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Digital Circuits Tutorial
This tutorial aims to instruct readers on how to construct and
evaluate sequential and combinational circuits
...
After
completing this session, you will be able to identify the type of
digital circuit that is suitable for a certain application
...
In that number system, there are 'r' total
numbers
...
In this chapter, let's discuss the many number representations
that are used, along with how a number is represented in each
system
...
Decimal Number system
• Binary Number system
• Octal Number system
• Hexadecimal Number system
Decimal Number System
•
The base or radix of Decimal number system is 10
...
The
part of the number that lies to the left of the decimal point is
known as integer part
...
In this number system, the successive positions to the left of the
decimal point having weights of 100, 101, 102, 103 and so on
...
That means,
each position has specific weight, which is power of base 10
Example
Consider the decimal number 1358
...
Integer part of this
number is 1358 and fractional part of this number is 0
...
The
digits 8, 5, 3 and 1 have weights of 100, 101, 102 and
103 respectively
...
Mathematically, we can write it as
1358
...
Binary Number System
All digital circuits and systems use this binary number system
...
So, the numbers 0
and 1 are used in this number system
...
Similarly, the part of the number,
which lies to the right of the binary point is known as fractional
part
...
Similarly,
the successive positions to the right of the binary point having
weights of 2-1, 2-2, 2-3 and so on
...
Example
Consider the binary number 1101
...
Integer part of this
number is 1101 and fractional part of this number is 0
...
The
digits 1, 0, 1 and 1 of integer part have weights of 20, 21, 22,
23 respectively
...
Mathematically, we can write it as
1101
...
Octal Number System
The basis, or radix, of the octal number system is eight
...
The
region to the left of the octal point is the integer portion of the
number
...
In this number system, the successive positions to the left of the
octal point having weights of 80, 81, 82, 83 and so on
...
That means, each position has
specific weight, which is power of base 8
...
236
...
236
...
Similarly, the digits 2, 3 and 6 have weights of 8-1, 8-2, 83
respectively
...
236 = (1 × 83) + (4 × 82) + (5 × 81) + (7 × 80) + (2 × 8-1) +
(3 × 8-2) + (6 × 8-3)
After simplifying the right hand side terms, we will get a decimal
number, which is an equivalent of octal number on left hand
side
...
As a
result, this number system uses the numerals 0 to 9 and the
letters A to F
...
The integer part of the number is the area to the left of the
hexadecimal point
...
In this number system, the successive positions to the left of the
Hexa-decimal point having weights of 160, 161, 162, 163 and so
on
...
That
means, each position has specific weight, which is power of base
16
...
2C4
...
2C4
...
Similarly, the digits 2, C and 4 have weights of
16-1, 16-2 and 16-3 respectively
...
2C4 = (1 × 163) + (10 × 162) + (0 × 161) + (5 × 160) + (2 × 161
)+
(12 × 16-2) + (4 × 16-3)
After simplifying the right hand side terms, we will get a decimal
number, which is an equivalent of Hexa-decimal number on left
hand side
...
Let's translate the numbers in
this chapter between two different number systems to
determine their equivalents
...
Follow these steps for converting the decimal
number into its equivalent number of any base ‘r’
...
Consider the
remainders in reverse order to get the integer part of
equivalent number of base ‘r’
...
• Do multiplication of fractional part of decimal number
and successive fractions with base ‘r’ and note down
the carry till the result is zero or the desired number of
equivalent digits is obtained
...
Decimal to Binary Conversion
•
The following two types of operations take place, while
converting decimal number into its equivalent binary number
...
Multiplication of fractional part and successive
fractions with base 2
...
25
...
25
...
Operation
Quotient
Remainder
58/2
29
0 LSBLSB
29/2
14
1
14/2
7
0
7/2
3
1
3/2
1
1
1/2
0
1MSBMSB
⇒585810 = 1110101110102
Therefore, the integer part of equivalent binary number
is 111010
...
25 and successive fractions with base
2
...
25 x 2
0
...
5 x 2
1
...
0
-
⇒
...
2510 =
...
012
Therefore, the fractional part of equivalent binary number is
...
2558
...
01111010
...
25 is
111010
...
Decimal to Octal Conversion
The following two types of operations take place, while
converting decimal number into its equivalent octal number
...
Multiplication of fractional part and successive
fractions with base 8
...
25
...
25
...
Operation
Quotient
Remainder
58/8
7
2
7/8
0
7
⇒585810 = 72728
Therefore, the integer part of equivalent octal number is 72
...
25 and successive fractions with base
8
...
25 x 8
2
...
00
-
⇒
...
2510 =
...
28
Therefore, the fractional part of equivalent octal number is
...
2558
...
272
...
25 is 72
...
Decimal to Hexa-Decimal Conversion
The following two types of operations take place, while
converting decimal number into its equivalent hexa-decimal
number
...
Multiplication of fractional part and successive
fractions with base 16
...
25
...
25
...
Operation
Quotient
Remainder
58/16
3
10=A
3/16
0
3
⇒ 585810 = 3A3A16
Therefore, the integer part of equivalent Hexa-decimal number
is 3A
...
25 and successive fractions with base
16
...
25 x 16
4
...
00
-
⇒
...
2510 =
...
416
Therefore, the fractional part of equivalent Hexa-decimal
number is
...
⇒58
...
2510 = 3A
...
416
Therefore, the Hexa-decimal equivalent of decimal number
58
...
4
...
Let's now talk about how to convert a binary
number to the decimal, octal, and hexadecimal number systems,
one at a time
...
Example
Consider the binary number 1101
...
Mathematically, we can write it as
1101
...
112 = (1 × 23) + (1 × 22) + (0 × 21) + (1 × 20) + (1 × 21
)+
(1 × 2-2)
⇒ 1101
...
112 = 8 + 4 + 0 + 1 + 0
...
25 = 13
...
111101
...
7513
...
11 is
13
...
Binary to Octal Conversion
We know that the bases of binary and octal number systems are
2 and 8 respectively
...
Follow these two steps for converting a binary number into its
equivalent octal number
...
If one or two bits are
less while making the group of 3 bits, then include
required number of zeros on extreme sides
...
Example
Consider the binary number 101110
...
Step 1 − Make the groups of 3 bits on both sides of binary point
...
011 01
Here, on right side of binary point, the last group is having only
2 bits
...
⇒ 101 110
...
⇒ 101110
...
0110102 = 56
...
328
Therefore, the octal equivalent of binary number 101110
...
32
...
Four bits of binary number is
equivalent to one Hexa-decimal digit, since 24 = 16
...
•
•
Start from the binary point and make the groups of 4
bits on both sides of binary point
...
Write the Hexa-decimal digits corresponding to each
group of 4 bits
...
01101
Step 1 − Make the groups of 4 bits on both sides of binary point
...
0110 1
Here, the first group is having only 2 bits
...
Similarly,
include three zeros on extreme side in order to make the last
group also as group of 4 bits
...
0110 1000
Step 2 − Write the Hexa-decimal digits corresponding to each
group of 4 bits
...
0110100000101110
...
682E
...
01101 is 2E
...
68
...
Now, let us discuss about the conversion of an octal
number to decimal, binary and Hexa-decimal number systems
one by one
...
Example
Consider the octal number 145
...
Mathematically, we can write it as
145
...
238 = (1 × 82) + (4 × 81) + (5 × 80) + (2 × 8-1) + (3 × 8-2)
⇒ 145
...
238 = 64 + 32 + 5 + 0
...
05 = 101
...
23145
...
3101
...
23 is
101
...
Octal to Binary Conversion
The process of converting an octal number to an equivalent
binary number is just opposite to that of binary to octal
conversion
...
Example
Consider the octal number 145
...
Represent each octal digit with 3 bits
...
23145
...
010011001100101
...
⇒ 145
...
238 = 1100101
...
0100112
Therefore, the binary equivalent of octal number 145
...
010011
...
•
•
Convert octal number into its equivalent binary
number
...
Example
Consider the octal number 145
...
23 as 1100101
...
By following the procedure of binary to Hexa-decimal
conversion, we will get
1100101
...
0100112 = 65
...
4C16
⇒145
...
238 = 65
...
4C16
Therefore, the Hexa-decimal equivalent of octal number 145
...
4C
...
Now, let us discuss about the
conversion of Hexa-decimal number to decimal, binary and octal
number systems one by one
...
Example
Consider the Hexa-decimal number 1A5
...
21A5
...
21A5
...
125 = 421
...
21A5
...
125421
...
2 is 421
...
Hexa-Decimal to Binary Conversion
The process of converting Hexa-decimal number into its
equivalent binary number is just opposite to that of binary to
Hexa-decimal conversion
...
Example
Consider the Hexa-decimal number 65
...
65
...
4C6 = 01100101
...
010011002
The value doesn’t change by removing the zeros, which are at
two extreme sides
...
4C65
...
0100111100101
...
4C
is 1100101
...
Hexa-Decimal to Octal Conversion
Follow these two steps for converting Hexa-decimal number into
its equivalent octal number
...
Convert the above binary number into its equivalent
octal number
...
4C
In previous example, we got the binary equivalent of Hexadecimal number 65
...
010011
...
0100111100101
...
23145
...
4C65
...
23145
...
4C is 145
...
Binary Numbers Representation
We can make the binary numbers into the following two groups
− Unsigned numbers and Signed numbers
...
They
don’t have any sign
...
As in decimal number system, the placing of positive
sign in front of the number is optional for representing positive
numbers
...
Signed Numbers
Signed numbers contain both sign and magnitude of the
number
...
So, we
have to consider the positive sign for positive numbers and
negative sign for negative numbers
...
If sign bit is zero, which indicates the binary number is positive
...
Representation of Un-Signed Binary Numbers
The bits present in the un-signed binary number holds
the magnitude of a number
...
Example
Consider the decimal number 108
...
This is the representation of unsigned binary
number
...
These 7 bits represent the magnitude of the
number 108
...
Hence, it is also called
as sign bit
...
Similarly, the negative sign is represented by placing ‘1’
in the sign bit
...
There are three types of representations for signed binary
numbers
•
•
•
Sign-Magnitude form
1’s complement form
2’s complement form
Representation of a positive number in all these 3 forms is same
...
Example
Consider the positive decimal number +108
...
These 7 bits
represent the magnitude of the number 108
...
+108+10810 = 01101100011011002
Therefore, the signed binary representation of positive decimal
number +108 is 𝟎𝟏𝟏𝟎𝟏𝟏𝟎𝟎
...
Sign-Magnitude form
In sign-magnitude form, the MSB is used for representing sign of
the number and the remaining bits represent the magnitude of
the number
...
This representation is similar to the
signed decimal numbers representation
...
The magnitude of
this number is 108
...
It is having 7 bits
...
Since the given number is negative, consider the sign bit as one,
which is placed on left most side of magnitude
...
sign-magnitude
representation
of
-108
1’s complement form
The 1’s complement of a number is obtained by complementing
all the bits of signed binary number
...
Similarly, 1’s
complement of negative number gives a positive number
...
Example
Consider the negative decimal number -108
...
We know the signed binary representation
of 108 is 01101100
...
The MSB of this number is zero, which indicates
positive number
...
So,
replace zeros by ones and ones by zeros in order to get the
negative number
...
2’s complement form
The 2’s complement of a binary number is obtained by adding
one to the 1’s complement of signed binary number
...
Similarly, 2’s complement of negative number gives a positive
number
...
Example
Consider the negative decimal number -108
...
= 10010011 + 1
= 10010100
Therefore,
the 2’s
of 10810810 is 10010100100101002
...
The basic arithmetic
operations are addition and subtraction
...
We can perform
the addition of these two numbers, which is similar to the
addition of two unsigned binary numbers
...
If resultant sum is positive, you can find the magnitude of it
directly
...
Example 1
Let us perform the addition of two decimal numbers +7 and
+4 using 2’s complement method
...
+7+710 = 00111001112
+4+410 = 00100001002
The addition of these two numbers is
+7+710 ++4+410 = 00111001112+00100001002
⇒+7+710 ++4+410 = 01011010112
...
So, there is no carry out from
sign bit
...
So, the magnitude of sum is 11 in decimal number
system
...
Example 2
Let us perform the addition of two decimal numbers -7 and 4 using 2’s complement method
...
−7−710 = 11001110012
−4−410 = 11100111002
The addition of these two numbers is
−7−710 + −4−410 = 11001110012 + 11100111002
⇒−7−710 + −4−410 = 1101011101012
...
In this case, carry is obtained
from sign bit
...
removing
carry
The sign bit ‘1’ indicates that the resultant sum is negative