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LOGO DKT 122/3

DIGITAL SYSTEM 1

BOOLEAN ALGEBRA (PART 2)

Boolean Algebra

Contents

Boolean Operations & Expression

Laws & Rules of Boolean algebra

DeMorgan’s Theorems

Boolean analysis of logic circuits

Simplification using Boolean Algebra

Standard forms of Boolean Expressions

Boolean Expressions & truth tables

The Karnaugh Map

Karnaugh Map SOP minimization

Karnaugh Map POS minimization

Programmable Logic

Boolean Algebra (Cont.)

Solve this..

B

A

X

(a) Write the equivalent Boolean expression of circuit shown above.

(b) Simplify the Boolean expression found in (a).

C

Solve this..

� Simplify the following Boolean expressions:

� (AB(C + BD) + AB)C

� ABC + ABC + ABC + ABC + ABC

� Write the Boolean expression of the following circuit.circuit.

RULE:Break the longest

bar first!

RULE:Break the longest

bar first!

� Sum-of-products (SOP)

Refer to two or more product terms, summed (added) by Boolean addition

Standard Forms of Boolean Expressions

summed (added) by Boolean addition

� Product-of-sums (POS)

Refer to multiplication of two or more sum terms

Sum-of-products (SOP)

� In an SOP expression, a single overbarcannot extend over more than one variable

The standard SOP form

Refer to expression in which all the variables in the domain appear in each product term in the expression


expression

Example:

ABCD + ABCD + ABCD (standard SOP expression)

ABC + ABD + ABCD (non-standard SOP expression)

� All variables appear in each product term.� Each of the product term in the expression is

called as minterm.

Example BCACABCBACBAf ++=),,(


The standard SOP form

Example BCACABCBACBAf ++=),,(

In compact form, f(A,B,C) may be written as

632),,( mmmCBAf ++=

)6,3,2(),,( mCBAf Σ=

Product-of-sums (POS)

� In a POS expression, a single overbarcannot extend over more than one variable

Refer to expression in which all the variables in the domain appear in each sum term in the expression

Example:


The standard POS form

Example:

(A + B + C)(B + C + D)(A + B + C + D) (non-standard POS expression)

(A + B + C + D)(A + B + C + D) (standard POS expression)

• All variables appear in each product term.• Each of the product term in the expression is called as maxterm.

Example: )()()(),,( CBACBACBACBAf ++•++•++=


The standard POS form

Example: )()()(),,( CBACBACBACBAf ++•++•++=

In compact form, f(A,B,C) may be written as

541),,( MMMCBAf =

)5,4,1(),,( MCBAf π=

Identify each of the following expressions as SOP, standard SOP, POS or standard POS:

(i) AB + ABD + ACD(ii) (A + B + C)(A + B + C)(iii) ABC + ABC

Solve this..

(iii) ABC + ABC(iv) A(A + C)(A + B)

Convert the following Boolean expressions to SOP form:

(i) (A + B)(B + C + D)(ii) AB + B(CD + EF)

Convert the following Boolean expressions into standard

SOP form:

AB + ABCD

Solution

Example (Standard SOP)

Solution

1st step: AB = AB (C + C) = ABC + ABC

2nd step: ABC (D + D) + ABC (D + D)

= ABCD + ABCD + ABCD + ABCD

3rd step: ABCD + ABCD + ABCD + ABCD + ABCD

Convert the following Boolean expressions into standard

POS form:

(A + B + C)(B + C + D)

Solution

Example (Standard POS)

Solution

1st step: A + B + C = A + B + C + DD

= (A + B + C + D) (A + B + C + D)

2nd step: B + C + D = B + C + D + AA

= (A + B + C + D)(A + B + C + D)

Final answer: Combine answer for 1st and 2nd step

Boolean Expressions & Truth Tables

Converting SOP to Truth Table

� Examine each of the products to determine where the product is equal to a 1.

� Set the remaining row outputs to 0.

Converting POS to Truth Table

�Opposite process from the SOP expressions.

� Each sum term results in a 0.

� Set the remaining row outputs to 1.


Determining Standard Expressions from Truth Table

INPUT OUTPUT

A B C X

0 0 0 0

SOP Expressions

X = ABC + ABC + ABC +


0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 1

X = ABC + ABC + ABC + ABC

POS Expressions

X = (A + B + C)(A + B + C)(A + B + C)(A + B + C)

Solve this..

� Convert the following SOP expression to an equivalent POS expression:

CBACBACABABCCBAf +++=),,(

� Develop a truth table for the expression:

)()()()(),,( CBACBACBACBACBAf ++•++•++•++=

CBACBACABABCCBAf +++=),,(

The Karnaugh Map (K-Map)

� Karnaugh Map (K-map) is an array of cells in which each cell represents a binary value of the input variables

� K-Mapping is used to minimize the number of logic gates that are required in a digital circuit. This will replace Boolean reduction when the This will replace Boolean reduction when the circuit is large.

� The number of cells in a K-map is equal to the total number of possible input variable combinations

� K-Map is similar to Truth Table because it present all possible values of input variables and the resulting output for each value

� The map is made up of a table of every possible SOP using the number of variables that are being used.

If 2 variables are used, then a 2X2 map is used,

The K-map

is used,



If 5 variables are used, then a 8X4 map is used

� Mapping Standard SOP Expression

For an SOP in standard form, a 1 is placed on the K-map for each product term in the expression

K-map SOP Minimization

Gray Code

Example of standard SOP form

� Mapping Non-standard SOP ExpressionExpand the non-standard expression

A + AB + ABC


A + AB + ABC

QuestionHow to expand this expression?

QuestionHow to expand this expression?

● Mapping directly from a truth table to a K-map


Thomas L. Floyd

Digital Fundamentals, 9e

Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

Notice that the map is going false to true, left to right and top to bottomA

B B

0 10 1

2 variables K-map


The upper right hand cell is A B if X= A B then put an X in that cell

This show the expression true when A = 0 and B = 0

A

A

B B

1

A 2 32 3

If X=AB + AB then put an X in both of these cells

A

A

B B

1

1

2 variables K-map


these cells

From Boolean reduction we know that A B + A B = B

From the Karnaugh map we can circle adjacent cell and find that X = B

A

A

B B

1

1

A 1

Gray Code 0 1

C C


3 variables K-map

00 A B

01 A B

11 A B

10 A B

0 10 1

2 32 3

6 76 7

4 54 5

0 1


X = A B C + A B C + A B C + A B C

3 variables K-map

00 A B

01 A B

11 A B

10 A B

0 1

C C

Each 3 variable term is one cell on a 4 X 2

Karnaugh map

Each 3 variable term is one cell on a 4 X 2

Karnaugh map

1 1

1 1

0 1

One

X = A B C + A B C + A B C + A B C


3 variables K-map

00 A B

01 A B

11 A B

10 A B

0 1

C COne simplification could be

X = A B + A B

1 1

1 1

X = A B C + A B C + A B C + A B C

0 1Another

X = A B C + A B C + A B C + A B C


3 variables K-map

00 A B

01 A B

11 A B

10 A B

C CAnother simplification could be

X = B C + B C

A Karnaugh Map does wrap around

1 1

1 1

X = A B C + A B C + A B C + A B C

0 1

X = A B C + A B C + A B C + A B C


3 variables K-map

00 A B

01 A B

11 A B

10 A B

0 1

C CThe bestsimplification would be

X = B

1 1

1 1

Conclusions

� One cell requires 3 variables


3 variables K-map

� One cell requires 3 variables

� Two adjacent cells require 2 variables

� Four adjacent cells require 1 variable

� Eight adjacent cells is a 1

Gray Code 0 0 0 1 1 1 1 0

C D C D C D C D


4 variables K-map

00 A B

01 A B

11 A B

10 A B

C D C D C D C D

0 1 3 20 1 3 2

4 5 7 64 5 7 6

12 13 15 1412 13 15 14

8 9 11 108 9 11 10

Gray Code 0 0 0 1 1 1 1 0

C D C D C D C D

Simplify:

X = A B C D + A B C D + A B C D + A B C D + A B C D + A B C D


00 A B

01 A B

11 A B

10 A B

C D C D C D C D

1

1

1

1

1

1

X = ABD + ABC + CD

Now, try it with Boolean reductions..


Conclusions

� One Cell requires 4 variables

4 variables K-map

� One Cell requires 4 variables

� Two adjacent cells require 3 variables

� Four adjacent cells require 2 variables

� Eight adjacent cells require 1 variable

� Sixteen adjacent cells give a 1 or true

Simplify

Z = B C D + B C D + C D + B C D + A B C

Gray Code 00 01 11 10

C D C D C D C D


00 A B

01 A B

11 A B

10 A B

C D C D C D C D

1 1

1 1

1 1

1 1

1

1

Z = BD + C

Example

Simplify the following circuit using K-map method.

Y = A + B + B C + ( A + B ) ( C + D)

Example (Cont.)

Y = A B + B C + A + B + ( C + D )

Y = A B + B C + (A + B ) + C D

Simplified SOP expression

Gray Code 00 01 11 10

C D C D C D C DY = 1

Then, map the SOP expression into the K-map & simplify the equation

Example (Cont.)

00 A B

01 A B

11 A B

10 A B

C D C D C D C D

1 1

1 1

1 1 1 1

1 1 1 1

1 1

1 1

K-map POS Minimization

Assume A, B, C, and D are variables.

3 variables 4 variables


3 variables K-map


4 variables K-map

Input Output

K-map Minimization – Don’t Cares

3 variables with output “don’t cares” (X)

K-map Minimization – Don’t Cares

4 variables with output “don’t cares” (X)

Determine the minimal SOP using K-Map:

14,15)D(5,12,13, 9,10)M(0,2,6,8, D)C,B,F(A, π=

Example

D refers to “don’t cares”

What is this?

AB

CD

00

00 01 11 10

0 1 1 00 1 3 2

Example (Cont.)

14,15)D(5,12,13, 9,10)M(0,2,6,8, D)C,B,F(A, π=

00

01

11

10

1 X 1 0

X X X X

0 0 1 0

0 1 3 2

4 5 7 6

12 13 15 14

8 9 11 10

DACBCDDCBAF ++=),,,(

Minimum SOP expression is:

CD

ADBC

Solve this..

Reduce (a), (b) and (c) using K-map:

(a) (b)

∑ ∑+= 13) 12, (6, d 9) 7, 5, 3, (1, m D) C, B, f(A, (c)

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