Binary Number System and Codes

Introduction to Number System and Codes
Presented by
Dr. Shirshendu Roy
Course - Digital Electronics
December 5, 2021
Dr. Shirshendu Roy Number System and Codes December 5, 2021 1 / 39

Radix Number System
A number system should be adopted to represent real numbers in hardware
platforms like computer. If a number system is weighted and weights are
calculated based on a parameter "Radix (r)" then it is called Radix Number
System. In a radix number system, a number X is represented as
X = {xn−1, xn−2, ..., x1, x0} (1)
Here, X has n number of digits and value of the digits can take any value
between 0 to (r-1). The value of the number can be evaluated as
X = Xn−1 × rm−1
+ Xn−2 × rm−2
+ ... + Xn−m × r0
+
Xn−m+1 × r−1
+ Xn−m+2 × r−2
+ ... + X0 × r−p
(2)
Here, total n digits are used, m digits are used for integer part and p digits
are used for fraction.

Decimal Number System
We are most familiar with Decimal Number System. In the Decimal Number
System, the value of the radix (r) is 10. Any real number can be represented
as
(1123.5652)10 = 1 × 103
+ 1 × 102
+ 2 × 101
+ 3 × 100
+
5 × 10−1
+ 6 × 10−2
+ 5 × 10−3
+ 2 × 10−4
(3)
Here, total n = 8 digits are used, the value of m is 4 and p = 4. In the
Decimal Number System, the digits can take any value from 0 to 9. So, this
is a weighted number system where weights are
{...1000, 100, 10, 1, 0.1, 0.01, 0.001, 0.0001, ....}

Binary Number System
Real numbers are represented using Binary Number System in the digital
systems implemented on computers, micro controllers or on FPGAs. Here,
the value of radix is 2 and thus value of any digit can be either ‘0’ or ‘1’. In
Binary Number System, a real number is represented as
X = 13.6875 = 1101.1011
In terms of weights the above representation can be written as
1101.1011 = 1 × 23
+ 1 × 22
+ 0 × 21
+ 1 × 20
+
1 × 2−1
+ 0 × 2−2
+ 1 × 2−3
+ 1 × 2−4
(4)
So, this is a weighted number system where weights are
{...8, 4, 2, 1, 0.5, 0.25, 0.125, 0.0625, ....}

Binary Number System
Each digit of a Binary number is called as a bit. Group of 8-bits called
as a byte.
The left most bit is called as Least Significant Bit (LSB) as the error
introduced by this bit is minimum.
The Right most bit is called as Most Significant Bit (MSB) as the
error introduced by this bit is maximum.
There is no separation between integer and fractional part in Binary
numbers. Real value of a number varies based on the number of bits
allocated for integer and fraction.
Unsigned value of the binary number 1101.1011 is 221. But real value
is 13.6875 as 4-bits are allocated for fraction. If two bits are allocated
for fraction, then the value will be 54.75.
Thus representation of any real numbers in binary depends on allocation
of bits and not fixed.

Binary to Decimal Conversion
A binary number B = B3B2B1B0.B−1B−2B−3B−4 is converted to decimal
number as
D = B3 × 23
+ B2 × 22
+ B1 × 21
+ B0 × 20
+ B−1 × 2−1
+ B−2 × 2−2
+ B−3 × 2−3
+ B−4 × 2−4
(5)
Example: Consider B = 1101.0101 then its equivalent decimal will be
D = 1 × 23
+ 1 × 22
+ 0 × 21
+ 1 × 20
+ 0 × 2−1
+ 1 × 2−2
+ 0 × 2−3
+ 1 × 2−4
= 8 + 4 + 0 + 1 + 0 + 0.25 + 0 + 0.0625
= 13.3125 (6)

Decimal to Binary Conversion
Decimal to binary conversion is done in two phases. In the first phase,
integer part is converted into binary. Then in the next phase fraction part is
converted.
Let’s say the decimal number is D = 53.456. In the first phase, integer part
is divided by 2 until the residue is less than 2. Conversion of the integer to
binary is shown below.
53
2
Remainder
1
2 26
2 13
2 6
2 3
1
0
1
0
1
MSB
Binary equivalent of the integer 53 is 110101

Decimal to Binary Conversion
Fraction is converted to binary by multiplying by 2. Multiplication is done
until the result is 1 or significant value is generated.
0.456 × 2 = 0.912
0.912 × 2 = 1.824
Integer
0
1
0.824 × 2 = 1.648 1
0.648 × 2 = 1.296 1
0.296 × 2 = 0.592 0 LSB
The binary value of 0.456 = 0.01110. Here, after 5-bits multiplication is
stopped.
Thus the overall binary equivalent of 53.456 is 110101.01110.

Other forms of Binary Number System
Machines can understand binary numbers. But for ease of understanding the big binary numbers
some forms binary numbers are adopted. They are Octal Number system and Hexadecimal
Number system.
Decimal Binary Octal Hexadecimal
0 0000 0 0
1 0001 1 1
2 0010 2 2
3 0011 3 3
4 0100 4 4
5 0101 5 5
6 0110 6 6
7 0111 7 7
8 1000 (10)8 8
9 1001 (11)8 9
10 1010 (12)8 A
11 1011 (13)8 B
12 1100 (14)8 C
13 1101 (15)8 D
14 1110 (16)8 E
15 1111 (17)8 F

Binary to Octal Conversion
To convert a Binary number to its Octal equivalent, first the binary number
is written as group of 3 binary digits and then these groups are replaced by
their Octal equivalents.
To make groups of 3-bits, additional zeros can be added in the MSB side for
the integer part and additional zeros can be added in the LSB side for the
fractional part
Example: Convert B = 1011101.01101 to its Hexadecimal equivalent.
To make group of 3-bits two zeros are padded in the MSB side and one zero
is padded in the LSB side. The modified number is
B = 001011101.011010
There are 5 groups of 3-bits. Writing their Octal equivalent, the Octal
number is
B = 001011101.011010 = (135.32)8

Octal to Binary Conversion
To convert an Octal number to Binary, Binary equivalent of Octal digits are
written in terms of 3-bit binary digits.
Example: Convert (73.62)8 to its equivalent binary representation
We know that (7)8 = 111, (3)8 = 011,(6)8 = 110 and (2)8 = 010 then
(73.62)8 = 111011.110010
111 011 . 110 010
7 3 6 2

Octal to Decimal Conversion
An Octal number O = O3O2O1O0.O−1O−2O−3O−4 is converted to decimal
number as
D = O3 × 23
+ O2 × 22
+ O1 × 21
+ O0 × 20
+ O−1 × 2−1
+ O−2 × 2−2
+ O−3 × 2−3
+ O−4 × 2−4
(7)
Example: Consider Octal number O = 1073.0721 then its equivalent deci-
mal will be
D = 1 × 83
+ 0 × 82
+ 7 × 81
+ 3 × 80
+ 0 × 8−1
+ 7 × 8−2
+ 2 × 8−3
+ 1 × 8−4
= 512 + 0 + 56 + 3 + 0 + 0.1093 + 0.0039 + 0.0002
= 571.1134 (8)

Decimal to Octal Conversion
Decimal to Octal conversion is done in the same way as decimal to binary
conversion is done. But here the decimal number is divided by 8 until the
residue is less than 8. Conversion is done in two phases.
Let’s consider the decimal number is 444.456. Octal equivalent of the integer
part is shown below.
444
8
Remainder
4
8 55
6
7
MSD
Octal equivalent of the decimal integer (444)10 is (674)8. Here Most Signif-
icant Digit (MSD) is 6.

Decimal to Octal Conversion
Conversion of fractional numbers to its Octal equivalent is done by repetitive
multiplication by 8 until the fraction part of the result is zero or significant
result is obtained. Conversion is shown below.
0.456 × 8 = 3.648
0.648 × 8 = 5.184
Integer
3
5
0.184 × 8 = 1.472 1
0.472 × 8 = 3.776 3
0.776 × 8 = 6.208 6 LSD
Octal equivalent of the fractional number (0.456)10 is (0.35136)8.
Thus Octal equivalent of (444.456)10 is (674.35136)8.

Binary to Hexadecimal Conversion
To convert a Binary number to its Hexadecimal equivalent, first the binary
number is written as group of 4 binary digits and then these groups are
replaced by their hexadecimal equivalents.
To make groups of 4-bits, additional zeros can be added in the MSB side for
the integer part and additional zeros can be added in the LSB side for the
fractional part
Example: Convert B = 1011101.011011 to its Hexadecimal equivalent.
To make group of 4-bits one zero is padded in the MSB side and two zeros
are padded in the LSB side. The modified number is
B = 01011101.01101100
There are 4 groups of 4-bits. Writing their hexadecimal equivalent, the
hexadecimal number is
B = 01011101.01101100 = 5D.6C H

Hexadecimal to Binary Conversion
To convert a Hexadecimal number to Binary, Binary equivalent of Hexadec-
imal digits are written in terms of 4-bit binary digits.
Example: Convert A3.B7H to its equivalent binary representation
We know that A = 1010, 3 = 0011,B = 1011 and 7 = 0111 then
A3.B7H = 10100011.10110111
1010 0011 . 1011 0111
A 3 B 7

Hexadecimal to Decimal Conversion
A Hexadecimal number H = H3H2H1H0.H−1H−2H−3H−4 is converted to
decimal number as
D = H3 × 23
+ H2 × 22
+ H1 × 21
+ H0 × 20
+ H−1 × 2−1
+ H−2 × 2−2
+ H−3 × 2−3
+ H−4 × 2−4
(9)
Example: Consider Octal number H = A05C.0DE1 then its equivalent
decimal will be
D = 10 × 163
+ 0 × 162
+ 5 × 161
+ 12 × 160
+ 0 × 16−1
+ 13 × 16−2
+ 14 × 16−3
+ 1 × 16−4
= 40960 + 0 + 80 + 12 + 0 + 0.05078 + 0.00341 + 0.00001
= 41052.0542 (10)

Decimal to Hexadecimal Conversion
Decimal to Hexadecimal conversion is done in the same way as decimal to
binary conversion is done. But here the decimal number is divided by 16
until the residue is less than 16. Conversion is done in two phases.
Let’s consider the decimal number is 165.456. Hexadecimal equivalent of
the integer part is shown below.
165
16
Remainder
5
10 MSD
Hexadecimal equivalent of the decimal integer (165)10 is (A5)16. Here Most
Significant Digit (MSD) is A → 10.

Decimal to Hexadecimal Conversion
Conversion of fractional numbers to its Hexadecimal equivalent is done by
repetitive multiplication by 16 until the fraction part of the result is zero or
significant result is obtained. Conversion is shown below.
0.456 × 16 = 7.296
0.296 × 16 = 4.736
Integer
7
4
0.736 × 16 = 11.776 11
0.776 × 16 = 12.416 12 LSD
Hexadecimal equivalent of the fractional number (0.456)10 is (0.74BC)16.
Thus Hexadecimal equivalent of (165.456)10 is (A5.74BC)16.

Hexadecimal to Octal Conversion
Hexadecimal to Octal conversion can be done by the following steps
Convert Hexadecimal number to its Binary.
Make groups of 3-bits starting from the LSB.
Write equivalent Octal digits for the groups.
Example: Convert (47)16 to its equivalent Octal.
(47)16 = (01000111)2
= (001000111)2
= (107)8

Octal to Hexadecimal Conversion
Octal to Hexadecimal conversion can be done by the following steps
Convert Octal number to its Binary.
Make groups of 4-bits starting from the LSB.
Write equivalent Hexadecimal digits for the groups.
Example: Convert (107)8 to its equivalent Hexadecimal.
(107)8 = (001000111)2
= (01000111)2
= (47)16

Binary Coded Decimal
Binary Coded Decimal (BCD) is a combination of four binary digits that
represent decimal numbers. BCD uses four bits to represent the decimal
digits from 0 to 9.
Decimal Binary BCD
0 0000 0000
1 0001 0001
2 0010 0010
3 0011 0011
4 0100 0100
5 0101 0101
6 0110 0110
7 0111 0111
8 1000 1000
9 1001 1001
10 1010 00010000
11 1011 00010001
12 1100 00010010
13 1101 00010011
14 1110 00010100
15 1111 00010101

Gray Reflected Code
Gray reflected code is a non-weighted code which is used in many applica-
tions. In this code, only one bit transition occurs between two consecutive
numbers.
Decimal Binary Gray Code
0 0000 0000
1 0001 0001
2 0010 0011
3 0011 0010
4 0100 0110
5 0101 0111
6 0110 0101
7 0111 0100
8 1000 1100
9 1001 1101
10 1010 1111
11 1011 1110
12 1100 1010
13 1101 1011
14 1110 1001
15 1111 1000

Representation of Both Negative and Positive Numbers
Till now we have discussed representation of unsigned (positive) numbers.
But the representation system should be able to represent both negative and
positive numbers means signed numbers. Three popular ways to represent
signed binary numbers are
Signed Binary Number System
One’s Complement Number System
Two’s Complement Number System

Signed Binary Number System
In Signed Binary Number system sign and magnitude of a number is repre-
sented separately. For an n-bit binary number, 1-bit is reserved for sign and
(n − 1) bits are reserved for magnitude. Generally MSB indicates sign.
MSB
(Sign)
(n-1)-bits
(Magnitude)
If n = 4 then, +7 is represented as 0111 and −7 is represented as 1111.
The maximum number that can be represented in this number system is
Xmax = ±(2n−1
− 1) (11)

One’s complement is used to represent signed numbers. One’s complement
of a n-bit binary number X is
Y = (2n
− 1) − X (12)
One’s complement of a number is obtained by taking bit-wise inversion of
that number. For example if X = 1010 then its one’s complement is Y =
0101.
The range of the one’s complement representation is
−2n−1
+ 1 ≥ X ≥ 2n−1
− 1 (13)
In this representation, zero does not have unique representation as in case of
signed magnitude representation. This representation also has the symmetric
range. The MSB differentiates the positive and negative numbers. The range
of one’s complement representation for n = 8 is −127 ≥ X ≥ 127.

Representation of different real numbers in One’s Complement Number sys-
tem using n = 4 is shown below.
Number Representation Decimal Number Representation Decimal
+0 0000 0 −0 1111 15
+1 0001 1 −1 1110 14
+2 0010 2 −2 1101 13
+3 0011 3 −3 1100 12
+4 0100 4 −4 1011 11
+5 0101 5 −5 1010 10
+6 0110 6 −6 1001 9
+7 0111 7 −7 1000 8

Two’s complement number system is also used to represent signed numbers.
Two’s complement of a n-bit binary number X is
Y = (2n
− 1) − X + 1 (14)
To obtain Two’s complement of a number, One’s complement of the number
is first obtained. Then ulp = 20
is added to the number. For example if
X = 1010 then its one’s complement is 0101. Then its Two’s complement
equivalent is Y = 0101 + 0001 = 0110.
The range of numbers in Two’s complement number system is
−2n−1
≥ X ≥ 2n−1
− 1 (15)
The range is asymmetric as the number −2n−1
(100...000) does not have
the positive counter part. If a complement operation is attempted on 2n−1
(100...000) then the result will be same. Thus in a design with fixed word-
length, this value is ignored and symmetric range is used. The usable range
of two’s complement representation for n = 8 is same as that of one’s
complement representation.

Representation of different real numbers in Two’s Complement Number sys-
tem using n = 4 is shown below.
Number Representation Decimal Number Representation Decimal
+0 0000 0 NA NA NA
+1 0001 1 −1 1111 15
+2 0010 2 −2 1110 14
+3 0011 3 −3 1101 13
+4 0100 4 −4 1100 12
+5 0101 5 −5 1011 11
+6 0110 6 −6 1010 10
+7 0111 7 −7 1001 9
NA NA NA −8 1000 8

Real Data Representation Techniques
In hardware platform, real data is represented by some specific formats.
Representation techniques as well as data-width also can vary. Two type of
techniques are there to represent real data and they are
Fixed Point Representation - In this technique, the number of bits
to represent fractional and integer part are fixed for a fixed data-width.
Floating Point Representation - This technique resembles the float-
ing point representation in decimal number system. Wide range of data
can be represented by this technique.

Fixed Point Representation
X = xm−12m−1
+...+x121
+x020
· x−12−1
+x−22−2
+...+x−(n−m)2−(n−m)
=
m−1
X
i=0
xi 2i
+
(n−m)
X
i=1
x−i 2−i
(16)
Here, m-bits are reserved to represent the integer part and (n − m)-bits are
reserved for fractional part. For example, if the data-length is 16-bit and
value of m is 6 then 6-bits are reserved for the integer part and rest of the
bits are reserved for the fractional part.
m-bits
(Integer Part)
(n-m)-bits
(Fractional Part)
Note: Practically (m − 1)-bits are used to represent the integer as the MSB
is used to denote the sign of the data.

Fixed Point Representation
Convert X = −9.875 in fixed point representation.
1 Let the number is X = −9.875, n = 16 and m = 6.
2 Binary representation of the integer part using 6-bits is 9 = 001001.
3 Binary representation of the fractional part using 10-bits is 0.875 =
2−1
+ 2−2
+ 2−3
= 1110000000.
4 Magnitude of the number in fixed point format is 001001_1110000000.
5 Perform two’s complement as the number is negative. Thus X =
1_10110_0010000000.
If 6-bits are reserved for integer part for n = 16, then maximum value
of positive integer that can be represented is 2m−1
− 1 = 31. Maximum
representable negative number is −2m−1
= −32. But to avoid confusion, the
range of negative and positive numbers must be kept same. The maximum
representable real number with this format is
011111_1111111111 = 31.999023438 (17)

Floating Point Representation
The concept of floating point data format comes from the representation of
real floating point numbers. For example, the fractional number −9.875 can
also be represented as
−9.875 = −1 × 9875 × 10−3
(18)
Other representations are also possible. Thus floating point representation
is not unique. The general format of the floating point representation is
X = S.M.rEb
(19)
1-bit
(Sign Bit)
4-bit
(Biased Exponent)
11-bits
(Mantissa)
A floating point number has three fields, viz. Sign (S), Mantissa (M) and
Biased Exponent (Eb). Here, r represents the radix and its value is 10 for
decimal numbers. Eb = E + bias where bias is included to differentiate
negative and positive numbers. Bias is calculated as bias = 2p−1
− 1 where
p is the number of bits used to represent the biased exponent.

Fixed Point to Floating Point Conversion
Represent the fractional number X = −9.875 in floating point data format
for 16-bit word-length.
1 Represent the magnitude of the fractional number in binary. abs(x) =
9.875 = 1001_111.
2 First decide the sign bit. The sign bit is 1 as X is negative.
3 In the mantissa part 11-bits are reserved. The first step to represent the
mantissa part is to normalize the binary representation. After normaliza-
tion and adding zeros the result is 1_00111100000. The normalization
is done to restrict the mantissa part between 1 and 2. Here, the number
is shifted three bits in the left side. The MSB may not be included in
the final version of the mantissa. The MSB is called as hidden bit and
this bit is ignored in the mantissa representation according to the IEEE
754 standard.
4 As the mantissa part is three bits left shifted then the value of exponent
is E = −3 and the value of the biased exponent is Eb = 1010 = 10102.
5 The floating point representation is 1_1010_00111100000.

Fixed Point to Floating Point Conversion Example
Convert a = 001000_0010000000 to floating point representation.
The input fixed point number is a = 001000_0010000000 whose deci-
mal value is 8.125 for m equal to 6-bits.
As this number is positive so no inversion is required
There are two leading zeros present in the number so it is left shifted
by two bits. The result of shifting is 100000_1000000000.
The value of the mantissa is M = 00000_100000 by choosing 11-bits
from MSB and excluding the MSB.
The leading zero count is subtracted from (m − 1) and the result is
(5 − 2) = 3.
The result is added to the bias to get the exponent (E = (7+3) = 10).
Thus the floating point number is 0_1010_00000100000.

Fixed Point to Floating Point Conversion Example
Convert a = 111111_0100000000 to floating point representation.
The input fixed point number is a = 111111_0100000000 whose deci-
mal value is -0.75 for m equal to 6-bits.
As this number is negative so inversion is required. After inversion,
a = 000000_1100000000
There are six leading zeros present in the number so it is left shifted by
six bits. The result of shifting is 110000_0000000000.
The value of the mantissa is M = 10000_000000 by choosing 11-bits
from MSB and excluding the MSB.
The leading zero count is subtracted from (m − 1) and the result is
(5 − 6) = −1.
The result is added to the bias to get the exponent (E = (7 − 1) = 6).
Thus the floating point number is 0_0110_10000000000.

Floating Point to Fixed Point Conversion Example
Convert floating point number a = 0_1011_01000000000 to fixed point
representation.
Input data is represented in floating point as a = 0_1011_01000000000.
Prepare the data as 0000_1_01000000000 for 16-bit fixed point rep-
resentation with 6 integer bits.
The difference between exponent and bias is 1011 − 0111 = 0100 and
exponent is greater than the bias.
Left shift the number by 4-bit. Result is 1010000000000000.
As the sign bit is 0, no need of inversion. Discard the LSB and concate-
nate the sign bit at the MSB side. The final output is 0101000000000000.

Floating Point to Fixed Point Conversion Example
Convert floating point number a = 1_0011_01000000000 to fixed point
representation.
Input data is represented in floating point as a = 1_0011_01000000000.
Prepare the data as 0000_1_01000000000 for 16-bit fixed point rep-
resentation with 6 integer bits.
The difference between exponent and bias is 7 − 3 = 4 and exponent is
lesser than the bias.
Right shift the number by 4-bit. Result is 0000_0_00010100000.
As sign bit is 1, inversion is needed. After inversion the finial result is
1111111101100000.

THANK YOU

Binary Number System and Codes

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