § 1
...
3(37)
modem, § 1
...
12(208)
modulation, § 1
...
11(207)
Morse code, § 6
...
35(236)
name servers, 236
negative, § 3
...
1(33)
network, 223, § 6
...
36(237)
network architecture, § 6
...
36(237)
networks, 203
node, § 3
...
15(61), 61
node voltages, 61
nodes, 37, 235

270

INDEX
noise, 125, § 6
...
8(204), 204,
§ 6
...
17(215), § 6
...
30(230)
nonlinear, § 3
...
7(45), § 3
...
2(145)
Nyquist frequency, 151, § 5
...
2(34)
Oliver Heaviside, § 1
...
17(66), 67, § 3
...
2(34), 35
operational ampliﬁer, § 3
...
19(69)
orthogonality, § 4
...
3(106), 106
output, § 2
...
9(122)

P packet, § 6
...
36(237)
packet size, § 6
...
34(235), 236
packets, 236
parallel, § 2
...
6(41), 42, 44
parity, § 6
...
29(230)
parity check matrix, 229
Parseval’s theorem, § 4
...
8(117),
119, § 5
...
2(101)
permutation, § 7
...
4(6), 6, § 2
...
2(15), 15
physical, § 3
...
33(234)
point to point communication, § 6
...
2(196), 196, 234,
§ 6
...
1(195), 195
point-wise equality, 113
polar form, § 2
...
2(145)
positive, § 3
...
33(234)
Power, 1, § 3
...
5(40), 40,

§ 3
...
16(65), § 4
...
1(261)
power factor, 83, 99
power spectrum, 109, § 6
...
2(262)
probability of error, § 6
...
21(76)
propagating wave, 199
propagation speed, 199
proportional, 162
protocol, § 6
...
37(239), 239
pulse, § 2
...
2(101), § 4
...
2(145)
quantization interval, § 5
...
4(152), 152

R random access, § 6
...
2(15)
real part, 11
real-valued, § 2
...
12(208)
receiver, § 1
...
16(213)
rectiﬁcation, § 4
...
5(202)
repeaters, 253
repetition code, § 6
...
2(34), 35
resistivity, 40
resistor, § 3
...
5(40)
reverse bias, § 3
...
5(7), 109
route, 235
routing, § 6
...
7(203)
satellite communication, § 6
...
16(213), 213

INDEX
self-synchronization, § 6
...
4(19)
series, § 3
...
3(4), § 6
...
20(218),
§ 6
...
30(230), § 6
...
14(173)
shift-invariant systems, § 5
...
2(34)
sign bit, § 5
...
2(2), 2, § 1
...
5(21)
signal decomposition, 18
signal set, 210, § 6
...
17(215),
§ 6
...
4(108)
signal-to-noise, § 5
...
10(206), 206,
§ 6
...
12(208)
signal-to-noise-ration, § 6
...
27(226)
signals, 1, § 2
...
21(76)
simple binary code, 220
sinc, 118
sine, § 1
...
2(15), § 2
...
29(230)
sink, § 1
...
4(6), 6, § 2
...
4(19),
§ 4
...
3(106)
SIR, § 6
...
9(205), § 6
...
12(208),
§ 6
...
27(226), § 6
...
3(4), 5, § 2
...
2(34)
source coding theorem, § 6
...
22(220)
space constant, 199
spectrograms, § 5
...
1(101), 101, § 4
...
10(124)
square wave, § 2
...
2(101), § 4
...
2(15)
superposition, § 2
...
1(101), § 5
...
4(19)
synchronization, § 6
...
5(21)
systems, § 2
...
33(234)
telephone, § 6
...
1(1)
Thévenin, § 3
...
12(56)
Thévenin equivalent circuit, 47
time constant, § 2
...
6(23)
time domain, § 3
...
12(169)
time invariant, § 4
...
6(23)
time-domain multiplexing, § 6
...
6(23)
total harmonic distortion, 109
transatlantic communication, § 6
...
36(237), 237
transfer function, § 3
...
14(59),
§ 5
...
14(173)
transforms, 119
transition diagrams, § 6
...
5(202), § 6
...
14(210),
§ 6
...
17(215)
transmission line, § 6
...
3(4), 5
twisted pair, § 6
...
4(19), 20, 155, 156
unit step, § 2
...
10(124), 126
Volta, § 3
...
1(33), 33, § 3
...
6(41), 41
voltage gain, 69

W WAN, § 6
...
4(201)
white noise, § 6
...
31(232)
wide area network, § 6
...
5(202), § 6
...
2(196), § 6
...
5(202)
wireline channel, § 6
...
3(196)
World Wide Web, § 6
...
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Elementary signal theory; time- and frequency-domain analysis; Sampling Theorem
...

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Title: electrical hand book
Description: basics of electrical engineering that will help you in your study

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