Module I - Digital Systems & Logic Gates.ppt

ENGIN112 L2: Number Systems September 5, 2003
Number Systems

Overview
° The design of computers
• It all starts with numbers
• Building circuits
• Building computing machines
° Digital systems
° Understanding decimal numbers
° Binary and octal numbers
• The basis of computers!
° Conversion between different number systems

Digital Computer Systems
° Digital systems consider discrete amounts of data.
° Examples
• 26 letters in the alphabet
• 10 decimal digits
° Larger quantities can be built from discrete values:
• Words made of letters
• Numbers made of decimal digits (e.g. 239875.32)
° Computers operate on binary values (0 and 1)
° Easy to represent binary values electrically
• Voltages and currents.
• Can be implemented using circuits
• Create the building blocks of modern computers

Understanding Decimal Numbers
° Decimal numbers are made of decimal digits:
(0,1,2,3,4,5,6,7,8,9)
° But how many items does a decimal number
represent?
• 8653 = 8x103 + 6x102 + 5x101 + 3x100
° What about fractions?
• 97654.35 = 9x104 + 7x103 + 6x102 + 5x101 + 4x100 + 3x10-1 + 5x10-2
• In formal notation -> (97654.35)10

Understanding Octal Numbers
° Octal numbers are made of octal digits:
(0,1,2,3,4,5,6,7)
° How many items does an octal number represent?
• (4536)8 = 4x83 + 5x82 + 3x81 + 6x80 = (1362)10
• (465.27)8 = 4x82 + 6x81 + 5x80 + 2x8-1 + 7x8-2
° Octal numbers don’t use digits 8 or 9

Understanding Binary Numbers
° Binary numbers are made of binary digits (bits):
• 0 and 1
° How many items does an binary number represent?
• (1011)2 = 1x23 + 0x22 + 1x21 + 1x20 = (11)10
• (110.10)2 = 1x22 + 1x21 + 0x20 + 1x2-1 + 0x2-2
° Groups of eight bits are called a byte
• (11001001) 2
° Groups of four bits are called a nibble.
• (1101) 2

Why Use Binary Numbers?
° Easy to represent 0 and 1 using
electrical values.
° Possible to tolerate noise.
° Easy to transmit data
° Easy to build binary circuits.
AND Gate
1
0
0

Conversion Between Number Bases
Decimal(base 10)
Octal(base 8)
Binary(base 2)
Hexadecimal
(base16)
° Learn to convert between bases.
° Already demonstrated how to convert
from binary to decimal.
° Hexadecimal described in next
lecture.

Convert an Integer from Decimal to Another Base
1. Divide decimal number by the base (e.g. 2)
2. The remainder is the lowest-order digit
3. Repeat first two steps until no divisor remains.
For each digit position:
Example for (13)10:
Integer
Quotient
13/2 = 6 + ½ a0 = 1
6/2 = 3 + 0 a1 = 0
3/2 = 1 + ½ a2 = 1
1/2 = 0 + ½ a3 = 1
Remainder Coefficient
Answer (13)10 = (a3 a2 a1 a0)2 = (1101)2

Convert a Fraction from Decimal to Another Base
1. Multiply decimal number by the base (e.g. 2)
2. The integer is the highest-order digit
3. Repeat first two steps until fraction becomes
zero.
Example for (0.625)10:
Integer
0.625 x 2 = 1 + 0.25 a-1 = 1
0.250 x 2 = 0 + 0.50 a-2 = 0
0.500 x 2 = 1 + 0 a-3 = 1
Fraction Coefficient
Answer (0.625)10 = (0.a-1 a-2 a-3 )2 = (0.101)2

The Growth of Binary Numbers
n 2n
0 20=1
1 21=2
2 22=4
3 23=8
4 24=16
5 25=32
6 26=64
7 27=128
n 2n
8 28=256
9 29=512
10 210=1024
11 211=2048
12 212=4096
20 220=1M
30 230=1G
40 240=1T
Mega
Giga
Tera

Binary Addition
° Example of adding two binary numbers…
1 1 1 1 0 1
+ 1 0 1 1 1
---------------------
0
1
0
1
1
1
1
1
1
1 1 0
0
carries

Binary Subtraction
° We can also perform subtraction (with borrows in place of
carries).
° Let’s subtract (10111)2 from (1001101)2…
1 10
0 10 10 0 0 10
1 0 0 1 1 0 1
- 1 0 1 1 1
------------------------
1 1 0 1 1 0
borrows

Binary Multiplication
° Binary multiplication is much the same as decimal
multiplication, except that the multiplication
operations are much simpler…
1 0 1 1 1
X 1 0 1 0
-----------------------
0 0 0 0 0
1 0 1 1 1
0 0 0 0 0
1 0 1 1 1
-----------------------
1 1 1 0 0 1 1 0

Convert an Integer from Decimal to Octal
1. Divide decimal number by the base (8)
2. The remainder is the lowest-order digit
3. Repeat first two steps until no divisor remains.
Example for (175)10:
Integer
Quotient
175/8 = 21 + 7/8 a0 = 7
21/8 = 2 + 5/8 a1 = 5
2/8 = 0 + 2/8 a2 = 2
Remainder Coefficient
Answer (175)10 = (a2 a1 a0)2 = (257)8

Convert an Fraction from Decimal to Octal
1. Multiply decimal number by the base (e.g. 8)
2. The integer is the highest-order digit
3. Repeat first two steps until fraction becomes
zero.
Example for (0.3125)10:
Integer
0.3125 x 8 = 2 + 5 a-1 = 2
0.5000 x 8 = 4 + 0 a-2 = 4
Fraction Coefficient
Answer (0.3125)10 = (0.24)8

Summary
° Binary numbers are made of binary digits (bits)
° Binary and octal number systems
° Conversion between number systems
° Addition, subtraction, and multiplication in binary

Overview
° Hexadecimal numbers
• Related to binary and octal numbers
° Conversion between hexadecimal, octal and binary
° Value ranges of numbers
° Representing positive and negative numbers
° Creating the complement of a number
• Make a positive number negative (and vice versa)

Understanding Hexadecimal Numbers
° Hexadecimal numbers are made of 16 digits:
• (0,1,2,3,4,5,6,7,8,9,A, B, C, D, E, F)
° How many items does an hex number represent?
• (3A9F)16 = 3x163 + 10x162 + 9x161 + 15x160 = 1499910
• (2D3.5)16 = 2x162 + 13x161 + 3x160 + 5x16-1 = 723.312510
° Note that each hexadecimal digit can be represented
with four bits.
• (1110) 2 = (E)16
° Groups of four bits are called a nibble.
• (1110) 2

Putting It All Together
° Binary, octal, and
hexadecimal similar
° Easy to build circuits to
operate on these
representations
° Possible to convert
between the three
formats

Converting Between Base 16 and Base 2
° Conversion is easy!
 Determine 4-bit value for each hex digit
° Note that there are 24 = 16 different values of four
bits
° Easier to read and write in hexadecimal.
° Representations are equivalent!
3A9F16 = 0011 1010 1001 11112
3 A 9 F

Converting Between Base 16 and Base 8
1. Convert from Base 16 to Base 2
2. Regroup bits into groups of three starting from right
3. Ignore leading zeros
4. Each group of three bits forms an octal digit.
352378 = 011 101 010 011 1112
5 2 3 7
3
3A9F16 = 0011 1010 1001 11112
3 A 9 F

How To Represent Signed Numbers
• Plus and minus sign used for decimal
numbers: 25 (or +25), -16, etc.
• For computers, desirable to represent
everything as bits.
• Three types of signed binary number
representations: signed magnitude, 1’s
complement, 2’s complement.
• In each case: left-most bit indicates sign:
positive (0) or negative (1).
Consider signed magnitude:
000011002 = 1210
Sign bit Magnitude
100011002 = -1210
Sign bit Magnitude

One’s Complement Representation
• The one’s complement of a binary number
involves inverting all bits.
• 1’s comp of 00110011 is 11001100
• 1’s comp of 10101010 is 01010101
• For an n bit number N the 1’s complement is
(2n-1) – N.
• Called diminished radix complement .since 1’s
complement for base (radix 2).
• To find negative of 1’s complement number
take the 1’s complement.
000011002 = 1210
Sign bit Magnitude
111100112 = -1210
Sign bit Magnitude

Two’s Complement Representation
• The two’s complement of a binary number
involves inverting all bits and adding 1.
• 2’s comp of 00110011 is 11001101
• 2’s comp of 10101010 is 01010110
• For an n bit number N the 2’s complement is
(2n-1) – N + 1.
• Called radix complement. since 2’s
complement for base (radix 2).
• To find negative of 2’s complement number
take the 2’s complement.
000011002 = 1210
Sign bit Magnitude
111101002 = -1210
Sign bit Magnitude

Two’s Complement Shortcuts
° Algorithm 1 – Simply complement each bit and
then add 1 to the result.
• Finding the 2’s complement of (01100101)2 and of its 2’s
complement…
N = 01100101 [N] = 10011011
10011010 01100100
+ 1 + 1
--------------- ---------------
10011011 01100101
° Algorithm 2 – Starting with the least significant bit,
copy all of the bits up to and including the first 1
bit and then complementing the remaining bits.
• N = 0 1 1 0 0 1 0 1
[N] = 1 0 0 1 1 0 1 1

Finite Number Representation
° Machines that use 2’s complement arithmetic can
represent integers in the range
-2n-1 <= N <= 2n-1-1
where n is the number of bits available for
representing N. Note that 2n-1-1 = (011..11)2 and
–2n-1 = (100..00)2
oFor 2’s complement more negative numbers than
positive.
oFor 1’s complement two representations for zero.
oFor an n bit number in base (radix) z there are zn
different unsigned values.
(0, 1, …zn-1)

1’s Complement Addition
° Using 1’s complement numbers, adding numbers
is easy.
° For example, suppose we wish to add +(1100)2
and +(0001)2.
° Let’s compute (12)10 + (1)10.
• (12)10 = +(1100)2 = 011002 in 1’s comp.
• (1)10 = +(0001)2 = 000012 in 1’s comp. 0 1 1 0 0
+ 0 0 0 0 1
--------------
0 0 1 1 0 1
0
--------------
0 1 1 0 1
Add carry
Final
Result
Step 1: Add binary numbers
Step 2: Add carry to low-order bit
Add

1’s Complement Subtraction
° Using 1’s complement numbers, subtracting
numbers is also easy.
° For example, suppose we wish to subtract
+(0001)2 from +(1100)2.
° Let’s compute (12)10 - (1)10.
• (12)10 = +(1100)2 = 011002 in 1’s comp.
• (-1)10 = -(0001)2 = 111102 in 1’s comp.
0 1 1 0 0
- 0 0 0 0 1
--------------
0 1 1 0 0
+ 1 1 1 1 0
--------------
1 0 1 0 1 0
1
--------------
0 1 0 1 1
Add carry
Final
Result
Step 1: Take 1’s complement of 2nd operand
Step 3: Add carry to low order bit
1’s comp
Add

2’s Complement Addition
° Using 2’s complement numbers, adding numbers
is easy.
° For example, suppose we wish to add +(1100)2
and +(0001)2.
° Let’s compute (12)10 + (1)10.
• (12)10 = +(1100)2 = 011002 in 2’s comp.
• (1)10 = +(0001)2 = 000012 in 2’s comp. 0 1 1 0 0
+ 0 0 0 0 1
--------------
0 0 1 1 0 1
Final
Result
Step 2: Ignore carry bit
Add
Ignore

° Using 2’s complement numbers, follow steps for
subtraction
° For example, suppose we wish to subtract
+(0001)2 from +(1100)2.
° Let’s compute (12)10 - (1)10.
• (12)10 = +(1100)2 = 011002 in 2’s comp.
• (-1)10 = -(0001)2 = 111112 in 2’s comp.
0 1 1 0 0
- 0 0 0 0 1
--------------
0 1 1 0 0
+ 1 1 1 1 1
--------------
1 0 1 0 1 1
Final
Result
Step 1: Take 2’s complement of 2nd operand
Step 3: Ignore carry bit
2’s comp
Add
Ignore
Carry

2’s Complement Subtraction: Example #2
° Let’s compute (13)10 – (5)10.
• (13)10 = +(1101)2 = (01101)2
• (-5)10 = -(0101)2 = (11011)2
° Adding these two 5-bit codes…
° Discarding the carry bit, the sign bit is seen to be
zero, indicating a correct result. Indeed,
(01000)2 = +(1000)2 = +(8)10.
0 1 1 0 1
+ 1 1 0 1 1
--------------
1 0 1 0 0 0
carry

2’s Complement Subtraction: Example #3
° Let’s compute (5)10 – (12)10.
• (-12)10 = -(1100)2 = (10100)2
• (5)10 = +(0101)2 = (00101)2
° Here, there is no carry bit and the sign bit is 1.
This indicates a negative result, which is what we
expect. (11001)2 = -(7)10.
0 0 1 0 1
+ 1 0 1 0 0
--------------
1 1 0 0 1

Summary
° Binary numbers can also be represented in octal and
hexadecimal
° Easy to convert between binary, octal, and hexadecimal
° Signed numbers represented in signed magnitude, 1’s
complement, and 2’s complement
° 2’s complement most important (only 1 representation
for zero).
° Important to understand treatment of sign bit for 1’s
and 2’s complement.

Number Codes

Overview
° 2’s complement numbers
• Addition and subtraction
° Binary coded decimal
° Gray codes for binary numbers
° ASCII characters
° Moving towards hardware
• Storing data
• Processing data

1’s complement
° The ones' complement of a binary number is
defined as the value obtained by inverting all the
bits in the binary representation of the number
(swapping 0s and 1s).
0 0000 1111
1 0001 1110
2 0010 1101
3 0011 1100
4 0100 1011
5 0101 1010
6 0110 1001
7 0111 1000

2’s Complement
To get 2’s complement of a binary number, simply
invert the given number and add 1 to the least
significant bit (LSB) of given result.
Binary number 1’s complement 2’s complement
000 111 000
001 110 111
010 101 110
011 100 101
100 011 100
101 010 011
110 001 010
111 000 001

° Let’s compute (13)10 - (5)10.
• (13)10 = +(1101)2 = (01101)2
• (-5)10 = -(0101)2 = (11011)2
° Discarding the carry bit, the sign bit is seen to be
zero, indicating a correct result.
0 1 1 0 1
+ 1 1 0 1 1
--------------
1 0 1 0 0 0
carry

° Let’s compute (5)10 – (12)10.
• (-12)10 = -(1100)2 = (10100)2
• (5)10 = +(0101)2 = (00101)2
° Here, there is no carry bit and the sign bit is 1.
This indicates a negative result, which is what we
expect. (11001)2 = -(7)10.
° Numbers in hexadecimal
0 0 1 0 1
+ 1 0 1 0 0
--------------
1 1 0 0 1

Binary codes

Binary Coded Decimal Code
° Binary coded decimal (BCD) represents each decimal
digit with four bits
• Ex. 0011 0010 1001 = 32910
° This is NOT the same as 0011001010012
° BCD code is a class of binary encodings of decimal
numbers where each decimal digit is represented by
four number of bits.
Digit BCD
Code
Digit BCD
Code
0 0000 5 0101
1 0001 6 0110
2 0010 7 0111
3 0011 8 1000
4 0100 9 1001
3 2 9

Putting It All Together
° BCD not very efficient
° Used in early
computers (40s, 50s)
° Used to encode
numbers for seven-
segment displays.

Gray Code
The gray code is an ordering of the binary numeral system
such that two successive values differ in only one bit(binary
digit).
The advantage of the Gray code over the straight binary
number sequence is that only one bit in the code group
changes in going from one number to the next.
For example,
in going from 7 to 8, the Gray code changes from 0100 to
1100. Only the first bit changes, from 0 to 1; the other three
bits remain the same. By contrast, with binary numbers the
change from 7 to 8 will be from 0111 to 1000, which causes all
four bits to change values.

Gray Code
° Gray code is not a number system.
• It is an alternate way to represent four
bit data
° Only one bit changes from one
decimal digit to the next
° Useful for reducing errors in
communication.
° Can be scaled to larger numbers.
° It is a non-weighted code and it is
not arithmetic codes. That means
there are no specific weights
assigned to the bit position.
Digit Binary Gray
Code
0 0000 0000
1 0001 0001
2 0010 0011
3 0011 0010
4 0100 0110
5 0101 0111
6 0110 0101
7 0111 0100
8 1000 1100
9 1001 1101
10 1010 1111
11 1011 1110
12 1100 1010
13 1101 1011
14 1110 1001
15 1111 1000

Application of the Gray code
° The reflected binary code or gray code was
originally designed to prevent spurious output from
electromechanical switches.
° Today, Gray codes are widely used to facilitate
error correction in digital communications such as
digital terrestrial television and some cable TV
systems.

ASCII Code
Many applications of digital computers require the
handling not only of numbers, but also of other characters
or symbols, such as the letters of the alphabet. For
instance, consider a high‐tech company with thousands of
employees. To represent the names and other pertinent
information, it is necessary to formulate a binary code for
the letters of the alphabet.
The standard binary code for the alphanumeric characters
is the American Standard Code for Information Interchange
(ASCII), which uses seven bits to code 128 characters

ASCII Code
° American Standard Code for Information
Interchange
° ASCII is a 7-bit code, frequently used with an 8th bit
for error detection (more about that in a bit).
Character ASCII (bin) ASCII (hex) Decimal Octal
A 1000001 41 65 101
B 1000010 42 66 102
C 1000011 43 67 103
…
Z
a
…
1
‘

ASCII Codes and Data Transmission
° ASCII Codes
° A – Z (26 codes), a – z (26 codes)
° 0-9 (10 codes), others (@#$%^&*….)
° Transmission susceptible to noise
° Typical transmission rates (1500 Kbps, 56.6 Kbps)
° How to keep data transmission accurate?

Parity bit
° A parity bit, or check bit, is a bit added to a string of
binary code to ensure that the total number of 1-
bits in the string is even or odd. Parity bits are used
as the simplest form of error detecting code.
° There are two variants of parity bits: even parity bit
and odd parity bit.

° In the case of even parity, for a given set of bits, the
occurrences of bits whose value is 1 is counted. If
that count is odd, the parity bit value is set to 1,
making the total count of occurrences of 1s in the
whole set (including the parity bit) an even number.
If the count of 1s in a given set of bits is already
even, the parity bit's value is 0.
° In the case of odd parity, the coding is reversed. For
a given set of bits, if the count of bits with a value of
1 is even, the parity bit value is set to 1 making the
total count of 1s in the whole set (including the
parity bit) an odd number. If the count of bits with a
value of 1 is odd, the count is already odd so the
parity bit's value is 0.

Parity Codes
° Parity codes are formed by concatenating a parity
bit, P to each code word of C.
° In an odd-parity code, the parity bit is specified so
that the total number of ones is odd.
° In an even-parity code, the parity bit is specified
so that the total number of ones is even.
Information Bits
P
1 1 0 0 0 0 1 1

Added even parity bit
0 1 0 0 0 0 1 1

Added odd parity bit

Parity Code Example
° Concatenate a parity bit to the ASCII code for the
characters 0, X, and = to produce both odd-parity
and even-parity codes.
Character ASCII Odd-Parity
ASCII
Even-Parity
ASCII
0 0110000 10110000 00110000
X 1011000 01011000 11011000
= 0111100 10111100 00111100

Binary Data Storage
• Binary cells store individual bits of data
• Multiple cells form a register.
• Data in registers can indicate different values
• Hex (decimal)
• BCD
• ASCII
Binary Cell
0 0 1 0 1 0 1 1

circuit that generates parity bit from four message bits

Basic Logic Gates
© Alan T. Pinck / Algonquin College; 2003

Logic Gates
° a logic gate is a physical device implementing a
Boolean function;
° that is, it performs a logical operation on one or
more binary inputs and produces a single binary
output.

° Logic operations include any operations that
manipulate Boolean values.
° Boolean values are either true or false. They are
named after English mathematician George Boole,
who invented Boolean algebra,

Describing Circuit Functionality: Inverter
° Basic logic functions have symbols.
° The same functionality can be represented with truth
tables.
• Truth table completely specifies outputs for all input combinations.
° The Output signal from a NOT gate is True (on, 1) if and
only if the Input signal is False.
° The Output signal from a NOT gate is False (off, 0) if
and only if the Input signal is True.
A Y
0 1
1 0
Input Output
A Y
Symbol
Truth Table

The AND Gate
° The Output signal from an AND
gate is True (on, 1) if and only if
both Input signals are True (on,
1).
° The Output signal from an AND
gate is False (off, 0), otherwise.
A B Y
0 0 0
0 1 0
1 0 0
1 1 1
A
B
Y
Truth Table

The OR Gate
° The Output signal from an OR
gate is True (on, 1) if either, or
both, Input signals are True (on,
1).
° The Output signal from an OR
gate is False (off, 0) if and only if
both Input signals are False (off,
0).
A B Y
0 0 0
0 1 1
1 0 1
1 1 1
A
B
Y

Describing Circuit Functionality: Waveforms
° Waveforms provide another approach for representing
functionality.
° Values are either high (logic 1) or low (logic 0).
° Can you create a truth table from the waveforms?
A B Y
0 0 0
0 1 0
1 0 0
1 1 1
AND Gate

Logic Gates : NOT combined with other gates
° Note that in the “classic: electronic
engineering form, it is really the “bubble”
that indicates the NOT activity.

Consider three-input gates
3 Input OR Gate

Ordering Boolean Functions
° How to interpret AB+C?
• Is it AB ORed with C ?
• Is it A ANDed with B+C ?
° Order of precedence for Boolean algebra: AND
before OR.
° Note that parentheses are needed here :

Boolean Algebra
° A Boolean algebra is defined as a closed algebraic
system containing a set K or two or more elements
and the two operators, . and +.
° Useful for identifying and minimizing circuit
functionality
° Identity elements
a + 0 = a
a . 1 = a
° 0 is the identity element for the + operation.
° 1 is the identity element for the . operation.

Commutativity and Associativity of the Operators
° The Commutative Property:
° a binary operation is commutative if changing the order of
the operands does not change the result.
For every a and b in K,
a + b = b + a
a . b = b . a
° The Associative Property:
° Within an expression containing two or more occurrences
in a row of the same associative operator, the order in
which the operations are performed does not matter as
long as the sequence of the operands is not changed.
For every a, b, and c in K,
a + (b + c) = (a + b) + c
a . (b . c) = (a . b) . c

Distributivity of the Operators and Complements
° The Distributive Property:
For every a, b, and c in K,
• a + ( b . c ) = ( a + b ) . ( a + c )
• a . ( b + c ) = ( a . b ) + ( a . c )
° The Existence of the Complement:
For every a in K there exists a unique element
called a’ (complement of a) such that,
• a + a’ = 1
• a . a’ = 0
° To simplify notation, the . operator is frequently
omitted. When two elements are written next to
each other, the AND (.) operator is implied…
• a + b . c = ( a + b ) . ( a + c )
• a + bc = ( a + b )( a + c )

Duality
° The principle of duality is an important concept.
This says that if an expression is valid in Boolean
algebra, the dual of that expression is also valid.
° To form the dual of an expression, replace all +
operators with . operators, all . operators with +
operators, all ones with zeros, and all zeros with
ones.
° Form the dual of the expression
a + (bc) = (a + b)(a + c)
° Following the replacement rules…
a(b + c) = ab + ac
° Take care not to alter the location of the
parentheses if they are present.

Involution
° This theorem states:
a’’ = a
° Remember that aa’ = 0 and a+a’=1.
• Therefore, a’ is the complement of a and a is also the
complement of a’.
• As the complement of a’ is unique, it follows that a’’=a.
° Taking the double inverse of a value will give the
initial value.

Absorption
° This theorem states:
a + ab = a a(a+b) = a
° To prove the first half of this theorem:
a + ab = a . 1 + ab
= a (1 + b)
= a (b + 1)
= a (1)
a + ab= a

DeMorgan’s Theorem
° A key theorem in simplifying Boolean algebra
expression is DeMorgan’s Theorem. It states:
(a + b)’ = a’b’ (ab)’ = a’ + b’
° Complement the expression
a(b + z(x + a’)) and simplify.
(a(b+z(x + a’)))’ = a’ + (b + z(x + a’))’
= a’ + b’(z(x + a’))’
= a’ + b’(z’ + (x + a’)’)
= a’ + b’(z’ + x’a’’)
= a’ + b’(z’ + x’a)

Basic theorems

Summary
° Basic logic functions can be made from AND, OR,
and NOT (invert) functions
° The behavior of digital circuits can be represented
with waveforms, truth tables, or symbols
° Primitive gates can be combined to form larger
circuits
° Boolean algebra defines how binary variables can
be combined
° Rules for associativity, commutativity, and
distribution are similar to algebra
° DeMorgan’s rules are important.
• Will allow us to reduce circuit sizes.

Overview
° Expressing Boolean functions
° Relationships between algebraic equations, symbols,
and truth tables
° Simplification of Boolean expressions
° Minterms and Maxterms
° AND-OR representations
• Product of sums
• Sum of products

Boolean Functions
° Boolean algebra deals with binary variables and
logic operations.
° Function results in binary 0 or 1
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z
0
1
0
1
0
1
0
1
F
0
0
0
0
1
0
1
1 F = x(y+z’)
x
y
z
z’
y+z’ F = x(y+z’)

Boolean Functions
° Boolean algebra deals with binary variables and
logic operations.
° Function results in binary 0 or 1
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z
0
1
0
1
0
1
0
1
xy
0
0
0
0
0
0
1
1
x
y
z
G = xy +yz
yz
xy
We will learn how to transition between equation,
symbols, and truth table.
yz
0
0
0
1
0
0
0
1
G
0
0
0
1
0
0
1
1

Representation Conversion
° Need to transition between boolean expression,
truth table, and circuit (symbols).
° Converting between truth table and expression is
easy.
° Converting between expression and circuit is
easy.
° More difficult to convert to truth table.
Truth
Table
Circuit Boolean
Expression

Truth Table to Expression
° Converting a truth table to an expression
• Each row with output of 1 becomes a product term
• Sum product terms together.
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z
0
1
0
1
0
1
0
1
G
0
0
0
1
0
0
1
1
xyz + xyz’ + x’yz
Any Boolean Expression can be
represented in sum of products form!

Equivalent Representations of Circuits
° All three formats are equivalent
° Number of 1’s in truth table output column equals AND
terms for Sum-of-Products (SOP)
x y z
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z
0
1
0
1
0
1
0
1
G
0
0
0
1
0
0
1
1
G = xyz + xyz’ + x’yz
G
x x
x
x
x
x
x
x
x

Reducing Boolean Expressions 1
° Is this the smallest possible implementation of
this expression? No!
° Use Boolean Algebra rules to reduce complexity
while preserving functionality.
° Step 1: Use Theorum 1 (a + a = a)
• So xyz + xyz’ + xy’z’ = xyz + xyz’ + xyz’ + xy’z’
° Step 2: Use distributive rule a(b + c) = ab + ac
• So xyz + xyz’ + xyz’ + xy’z’ = xy(z + z’) + xz’(y + y’)
° Step 3: Use Postulate 3 (a + a’ = 1)
• So xy(z + z’) + xz’(y + y’) = xy.1 + xz’.1
° Step 4: Use Postulate 2 (a . 1 = a)
• So xy.1 + xz’.1 = xy + xz’ = xyz + xyz’ + xy’z’
G = xyz + xyz’ + xy’z’

Reducing Boolean Expressions 2
° Is this the smallest possible implementation of
this expression? No!
° Use Boolean Algebra rules to reduce complexity
while preserving functionality.
° Step 1: Use Theorum 1 (a + a = a)
• So xyz + xyz’ + x’yz = xyz + xyz + xyz’ + x’yz
° Step 2: Use distributive rule a(b + c) = ab + ac
• So xyz + xyz + xyz’ + x’yz = xy(z + z’) + yz(x + x’)
° Step 3: Use Postulate 3 (a + a’ = 1)
• So xy(z + z’) + yz(x + x’) = xy.1 + yz.1
° Step 4: Use Postulate 2 (a . 1 = a)
• So xy.1 + yz.1 = xy + yz = xyz + xyz’ + x’yz

Reduced Hardware Implementation
° Reduced equation requires less hardware!
° Same function implemented!
x y z
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z
0
1
0
1
0
1
0
1
G
0
0
0
1
0
0
1
1
G = xyz + xyz’ + x’yz = xy + yz
G
x x
x
x

Minterms and Maxterms (POS AND SOP Forms
° Each variable in a Boolean expression is a literal
° Boolean variables can appear in normal (x) or
complement form (x’)
° Each AND combination of terms is a minterm
° Each OR combination of terms is a maxterm
For example:
Minterms
x y z Minterm
0 0 0 x’y’z’ m0
0 0 1 x’y’z m1
…
1 0 0 xy’z’ m4
…
1 1 1 xyz m7
For example:
Maxterms
x y z Maxterm
0 0 0 x+y+z M0
0 0 1 x+y+z’ M1
…
1 0 0 x’+y+z M4
…
1 1 1 x’+y’+z’ M7

Representing Functions with Minterms
° Minterm number same as row position in truth table
(starting from top from 0)
° Shorthand way to represent functions
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z
0
1
0
1
0
1
0
1
G
0
0
0
1
0
0
1
1
G = m7 + m6 + m3 = Σ(3, 6, 7)

Complementing Functions
° Minterm number same as row position in truth table
(starting from top from 0)
° Shorthand way to represent functions
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z
0
1
0
1
0
1
0
1
G
0
0
0
1
0
0
1
1
G’ = (xyz + xyz’ + x’yz)’ =
G’
1
1
1
0
1
1
0
0

Truth table representing minterm and maxterm

Canonical Form
° In Boolean algebra,Boolean function can be
expressed as Canonical Disjunctive Normal Form
known as minterm and some are expressed as
Canonical Conjunctive Normal Form known as maxterm
.
° In Minterm, we look for the functions where the output
results in “1” while in Maxterm we look for function
where the output results in “0”.
° We perform Sum of minterm also known as Sum of
products (SOP) .
° We perform Product of Maxterm also known as Product
of sum (POS).
° Boolean functions expressed as a sum of product
form(minterm) or product of sum form(maxterm) are
said to be in canonical form.

Standard Form
A Boolean variable can be expressed in either true form or
complemented form.
In standard form Boolean function will contain all the
variables in either true form or complemented form
while in canonical form number of variables depends on
the output of SOP or POS.

A Boolean function can be expressed algebraically
from a given truth table by forming a :
minterm for each combination of the variables that
produces a 1 in the function and then taking the OR
of all those terms.
maxterm for each combination of the variables that
produces a 0 in the function and then taking the AND
of all those terms

Complementing Functions
° Step 1: assign temporary names
• b + c -> z
• (a + z)’ = G’
° Step 2: Use DeMorgans’ Law
• (a + z)’ = a’ . z’
° Step 3: Resubstitute (b+c) for z
• a’ . z’ = a’ . (b + c)’
° Step 4: Use DeMorgans’ Law
• a’ . (b + c)’ = a’ . (b’. c’)
° Step 5: Associative rule
• a’ . (b’. c’) = a’ . b’ . c’
G’ = (a + b + c)’
G = a + b + c
G’ = a’ . b’ . c’ = a’b’c’
G = a + b + c

Complementation Example
° Find complement of F = x’z + yz
• F’ = (x’z + yz)’
° DeMorgan’s
• F’ = (x’z)’ (yz)’
° DeMorgan’s
• F’ = (x’’+z’)(y’+z’)
° Reduction -> eliminate double negation on x
• F’ = (x+z’)(y’+z’)
This format is called product of sums

Conversion Between Canonical Forms
° Easy to convert between minterm and maxterm
representations
° For maxterm representation, select rows with 0’s
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z
0
1
0
1
0
1
0
1
G
0
0
0
1
0
0
1
1
G = m7 + m6 + m3 = Σ(3, 6, 7)
G = M0M1M2M4M5 = Π(0,1,2,4,5)
G = (x+y+z)(x+y+z’)(x+y’+z)(x’+y+z)(x’+y+z’)

Representation of Circuits
° All logic expressions can be represented in 2-
level format
° Circuits can be reduced to minimal 2-level
representation
° Sum of products representation most common in
industry.

Summary
° Truth table, circuit, and boolean expression formats
are equivalent
° Easy to translate truth table to SOP and POS
representation
° Boolean algebra rules can be used to reduce circuit
size while maintaining function
° All logic functions can be made from AND, OR, and
NOT
° Easiest way to understand: Do examples!
° Next time: More logic gates!

The NAND Gate
° This is a NAND gate. It is a combination of an
AND gate followed by an inverter. Its truth table
shows this…
° NAND gates have several interesting properties…
• NAND(a,a)=(aa)’ = a’ = NOT(a)
• NAND’(a,b)=(ab)’’ = ab = AND(a,b)
• NAND(a’,b’)=(a’b’)’ = a+b = OR(a,b)
A B Y
0 0 1
0 1 1
1 0 1
1 1 0
A
B
Y

The NAND Gate
° These three properties show that a NAND gate with both
of its inputs driven by the same signal is equivalent to a
NOT gate
° A NAND gate whose output is complemented is
equivalent to an AND gate, and a NAND gate with
complemented inputs acts as an OR gate.
° Therefore, we can use a NAND gate to implement all three
of the elementary operators (AND,OR,NOT).
° Therefore, ANY switching function can be constructed using
only NAND gates. Such a gate is said to be primitive or
functionally complete or UNIVERSAL GATE.

A
Y
A
B
Y
Y
A
B
NOT Gate
AND Gate
OR Gate
NAND Gates into Other Gates
(what are these circuits?)

The NOR Gate
° This is a NOR gate. It is a combination of an OR
gate followed by an inverter. It’s truth table shows
this…
° NOR gates also have several
interesting properties…
• NOR(a,a)=(a+a)’ = a’ = NOT(a)
• NOR’(a,b)=(a+b)’’ = a+b = OR(a,b)
• NOR(a’,b’)=(a’+b’)’ = ab = AND(a,b)
A B Y
0 0 1
0 1 0
1 0 0
1 1 0
A
B
Y

Functionally Complete Gates
° Just like the NAND gate, the NOR gate is
functionally complete…any logic function can be
implemented using just NOR gates.
° Both NAND and NOR gates are very valuable as
any design can be realized using either one.
° It is easier to build an IC chip using all NAND or
NOR gates than to combine AND,OR, and NOT
gates.
° NAND/NOR gates are typically faster at switching
and cheaper to produce.

NOT Gate
OR Gate
AND Gate
NOR Gates into Other Gates
A
Y
Y
A
B
A
B
Y

The XOR Gate (Exclusive-OR)
° This is a XOR gate.
° XOR gates assert their output
when exactly one of the inputs
is asserted, hence the name.
° The switching algebra symbol
for this operation is , i.e.
1  1 = 0 and 1  0 = 1.
A B Y
0 0 0
0 1 1
1 0 1
1 1 0
A
B
Y

The XNOR Gate
° This is a XNOR gate.
° This functions as an
exclusive-NOR gate, or
simply the complement of
the XOR gate.
° The switching algebra symbol
for this operation is , i.e.
1  1 = 1 and 1  0 = 0.
A B Y
0 0 1
0 1 0
1 0 0
1 1 1
A
B
Y

NOR Gate Equivalence
° NOR Symbol, Equivalent Circuit, Truth Table

° The theorem explains that the complement of the product
of all the terms is equal to the sum of the complement of
each term. Likewise, the complement of the sum of all the
terms is equal to the product of the complement of each
term. ie
(a + b)’ = a’b’ (ab)’ = a’ + b’
° Complement the expression
a(b + z(x + a’)) and simplify.
(a(b+z(x + a’)))’ = a’ + (b + z(x + a’))’
= a’ + b’(z(x + a’))’
= a’ + b’(z’ + (x + a’)’)
= a’ + b’(z’ + x’a’’)
= a’ + b’(z’ + x’a)

Example
° Determine the output expression for the below
circuit and simplify it using DeMorgan’s Theorem

A universal gate is a gate which can implement any Boolean
function without need to use any other gate type.
The NAND and NOR gates are universal gates.
In practice, this is advantageous since NAND and NOR gates
are economical and easier to fabricate and are the basic gates
used in all IC digital logic families.
universal gate

Universality of NAND and NOR gates

Universality of NOR gate
° Equivalent representations of the AND, OR, and
NOT gates

Example

Interpretation of the two NAND gate symbols
° Determine the output expression for circuit via

Interpretation of the two OR gate symbols
° Determine the output expression for circuit via

Summary
° Basic logic functions can be made from NAND, and
NOR functions
° The behavior of digital circuits can be represented
with waveforms, truth tables, or symbols
° Primitive gates can be combined to form larger
circuits
° Boolean algebra defines how binary variables with
NAND, NOR can be combined
° DeMorgan’s rules are important.
• Allow conversion to NAND/NOR representations

Module I - Digital Systems & Logic Gates.ppt

More Related Content

Similar to Module I - Digital Systems & Logic Gates.ppt (20)

Recently uploaded (20)

Module I - Digital Systems & Logic Gates.ppt