cs121 : digital design
DESCRIPTION
CS121 : Digital Design. Basic Information. Title : Digital Design Code : CS121 Lecture : 3 Tutorial : 1 Pre-Requisite : Computer Introduction. Overall Aims of Course. By the end of the course the students will be able to: - PowerPoint PPT PresentationTRANSCRIPT
CS121 : Digital Design
Basic Information
Title: Digital Design Code: CS121 Lecture: 3 Tutorial: 1 Pre-Requisite: Computer Introduction
Overall Aims of Course
By the end of the course the students will be able to: Grasp basic principles of combinational and sequential
logic design. Determine the behavior of a digital logic circuit
(analysis) and translate description of logical problems to efficient digital logic circuits (synthesis).
Understanding of how to design a general-purpose computer, starting with simple logic gates.
Contents
TopicsContactHours
No. ofWeeks
-Introduction to the course content, text book(s), reference(s) and course plane.
- Digital Systems and Binary numbers9 3
- Boolean Algebra and Logic Gates 6 2
- Gate Level Minimization 6 2
- Combinational Logic 12 4
- Synchronous Sequential Logic 12 4
Total 45 15
Assessment schedule
Assessment Methods Week Weighting of Assessments
First Midterm Exam 6 10%
Second Midterm Exam 12 15%
Quiz 5 & 10 5%
Lab-project 11 5%
Lab- Evaluation 2,4,6,8,10,12 5%
Final Lab 15 20%
Final Exam After week15 40%
Total 100%
List of References
Essential Books “DIGITAL DESIGN”, by Mano M. Morris, 4th edition, Prentice- Hall.
Recommended Books “FUNDAMENTALS OF LOGIC DESIGN”, by Charles H. Roth,
Brooks/Cole Thomson Learning. “INTRODUCTION TO DIGITAL SYSTEMS”, by M.D. ERCEGOVAC, T.
Lang, and J.H. Moreno, Wiley and Sons. 1998. “DIGITAL DESIGN, PRINCIPLES AND PRACTICES”, by John F.Wakely,
Latest Edition, Prentice Hall, Eaglewood Cliffs, NJ. “FUNDMENTALS OF DIGITAL LOGIC WITH VHDL DESIGN”, by
Stephen Brown and Zvonko Vranesic, McGraw Hill. “INTRODUCTION TO DIGITAL LOGIC DESIGN”, by John Hayes,
Addison Wesley, Reading, MA.
1. Digital Systems and Binary Numbers
1.1 Digital Systems1.2 Binary Numbers1.3 Number-Base Conversions
1.4 Octal and Hexadecimal Numbers
1.5 Complements
1.6 Signed Binary Numbers
1.7 Binary Codes
1.1 Digital Systems
1.2 Binary Numbers
In general, a number expressed in a base-r system has coefficients multiplied by powers of r:
r is called base or radix.
1.3 Number-Base Conversions (Integer Part)
Example:
1.3 Number-Base Conversions (Fraction Part)
Example:
Binary-to-Decimal Conversion
Example:
Example:
1.4 Octal and Hexadecimal Numbers
Decimal-to-Octal Conversion
Example:
Decimal-to-Hexadecimal Conversion
Example:
Octal-to-Decimal Conversion
Example:
Example:
Hexadecimal-to-Decimal Conversion
Example:
Example:
Binary–Octal and Octal–Binary Conversions
Example:
Example:
Hex–Binary and Binary–Hex Conversions
Example:
Example:
Hex–Octal and Octal–Hex Conversions
For Hexadecimal–Octal conversion, the given hex number is firstly converted into its binary equivalent which is further converted into its octal equivalent.
An alternative approach is firstly to convert the given hexadecimal number into its decimal equivalent and then convert the decimal number into an equivalent octal number. The former method is definitely more convenient and straightforward.
For Octal–Hexadecimal conversion, the octal number may first be converted into an equivalent binary number and then the binary number transformed into its hex equivalent.
The other option is firstly to convert the given octal number into its decimal equivalent and then convert the decimal number into its hex equivalent. The former approach is definitely the preferred one.
Example
Arithmetic Operation
augend 101101Added: + 100111 ----------
Sum: 1010100
Addition
Subtraction
minuend: 101101subtrahend: - 100111 -------------difference: 000110
Multiplication
Diminished Radix Complement ((r-1)‘s complement)
Given a number N in base r having n digits, the (r - 1)’sComplement of N is defined as (rn- 1) -N.
the 9’s complement of 546700 is 999999 – 46700=453299the 1’s complement of 1011000 is 0100111
Note: The (r-1)’s complement of octal or hexadecimal numbers is obtained by subtracting each digit from 7 or F (decimal 15), respectively
1.5 Complements
Radix Complement
Given a number N in base r having n digit, the r’s complement of Nis defined as (rn -N) for N ≠0 and as 0 for N =0 .
The 10’s complement of 012398 is 987602The 10’s complement of 246700 is 753300The 2’s complement of 1011000 is 0101000
Subtraction with Complement
The subtraction of two n-digit unsigned numbers M – N in base r can be done as follows:
M + (rn - N), note that (rn - N) is r’s complement of N. If M N, the sum will produce an end carry x,
which can be discarded; what is left is the result M- N. If M < N, the sum does not produce an end carry
and is (N - M). Take the r’x complement of the sum and place a negative sign in front.
Example:
Using 10’s complement subtract 72532 – 3250
M = 72532
10’s complement of N = 96750
sum = 169282
Discarded end carry 105 = -100000 answer: 69282
Example:
Using 10’s complement subtract 3250 - 72532
M = 03250
10’s complement of N = 27468
sum = 30718
Discarded end carry 105 = -100000 answer: -(100000 - 30718) = -69282
The answer is –(10’s complement of 30718) = -69282
Example
Using 2’s complement subtract (a) 1010100 – 1000011
M = 1010100
N = 1000011, 2’s complement of N = 0111101
1010100
0111101 sum = 10010001
Discarded end carry 27=-10000000answer: 0010001
Example
(b) 1000011 – 1010100M = 1000011
N = 1010100, 2’s complement of N = 0101100
1000011 0101100sum = 1101111
answer: - (10000000 - 1101111) = -0010001
The answer is –(2’s complement of 1101111) = - 0010001
Using 1’s complement, subtract 1010100 - 1000011
M = 1010100
N = 1000011, 1’s complement of N = 0111100
answer: 0010001
1010100 0111100
10010000
33
Example
end-around carry = + 1
Using 1’s complement, subtract 1000011 - 1010100
M = 1000011
N = 1010100, 1’s complement of N = 0101011
1000011
0101011
1101110
34
Example
Answer: -0010001
1.6 Signed Binary Numbers
1.6 Signed Binary Numbers
1 - Sign and Magnitude representation
2 - 1’s Complement Representation
3 - 2’s Complement Representation
1 - The previous representation are the same for positive numbers and different for negative numbers
2 - For a signed binary number the most significant bit is used for representing the sign of the number
We use 0 for positive numbers and 1 for negative numbers
Notes
Example : represent +76
10 2
10 2
10 2
76 1001100 &
76 1001100 1'
0
76 1
0
0 001100 2 '
Sign Magnitude
s Complement
s Complement
Representing negative numbers in the previous three systems
1’s Complement of a negative number can be obtained by flipping all bits of the positive binary number
2’s Complement of a negative number can be obtained by adding 1 to the 1’s Complement or by flipping bits of the positive binary number after the first one from the right
Example : represent -76
10 2
10 2
10 2
76 1001100 &
76 0110011 1'
1
76 0
1
1 110100 2 '
Sign Magnitude
s Complement
s Complement
Arithmetic Addition with Comparison
The addition of two numbers in the signed mgnitude system follow the rules of ordinary arithmetic.
If the signed are the same, we add the two magnitudes and give the sum the common sign.
If the signed are different, we subtract the smaller magnitudefrom the larger and give the difference the sign of the largermagnitude. EX. (+25) + (-38) = -(38 - 25) = -13
Arithmetic Addition
Arithmetic Addition without Comparison
The addition of two signed binary number with negative numbers represented in signed 2’s complement form is obtained from the addition of the two numbers, including their signed bits. A carry out of the signed bit position is discarded (note that the 4th case).
Arithmetic Addition without Comparison
19 1110110113 1111001106 11111010
07 1111100113 1111001106 00000110
07 00000111
13 0000110106 11111010
19 00010011
13 0000110106 00000110
40
Arithmetic Subtraction
(+/-) A – (+B)= (+/-) A + (-B) (+/-) A – (-B)= (+/-) A + (+B)
Example (-6) – (-13)= +7In binary: (1111010 – 11110011)= (1111010 +
00001101) =100000111 after removing the carry out the result will be : 00000111
1.7 Binary Codes
Binary Coded Decimal (BCD)
Binary Coded Decimal (BCD)
in this system each digit is represented in 4 bits
For example : to represent in BCD
9 54
1001
10945
0100 0101
10 BCD10010100010 9 5 14
BCD Addition
Example : Evaluate the following operations in BCD System
1 – 3 + 4
2 – 4 + 8
3 - 148 + 576
3
4
7
01000111
0011BCD BCD Decimal
BCD Addition
Example : Evaluate the following operations in BCD System
1 – 3 + 4
2 – 4 + 8
3 - 148 + 5764
8
12
10001100
0100BCD BCD
Decimal
Error
01100001001012
We must add 6 (0110) to the result
BCD
BCD Addition
Example : Evaluate the following operations in BCD System
1 – 3 + 4
2 – 4 + 8
3 - 184 + 576
1846 57
1
0001BCD 1000 0100
0101 0111 0110
Decimal
0111 0000 1010
0110 0110
0111 0110 00001 1
1
760 1
760
1 – In BCD Addition , we add (0110)=(6) if the result value was greater than (1001)=(9) or if the result was more than 4 digits
Notes
In previous Example we added 0110 when the result was
1 - greater than 9 (1001)
2 - more than 4 digits (10000)
Note : result more than 4 digit is greater than 9(1001)
Decimal ArithmeticAddition for signed numbers
Example: (+375) + (- 240) = + 135 in BCD
Apply 10‘s complement to the negative number only. Addition is done by summing all digits,including the
sign digit,and discarding the end carry 0 375 +9 760 ------------ 0 135
Decimal Arithmetic
Subtraction for signed and unsigned numbers
Apply 10‘s complement to the subtrahend and apply addition (same as binary case)
(ex-3) is like (BCD) in the way of representing number
i.e. each digit is represented in 4 bits
Except that : each digit is firstly incremented by three
Excess-3 (ex-3)Excess-three (ex-3)is another system to represent a number
For example : to represent in ex-3
12 87
1100
10945
0111 1000
10 311000111100 94 05
ex
9 54
Gray Code
ASCII code is used to represent characters , Symbols , …
ASCII code consists of 7-bits (to represent 128 character)
ASCII character code
ASCII : American Standard Code for Information Interchange
Upper case Letters are represented by ASCII (65 : 90)
Lower case Letters are represented by ASCII (97 : 122)
# ASCII Ch
65 1000001 A
66 1000010 B
90 1011010 Z
97 1100001 a
98 1100010 b
122 1111001 z
Error Detecting Code
with even parity with odd parityASCII A 1000001 01000001 11000001ASCII T 1010100 11010100 01010100
For more information about Number Systems and Conversations between them
Check these
1 – Our Logic Book
2 - Computer Organization's Lectures
3 – Any other References