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Electrical Engineering
Principles & Applications
Chapter 5
Steady-State Sinusoidal
Analysis
Slide 1

Outline
1
...
Solve steady-state AC circuits using phasors
and complex impedances

Slide 2

Importance of Sinusoidal Sources
• Appear in many practical applications
– Electric power is distributed by sinusoidal currents and voltages
– Sinusoidal signals are used widely in radio communications

• Any signal can be represented by a sum of sinusoidal
components (Fourier Analysis)
• Sinusoids have good mathematical properties
– Derivative is a sinusoid
– Integral is a sinusoid

Sinusoidal Steady-State
• Whenever the forced input to the circuit is sinusoidal the
response will be sinusoidal
• If the input persists, the response will persist and we call
it steady-state response
Slide 3

Sinusoidal Currents or Voltages

Slide 4

Sinusoidal Currents and Voltages
v(t ) = Vm cos(ωt + θ )

Vm is the peak value
is the angular frequency in radians per second
is the phase angle

1
T is the period , where f =
T

2π
ω =
T

Slide 5

is the frequency

is the angular frequency, where: ω = 2πf

(

sin (z ) = cos z − 90

o

)

Root-Mean-Square (RMS) Values of a Sinusoid
V

rms

=

P avg

1
T

=

T

∫

v

2

0

V

2
rms

R

(t ) dt

I

rms

=

1
T

T

∫

i

2

(t ) dt

0

2
Pavg = I rms R

Note: for DC sources, the power is V2/R = I2R
The RMS value for a sinusoid is the peak value
divided by the square root of 2
Vm
V rms =
2
This is NOT true for other periodic waveforms such as
square waves or triangular waves
Slide 6

Voltage Applied to Resistors
Voltage applied to a 50-

resistance

Voltage

Pavg

Power

2
Vrms
=
R
(100 / 2 ) 2
=
= 100W
50

v 2 (t )
p(t ) =
R
= 200 cos (100πt ) W
2

Slide 7

Real and Complex Signals
A complex number is given by:

Z = x + jy

Real part

−1

Rectangular form
Imaginary part

It can also be represented by a POINT in the complex Plane

Note that Z can be written as :

| Z | ∠θ

y

Polar form

Complex conjugate of z is:

z* = x - jy
Slide 8

x

Phasor Definition
Phasors are complex numbers that can be used to represent sinusoidal
signals
The magnitude of the phasor = Peak value
Angle of the phasor = phase of the sinusoid (written as a cosine)

Consider v1 (t ) = V1 cos(ωt + θ1 )

The phasor is V1 = V1∠θ1

and i1(t ) = I1 sin(ωt +θ2 )
The phasor is I1 = I1∠θ 2 − 90

The steady state analysis of sinusoidal signals is easily done
if signals are represented as phasors (vectors)
Slide 9

Complex Numbers
z = x + jy
Real part

−1

Rectangular form

Imaginary part

A complex number can be represented as a point in the complex Plane

Imaginary

Complex conjugate of z is:

z* = x - jy

y

z

x
Slide 10

Real

Complex Numbers in Polar Form
• Represent the complex number by the length of the
arrow and the angle between the arrow and the
positive real axis
• We write complex numbers in polar form as:

z = z ∠θ
z

2

= x 2 + y 2 ==> z =

y
tan θ =
or θ = tan
x
Polar to Rectangular:

−1

x2 + y2

Imaginary

y
x

1

x = z cos θ

x

y = z sin θ
In exponential form:
Slide 11

θ

=| Z | e

jθ

y
Real

Euler’s Identities
e jθ = 1∠θ = cos θ + j sin θ
Think of it as a complex variable in polar form

e − jθ = 1∠ − θ = cos θ − j sin θ

Imaginary

e jθ = cos 2 θ + sin 2 θ = 1
jθ

e +e

− jθ

e jθ − e − jθ

e jθ + e − jθ
= 2 cos θ  cos θ =
→
2
e jθ − e − jθ
= j 2 sin θ  sin θ =
→
2j

1

θ

x

y
Real

cos(θ ) = Re (e jθ ) = Re[cosθ + j sin θ ]
Real part of

Ae
Slide 12

jθ

= A∠θ

Exponential form of a complex number

Complex Arithmetic
z1 = 5 + j 5, and

Let

z2 = 3 − j 4

z1 + z2 = (5 + j 5) + (3 − j 4) = 8 + j1
z1 − z2 = (5 + j 5) − (3 − j 4) = 2 + j 9
Multiplication

z1 z 2 = (5 + j 5)(3 − j 4)

Division
*
z1
z2
5 + j5
=
× *
z2
z2
3 − j4

=
=
=

= 15 − j 20 + j15 − j 2 20

=

= 15 − j 20 + j15 + 20

=

= 35 − j 5
Slide 13

=

5 + j5
3 + j4
×
3 − j4
3 + j4
15 + j 20 + j 15
9 + j 12 − j 12
15 + j 20 + j 15
9 + j 12 − j 12
− 5 + j 35
25
5
35
−
+ j
25
25
0
...
4

+ j 2 20
− j 2 16
− 20
+ 16

Examples
Phasor

Rectangular

z1 = 5∠30o
z1 = 5 cos(30o ) + j 5 sin(30 o )
= 4
...
5 = x + jy

Rectangular

Phasor

z2 =10+ j5

z3 =

(10 ) + ( 5 ) ∠ tan
2

2

−1

= 11
...
57 o
Wrong angle since real part
is negative; the true angle is:

θ = tan−1( y / x) ±180o

2

−1

Be careful when using your calculator
5

z3

z2
153
...
57
-10
Slide 14

5
(
)
− 10

= −26
...
43o

5
z2 = (10) + (5) ∠tan ( )
10
=11
...
57o
2

z 3 = − 10 + j 5

10

Forms of a Complex Number

z2 = 10 + j5

Rectangular form

5
z2 = (10) + (5) ∠ tan ( )
10
o
Polar form
= 11
...
57
2

= 11
...
57o
Exponential form

1∠90o = cos 90 + j sin 90 = j

Arithmetic Operations in Polar and Complex
Form
• To add or subtract two numbers you must
convert them to rectangular first and then do
the operation
• To multiply two complex numbers in polar form

z1 z 2 = z1 ∠ θ1 × z 2 ∠ θ 2
= z1 z2 ∠ (θ1 + θ 2 )
• To multiply two numbers in exponential form

z1 z 2 = z1 e jθ1 × z 2 e jθ 2
= z1 z 2 e j (θ1 +θ 2 )
Slide 16

Division in Polar and Complex Form
• To divide two complex numbers in polar form
z1 ∠ θ 1
z1
=
z2
z2 ∠θ 2
=

z1
z2

∠ (θ 1 − θ 2 )

• To divide two numbers in exponential form
z 1 e jθ 1
z1
=
z2
z 2 e jθ 2
=
Slide 17

z1
z2

e

j (θ 1 − θ 2 )

Adding Sinusoids Using Phasors
Step 1: Determine the phasor for each term
Step 2: Add the phasors using complex arithmetic
Step 3: Convert the sum to polar form
Step 4: Write the result as a time function

Slide 18

Using Phasors to Add Sinusoids
v1 (t ) = 20 cos(ωt − 45o )

(

v2 (t ) = 10 sin ωt + 60

V1 = 20∠ − 45

o

o

)

V 2 = 10 ∠ − 30 o

V s = V1 + V 2
= 20 ∠ − 45 o + 10 ∠ − 30 o
= 14
...
14 + 8
...
06 − j19
...
97 ∠ − 39
...
97 cos ω t − 39
...
Then
when standing at a fixed point, if V1 arrives first followed
by V2 after a rotation of , we say that V1 leads V2 by
...
(Usually,
we take as the smaller angle between the two phasors)

• To determine phase relationships between sinusoids from
their plots versus time:
– Find the shortest time interval tp between positive peaks of the
two waveforms
...
Replace the time descriptions of the voltage and
current sources with the corresponding phasors
...
Express components by their complex impedances:
Replace inductances by their complex impedances

ZL = j L
Replace capacitances by their complex impedances

ZC = 1/(j C)
Resistances have impedances equal to their resistances

ZR = R
1
...
3 = j150 Ω
1
1
=−j
= − j 50 Ω
−6
ωC
500 × 40 × 10
= 100 + j150 − j 50 = 100 + j100 = 141
...
707 ∠ − 15 o
Z eq 141
...
707 cos( 500 t − 15 o )

Inductive Circuit

Example (cont
...
707∠ − 15o = 70
...
707∠ − 15o = 106
...
707∠ − 15o = 35
...
71 − 45o
∠
= 10∠ − 90
j100+ 50− j50
o

70
...
71∠45o
= 10∠ −180o = −10

vc (t ) = −10 cos(1000t )
Slide 32

Z

RC

=
=
=

1
1/ R +1/Z

=
c

1
0
...
01

1
0
...
71 ∠ − 45
= 50 − j 50

o
o

Example (cont
...
1∠ − 180
= −0
...
141∠ − 135o
Vc
Ic =
Zc
10 ∠ − 180
=
− j 100

Slide 33

10 ∠ −180o
100

o

10 ∠ − 180
=
100 ∠ − 90
= 0
...
1

o
o

o

Example
Use nodal analysis to find v1(t) in steady state

V1 V1 − V2
+
= 2∠ − 90o
− j5
10
V2 V2 − V1
+
= 1
...
1∠29
...
1 cos(100t + 29
...
1 + j 0
...
2V2 = − j 2
− j 0
...
1V2 = 1
...
28∠ − 135o mA
Z 250 + j 250

i(t) = 28
...
07∠ − 135o

VL = jωLI = 7
...
414∠ − 45o A and I 2 = 1∠0o A

Thus we have

i1 ( t ) = 1
...
I rms ∠ −

= Vrms I rms ∠(

v

−

i

= Vrms I rms cos (

v

−

i

)
i

) + jVrms I rms sin (

v

= P + jQ
2
S = I rms Z
Slide 48

S is measured in VA

−

i

)

Example
Compute the power and reactive power for each element
...


Source:
Vrms =

θ = θv − θi
= −90° − (−135°)
= 45° i
...


is +ve

PF = cos(45)
= 0
...
071V

0
...
1A

Current lags voltage

P = Vsrms I rms cos(θ )
= 7
...
1× cos(45°)
= 0
...
071× 0
...
5 VAR

Example (cont
...
1) 2100 = 1VAR

Capacitor:
2
QC = I Crms X C = (

0
...
5VAR

Resistor:
PR =

2
I Rrms R

=(

0
...
5W

It is important to note that:

PL = 0
PC = 0

Check:
Q = QL + QC

P = PR

In real life, the values are much bigger (kW, MW, kVA)
Slide 50

Example
Find the power, reactive power, and power factor of the
source
...
)

given

θ A = θ vA − θ iA = cos −1 ( 0
...
5 = 5kW
Q A = V rms I Arms sin( θ A ) = 10 ,000 × 0
...
66 kVAR

Q A = −8
...
7 ) = 45
...
57 °
PA = V rms I Brms cos( θ B ) =
B

Apparent
power of B

1414
2

given

× I Brms × 0
...
1439 A
999
...
7
= V rms I Brms sin( θ B ) = 999
...
1439 × 0
...
101 kVAR

Q B = PB tan(θ B )
= 5000 tan( 45
...
101kVAR

Slide 52

Example (cont
...
66 + 5
...
559kVAR

Apparent Power at source:

Vrms I rms = P 2 + Q 2 = (10) 2 + (−3
...
61kVA
PF at source:
−1 Q
−1 −3
...
59°
P
10
94
...
9421
Slide 53

Example (cont
...
61kVA
=
=
= 10
...
8489V

θ = θv − θi

= source PF

− 19
...
59°
I = 2 × 10
...
59°
= 15∠49
...
71∠ − 45°
= 50 − j 50Ω
I sc = I R − I s
100∠0°
=
− 1∠90°
100
= 1 − j = 1
...
)

Slide 58

Maximum Average Power Transfer
• If the load can take on any complex value, maximum
power transfer is attained for a load impedance equal to
the complex conjugate of the Thévenin impedance
• If the load is required to be a pure resistance,
maximum power transfer is attained for a load
resistance equal to the magnitude of the Thévenin
impedance

Slide 59

Example
Determine the maximum power that can be delivered to
a load by the previous circuit
1
...
load is a pure resistance

magnitude of
the Thévenin
impedance

complex
conjugate of
the Thévenin
impedance

2
P = I arms Rload

1 2
=(
) × 50 = 25W
2
Slide 60

2
P = I brms Rload

0
...
71 = 20

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