1. Courtesy of Dr Radwan E Abdel-Aal
COE 202: Digital Logic Design
Courtesy of Dr Radwan E Abdel-Aal
Unit 1
Number Systems & Codes

2. Unit 1: Number Systems and Codes Contents
Introduction: Information representation & processing and 1.1
Number Systems: Binary, Octal and Hexadecimal
Number Base Conversion: Dec, Bin, Oct, and Hex
Digital Codes for Numbers and Symbols , 1.5, 1.6
Numbers: Binary, BCD, Excess-3, Gray, Parity , 1.5
Alphanumeric (symbols): ASCII, Unicode
Arithmetic on Unsigned Numbers:
Basic Operations in binary: Add, Subtract, Multiply
Arithmetic in Octal and Hexadecimal, BCD addition
Binary Signed Number Representation and Arithmetic:
Subtraction, complement notation
Signed representation, Addition/Subtraction, Overflow

3. We live in a predominantly Analog world
1. Introduction to Digital Systems Continuous Vs Discrete (Analog Vs Digital)
We live in a predominantly Analog world
Analog means Continuous (both in time and amplitude)
Analog information exhibit smooth, gradual changes over time and assume a continuous (infinite) range of amplitudes
Examples:
Earth’s movement
Body temperature
Our speech
Analog
Signal

4. Digital Information Digital  Discrete, Not continuous
Digital information assume a limited (finite) set of “Discrete” values, not a continuous range of values
Values change abruptly (not smoothly) by “Jumping” between values
Examples:
The Alphabet
Position of a switch
DNA sequences (TAGC)
Energy levels of
electrons in atoms
Digital
Signal
Only 4 allowed
Signal levels
If we have only two signal levels  Binary signal
… So binary is a special case of digital

5. Analog Vs Digital
It is a lot easier to design digital systems than analog ones, Why?
Much simpler to deal with a limited set of values as inputs and outputs for the circuits
Greater tolerance to drift, noise  low error rates
Digital circuits are more available, simpler and cheaper (VLSI)
It is also more advantageous to store and communicate information digitally
Dilemma here: Our natural world is mainly analog… but it is easier to process it digitally!
Also

6. Digital Processing in the Real World
Analog-to-digital-converters (ADC) are used to digitize raw analog inputs.
Digital-to-analog-converters (DAC) are used to regenerate analog signals from their digitized form

7. Digitization of Analog Signals
Nowadays most processing, computing, and communication are done digitally
Solution: Convert analog signals to digital parameters and process them digitally
This requires two steps:
1. Sampling in time (impossible to handle the  number of values existing on the time axis!). Ignore signal between samples
2. Quantization in amplitude (impossible to handle the  number of values existing on the amplitude axis!). Approximate sample value to the nearest quantization level
2. Quantization to discrete
levels in amplitude
Ignore
Ignore
1. Sampling at discrete points in time

8. Amplitude Quantization: 4 discrete levels
Example:
Amplitude Quantization: 4 discrete levels
Using a larger number of discrete levels
We can reduce the quantization errors
(noise) we introduced!
Quantization
Errors
Analog Signal levels are mapped to the nearest value among the set of discrete voltages  {V1, V2, V3, V4} allowed for the digital signal

9. Information Representation How to represent 10 discrete levels (0 to 9)?
Assume a 0 to 5 V range to represent the discrete quantization levels
Direct 10-level Representation
Using Binary (2-level) Representation
Our circuits deal with:
Ten Signal levels
Two Signal levels (ON/OFF)
Simpler, reliable Circuits
Noise Margin:
(5/9)/2  V
(5/1)/2 = 2.5 V
Larger (better)
Number of steps
1 variable takes
1 of 10 values
Use n variables, each takes
1 of 2 values {0,1}
 n binary digits (bits)
e.g. with n = 4 bits
 6 is represented as 0110
Noise Margin

10. Binary Values: Abstract and Physical Representations
Abstract Representations
Digit Values: 0, 1
Logic Levels: True, False
Signal Levels: Low, High
States: ON, OFF
Physical Representations
In an IC (e.g. a microprocessor): Voltage or Current
On a Hard Disk: Magnetic Field Direction
On a CD: Surface pits/Light interference
In a Dynamic RAM: Electrical Charge

11. Signal Voltage Levels: Noise Considerations
1 Logic Level
Min O/P
representing 1
Max allowed
noise pickup
on Level 1
Forbidden
Region
Min I/P
Accepted as 1
0 Logic Level
The HIGH range typically corresponds to binary 1 and LOW range to binary 0. The threshold region is a range of voltages
for which the input voltage value cannot be interpreted reliably as either a 0 or a 1.
Output
Input
Circuit
Circuit
Noise pickup on connecting medium

12. Types of Digital Systems
No State Present
Response determined only by present inputs
i.e. Output = Function (Inputs only)
Combinational Logic Circuits, e.g. adder
A State Present
Response determined by present inputs and the present state
of the system
i.e. Output = Function (Present State, Inputs)
and State (new) = Function (Present State, Inputs)
Sequential Logic Circuits, e.g. Counter
Two Types of state-based systems:
State updated at discrete times (e.g. at clock pulses)
 Synchronous Sequential System
State updated at any arbitrary time
 Asynchronous Sequential System

13. Digital Processing System
Takes a set of discrete information inputs and discrete internal system information (system state) and generates a set of discrete information outputs.
Data Path
System State
Discrete
Information
Processing
System
Inputs
Outputs
Processed
Digital
Data
Row
Digital
Data
Control
Outputs
Results can
affect next
Processing
Sequence of system states determines the
processing performed
Same hardware can do different processing by simply changing the stored program!
Control
Path
Stored Program

14. A Generic Digital Computer
CPU: Central Processing Unit
FPU: Floating Point Unit
MMU: Memory Management Unit
Program
Data
Processor
FPU
MMU
Cache
Input Devices:
Keyboard, Mouse, Scanner, Microphone
Output Devices:
Monitor, Printer, Speaker
Answer: Part of specification for a PC is in MHz. What does that imply? A clock which defines the discrete times for update of state for a synchronous system. Not all of the computer may be synchronous, however.
Input/Output
Hard disk, Modem, Network Card

15. 2. Number Systems Representation
Positive radix, positional number systems
A number with radix (base) r (or b) is represented by a string of digits (n in integer and m in fraction):
An - 1An - 2 … A1A0 . A- 1 A- 2 … A- m in which 0 £ Ai < r and “.” is the radix point.
The string of digits represents the power series:
Integer Part
Fraction Part ( ) ( )
i = n - 1
j = - m å å
(Number)r = A r i + A r j i j
i = 0
j = - 1
+ ive exponents
- ive exponents
(Integer Portion) +
(Fraction Portion)

16. Examples Decimal (base 10):
(724.56)10= 7 x x x x x 10-2
= =
(base 5):
(312.4)5= 3 x x x x 5-1
= = 82.8
Radix Powers
Radix Powers

17. Number Systems – Examples
For these 3 systems, Radix is a Power of 2
Number Systems – Examples
System
General
Decimal
Binary
Octal
Hexadecimal
Radix (Base) r 10 2 8 16
Digits: 0 to (r-1)
0, 1, …, (r – 1)
0, 1, …, 9
0, 1
0,…, 7
0, …, 15 1 3
Power of 4
Radix -1 -2 -3 -4 -5 r0 r1 r2 r3 r4 r5
r -1
r -2
r -3
r -4
r -5
100
1000
10,000
100,000
0.1
0.01
0.001
0.0001 4 32
0.5
0.25
0.125
0.0625
Integer
Fraction

18. Commonly Used Bases Name Radix Digit takes values: Binary (b) 2 0,1*
Octal 8
0,1,2,3,4,5,6,7 10
0,1,2,3,4,5,6,7,8,9
Hexa-
Decimal (H) 16
0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
10,11,12,13,14,15
Example
57AC.F4D
Decimal (d)
Answer: The six letters A, B, C, D, E, and F represent the digits for values
10, 11, 12, 13, 14, 15 (given in decimal), respectively, in hexadecimal.
Alternatively, a, b, c, d, e, f are used.
*The Binary digit is called a bit
Bits take the values 0,1 only (OFF/ON)  can be manipulated
with Simple, Reliable Circuits

19. Integers (0 – 16) Represented in Various Bases
More Concise
0-15 : digits
The larger the
Radix the fewer the digits needed to represent a given number
LS digit goes
through
all possible
Combinations, then the more significant digit, etc.
Octal & Hexa
offer a more
concise way
of representing
long bit (binary)
sequences
in computer
systems

20. Binary Numbers: Special Powers of 2
210 (1024) is Kilo, denoted "K"
220 (1,048,576) is Mega, denoted "M"
230 (1,073, 741,824) is Giga, denoted "G"
240 (1,099,511,628) is Tera, denoted “T"

21. Positive Powers of 2 Powers of 2 Exponent Value Exponent Value 1 111 11
2,048
2 K 1 2 12
4,096
4 K 2 4 13
8,192
8 K 3 8 14
16,384
16 K 4 16 15
32,768
= 215 = 25 x 210 = 32 K 5 32 16
65,536
64 K 6 64 17
131,072
128 K 7
128 18
262,144
256 K 8
256 19
524,288
512 K 9
512 20
1,048,576
1 M 10
1024
1 K 21
2,097,152
2 M

22. Conversion Between Number Systems
We will consider: (N may contain a fraction part)
Conversion from base b to decimal: (N)b  (?)10
Conversion from decimal to base b: (N)10  (?)b
Conversion between 2 non-decimal bases b1 and b2: (N)b1  (?)b2:
In general: Go through decimal. i.e.
(N)b1  (?)10  (?) b2
Special case when b1 and b2 are powers of the same number: e.g. between binary (b = 21), octal (b = 8 = 23), and hexadecimal (b = 16 = 24):
Here direct conversion is easy…
Powers of 2: 43210
=>
1 X = 16
+ 1 X = 8
+ 0 X = 0
+ 1 X 21 = 2
+ 0 X 20 = 0
2610

23. Conversion from base b to decimal: (N)b  (?)10
Simply substitute in the power series expansion
Examples: compare lengths
b = 2: ( )2  (D)10:  
D = 1 x x x x x x 2-2 = = 53.75
Notice: smaller bases give longer Number representation
 Binary is the longest representation!
Powers of 2: 43210
=>
1 X = 16
+ 1 X = 8
+ 0 X = 0
+ 1 X 21 = 2
+ 0 X 20 = 0
2610
b = 16: (E9C4.B6)16  (D)10:  
D =14 x x x x x x = =

24. Conversion from decimal to base b: (N)10  (?)b
Procedure:
Convert the Integer Part
Convert the Fraction Part
Join the two results with the new radix point
625 – 512 = 113 => 9
113 – 64 = 49 => 6
49 – 32 = 17 => 5
17 – 16 = => 4
1 – = => 0
Placing 1’s in the result for the positions recorded and 0’s elsewhere,

25. Converting integer from decimal to base b:
Repeatedly divide the integer by the new base b and save the remainders, until you get a 0 quotient.
The digits for the new base are the remainders with the LS digit being the first remainder
(41)10 = (??)2 
+0/2
radix point
+0/2
Integer in new base
Continue until
you get a 0
quotient
+0/2 
Verify result!

26. Converting fraction from decimal to base b:
Repeatedly multiply the integer by the new base b and save the integer part of the result (will never exceed b) until 0 fraction or enough digits
 The digits for the new base are those integers with the MS digit being that computed first
(0.6875)10 = (??)2 
radix point
Fraction in new base
Continue until
you get either:
Zero fraction, or
Enough # of
Significant fraction digits 
Verify Result

27. Comment on the Fractional Part
In the above example the fractional part became 0 as a result of the repeated multiplications and exact conversion was achieved
In general it may take many bits to get this to happen or it may never happen
Example: Convert to N2
= …2
The fractional part begins repeating every 4 steps, thus repeating 1001 forever!
Solution: Specify number of bits to right of radix point and truncate or round to this number of bits, e.g = or = (to 3 bits)

28. 1/2 = 0 remainder = 1 Reading off in the forward direction: 0.10112
Example: Convert to Base 8, rounding the resulting fraction to 3 octal digits
1. Convert Integer Part: 153 to Base 8
2. Convert Fraction Part: to Base 8:   
Answer 1: Converting 46 as integral part: Answer 2: Converting as fractional part:
46/2 = 23 rem = * 2 = int = 1
23/2 = 11 rem = * 2 = int = 0
11/2 = 5 remainder = * 2 = int = 1
5/2 = 2 remainder = * 2 = int = 1
2/2 = 1 remainder =
1/2 = 0 remainder = Reading off in the forward direction:
Reading off in the reverse direction:
Answer 3: Combining Integral and Fractional Parts:
Get 4 octal digits in the
Fraction for rounding to
3 significant digits
(0.513)10 = (0.4065)8  (0.407)8 after rounding, since 5 is >4 we add 1 to fraction 
So: =

29. Contd. Example: Convert 153.51310 to Base 8
3. Join the results together with the new radix point:
4. Verify the result:
( )8  (D)10
= 2 x x x x x x 8-3
= 2 x x x 8-3
= ( )10
 ( )10
(Difference is due to rounding error)
( )10 = ( )8

30. Conversion between 2 non-decimal bases b1 and b2: (N)b1  (?)b2:
General Case: bases are not powers of the same number:
Go through the intermediate step of decimal
e.g. (231)5  (??)8 (231)5  (??)10  (??)8
Special (and more important case with computers):
Bases are powers of the same number (2), e.g.
System Base Digit Range # of Binary bits/digit
Binary =
Octal =
Hexadecimal =
Exact number
of binary bits
needed to store a digit!
Range of
Digit values
Number of
binary bits needed to
represent a digit
Here binary is a better intermediate step!

31. Binary, Octal, & Hexadecimal Conversions
Octal (or Hexadecimal) to Binary:
Express each octal (hexadecimal) digit as three (four) binary bits starting at the radix point and going both ways.
Binary to Octal (or Hexadecimal):
Group the binary bits into three (four) bit groups starting at the radix point and going both ways, padding LH zeros in the integer part and RH zeros in the fractional part as needed
Replace each group of three (four) bits with the equivalent octal (hexadecimal) digit
Octal Hexadecimal
Go through binary as an intermediate step (use above)
e.g. Octal  Binary  Hexadecimal

32. Examples: Octal (Hexadecimal) Binary
Example: Hexadecimal to Binary
(F C . B D)16 =
= ( )2
Example: Binary to Octal
( )2
= ( )2
= ( )2
= ( )8
Always start at the radix point
Working your way away from it
in both directions
Answer 1:
110|011| |111|111 2
Regroup:
1|1001| |1111|1(000)2
Convert:
D F
Answer 2: Marking off in groups of three (four) bits corresponds to dividing or multiplying by 23 = 8 (24 = 16) in the binary system.
Appended 0’s

33. Example: Octal to Hexadecimal via Binary
1. Convert octal to binary
2. Use groups of four binary bits and express them as hexadecimal digits
Example: Octal  Binary  Hexadecimal
( )8
= ( D E 8)16
Represent Octal in binary
Group into 4 bit groups for both the integer and fraction parts, starting at the radix point
Append leading 0’s to the left of integer part and trailing 0’s to the right of the fraction part as needed
Express each group of 4 bits in hex
Appended 0’s
Answer 1:
110|011| |111|111 2
Regroup:
1|1001| |1111|1(000)2
Convert:
D F
Answer 2: Marking off in groups of three (four) bits corresponds to dividing or multiplying by 23 = 8 (24 = 16) in the binary system.

34. Using Binary Bits to Represent Information: Numbers and Codes
We can assign any combination of binary bits (called a code word) to represent any information item as long as data is uniquely represented
Two Basic Types of Information:
Numeric (e.g. signed or unsigned integers)
Code tied to binary numbers
Better use codes that allow simple implementation of common arithmetic operations
Non-numeric (e.g. symbols, colors, the alphabet)
Code not tied to binary numbers
Greater flexibility: No arithmetic operations involved

35. If you represent only a subset of m items, m < rn
Q 1
Given n digits of radix r, what is the maximum number of different information items that can be represented?
With n digits in radix r, there are rn distinct combinations (codes), i.e. a maximum of rn items can be uniquely represented
If you represent only a subset of m items, m < rn
, this leaves (rn – m) codes unused
Examples:
2 decimal digits (r = 10) can represent up to 102 = 100 elements: (00, 01, 02, …, 98, 99)
3 binary bits (r = 2) can represent up to 23 = 8 elements: (000, 001, 010, 011, 100, 101, 110, 111)
If only the blue 6 codes are used, the remaining 2 codes are unused and maybe we don’t care what the output would be for them- this simplifies the circuits needed, see Unit 2

36. Q 2
Given M different information items to represent, what is the minimum number (n) of radix r digits do we need?
rn  M > r(n – 1)
i.e. n = the smallest integer that is ≥ [logr M]
Example: What is the minimum # of binary bits are required to represent the 10 decimal digits with a binary code?
n = smallest integer that is ≥ log2 (10)
≥ 3.33  n = 4
 need at least 4 bits:
24 > 10 > 23
M = 10
Therefore n = 4 since:
24 =16 is  10 > 23 = 8
and the ceiling function for log2 10 is 4.
How many spare codes do we have here?

37. Using Binary to represent Symbols: 1. ASCII Character Codes
American Standard Code for Information Interchange (ASCII)
Popular for representing information as character-based textual data
Uses 7-bits (i.e. 27 = 128 combinations) to represent:
94 Graphic printing characters (alphanumeric, symbols,…)
34 Non-printable characters
Non-printable characters are used for:
Text formatting (e.g. BS = Backspace, CR = carriage return)
Record marking and flow control (e.g. STX and ETX start and end text areas)
Communication handshaking, e.g. ACK

38. ASCII Character Codes A = 100 0001 = 41 Hex a = 110 0001 = 61 Hex
Non-Printable Characters
MSB
8 Columns
LSB 16
Rows
16 x 8
= 128

39. Extends ASCII to 65,536 universal characters codes
2. UNICODE
Extends ASCII to 65,536 universal characters codes
2 byte (16-bit) code words
For encoding characters in world languages
Used in many modern applications
Arabic codes: from 0600 to 06FF (hex)
How many codes are used for Arabic? Answer after we cover hexadecimal arithmetic

40. Using Binary to represent Numbers: e.g. 0-9 using 4 bits (16 codes)10 16
Using Binary to represent Numbers: e.g. 0-9 using 4 bits (16 codes)
0000
0001 1 9
1111
There are different ways to map the 10 decimal elements to the 16 binary codes of 4 bits. 4 of those are shown below:
Straight Binary
4 coding
schemes
1’s Complementing
G2 G1 G0
Gray:
Always
only 1-bit
changes
between
adjacent
codes
All 4 bits
Change

41. Gray Code: Advantages for Optical Encoding of Angular Shaft Position
The 3 optical sources/sensors (or sector marks) may not be perfectly aligned
Some may change before the others at sector crossings
For example, at the crossing between positions 3 and 4:
Gray: Only bit G0 changes  code can change only from 111 to (i.e. correct new code is the only possibility)
Binary: All 3 bits change  several wrong intermediate codes can be generated (e.g. 110). If read, they give wrong information on shaft position
111
000
100
000 B
White = 1 B 1
110
001
101
001 B 1 2
Yes, as illustrated by the example that followed. 3 2
111
101 4 3
010 4
011
G2 G1 G0
Shaft
Rotation
3 sensors
100
3 sensors
011
110
010
(a) Binary Code for Positions 0 through 7
(b) Gray Code for Positions 0 through 7

42. Binary-Coded Decimal (BCD)
The BCD code consists of the first ten values (0 to 9) of the bit (8,4,2,1) binary codes
Can be used to encode
the digits of a decimal number separately in in binary, e.g.
(863)10 = ( )BCD
Note this is different from converting the decimal number to binary:
(863)10 = ( )2
Which is a more concise representation?
BCD is Basis for BCD computer arithmetic: e.g. BCD addition,
subtraction, etc. See later
10 BCD Codes
Answer 1: 6
Answer 2: 1010, 1011, 1100, 1101, 1110, 1111
6 Unused
Binary Codes
(don’t care conditions)

43. Error-Detection Codes
Redundancy (additional information), in the form of extra bits, can be appended to the binary code for the data to allow error detection, and possibly error correction
A simple form of redundancy is the parity bit, an extra bit appended to the code word to make the total number of 1’s odd or even. Parity can detect all single-bit errors and some multiple-bit errors
Even parity bit: Makes total number of 1’s even
Odd parity: Makes total number of 1’s odd

44. Example: 4-Bit Parity Code (3 bits of data + 1 parity bit)
Even Parity
Odd Parity
Message
Parity
Message
Parity - -
0000
0001
0011
0010
0101
0100
0110
0111
1001
1000
1010
1011
1100
1101
1111
1110
Even Parity Bits: 0, 1, 1, 0, 1, 0, 0, 1
Odd Parity Bits: 1, 0, 0, 1, 0, 1, 1, 0
 Note: Odd parity is the complement (inversion) of even parity
Question: can the parity scheme detect 1, 2, or 3 bit errors?
We will design circuits that generate parity bits in Unit 3

45. e.g. for n = 8 bits (a byte): 0 to [28-1]/28 = 0 to 0.9961
Range of values (smallest to largest) of unsigned integers and fractions represented using n bits
Unsigned integer values range: 0 to 2n-1
00……0 = 0
to 11……1 = 1 x x 21 + …. + 1 x 2(n-1) = 2n-1
e.g. for n = 8 bits (a byte): Range: 0 to 28-1 = 0 to 255
Unsigned fraction values range: 0.0 to [2n-1]/2n
0.00….0 = 0
to …1 = 1 x x …. + 1 x 2-n = 2n [1 x x …. + 1 x 2-n]/2n = [1 x 2n x 2(n-2) + …. + 1 x 20]/2n = [2n-1]/2n (an n-bit integer shifted right 8 places!)
e.g. for n = 8 bits (a byte): 0 to [28-1]/28 = 0 to
(n-1)
-1 ……. -n
M = 10
Therefore n = 4 since:
24 =16 is  10 > 23 = 8
and the ceiling function for log2 10 is 4.

46. 3. Basic Computer Arithmetic
1. Single Bit Arithmetic:
Addition with Carry, Subtraction with Borrow, and Multiplication
2. Multiple Bits Arithmetic:
On Unsigned Binary Integers
Binary Addition, Subtraction, and Multiplication
BCD Addition
b. On Signed Binary Integers
Signed-magnitude
1’s and 2’s complement
c. Floating Point Binary Arithmetic: e.g x 106
Not covered in this course
3. Octal and Hex arithmetic: Unsigned addition, subtraction and multiplication

47. 1. Single Bit Binary Addition with CarryX
1-bit
Adder Z C S

48. Single Bit Binary Subtraction with BorrowX
1-bit
Subtractor Z B D

49. Single Bit Binary Multiplication

50. 2. Unsigned Numbers (Multiple Bit) Binary Addition
Extending bit operation to multiple bit:
Generated carries to the bit
Augend
Addend
Sum
Notes:
Assume 0 default Carry-In to the least significant bit
The red carry is end-carry from the most significant bit
In unsigned addition, an end carry represents an overflow- Result is larger than can be represented in the used number of bits (here 5 bits)
Always 0 for 1st bit (1st stage)
End Carry
Verify results by doing the operation on the equivalent decimal values
= 29
< 33, No overflow
= 45
> 33, overflow
Result too large for 5 bits!
Carries
Augend
Addend
Sum

51. Unsigned Numbers Binary Subtraction with borrow
M – N
First ensure that M  N
(by swapping numbers if necessary)
Extending bit operation to multiple bits: (here 5 bits)
Borrows from bit
Minuend
Subtrahend
Difference
Notes:
If N is larger than M, swap the two, do the operation, and consider the result negative (we do not have a sign bit yet!)
Assume 0 default Borrow-In to the least significant bit
The red borrow is Borrow-out from the most significant bit (End Borrow)
Borrow and subtraction mechanisms are cumbersome- Will seek better solutions for subtraction (through addition)
Always 0 for 1st bit in 1st stage
Verify results by doing the operation on the equivalent decimal values
= 4
Carries
Augend
Addend
Sum

52. Unsigned Numbers Binary Subtraction with borrow
M – N
No restrictions, M >=< N
No Prior Comparison of M. N
Algorithm:
Subtract N from M without prior comparison or swapping
If no end borrow occurs, then M ³ N, and the result obtained is non-negative and correct.
If an end borrow occurs, then N > M. the result is (M + 2n) - N. To get N-M, we subtract the result from 2n , and consider it negative.
Examples:
(-)
End borrow caused by last stage
It amounts to adding 2n to M 9 M 4 7 -N 7
What is the problem on the RHS?
How we discover it?
How we handle (correct) it? 2 13
This is (M + 2n ) – N (Wrong)
To get N-M from this: do
2n – (M+2n-N)
i.e. subtract the result from 2n
and call it - ive 16
Correction
Step 13
2n – number = its 2’s complement
So, 2’s complementing can help! 3

53. The Complement of an integer
Depends not only on the
Integer N but also on the number
of digits (bits) n used to represent it
The Complement of an integer
For a number N of n digits in radix r we define:
The Radix (r’s) Complement
Defined as rn - N
The Diminished Radix [(r - 1)’s] Complement
Defined as (rn - 1) - N
For binary number, r = 2, we have:
The 2’s Complement
Defined as 2n - N
The 1’s Complement
Defined as (2n -1) - N
Decimal number system:
r = 10, 2 digits # e.g. 57
10’s comp. = = 43
9’s comp. = (102-1)-57
= = 42
For n = 4 bits
2n = 24 = 16
2n-1 = 16-1 = 15 = (n 1’s)
Let N = 1001, 1’s comp of N
- 1001
0110
1’s comp. is easier to generate!
Then 2’s comp = 1’s comp. + 1
1  0 , 0  1 (NOT gates)

54. Binary 1's Complement For r = 2, N = 011100112, n = 8 (8 digits):
(rn – 1) = = or
The 1's complement of is then:
(8 1’s) –
Since the 2n – 1 factor consists of n 1's and since 1 – 0 = 1 and 1 – 1 = 0, the one's complement is obtained by complementing each individual bit (bitwise NOT).

55. Binary 2's Complement
For r = 2, N = , n = 8 (8 digits), we have:
(2n ) = or (1 in the (n+1)th position)
The 2's complement of is then: –
Note this is the 1's complement plus 1
 Can design simple hardware for generating the 2’s complement from the easier-to-generate 1’s complement
Eight 0’s
2n-N = (2n-1-N) +1
2’s Complement
1’s Complement

56. Alternative 2’s Complement Method
Given: an n-bit binary number, beginning at the least significant bit and proceeding to higher bits:
Starting at the LS bit, Copy all least significant 0’s
Copy also the first 1
Complement all remaining bits thereafter.
2’s Complement Example:
Copy underlined bits:
and complement all remaining bits to the left:

57. Important Complement Remarks
Taking the complement of a complement returns you back to the original number,
Example: 2’s complement for an n-bit number N
2’s comp. of N = 2n-N
2’s comp. of the 2’s comp. of N = 2n- (2n-N) = N
Complements of 0 (represented as n bits):
1’s comp of 0 = n 1’s
2’s comp of 0 = 1’s comp of = Carry + n 0’s =
1’s Complementing
0000
1111
Ignored-
Outside the n bits +1
2’s Complementing
(also by the rule on previous slide!)
0000

58. Subtraction of two Unsigned Numbers (M-N) Using the 2’s Complement
For n-bit, unsigned numbers M and N, find M - N:
Add the 2's complement of the subtrahend N to the minuend M: M + (2n  N) = M  N + 2n
If M  N, the sum produces end carry (rn) which is discarded; leaving the correct answer M - N.
If M < N, the sum does not produce an end carry and, from above, is equal to 2n  ( N  M ), i.e. the 2's complement of ( N  M ).
To obtain the result, i.e. negative (N – M):
Take the 2's complement of the sum  (N-M)
and place a  sign to its left!. M
- N
i.e. we are doing subtraction by addition!
(Simpler)

59. Unsigned Subtraction with 2’s Complement Example 1
Find – –
= 17
The end carry of 1 indicates that no correction is required on the result .
No need to compare
amplitudes
or re-arrange! 84 1
- 67
2’s comp 17

60. Unsigned Subtraction with 2’s Complement Example 2
Find – –
The 0 end carry indicates that a correction of the result is required (taking the 2’s comp of the result and considering it negative)
Result = – ( ) = -17 67
- 84
2’s comp
- 17
Take
2’s comp

61. Subtraction of two Unsigned Numbers (M-N) Using the 1’s Complement
For n-bit, unsigned numbers M and N, find M  N:
Add the 1's complement of the subtrahend N to the minuend M: M + (2n  1  N) = M  N  1 + 2n
If M  N, the result is excess by 2n  1 and a carry is generated. Ignoring the carry removes 2n, leaving the result short by 1. To fix this shortage, whenever and end carry occurs, add it in the LSB position. This is called the end-around carry.
If M < N, the sum does not produce an end carry. The sum is equal to 2n  1  ( N  M ), the 1's complement of ( N  M ).
So, to obtain the final result,  (N – M), take the 1's complement of the sum and place a  to its left.

62. End-around carry occurs. 1
Subtraction of two Unsigned Numbers (M-N) Using the 1’s Complement Example 1
Find – – +1
End-around carry occurs. 1 84
- 67
1’s comp 17

63. Subtraction of two Unsigned Numbers (M-N) Using the 1’s Complement Example 2
Find – –
The end carry of 0 indicates that a correction of the result is required (1’s comp and consider negative)
Result = – ( ) 67
- 84
1’s comp
-17
Take
1’s comp

64. Unsigned Numbers (Multiple Bits) Binary Multiplication11 5 55
Verify the result by doing the operation on the equivalent decimal values

65. Unsigned Numbers BCD Addition
Use binary arithmetic to add the BCD digits: 3 8
1000
Eight +5 +5
+0101
Plus 5 8
OK (< 9) 13
1101
is 13 (> 9)
If result is > 9, it must generate a carry and be corrected!
To correct the digit, subtract 10 by adding 6 and ignoring
the carry: -10 = -(16-6) = (the carry ignored within digit) 8
1000
Eight
We try to avoid subtraction!
Replacing it with addition! +5
+0101
Plus 5 13
1101
13 ( is > 9)
+0110
so add 6 (always, for results > 9)
carry = 1
0011
giving 3 + carry
0001 | 0011
Final answer (two digits)
The adder circuit utilizes the resulting carry bit by sending it as
carry-in to the next digit

66. Add 2905BCD to 1857BCD showing carries and digit corrections.
Unsigned Numbers BCD Addition: Example
Add 2905BCD to 1857BCD showing carries and digit corrections.
1010 1 1
17>9
+6
6<9
+0
12>9
+6
4<9
+0
Add 6 or 0
4762

67. Unsigned Numbers Arithmetic in Octal & Hexadecimal
Perform digit calculations in decimal then express results in the appropriate base system
Start here
At the LS
Digit
Base 16 1
Carry to next digit 8 10 8
Remains in this digit 2
Base 8

68. Integer Representations using n bits
Unsigned Integers
The full n bits are the magnitude of the number. The sign is not represented as part of the number
Signed Integers
The MS bit indicates the sign: 0 = Positive, 1 = Negative
Signed-Magnitude The remaining (n – 1) bits represent the positive magnitude of the number (sign and magnitude are treated separately by the hardware)
Signed-Complement (sign is part of the number)
- Positive integer: Exactly as in signed-magnitude,
- Negative integer: As the complement of the positive value of the number
Two Types:
Signed 1's Complement
Uses 1's Complement Arithmetic
Signed 2's Complement
Uses 2's Complement Arithmetic
Magnitude
n bits
Sign is
separate
Magnitude
Sign
1 bit
n-1 bits
Sign is
part of the
number
Sign
1 bit
Magnitude only for
+ ive integers
Example: Interpret as : a. Unsigned number b. Signed-Magnitude number
c. Signed-1’s Complement c. Signed-2’s Complement

69. Signed Integer Representation in 3 notations Examples for n = 4
4-bits  16 different codes
Number Range (4 bits16 values)
(2n-1-1)
To +(2n-1-1)
(2n-1)
To +(2n-1-1)
(2n-1-1)
To +(2n-1-1)
Positive numbers
have identical
Representation
in all 3 notations
Dual representations
of the 0 is a disadvantage
0000
2’s Complementing
1’s Complementing
Unique
Codes only:
A code
Should not
Represent
2 numbers
All Negative #s
Represented
Have MSB = 1
(Use as sign bit)
How to negate
A number?
16th code
Used for -8
16 codes
already used
16 codes
already used

70. Signed-Magnitude Addition/Subtraction MN (each of M and B can be + ive or – ive)
Follow rules of ordinary arithmetic
We compare only the signs of the two operands and conclude the sign of the result based on these signs and the outcome of operation
Addition/Subtraction of magnitudes is similar to using unsigned integers
Addition: (operations done on magnitudes)
a. If the two signs are equal: (+A) + (+B) = +(A+B) or (-A) + (-B) = -(A+B)
1. Add the two (unsigned) magnitudes, overflow can occur
2. Check that there is no overflow (an end carry out of the mag.)
3. The sign of the result is the same as the common sign
b. If the two signs are different: (+A) + (-B) = +(A-B) or (-A) + (+B) = -(A-B)
Subtract the second magnitude from the first
(Since we subtract 2 unsigned #s, Overflow can not occur)
2. If no borrow occurs (A>B), result is correct and has the sign of the first operand
If a borrow occurs (A<B): Result needs correction
 Correct it by taking its two’s complement and give it the sign of the second operand.

71. Signed-Magnitude Addition/Subtraction MN (each of M and B can be + ive or – ive)
Follow rules of ordinary arithmetic
We compare only the signs of the two operands and conclude the sign of the result based on these signs and the outcome of operation
Addition/Subtraction of magnitudes is similar to unsigned integers
Subtraction: (operations done on magnitudes)
a. If the two signs are different: (+A) - (-B) = +(A+B) or (-A) - (+B) = -(A+B)
1. Add the two (unsigned) magnitudes, Overflow can occur
2. Check that there is no overflow (an end carry out of the mag.)
3. The sign of the result is the same as the sign of the first operand
b. If the two signs are equal: (+A) - (+B) = +(A-B) or (-A) - (-B) = -(A-B)
Subtract the second magnitude from the first
(Since we subtract 2 unsigned #s, Overflow can not occur)
2. If no borrow occurs (A>B), results is correct and has the common sign
If a borrow occurs (A<B): Result needs correction
 Correct it by taking its two’s complement and give it the opposite of the common sign.

72. Signed-Magnitude Arithmetic- Examples Operations are performed on magnitudes only, Get Sign separately
Example 1:
Example 2:
Example 3:
Example 3: +2 +5
Addition (a), overflow did not occur
0111 +2
- 5
Addition (b, borrow  correction)
Overflow not possible
1011 +2
- 5
Subtraction (a), overflow did not occur
Example 1: Since the parity of the signs is 0: ADD magnitudes giving 111
Append the sign of the first operand to obtain 0111
Example 2: Since the parity of the signs is 1: SUBTRACT 010
- 101
101 Borrow of 1
Taking the 2’s complement and appending the complement of the sign: 1011
Example 3: Since the parity of the signs is 0, ADD the magnitudes
101
111
Appending the sign if the top operand:
0111
- 2
- 5
Subtraction (b, borrow  correction)
Overflow not possible
0011

73. 1. Add the two numbers including the sign bits,
Signed-Complement Arithmetic Both addition and subtraction use addition
Start with both numbers A, B represented in the complement notation:
For Addition:
1. Add the two numbers including the sign bits,
 Discarding any carry out of the sign bits (for 2's Complement representations),
 or using it as end-around carry (for 1's Complement representation)
This automatically determines the sign of the result
2. Overflow is possible only if both numbers added have the same sign. In this case if the sign of the result is different from this common sign, an overflow has occurred (result is wrong)
For Subtraction: A – B = A + (-B) = A + Comp of B
Form the complement of the number you are subtracting and then perform addition- applying the same addition rules above
What does taking the 2’s complement of a number do to it?

74. Signed Complement Arithmetic Examples, n = 4 bits
Carry ignored
Carry, End around
Example 1:
0000
Example 2:
Example 1:
0000
Example 2: -3 +3 -2 +3
An overflow?
Using 1’s Complement
Notation 1
Using 2’s Complement
Notation
Sign bit
0001 -3 +3 -2 +3
Carry ignored
Carry, End around
1101
+1101
1010
1101
+1100
1001
Example 1: Result is The carry out of the MSB is discarded.
Example 2: Complement 0011 to 1101 and add. Result is The carry out of the MSB is discarded.
An overflow?
Last 2 carries are
identical 1 1
- ive, =
- ive, = 1
1010
Judge Overflow
from addition signs

75. Signed Complement Arithmetic Examples, n = 4 bits
Carry ignored
Example 3:
0110 -6 -4
 Largest negative number that can be
Represented in 4 bits is -8
(-6) + (-4) = -10 , i.e. overflow expected
Signs of overflow from addition operation:
- Sign of result is different from that of the two
numbers added
- Last 2 carries are different
- Result is wrong
An overflow? 1
+4!
Sign bit
Using 2’s Complement
Notation
Example 1: Result is The carry out of the MSB is discarded.
Example 2: Complement 0011 to 1101 and add. Result is The carry out of the MSB is discarded.

76. Overflow and its Detection
Overflow occurs if the result from an n-bit addition or subtraction can not be accommodated in the n-bit hardware used and an (n+1) th bit is needed
With Unsigned numbers:
 Overflow is possible when adding two number
 Overflow is not possible when subtracting two number (since the results is  the largest of the two, hence fits into the n-bits)
With Signed Numbers:
Overflow can occur when:
Adding two operands of the same sign
Subtracting two operands of different signs
How to detect it:
When the sign bit is not part of the binary operation, we could rely on the end carry as an indication of overflow
But with 2’s complement or 1’s complement the sign bit is used as part of the number and we can not rely on end carry to indicate overflow
In this case, overflow is detected if the two input numbers that are added have the same sign while the result has the opposite sign
Another way is overflow V = cn XOR cn-1. (different carries from last 2 stages). Works fine for 1’s complement notation and with limitations for the 2’s Complement notation
In both cases the two numbers
are added as unsigned numbers
 overflow is possible

77. Overflow Detection Example, Adding with 2’s complement notation:
Carries
___________________ 1
Signs
With 8 bits , max + ive number is +127
+150 > +127  Overflow
Last 2 carries are different

78. Unit 1: Number Systems and Codes Overview
Introduction: Information processing and representation
Number Systems: Binary, Octal and Hexadecimal
Number Base Conversion: Dec, Bin, Oct, and Hex
Digital Codes for Numbers and Symbols
Numbers: Binary, BCD, Excess-3, Gray, Parity
Alphanumeric (symbols) :ASCII, Unicode
Computer Arithmetic on Unsigned Numbers:
Basic Operations on binary: Add, Subtract, Multiply
Arithmetic in Octal and Hexadecimal, BCD addition
Binary Signed Number Representation and Arithmetic:
Subtraction, complement notation
Signed representation, Addition/Subtraction, Overflow