Digital Design Digital Sytems Number Systems

19Z204- Digital Design
Dr.S.Sivaranjani,AP-CSE

Course Topics(Unit –I)
• Introduction to Digital Systems
• Number systems
• Number-Base conversion
• Complements of Numbers (Diminished Radix
complement ,Radix Complement)
• Signed Binary Numbers
• Arithmetic operation with the Binary Numbers
• Binary Codes ( BCD, 8421 code,Gray code,ASCII)

• Digital Systems are part of your daily activities.
• Everything in Digital(High Resolution Display) : Camera,
Laptop, Mobile phones, Tab- we are using Digital Electronics
Data processing
Data transmission
Device control
Digital
Systems
Made or build using
Introduction to Digital Systems

• Digital Devices have GUI [ Graphical User
Interface]
• Enable the devices to execute commands that
appear to the user to be simple but which
involves precise execution of a sequence of
complex internal instructions
• Digital Systems are fast, accurate and
consume less power.

Digital Systems
• Stemming from word “Digit” [means 0,1,2,3…]
• Use digital signals or information to operate
• Digital Signals : are discrete signals that
represent information in the form of binary
digits (bits) which can take only one of two
values (0&1)

Processed
Stored
Transmitted
Digital
Signals
Can be
Digital
Devices
Using
Smartphones,
Computers,Tablets &
other Electronic devices

Components of Digital Systems
• These components work together to process, store and transmit digital
information
Digital
System
I/P Devices
[used to provide input to
the system in the form of
digital signals, such as
switches, sensors, and
keyboards]
Output Devices
[used to display or transmit
the processed information,
such as monitors, printers,
and speakers]
Processors
[The processing devices
perform the required logic
and arithmetic operations
on the input signals, such
as microprocessors and
microcontrollers. ]
Memory
Storage Devices
[storage devices are used
to store data and
instructions, such as
memory chips and hard
drives]

What is Digital System?
Digital System
Digital Input Digital output
Discrete numbers
Process Digital information.
Processing will happen using Digital Logic

• A digital system is a system that processes
information in a digital form.
• It consists of electronic devices that
manipulate digital signals, such as binary
code, using a set of logic gates and algorithms.
• These devices include microprocessors,
microcontrollers, digital signal processors, and
other digital integrated circuits.

Applications of digital system
Digital systems have a wide range of applications in various fields, including:
Communication systems:
• Digital systems are used in communication systems, such as cellular
phones, satellites, and the internet.
• They enable reliable transmission and reception of data and voice signals
over long distances.
Control systems:
• Digital systems are used in control systems, such as robotics, process
control, and automotive control systems.
• They allow for precise control of devices and processes, ensuring optimal
performance.
Digital signal processing:
• Digital systems are used in digital signal processing applications, such as
audio and video processing, image and speech recognition, and radar and
sonar systems.

Medical devices:
• Digital systems are used in medical devices, such
as pacemakers, MRI machines, and digital X-ray
machines.
• They enable accurate and reliable diagnostic and
treatment procedures.
Consumer electronics:
• Digital systems are used in consumer electronics,
such as smartphones, tablets, laptops, and smart
TVs.
• They provide a wide range of functions and
capabilities, such as multimedia playback,
internet access, and gaming.

Industrial automation:
• Digital systems are used in industrial automation
systems, such as manufacturing plants, assembly
lines, and logistics systems.
• They enable efficient and automated control of
production and logistics processes.
Defense and aerospace:
• Digital systems are used in defense and
aerospace applications, such as military
communication systems, guidance and control
systems for aircraft and missiles, and satellite
communication and navigation systems.

Number System

General Representation
• A number is represented as
a5a4a3a2a1a0.a-1a-2a-3a-4
• . - radix point
• aj – coefficients(symbols used in a number representation)
• j – place value
• Example:
– Decimal number: 567.28

Radix or Base
• The radix of a number system is also known as
its base or its numerical base.
• It refers to the number of unique digits or
symbols used in the system to represent
numbers.
• For example, the decimal system that we use
in everyday life has a radix of 10 because it
uses 10 digits (0-9) to represent numbers.

• The radix of a number system is an integer greater than
1.
• The value of the radix determines the range of values
that can be represented using the system, as well as
the way in which numbers are written and
manipulated.
• For example, in a binary system with a radix of 2, there
are only two possible digits (0 and 1), which means
that all numbers are represented using only these two
digits.
• Different number systems are used in different
contexts, and each system has its own radix.

Common number systems and their
radices
• Binary system (radix 2): Uses two digits (0 and 1) to
represent numbers.
• Octal system (radix 8): Uses eight digits (0-7) to
represent numbers.
• Decimal system (radix 10): Uses ten digits (0-9) to
represent numbers.
• Hexadecimal system (radix 16): Uses sixteen digits (0-9
and A-F) to represent numbers.
Understanding the radix of a number system is
important for converting between different systems
and for understanding how computers represent and
manipulate numbers.

Radix or Base
• In a positional numeral system(value of each
symbol depends on the position),
the radix or base is the number of
unique digits, including the digit zero, used to
represent numbers.
• Eg: Decimal number system:
– Uses 10 symbols (0,1,2,3,4,5,6,7,8,9)
– Hence radix or base = 10

Positional Number System
• Another representation of a number:
anrn+an-1rn-1+…+a1r1+a0r0+a-1r-1+a-2r-2+…+a-mr-m
• aj – coefficients
• j – place value/ positional value
• r – radix or base
• Eg: 567.28(decimal number)
– 567.28 = 5x102+6x101+7x100+2x10-1+8x10-2

Radix - 2 (Binary Numbers)
• Symbols Used: 0,1
• Base: 2

Radix - 8 (Octal Numbers)
• Symbols Used: 0,1,2,3,4,5,6,7
• Base: 8

Radix – 10 (Decimal Numbers )
• Symbols Used: 0,1,2,3,4,5,6,7,8,9
• Base: 10

Radix – 16 (Hexadecimal Numbers)
• Symbols Used: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
• Base: 16

Example
• The number “twenty-seven” can be represented in
different ways :
– IIIII IIIII IIIII IIIII IIIII II (sticks or unary code)
– 27 (radix-10 or decimal code)
– 11011 (radix-2 or binary code)
– XXVII (roman numerals)
• The use of radix-2 ( binary) numbers became popular with
the onset of electronic computers,
– binary digits or bits, having only two possible values 0 and 1, is
compatible with electronic signals
• Radix-8 (octal) and radix-16 (hexadecimal) numbers have
been used as shorthand notation for binary numbers.

General representation
• Usually first 10 symbols in a number system of radix
r is represented by the symbols of decimal number
system and for the 11th symbol it starts with the
symbols of English alphabets.
– Eg:
• Radix-6 , Symbols: 0,1,2,3,4,5
• Radix-19, Symbols:0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I.
• To distinguish a number represented in a particular
number system, the number is usually written by
enclosing it in a parenthesis with a subscript of r.
– Eg:
• (56)8 – octal
• (56)10 – decimal
• (1011)2 - binary

Conversion between Number Systems
(Conversion from base-r to decimal)
• Conversion of any base-r number to decimal:
– Multiply each digit with its weight(radix raised to
its positional value) to get the resultant value of
each symbol.
– Add all the resultant symbol values.

(Binary to decimal number)
• 100.111
(100.111)2=1x22+0x21+0x20+1x2-1+1x2-2+1x2-3
= 4+0+0+0.5+0.25+0.125
= (4.875)10

To convert the binary number 11011.11 to decimal,
we can use the following method:
• Separate the integer and fractional parts of the binary number:
• Integer part: 11011
• Fractional part: 0.11
1.Convert the integer part to decimal by using the positional notation
of the binary system:11011 in binary equals
1 x 2^4 + 1 x 2^3 + 0 x 2^2 + 1 x 2^1 + 1 x 2^0= 16 + 8 + 0 + 2 + 1= 27
in decimal
2.Convert the fractional part to decimal by using the positional
notation of the binary system:
0.11 in binary equals 1 x 2^-1 + 1 x 2^-2= 0.5 + 0.25= 0.75 in decimal
3.Add the decimal values of the integer and fractional parts to get
the final decimal value:27 + 0.75= 27.75 in decimal
Therefore, the binary number 11011.11 is equal to the decimal
number 27.75.

(Octal to decimal number)
• 517.35
(517.35)8=5x82+1x81+7x80+3x8-1+5x8-2
= 320+8+7+0.375+0.078125
= (335.453125)10

(Hexadecimal to decimal number)
• 786.BC
(786.BC)16=7x162+8x161+6x160+Bx16-1+Cx16-2
= 1792+128+6+0.6875+0.046875
= (1926.734375)10

(Conversion from decimal to base-r)
• Conversion of a decimal integer to a number in
base r:
– Divide the integer and all its successive quotients by r
– accumulate the remainder in reverse order
– Range of remainder: 0 to r-1
• Conversion of a decimal fraction to a number in
base r:
– Multiply the fraction by r and its successive remainder
by r
– accumulate the quotients in the same order
– Range of coefficients: 0 to r-1

(decimal to binary number)
• 343.392
(343)10 = (101010111)2
0.392 x 2 = 0 .784
0.784 x 2 = 1 .568
0.568 x 2 = 1 .136
0.136 x 2 = 0 .272
0.272 x 2 = 0 .544
0.544 x 2 = 1 .088
2 343
2 171 - 1
2 85 - 1
2 42 - 1
2 21 - 0
2 10 - 1
2 5 - 0
2 2 - 1
1 - 0
(0.392)10 = (0.011001…)2
Ans: (343.392)10 = (101010111.011001)2

(decimal to octal number)
• 153.513
(153)10 = (231)8
0.513 x 8 = 4 .104
0.104 x 8 = 0 .832
0.832 x 8 = 6 .656
0.656 x 8 = 5 .248
0.248 x 8 = 1 .984
0.984 x 8 = 7 .872
8 153
8 19 - 1
2 - 3
(0.513)10 = (0.406517…)8
Ans: (153.513)10 = (231.406517)8

(decimal to hexadecimal number)
• 487.365
(487)10 = (1E7)16
0.365 x 16 = 5 .84
0.84 x 16 = 13 .44
0.44 x 16 = 7 .04
0.04 x 16 = 0 .64
0.64 x 16 = 10 .24
0.24 x 16 = 3 .84
16 487
16 30 - 7
1 - E
(0.365)10 = (0.5D70A3…)16
Ans: (487.365)10 = (1E7.5D70A3)16

Conversion from Binary to Octal and
Octal to Binary
• Conversion of a binary number to a octal number:
– Keep splitting the binary number into 3 bits from right to
left before radix point and left to right after radix point.
– If the leftmost part before radix point has lesser than 3
bits, add 0s to fill the places.
– If the rightmost part after radix point has lesser than 3 bits,
add 0s to fill the places.
– Write the corresponding octal symbol for each 3 bits and
accumulate them.
• Conversion of a octal number to a binary number:
– Write the 3 bit binary equivalent of each octal number and
accumulate them.

Conversion from Binary to Octal
• 10110001101011.111100000110
– Splitting the numbers into 3 bits:
010 110 001 101 011 . 111 100 000 110
2 6 1 5 3 . 7 4 0 6
– Answer:
(10110001101011.111100000110)2=(26153.7406)8

Conversion from Octal to Binary
• 673.124
– Writing 3-bit binary equivalent for each number.
6 7 3 . 1 2 4
110 111 011 001 010 100
– Answer: (673.124)8=(110111011.001010100)2

Conversion from Binary to Hexadecimal
and Hexadecimal to Binary
• Conversion of a binary number to a hexadecimal number:
– Keep splitting the binary number into 4 bits from right to left
before radix point and left to right after radix point.
– If the leftmost part has lesser than 4 bits, add 0s to fill the
places.
– If the rightmost part after radix point has lesser than 4 bits, add
0s to fill the places.
– Write the corresponding hexadecimal symbol for each 4 bits and
accumulate them.
• Conversion of a hexadecimal number to a binary number:
– Write the 4 bit binary equivalent of each hexadecimal number
and accumulate them.

Conversion from Binary to Hexadecimal
• 10110001101011.111100000110
– Splitting the numbers into 4 bits:
0010 1100 0110 1011 . 1111 0000 0110
2 C 6 B . F 0 6
– Answer:
(10110001101011.111100000110)2=(2C6B.F06)16

Conversion from Hexadecimal to Binary
• 306.D
– Writing 4-bit binary equivalent for each number.
3 0 6 . D
0011 0000 0110 . 1101
– Answer: (306.D)16=(001100000110.1101)2

Complements of Numbers
(Diminished Radix complement ,Radix
Complement)

Complement
• There are two types of complements for each
base-r number system:
– Radix complement (or) r’s complement
– Diminished radix complement (or) (r-1)’s
complement
• Complements are used in digital computers
for simplifying the subtraction operation.

r-1’s Complement/ Diminished Radix
Complement
• The (r-1)’s complement of N:
(rn-1)-N
 N – Number
 r – radix /base
 n – number of digits in N
 (rn-1) is the largest number with n digits in base r
 Hence, subtraction is between the number N from the largest number
with n digits.

r-1’s Complement of Binary Number/
1’s Complement
• N = 1011000 ; r=2
n = 7 ; r – 1 = 1 (1’s complement)
1’s complement = (27-1)10-(1011000)2
= ( 128 -1)10 – (1011000)2
= ( 127)10 – (1011000)2
= (1111111)2– (1011000)2
= 0100111
• N = 0101101 ; r=2
n =7
1’s complement = 1111111 – 0101101
= 1010010
Largest 7 digit binary number

1’s complement – Another Method
• 1’s complement of a binary number is formed
by
– changing 1’s to 0’s and 0’s to 1’s
• Eg: N = 1011000
– 1’s complement is 0100111

r-1’s Complement of Octal Number/
7’s Complement
• N = 563 ; r = 8
n= 3; r-1 = 7 (7’s complement)
7’s complement = (83-1)10-(563)8
= (512-1)10-(563)8
= (511)10-(563)8
= (777)8-(563)8
= 214 Largest 3 digit octal number

r-1’s Complement of Decimal Number/
9’s Complement
• N = 546700 ; r = 10
9’s complement = (106-1)10 – (546700)10
= (1000000-1)10 – (546700)10
= (999999)10 – (546700)10
= 453299 Largest 6 digit decimal
number

r-1’s Complement of Hexdecimal Number/
15’s Complement
• N = C3DF ; r= 16
15’s complement = (164-1)10 – (C3DF)16
= (65536-1)10 – (C3DF)16
= (65535)10 – (C3DF)16
= (FFFF)16 – (C3DF)16
= 3C20 Largest 4 digit hexadecimal
number

r’s Complement/Radix Complement
• The r’s complement of N:
rn-N for N ≠ 0
0 for N = 0
 N – Number
 r – radix /base
 n – number of digits in N
• r’s complement is obtained by adding 1 to (r-1)’s complement:
rn-N = [(rn-1)-N]+1
• Note: It is better to do (r-1)’s complement first for r’s complement since it is
easy to do the subtraction in (r-1)’s complement(no borrow problem!).

r’s Complement of Binary Number/ 2’s
Complement
• N = 1011000 ; r=2
n = 7 ; r = 2 (2’s complement)
1’s complement = (1111111)2– (1011000)2
= 0100111
2’s complement = 1’s complement + 1
= 0100111 + 1
= 0101000

2’s complement – Another Method
• 2’s complement of a binary number is formed by
– Scan the numbers from right to left
– Till the first ‘1’ is found write the digits as such.
– After the first ‘1’, invert all the digits.
• Eg: N = 1011000
– Writing the digits till 1st ‘1’ from right to left :
- - - 1 0 0 0
– Inverting the rest of the numbers
0 1 0 1 0 0 0
– Hence 2’s complement is : 0101000
• Eg: N = 1011001
– Writing the digits till 1st ‘1’ from right to left :
- - - - - - 1
– Inverting the rest of the numbers
0 1 0 0 1 1 1
– Hence 2’s complement is : 0100111

r’s Complement of Octal Number/
8’s Complement
• N = 563 ; r = 8
n= 3; r = 8 (8’s complement)
7’s complement = (777)8-(563)8
= 214
= 214 +1
= 215

r’s Complement of Decimal Number/
10’s Complement
• N = 546700 ; r = 10
9’s complement = (999999)10 – (546700)10
= 453299
= 453299 + 1
= 453300

r’s Complement of Hexadecimal Number/
16’s Complement
• N = C3DF ; r= 16
15’s complement = (FFFF)16 – (C3DF)16
= 3C20
= 3C20 + 1
= 3C21

Complement of a number with radix
point
• If the original number N contains a radix point,
– Temporarily remove the point to perform
complement.
– The radix point is then restored to the
complemented number in the same relative
position.

Complement with radix point (Base- 2)
• N = 1101.011
1101.011= 1101011 x 2-3
1’s complement of 1101011=1111111-1101011
= 0010100
2’s complement of 1101011 = 0010100 + 1
= 0010101
1’s complement of 1101.011= 0010100 x 2 -3
= 0010.100
2’s complement of 1101.011 = 0010101 x2 -3
= 0010.101

• N = 323.64
323.64= 32364x 8-2
7’s complement of 32364 =77777-32364
= 45413
8’s complement of 32364 = 45413 + 1
= 45414
7’s complement of 323.64 = 45413 x 8 -2
= 454.13
8’s complement of 323.64 = 45414x8 -2
= 454.14

• N = 325.93
325.93= 32593x 10-2
9’s complement of 32593 =99999-32593
=67406
10’s complement of 32593 = 67406 + 1
= 67407
9’s complement of 325.93 = 67406 x 10 -2
= 674.06
10’s complement of 325.93 = 67407 x10 -2
= 674.07

• N = ABC.3E2
ABC.3E2 = ABC3E2 x 16-3
15’s complement of ABC3E2 =FFFFFF- ABC3E2
= 543C1D
16’s complement of ABC3E2 = 543C1D + 1
= 543C1E
15’s complement of ABC.3E2 = 543C1D x 16 -3
= 543.C1D
16’s complement of ABC.3E2 = 543C1E x16 -3
= 543.C1E

Subtraction with Complements
r’s Complement Subtraction
• The subtraction of 2 n-digit unsigned numbers
M-N in base r can be done as follows:
– Add the minuend M to the r’s complement of the
subtrahend N. This performs: M+(rn-N)=M-N+rn
– If M≥ N, the sum will produce an end carry,rn,
which can be discarded. Hence the result is M-N
– If M< N, the sum does not produce an end carry
and is equal to rn-(N-M), which is the r’s
complement of N-M

r’s Complement Subtraction(Base-2)
• 1010100-1000011
2’s complement of 1000011=1111111-1000011+1 = 0111101
Answer =0010001
• 1000011-1010100
2’s complement of 1010100=1111111-1010100+1 = 0101100
Answer =1101111 (or) - 0010001
1 0 1 0 1 0 0
0 1 1 1 1 0 1
1 0 0 1 0 0 0 1
1 0 0 0 0 1 1
0 1 0 1 1 0 0
1 1 0 1 1 1 1

• 342-614
8’s complement of 614=777-614+1 = 164
Answer =526 (or) -252
• 614-342
8’s complement of 342=777-342+1 = 436
Answer =252
3 4 2
1 6 4
5 2 6
6 1 4
4 3 6
1 2 5 2

• 72532-3250
10’s complement of 03250=99999-03250+1 = 96750
Answer =69282
• 3250-72532
10’s complement of 72532=99999-72532+1 = 27468
Answer =30718 (or) -69282
7 2 5 3 2
9 6 7 5 0
1 6 9 2 8 2
0 3 2 5 0
2 7 4 6 8
3 0 7 1 8

• CB2-672
16’s complement of 672=FFF-672+1 =98E
Answer =640
• 672-CB2
16’s complement of CB2=FFF-CB2+1 = 34E
Answer =9C0 (or) - 640
C B 2
9 8 E
1 6 4 0
6 7 2
3 4 E
9 C 0

Subtraction with Complements
r-1’s Complement Subtraction
• The subtraction of 2 n-digit unsigned numbers M-N
in base r can be done as follows:
– Add the minuend M to the r’s complement of the
subtrahend N.
– If M≥ N, the sum will produce an end carry, which is added
to the result since it produces a sum that is 1 less than the
correct difference(only if carry is generated).
• Removing the end carry and adding 1 to the sum is referred to as
an end-around carry.
– If M< N, the sum does not produce an end carry,which is
the r-1’s complement of N-M

r-1’s Complement Subtraction(Base-2)
• 1010100-1000011
1’s complement of 1000011=1111111-1000011 = 0111100
Answer =0010001
• 1000011-1010100
1’s complement of 1010100=1111111-1010100 = 0101011
Answer =1101110 (or) -0010001
1 0 0 0 0 1 1
0 1 0 1 0 1 1
1 1 0 1 1 1 0
1 0 1 0 1 0 0
0 1 1 1 1 0 0
1 0 0 1 0 0 0 0
1
0 0 1 0 0 0 1

• 402-314
7’s complement of 314=777-314 = 463
Answer =066
• 314-402
7’s complement of 402=777-402 = 375
Answer =711 (or) - 066
3 1 4
3 7 5
7 1 1
4 0 2
4 6 3
1 0 6 5
1
0 6 6

• 4567-1234
9’s complement of 1234=99999-1234= 8765
Answer =3333
• 1234-4567
9’s complement of 4567=99999-4567= 5432
Answer =6666 (or) -3333
1 2 3 4
5 4 3 2
6 6 6 6
4 5 6 7
8 7 6 5
1 3 3 3 2
1
3 3 3 3

• B06-C7C
15’s complement of C7C=FFF-C7C = 383
Answer =E89 (or) -176
• C7C-B06
15’s complement of B06=FFF-B06 = 4F9
Answer =176
B 0 6
3 8 3
E 8 9
C 7 C
4 F 9
1 1 7 5
1
1 7 6 Dr.S.Sivaranjani,AP-CSE

Signed Binary Numbers

Signed Binary Number Representations
• Both signed and unsigned binary numbers
consists of a string of bits when represented in
computer.
• User determines whether a number is signed or
not.
• Representation Types:
– Signed-magnitude representation
– Signed- Complement representation
• 1’s Complement
• 2’s Complement

Sign-Magnitude Representation
• The number consists of two parts:
 Sign bit (leftmost bit)
 Magnitude bits (other than leftmost bit)
• If the leftmost bit is
 0 – positive number
 1 – negative number
• The negative number has the same magnitude bits as the
corresponding positive number but the sign bit is 1 rather than 0.
• Eg: 8-bit representation of ‘fifteen’
 +15 – 0 0001111
 -15 – 1 0001111
 It is used in ordinary arithmetic but usually not in computer
arithmetic, since sign and magnitude bits must be handled
separately.

1’s Complement Representation
• The negative number is the 1’s complement of
the corresponding positive number.
• Has some difficulties while used for arithmetic
operations.
• It is used in logical operations.
• There are two different representations for
zero.(i.e) 0000 and 1111 (4 bit +0 and -0).
 +15 – 00001111
 -15 – 11110000

• The negative number is the 2’s complement of
the corresponding positive number.
• This is the most common representation used
in computer arithmetic
 +15 – 0 0001111
 -15 – 11110001

Note: leftmost bit of the
representation acts a the sign bit (0
for positive values, 1 for negative
ones)

Conversion of decimal numbers to
signed binary numbers
• Express decimal number -39 as 8-bit number in (a)sign-
magnitude (b)1’s complement and (c)2’s complement
representations.
– 8-bit representation for +39
00100111
– (a)8-bit Sign magnitude representation for -39:
10100111
– (b) 8-bit 1’s complement representation for -39:
11011000
– (c) 8-bit 2’s complement representation for -39:
11011001
First represent the corresponding
positive number in the given number
of bits. Else the minimum number of
bits required to represent that
particular number should be taken.
Then use that
number
represented in
the required
number of bits to
find the negative
representation

Conversion of a signed binary number
to decimal number
Determine the decimal value of signed binary number expressed
in sign-magnitude representation.
• 10010101
– Computing the weights of rightmost 7 bits:
0x26+0x25+1x24+0x23+1x22+0x21+1x20 = 16+4+1=21
– Sign bit(leftmost bit) is 1.Hence it’s a negative number
– Therefore, the decimal number is -21

to decimal number
in 1’s complement representation.
• 00010111
– Computing the weights of the bits with the weight of the leftmost bit
as negative:
-0x27+0x26+0x25+1x24+0x23+1x22+1x21+1x20 = 16+4+2+1= +23
- Therefore, the decimal number is +23
• 11101000 (complement of the previous question)
as negative:
-1x27+1x26+1x25+0x24+1x23+0x22+0x21+0x20 = -128+64+32+8=-24
- Adding 1 to the result (i.e)-24+1=-23
- Therefore, the decimal number is -23
Negative numbers alone add
1 if 1’s complement
representation

to decimal number
in 2’s complement representation.
• 01010110
as negative:
-0x27+1x26+0x25+1x24+0x23+1x22+1x21+0x20 = 64+16+4+2= +86
- Therefore, the decimal number is +86
• 10101010 (complement of the previous question)
as negative:
-1x27+0x26+1x25+0x24+1x23+0x22+1x21+0x20 = -128+32+8+2=-86
- Therefore, the decimal number is -86
Need not add 1 like 1’s
complement representation

3-bit representation of signed numbers
No Possible
3-bit
represent
ations
If only
positive
numbers
represented
If negative
numbers also
should be
represented
(sign-
magnitude)
If negative
numbers also
should be
represented
(1’s
complement)
If negative
numbers also
should be
represented
(2’s
complement)
1 000 0 +0 +0 0
2 001 1 +1 +1 +1
3 010 2 +2 +2 +2
4 011 3 +3 +3 +3
5 100 4 -0 -3 -4
6 101 5 -1 -2 -3
7 110 6 -2 -1 -2
8 111 7 -3 -0 -1
With the available combinations of binary numbers for a given number of bits,
positive and negative numbers must be represented(FOR SIGNED NUMBERS)!

Arithmetic operation with the Binary Numbers

Binary Addition
• Addition
– The two numbers in addition:
• Augend
• Addend
– Result is:
• Sum
– Eg:
– Note: Generally, negative numbers are considered in 2’s
complement representation
0 0 0 0 0 1 1 1 Augend
+ 0 0 0 0 0 1 0 0 Addend
0 0 0 0 1 0 1 1 Sum

Binary Addition
• Both numbers are positive (consider 8 bit)
• Positive number with magnitude larger than
negative number
0 0 0 0 0 1 1 1 (+7)
+ 0 0 0 0 0 1 0 0 + (+4)
0 0 0 0 1 0 1 1 11
0 0 0 0 1 1 1 1 (+15)
+ 1 1 1 1 1 0 1 0 + (-6)
1 0 0 0 0 1 0 0 1 9
Discard Carry Dr.S.Sivaranjani,AP-CSE

Binary Addition
• Negative number with magnitude larger than
positive number
• Both numbers are negative
0 0 0 1 0 0 0 0 (+16)
+ 1 1 1 0 1 0 0 0 + (-24)
1 1 1 1 1 0 0 0 -8
1 1 1 1 1 0 1 1 (-5)
+ 1 1 1 1 0 1 1 1 + (-9)
1 1 1 1 1 0 0 1 0 -14
Discard Carry

Binary Addition
• Overflow condition
– When two numbers are added and the number of bits required to
represent that sum exceeds the number of bits in the two numbers, an
overflow condition occurs.
– It can occur only if both numbers are positive or both numbers are
negative.
– If the sign bit of the result is different than the sign bit of the numbers
that are added, overflow is indicated.
0 1 1 1 1 1 0 1 (+125)
+ 0 0 1 1 1 0 1 0 + (+58)
1 0 1 1 0 1 1 1 183
Incorrect Sign bit

Binary Subtraction
• Special case of addition is subtraction
• Subtraction
– The two numbers in subtraction:
• Minuend
• Subtrahend
– Result is:
• Difference
• Eg:
1 0 1 1 0 1 Minuend
+ 1 0 0 1 1 1 Subtrahend
0 0 0 1 1 0 Difference

Binary Subtraction
• The sign of the number is changed by taking
2’s complement
• To subtract 2 numbers
– take the 2’s complement of the subtrahend and
add.
– Discard any final carry

Binary Subtraction
• 00001000-00000011
– 2’s complement of 00000011=11111101
0 0 0 0 1 0 0 0 (+8)
+ 1 1 1 1 1 1 0 1 + (-3)
1 0 0 0 0 0 1 0 1 +5

Binary Multiplication
• Multiplication
– The numbers in multiplication:
• Multiplicand
• Multiplier
– Result is:
• Product
• (Partial Product)

Binary Multiplication
• 1011 x 101
1 0 1 1 Multiplicand(11)
x 1 0 1 Multiplier(5)
1 0 1 1 Partial Product
1 1 0 1 1 1 Product(55)

Binary Division
• Division
– The numbers in division:
• Dividend
• Divisor
– Result is:
• Quotient
• Remainder

Binary Division
• 11000101 ÷ 1010
1 0 0 1 1 Quotient(19)
1010 1 1 0 0 0 1 0 1 Divident(197)
1 0 1 0
0 1 0 0
0 0 0 0
1 0 0 1
0 0 0 0
1 0 0 1 0
1 0 1 0
1 0 0 0 1
1 0 1 0
1 1 1 Remainder(7)
Divisor(10)

Binary Codes
( BCD, 8421 code,Gray code,ASCII)

Binary Codes
• Any discrete elements of information that is distinct
among a group of quantities can be represented with
binary codes
• Sample Binary Codes:
– Binary Coded Decimal(BCD)/8421
– Gray Code
– Excess-3 Code
– 2421 Code
– ASCII Code
.
.
.

Binary Coded Decimal(BCD)/8421
• Straight binary assignment of the decimal numbers.
• It is a weighted code (codes which obey the
positional weight principle.)
• 10 decimal digits requires 4 bits for representation. But 6
out of 16 4-bit possible combination remains unassigned.
– A number with k decimal digits will require 4k bits in BCD
• Eg:
– (185)10 = (0001 1000 0101)BCD = (10111001)2
• Applications: Digital clocks, digital meters, Seven segment
display etc…(simplify the display of decimal numbers)
• This code is not very efficient but useful if only limited
processing is required. Dr.S.Sivaranjani,AP-CSE

Decimal & BCD
DECIMAL BCD
0 0 0 0 0
1 0 0 0 1
2 0 0 1 0
3 0 0 1 1
4 0 1 0 0
5 0 1 0 1
6 0 1 1 0
7 0 1 1 1
8 1 0 0 0
9 1 0 0 1
10 0 0 0 1 0 0 0 0

BCD Addition
• Add 2 BCD numbers using the rules for binary
addition
• If a 4-bit sum is equal or less than 9, it’s a valid
BCD number
• If a 4-bit sum is greater than 9 or if a carry out of
the 4-bit group is generated, it is an invalid result.
– Add 6(0110) to the 4-bit sum in order to skip the
invalid states.
– If a carry results when 6 is added, simply add the carry
to the next 4-bit group

BCD Addition
• 0110 0111 + 0101 0011
0 1 1 0 0 1 1 1 67
+ 0 1 0 1 0 0 1 1 + 53
1 0 1 1 1 0 1 0
0 1 1 0 0 1 1 0
0 0 0 1 0 0 1 0 0 0 0 0 120
Invalid BCD
Invalid BCD

BCD Addition
• The sign of the decimal number is represented
using 4 bits
– 0000 represents positive
– 1001 represents negative
• Sign-magnitude is seldom used in computers
• Sign-complement uses 9’s or 10’s complement

BCD Addition (10’s complement)
• (+ 0011 0111 0101) + (- 0010 0100 0000)
– 0011 0111 0101 –> 375
– 0010 0100 0000 –> 240
– 10’s complement of 0240 = 9999-0240+1= 9760
– Answer:
(+0011 0111 0101) + (-0010 0100 0000) = (+0001 0011 0101)
0 3 7 5
9 7 6 0
1 0 1 3 5
Discard Carry
Sign

BCD Subtraction
• At first the decimal equivalent of the given
Binary Coded Decimal (BCD) codes are found
out.
• Then the 10’s compliment of the subtrahend is
done and then that result is added to the
number from which the subtraction is to be
done.
• Discard Carry if generated
• Note: If 9’s complement is used, carry is added to the
result of subtraction!

BCD Subtraction (9’s complement)
• 0101 0001 − 0010 0001
– 0101 0001 -> 51
– 0010 0001 -> 21
– 9’s complement of 21 = 99-21 = 78
– 30 -> 0011 0000
– Answer: 0101 0001 − 0010 0001 = 0011 0000
5 1
+ 7 8
1 2 9
1
3 0
Carry
Carry generated added to
result

Gray Code
• Un-weighted code
• Important Feature:
– It exhibits one a single
bit change from one
code word to the next
in sequence
Decimal Binary Gray
0 0 0 0 0 0 0 0 0
1 0 0 0 1 0 0 0 1
2 0 0 1 0 0 0 1 1
3 0 0 1 1 0 0 1 0
4 0 1 0 0 0 1 1 0
5 0 1 0 1 0 1 1 1
6 0 1 1 0 0 1 0 1
7 0 1 1 1 0 1 0 0
8 1 0 0 0 1 1 0 0
9 1 0 0 1 1 1 0 1
10 1 0 1 0 1 1 1 1
11 1 0 1 1 1 1 1 0
12 1 1 0 0 1 0 1 0
12 1 1 0 1 1 0 1 1
14 1 1 1 0 1 0 0 1
15 1 1 1 1 1 0 0 0

Binary to Gray Code Conversion
• The MSB in the gray code is the same as the
corresponding MSB in the binary number
• Going from left to right, add each adjacent
pair of binary code bits to get the next gray
code bit.
• Discard Carries.

Binary to Gray Code Conversion
• Binary number = 10110
• (10110)2=(11101)Gray
• Note: Can perform XOR operation instead of addition. Hence
need not think about carry!
1 0 1 1 0 Binary
+ + + +
1 1 1 0 1 Gray
1+0 0+1 1+1 1+0

Gray to Binary Code Conversion
• MSB in the binary code is the same as the
corresponding bit in the gray code.
• Add each binary code bit generated to the
next gray code bit in the next adjacent
position.
• Discard carries.

Gray to Binary Code Conversion
• Gray code = 11101
• (11101)Gray= (10110)2
• Note: Can perform XOR operation instead of addition. Hence
need not think about carry!
1 1 1 0 1 Gray
+ + + +
1 0 1 1 0 Binary
1+1 0+1 1+0 1+1

Excess-3 Code
• Un-weighted Code
• Codes are obtained from
the corresponding decimal
value plus 3 in 4-bit binary
• Self complementing
code(9’s complement of a
number is directly obtained
by changing 1s to 0s and 0s
to 1s).
– 9’s complement of 4 is 5.
– 9’s complement of 3 is 6
Decimal BCD Excess-3
[(BCD+3) in
binary]
0 0 0 0 0 0 0 1 1
1 0 0 0 1 0 1 0 0
2 0 0 1 0 0 1 0 1
3 0 0 1 1 0 1 1 0
4 0 1 0 0 0 1 1 1
5 0 1 0 1 1 0 0 0
6 0 1 1 0 1 0 0 1
7 0 1 1 1 1 0 1 0
8 1 0 0 0 1 0 1 1
9 1 0 0 1 1 1 0 0
10 0001 0000 0100 0011

Converting Decimal to Excess-3
• (23)10
– Add both the digits separately by 3
• 2+3 =5
• 3+3 = 6
– Convert each corresponding decimal number to
equivalent 4 bit binary code
• 5 – 0101
• 6 – 0110
– Answer: (23)10 =(0101 0110)XS-3

Converting Decimal to Excess-3
• (359.8)10
– Add both the digits separately by 3
• 3+3 =6
• 5+3 = 8
• 9+3 =12
• 8 +3=11
– Convert each corresponding decimal number to equivalent
4 bit binary code
• 6 – 0110
• 8 – 1000
• 12 – 1100
• 11 – 1011
– Answer: (359.8)10 =(0110 1000 1100.1011)XS-3

American Standard Code for
Information Interchange (ASCII)
• An alphanumeric character set is a set of
elements that includes
– 10 decimal digits
– 26 letters of alphabets
– a number of special characters
• Uses 7 bit code and 128 characters
– Eg: A - 1000001

ASCII
• It contains
– 94 graphic characters that can be printed
• 26 uppercase letters( A – Z)
• 26 lowercase letters( a – z)
• 10 numerals ( 0 – 9 )
• 32 special printable characters such as %,*,$ …
– 34 non-printable characters used for various
control functions.
Totally 128
(94+34) characters

ASCII
• Control Characters
– Used for routing data and arranging printed text into prescribed
format
– Three types of control characters
• Format Effectors
– Characters that affect the layout of printing
– They include word processor and type writer controls:
» Backspace (BS)
» Horizontal tabulation (HT)
» Carriage return (CR)
• Information Separators
– Separate data into divisions such as paragraphs and pages
– They include:
» Record separator(RS)
» File separator(FS)

ASCII
• Communication Control Characters
– Useful during the transmission of text between remote
devices so that it can be distinguished from other messages
using the same communication channel before it and after it
» Start of Text (STX)
» End of Text (ETX)
• Mostly computers manipulate an 8-bit quantity as a
single quantity. Hence it is stored as one ASCII
character per byte
– Extra bit is used for other purposes depending on the
application

ASCII TABLE

Digital Design Digital Sytems Number Systems

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