Download CSC 331: DIGITAL LOGIC DESIGN

CSC 331: DIGITAL LOGIC DESIGN
COURSE LECTURER: E. Y. BAAGYERE.
CONTACT: 0249990362.
LECTURE TIME: 15:40 – 17:45 hrs.
VENUE: SP-LAB.
Course Reference Materials
1. Digital Fundamentals with PLD Programming
by Thomas L. FLOYD.
2. Logic and Computer Design Fundamentals, 2nd
Edition.
by M. Morris MANO and Charles R.
KIME.
3. Fundamentals of Digital Logic With VHDL Design.
by Stephen BROWN and Zvonko
VRANESIC.
4. Fundamentals of Logic Design, 5th Edition
by Charles H. ROTH, Jr.
COURSE OUTLINE
• INTRODUCTION TO DIGITAL LOGIC SYSTEMS
• NUMBER SYSTEM AND CONVERSION
•
•
•
•
•
BINARY TO DECIMAL
DECIMAL TO BINARY
BINARY ARITHMETIC
REPRESENTATION OF NEGATIVE NUMBERS
BINARY CODES
• BOOLEAN ALGEBRA
• APPLICATION OF BOOLEAN ALGEBRA
• KARNAUGH MAPS
COURSE OUTLINE
• QUINE – McCLUSKEY METHOD
• COMBINATIONAL CIRCUIT DESIGN AND
SIMULATION USING GATES
• MULTIPLEXERS, DECODERS
• SEQUENTIAL LOGIC DESIGN
• MINI PROJECT
CHAPTER ONE
NUMBER SYSTEM AND CONVERSION
INTRODUCTION
Digital systems are used extensively in
computation and data processing, control
systems, communications, and measurement.
REASONS
1. Digital systems are capable of greater
accuracy than analog systems
2. Reliability is greater than analog.
3. More data can be stored in storage device
when digitized than when they are not.
Design of Digital Systems
Digital System Design is in three parts
- System Design
- Logic Design
- Circuit Design
Design of Digital Systems
SYSTEM DESIGN: Breaking the overall
system into subsystems and specifying the
characteristics of each subsystem.
LOGIC DESIGN: Determining how to
interconnect basic logic building blocks to
perform a specific function.
CIRCUIT DESIGN: Specifying the
interconnection of specific components such as
resistors, diodes, and transistors to form a gate,
flip – flop, or other logic building block.
Switching Circuits
• A switching circuit has one or more inputs and
one or more outputs which take on discrete
values.
• In this course we shall study two types of
switching circuits – combinational and
sequential.
Switching Circuits
Combinational Circuit:The output values
depend only on the present value of the inputs
and not on the past values.
Sequential Circuit: The outputs depend on
both the present and past input values.
A block Diagram of a Switching
Circuit
Inputs
X1
X2
Xn
Z1
:
.
Switching
Circuit
:
.
Z1
Zn
Outputs
Number System and Conversion
The decimal (base 10) numbers are called
positional numbers, each position is assigned a
weight.
For example,
935.8710 = 9 X 102 + 3 X 101 + 5 X 100 +
8 X10-1+ 7 X 10-2.
1101.102 = 1 X 23 + 1 X 22 + 0 X 21 +
1 X 20 + 1 X 2-1 + 0 X 2-2
= 8 + 4 + 0 + 1 + ½ + 0 = 13.510.
Number Conversion
Any positive integer R (R > 1) can be chosen
as the radix or base of a number system.
If the base is R, then R digits (0, 1, ..., R – 1)
are used.
Number Conversion
A number written in positional notation can be
expanded in power series in R. For example,
N = (a4a3a2a1a0.a-1a-2a-3)R
= a4 X R4 + a3 X R3 + a2 X R2 + a1 X R1 + a0
X R0 + a-1 X R-1 + a-2 X R-2 + a-3 X R-3
Where ai is the coefficient of Ri and 0≤ ai ≥ R-1.
Number Conversion
For bases greater than 10, more than 10
symbols are needed to represent the digits.
For example,
In hexadecimal (base 16), A represents 1010, B
represents 1110, C represents 1210, D represents
1310, E represents 1410, and F represents 1510.
Thus;
A3F16 = 10 X 162 + 3 X 161 + 15 X 160 = 2560
+ 48 + 15 =262310.
NUMBER CONVERSION TYPES
Binary – to - decimal conversion
Decimal - to - binary conversion
-Repeated division – by – R method.
(R = Radix or Base)
-Sum – of – weights method.
REPEATED DIVISION - BY - R METHOD
The base R equivalent of a decimal integer N
can be represented as:
N = (anan-1...a2a1a0) = anRn + an-1Rn-1 + ...+
a2R2 + a1R1 + a0.
If we divide N by R, the remainder is a0:
N/R = anRn-1 + an-1Rn-2 + ...+ a2R1 + a1 = Q1 ,
remainder a0
REPEATED DIVISION - BY - R METHOD
Then we divide the quotient Q1 by R:
Q1/R = anRn-2 + an-1Rn-3 + ...+ a3R1 + a2 = Q2,
remainder a1
This process is continued until we finally
obtain an.
Note that the remainder obtained at each
division step is one of the desired digits and
the least significant digit (LSB) is obtained
first.
SUM – OF – WEIGHT METHOD

Determine the set of binary weights whose
sum is equal to the decimal number.
The lowest weight is 1 , which is 20.
 Doubling any weight, you get the next
higher weight.
Example: 9 = 8 + 1 or 9 = 23 + 20
Thus 9 = 1001
Conversion Decimal Fractions to
Binary
 There are two ways of converting Decimal
fractions to Binary
• Sum – of – weight method
• Repeated Multiplication by R Method
(where R is the Radix or Base).
Sum – of – Weight Method

The most significant weight is 0.5, which is
2-1
 Halving any weight, get you next lower
weight.
Example 0.5, 0.25, 0.125, 0.0625.
0.625 = 0.5 + 0.125 = 2-1 + 2-3 = 0.101.
There is a 1 in the 2-1 position, a 0 in the 2-2
position, and a 1 in the 2-3 position.
REPEATED MULTIPLICATION BY R
METHOD
 Conversion of a decimal fraction to base R
can be done using successive multiplications
by R. A decimal fraction F can be represented
as
 F = (.a-1a-2a-3...a-m)R = a-1R-1 + a-2R-2 + a-3R-3 +
... +a-mR-m
 Multiplying by R yields
FR = a-1 + a-2R-1 + a-3R-3 + ... + a-mR-m+1 = a-1 +
F1
REPEATED MULTIPLICATION BY R
METHOD
Where F1 represents the fractional part of the
result and a-1 is the integer part.
Multiplying F1 by R yields
F1R = a-2 + a-3R-1 + ... + a-mR-m+2 = a-2 + F2
This process is continued until we have
obtained a sufficient number of digits
Note that the integer part obtained at each step
is one of the desired digits and the most
significant digit (MSB) is obtained first.
Example: Convert 0.62510 to binary.
F = 0.625 X 2 = 1.250 (a-1 = 1), F1 = .250 X 2 =
0.500 (a-2 = 0), F2 =0.500 X 2 = 1.000 (a-3 = 1).
0.62510 = .1012.
BINARY ARITHMETIC
Arithmetic operations in digital systems are
usually done in binary because design of logic
circuits to perform binary arithmetic is much
easier than for decimal.
Binary arithmetic operations:
Addition
Multiplication
Subtraction
Division
BINARY ADDITION
The addition table for binary numbers is
0+0= 0
0+1=1
1+ 0=1
1 + 1 = 0 and carry 1 to the next column.
Carrying 1 to a column is equivalent to adding
1 to that column.
BINARY ADDITION
Add 1310 and 1110 in binary.
1111
carries
1310 = 1101
1110 = 1011
11000
= 2410
BINARY SUBTRACTION
The subtraction table for binary number is
0–0=0
0 – 1 = 1 and borrow 1 from next column
1–0=1
1–1–0
Borrowing 1 from a column is equivalent to
subtracting 1 from that column.
BINARY SUBTRACTION
Example of binary subtraction
A borrow from 3 position
1
11101
- 10011
1010
rd
BINARY SUBTRACTION
Exercise
1.
10000
11
2. 111001
- 1011
BINARY MULTIPLICATION
The multiplication table for binary numbers is
0 X 0 =0
0X1=0
1X0=0
1X1=1
Exercise
Find 1310 X 1410 in binary.
Binary Multiplication
Example (2): Find 101 X 111
101
X
111
111
000
111
100011
BINARY DIVISION
ASSIGNMENT (1)
Read on binary division
Representation of Signed Numbers
In most computers, in order to represent both
positive and negative numbers the first bit in a
word is used as a sign bit, with 0 used for plus
and 1 used for minus.
For an n – bit word, the first bit is the sign and
the remaining n -1 bits represent the magnitude
of the number. Thus an n – bit word can
represent any one of 2n-1 positive integers or
2n-1 negative integers.
The 1’s complement and 2’s
complement
The 1’s complement and 2’s complement are
commonly used because arithmetic units are
easy to design using these systems
Finding the 1’s complement
• For the 1’s complement system a negative
number, -N, is represented by its 1’s
complement, N. The 1’s complement of a
positive integer N is defined as
• N = (2n – 1) – N
• An alternative way of finding the 1’s
complement is to simply complement N bit –
by – bit by replacing 0’s with 1’s and 1’s with
0’s.
Finding the 2’s complement
For the 2’s complement number system, a
positive number, N, is represented by a 0
followed by the magnitude as in the sign and
magnitude system; however, a negative
number, -N, is represented by its 2’s
complement, Ѝ.
Finding the 2’s complement
Ѝ = 2n – N
Note that N = (2n – 1) – N.
This implies Ѝ = 2n – N = (2n – 1 – N) + 1 = N +
1
Another way of finding the 2’s complement of
a binary number is as follows:
 Start at the right with the LSB and write the
bits as they are up to and including the first 1.
Take the 1’s complement of the remaining bits.
Back to True Binary Form
To convert from 1’s complement back to the
true binary, reverse all the bits. To go from 2’s
complement form back to true binary, take the
1’s complement of the 2’s complement number
and add 1 to the least significant bit.
ARITHMETIC OPERATION WITH
SIGNED NUMBERS
 Addition of Signed Numbers
The two numbers in an addition are the
addend and augend. The result is the sum.
There are four cases that can occur when two
signed binary numbers are added.
Four Cases of Signed Binary Addition
• Both numbers positive
• Positive number with magnitude larger than
negative number
• Negative number with magnitude larger than
positive number
• Both numbers negative.
Both Numbers Positive
00000111
00000100
00001011
7
+
4
11
The sum is positive and is therefore in true
(uncomplemented) binary.
Positive number magnitude larger
than negative number:
00001111
11111010
1 00001001
15
6
9
+
The final carry bit is discarded. The sum is
positive and therefore in true
(uncomplemented) binary.
Negative number with magnitude
larger than positive number
00010000
15
+11101000
24
- 8
11111000
The sum is negative and therefore in 2’s
complement form.
Both numbers negative
11111011
+
-5
+
11110111
-9
+ 14
1 11110010
The final carry bit is discarded. The sum is
negative and therefore in 2’s complement
form.
Over flow:
When two numbers are added and the number of bits required to represent
the sum exceeds the number of bits in the two numbers, an overflow results
as indicated by an incorrect sign bit. An overflow occurs only when both
numbers are negative or both numbers are positive.
Example:
01111101
00111010
10110111
125
58
183
In this example the sum of 183 requires eight magnitude bits. Since there
are seven magnitude bits in the numbers (one bit is the sign bit), there is a
carry into the sign bit which produces the overflow indication.