1. Chapter 1 Digital Systems and Numbers System
وزارة التعليم العالي والبحث العلمي
جامعة الكوفة - كلية التربية – قسم علوم الحاسوب
Digital Logic Design I
Chapter 1 Digital Systems and Numbers System
Dr. Wissam Hasan Mahdi Alagele

2. Outline of Chapter 1 1.1 Digital Systems 1.2 Number System
1.2.1 Decimal Numbers
1.2.2 Binary Numbers
1.2.3 Octal Numbers
1.2.4 Hexadecimal Numbers
1.3 Number-base Conversions

3. Analog and Digital Signal
Analog system
The physical quantities or signals may vary continuously over a specified range.
Digital system
The physical quantities or signals can assume only discrete values.
Greater accuracy t
X(t)
Analog signal t
X(t)
Digital signal

4. Binary Digital Signal
An information variable represented by physical quantity.
For digital systems, the variable takes on discrete values.
Two level, or binary values are the most prevalent values.
Binary values are represented abstractly by:
Digits 0 and 1
Words (symbols) False (F) and True (T)
Words (symbols) Low (L) and High (H)
And words On and Off
Binary values are represented by values
or ranges of values of physical quantities. t
V(t)
Binary digital signal
Logic 1
Logic 0
undefine

5. Number System
Integers are normally written using positional number system, in which each digit represents the coefficient in a power series
𝑁= 𝑎 𝑛−1 𝑟 𝑛−1 + 𝑎 𝑛−2 𝑟 𝑛−2 +…+ 𝑎 2 𝑟 2 + 𝑎 1 𝑟 1 + 𝑎 0
Where 𝑛 is the number of digit, 𝑟 is the radix or base and 𝑎 𝑖 is the coefficient
0≤ 𝑎 𝑖 <𝑟
Ex.
There are four systems of arithmetic which are often used in digital circuits. These systems are:
decimal: it has a base (𝑟=10) and coefficients (𝑎) are in the range 0 to 9
binary: it has a base (𝑟=2) and coefficients (𝑎) are all either 0 or 1
octal : it has a base (𝑟=8) and coefficients (𝑎) are in the range 0 to 7
Hexadecimal: it has a base (𝑟=16) and coefficients (𝑎) are in the range
{ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F }
97142= 9*104+7 *103+1 *102+4*101+2 *100

6. Decimal Number System 5 1 2 7 4 Base (also called radix) = 10
10 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
Digit Position
Integer & fraction
Digit Weight
Weight = (Base) Position
Magnitude
Sum of “Digit x Weight”
Formal Notation 1 -1 2 -2 5 1 2 7 4 10 1
0.1
100
0.01
500 10 2
0.7
0.04
d2*B2+d1*B1+d0*B0+d-1*B-1+d-2*B-2
5 *102+1 *101+2 *100+7 * *10-2
(512.74)10

7. Binary Number System 1 1 1 Base = 2
2 digits { 0, 1 }, called binary digits or “bits”
Weights
Weight = (Base) Position
Magnitude
Sum of “Bit x Weight”
Formal Notation
Groups of bits bits = Nibble
8 bits = Byte 1 -1 2 -2
1/2 4
1/4 1 1 1
1 *22+0 *21+1 *20+0 *2-1+1 *2-2
=(5.25)10
(101.01)2

8. Octal Number System 5 1 2 7 4 Base = 8
8 digits { 0, 1, 2, 3, 4, 5, 6, 7 }
Weights
Weight = (Base) Position
Magnitude
Sum of “Digit x Weight”
Formal Notation 1 -1 2 -2 8
1/8 64
1/64 5 1 2 7 4
5 *82+1 *81+2 *80+7 *8-1+4 *8-2
=( )10
(512.74)8

9. Hexadecimal Number System
Base = 16
16 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F }
Weights
Weight = (Base) Position
Magnitude
Sum of “Digit x Weight”
Formal Notation 1 -1 2 -2 16
1/16
256
1/256 1 E 5 7 A
1 * *161+5 *160+7 * *16-2
=( )10
(1E5.7A)16

10. Number Base Conversions
Evaluate Magnitude
Octal
(Base 8)
Evaluate Magnitude
Decimal
(Base 10)
Binary
(Base 2)
Hexadecimal
(Base 16)
Evaluate Magnitude

11. Decimal (Integer) to Binary Conversion
Divide the number by the ‘Base’ (=2)
Take the remainder (either 0 or 1) as a coefficient
Take the quotient and repeat the division
Example: (13)10
Quotient
Remainder
Coefficient 13
/ 2 =
a0 = 1 6
/ 2 =
a1 = 0 3
/ 2 =
a2 = 1 1
/ 2 =
a3 = 1
Answer: (13)10 = (a3 a2 a1 a0)2 = (1101)2
MSB LSB

12. Decimal (Integer) to Binary Conversion

13. Decimal (Fraction) to Binary Conversion
Multiply the number by the ‘Base’ (=2)
Take the integer (either 0 or 1) as a coefficient
Take the resultant fraction and repeat the division
Example: (0.625)10
Integer
Fraction
Coefficient
0.625
* 2 =
a-1 = 1
0.25
* 2 = a-2 = 0
0.5
* 2 = a-3 = 1
Answer: (0.625)10 = (0.a-1 a-2 a-3)2 = (0.101)2
MSB LSB

14. Decimal to Binary Conversion

15. Table of binary equivalent decimal numbers1 11
1011 21
10101 2 10 12
1100 22
10110 3 13
1101 23
10111 4
100 14
1110 24
11000 5
101 15
1111 25
11001 6
110 16
10000 26
11010 7
111 17
10001 27
11011 8
1000 18
10010 28
11100 9
1001 19
10011 29
11101
1010 20
10100 30
11110

16. Decimal to Octal Conversion
Example: (175)10
Quotient
Remainder
Coefficient
175
/ 8 =
a0 = 7 21
/ 8 =
a1 = 5 2
/ 8 =
a2 = 2
Answer: (175)10 = (a2 a1 a0)8 = (257)8
Example: (0.3125)10
Integer
Fraction
Coefficient
0.3125
* 8 =
a-1 = 2
0.5
* 8 = a-2 = 4
Answer: (0.3125)10 = (0.a-1 a-2 a-3)8 = (0.24)8

17. Decimal to Hexadecimal Conversion
Example: (1983)10
Quotient
Remainder
Coefficient
1983
/ 16 = 123
a0 = F
123
/ 16 =
a1 = B 7
/ 16 =
a2 = 7
Answer: (1983)10 = (a2 a1 a0)16 = (7BF)16
Example: (0.5625)10
Integer
Fraction
Coefficient
0.5625
*16 =
a-1 = 9
Answer: (0.5625)10 = (0.a-1 a-2 a-3)16 = (0.9)16

18. Binary − Octal Conversion
0 0 0 1
0 0 1 2
0 1 0 3
0 1 1 4
1 0 0 5
1 0 1 6
1 1 0 7
1 1 1
8 = 23
Each group of 3 bits represents an octal digit
Example:
Assume Zeros
( )2
( )8
Works both ways (Binary to Octal & Octal to Binary)

19. Binary − Hexadecimal Conversion 1 2 3 4 5 6 7 8 9 A B C D E F
16 = 24
Each group of 4 bits represents a hexadecimal digit
Example:
Assume Zeros
( )2
( )16
Works both ways (Binary to Hex & Hex to Binary)

20. Octal − Hexadecimal Conversion
Convert to Binary as an intermediate step
Example:
( )8
Assume Zeros
Assume Zeros
( )2
( )16
Works both ways (Octal to Hex & Hex to Octal)

21. Decimal, Binary, Octal and Hexadecimal00
0000 01
0001 1 02
0010 2 03
0011 3 04
0100 4 05
0101 5 06
0110 6 07
0111 7 08
1000 10 8 09
1001 11 9
1010 12 A
1011 13 B
1100 14 C
1101 15 D
1110 16 E
1111 17 F

22. Tutorial Problems
Find the decimal equivalents of the following binary numbers:
(a) (101) (b) (1001) (c) (10.011)2
[(a) (5)10 (b) (9)10 (c) (3.0375)10]
What are the decimal equivalents of the following binary numbers ?
(a) (1111) (b) (10100) (c) ( ) (d) ( )2
[(a) (15)10 (b) (20)10 (c) (109)10 (d) (153)10]
Express the following binary numbers into their equivalent decimal numbers :
(a) (11.01) (b) (101.11) (c) (110.01)2
[(a) (3.25)10 (b) (5.75)10 (c) (6.25)10]
Convert the following decimal numbers into their binary equivalents:
(a) (25) (b) (125) (c) (0.85)10
[ (a) (11001)2 (b) ( )2 (c) ( )2]
What are the binary equivalents of the following decimal numbers ?
(a) (27) (b) (92) (c) (64)10
[(a) (11011)2 (b) ( )2 (c) ( )2]
Convert the following real numbers to the binary numbers:
(i) (12.0) (ii) (25.0) (iii) (0.125)10
[(i) (1100)2 (ii) (11001)2 (iii) (0.001)2]

23. Tutorial Problems Convert the following numbers :
(a) (35)8 to decimal (b) (6421)8 to decimal
(c) (1359)10 to octal (d) (7777)10 to octal
[(a) (239)10 (b) (3345)10 (c) (2517)8 (d) (17141)8]

24. Addition Decimal Addition 1 1 Carry 5 5 + 5 5 1 1 = Ten ≥ Base
= Ten ≥ Base
 Subtract a Base

25. Binary Addition Column Addition 1 1 1 1 1 1 Carry 1 1 1 1 1 + 1 1 1 11
= 61
= 23 + 1 1 1 1 1 1 1
= 84
Decimal
Binary 1 2 10 3 11
≥ (2)10
0 + 0= =1
1 + 0= =0 with a carry of 1 or =10

26. Binary Subtraction Borrow a “Base” when needed 1 2 = (10)2 2 2 2 1 1 12 2 2 1 1 1 1
= 77
= 23 − 1 1 1 1 1 1 1 1
= 54

27. Binary Multiplication
Bit by bit 1 1 1 1 x 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

28. 1.5 Complements
There are two types of complements for each base-r system: the radix complement and diminished radix complement.
Diminished Radix Complement - (r-1)’s Complement
Given a number N in base r having n digits, the (r–1)’s complement of N is defined as:
(rn –1) – N
Example for 6-digit decimal numbers:
9’s complement is (rn – 1)–N = (106–1)–N = –N
9’s complement of is – =
Example for 7-digit binary numbers:
1’s complement is (rn – 1) – N = (27–1)–N = –N
1’s complement of is – =
Observation:
Subtraction from (rn – 1) will never require a borrow
Diminished radix complement can be computed digit-by-digit
For binary: 1 – 0 = 1 and 1 – 1 = 0

29. Complements 1’s Complement (Diminished Radix Complement)
All ‘0’s become ‘1’s
All ‘1’s become ‘0’s
Example ( )2
 ( )2
If you add a number and its 1’s complement …

30. Complements Radix Complement Example: Base-10 Example: Base-2
The r's complement of an n-digit number N in base r is defined as
rn – N for N ≠ 0 and as 0 for N = 0. Comparing with the (r  1) 's complement, we note that the r's complement is obtained by adding 1 to the (r  1) 's complement, since rn – N = [(rn  1) – N] + 1.
The 10's complement of is
The 10's complement of is
The 2's complement of is
The 2's complement of is

31. Complements 2’s Complement (Radix Complement)
Take 1’s complement then add 1
Toggle all bits to the left of the first ‘1’ from the right
Example:
Number:
1’s Comp.: OR OR

32. Complements 1’s Complemental Subtraction as follows:
In digital work, both types of complements of a binary number are used for complemental subtraction :
as follows:
1’s Complemental Subtraction
compute the 1’s complement of the subtrahend by changing all its 1s to 0s and all its 0s to 1s.
add this complement to the minuend
perform the end-around carry of the last 1 or 0
Suppose we want to subtract (101)2 from (111)2 .
1 1 1
0 1 0
010
← 1’s complement of subtrahend (101)2
← end-around carry
As seen, we have removed from the addition sum the 1 carry in the last position and added it onto the remainder. It is called end-around carry.

33. Complements
4. if there is no end-around carry (i.e. 0 carry), then the answer must be re-complemented and a negative sign attached to it.
Example. Subtract (1101)2 from (1010)2 . 1
← 1’s complement of subtrahend (1101)2
← no end-around carry
re-complementing with negative sign

34. Complements 2’s Complemental Subtraction
In this case, the procedure is as under :
1.find the 2’s complement of the subtrahend,
2.add this complement to the minuend,
3.drop the final carry,
4.if the carry is 1, the answer is positive and needs no re-complementing,
5.if there is no carry, re-complement the answer and attach minus sign.
Example. Using 2’s complement, subtract: (A) (1010)2 from (1101)2 , (B) (1101)2 from (1010)2.
Solution. The 1’s complement of 1010 is The 2’s complement is = We will add it to 1101.
DROP
← 2’s complement of (1010)2
← The final answer is (0011)2 .

35. Complements Example D Drop final cary
Given the two binary numbers X = and Y = , perform the subtraction (a) X – Y ; and (b) Y  X, by using 2's complement.
D Drop final cary
There is no end carry. Therefore, the answer is Y – X =  (2's complement of ) = 

36. Complements
Example
Repeat Example, but this time using 1's complement.
There is no end carry, Therefore, the answer is Y – X =  (1's complement of ) = 

37. Digital Coding
In digital logic circuits, each number or piece of information is defined by an equivalent combination of binary digits. A complete group of these combinations which represents numbers, letters or symbols is called a digital code
Codes have been used for security reasons so that others may not be able to read the message even if it is intercepted.
The choice of a code depends on the function or purpose it has to serve.
There are many types of coding
Binary codes
BCD code
Gray code
ASCII code

38. Binary Codes
BCD Code
It is a binary code in which each decimal digit is represented by a group of four bits. it is also called an 8421 code. The indicates the
binary weights of the four bits (2 3 , 2 2 , 2 1 , 2 0 ).
A number with N decimal digits will require 4N bits in BCD.
Decimal 396 is represented in BCD with 12bits as , with each group of 4 bits representing one decimal digit.
A decimal number in BCD is the same as its equivalent binary number only when the number is between 0 and 9.

39. Binary Code
Example:
Consider decimal 185 and its corresponding value in BCD and binary:
Other Decimal Codes

40. Binary Codes) Gray Code
The advantage is that only bit in the code group changes in going from one number to the next.
Error detection.
Representation of analog data.
Low power design.
000
001
010
100
110
111
101
011

41. ASCII Character Codes
A popular code used to represent information sent as character- based data.
The ASCII code is a seven-bit code, and so it has 27 (=128) possible code groups. This is more than enough to represent all of the standard keyboard characters.