Lecture Digital logic design - Lecture 5: More boolean algebra

The main contents of the chapter consist of the following: Expressing boolean functions; relationships between algebraic equations, symbols, and truth tables; simplification of boolean expressions; minterms and maxterms; AND-OR representations.

Lecture 5

More Boolean Algebra

A B

w

° 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

Axioms and Graphical representation of DeMorgan's Law

Y X Y X

14B)

Y X Y

X

14A)

Y X Y X X

13D)

Y X Y X X

13C)

Y X XY X

13B)

Y X Y X X

13A)

YZ YW

XZ XW

Z W Y X

12B)

XZ XY

Z Y X

12A)

Z Y X Z

Y X

11B)

Z XY YZ

10A)

Commutative Law

Associative Law

Distributiv

e Law

Consensus Theorem

Simplification Using the Laws

x y z

z’ y+z’ F = x(y+z’)

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

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.

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)

z 0 1 0 1 0 1 0 1

G 0 0 0 1 0 0 1 1

G = xyz + xyz’ + x’yz

x x

° Step 1: Use Theorem 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

G = xyz + xyz’ + x’yz

Reducing Boolean Expressions

xy + yz two AND (Gate) operations and one OR (Gate)

y z

Reduced Hardware

Implementation

° Reduced equation requires less hardware!

° Same function implemented!

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

° 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

Representing Functions with

Minterms

° Minterm number same as row position in truth table

(starting from top from 0)

° Shorthand way to represent functions

z 0 1 0 1 0 1 0 1

G 0 0 0 1 0 0 1 1

G = xyz + xyz’ + x’yz

Complementing

Functions

° Minterm number same as row position in truth table

(starting from top from 0)

° Shorthand way to represent functions

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

Can we find a simpler representation?

Conversion Between Canonical

G 0 0 0 1 0 0 1 1

G = xyz + xyz’ + x’yz

G = (x+y+z)(x+y+z’)(x+y’+z)(x’+y+z)(x’+y+z’)

Simplification Using Boolean Algebra.

Example 1: AB + A(B+C) +B(B+C) =

Solution:

AB + A(B+C) +B(B+C) = AB+AB+AC+BB+BC

= AB+AB+AC+B+BC = AB+AC+B+BC

= AB+AC+B = B+AC

Boolean Algebra and Logic Simplification

Simplification Using Boolean Algebra.

Gate Network for Example 1:

ABAB

Boolean Algebra and Logic Simplification

Simplification Using Boolean Algebra.

Gate Network for Example 1:

B+AC

C

B+AC

ACA

B

Boolean Algebra and Logic Simplification

Simplification Using Boolean Algebra.

Gate Network for Example 1:

Simplification Using Boolean Algebra.

Simplification Using Boolean Algebra.

Example 2:

Solution:

C B A BD

B A C

B A C

B A BD

C B

[

C B A D

A C

B

(

C B

A C

B

(

C B

A C

C B A CC

B

A

C B A C

C

B

C B

Simplification Using Boolean Algebra.

Example 2:

Solution:

C B

A C

Simplification Using Boolean Algebra.

Example 3:

Using Boolean algebra techniques, simplify the following expression:

Boolean Algebra and Logic Simplification

ABC + ABC + ABC + ABC + ABC + ABC

ABC + ABC + ABC + ABC + ABC + ABC

BC(A + A) + BC(A + A) + AC(B + B)

BC + BC + AC

C(B + A) + BC

Simplification Using Boolean Algebra.

Example 4:

Using Boolean algebra techniques, simplify the following expression:

C B A AC

Boolean Algebra and Logic Simplification

Example 4:

Simplify the following Boolean

functions T1 and T2 to a minimum

number of literals:

° Truth table, circuit, and boolean expression

formats are equivalent

° Easy to translate truth table to SOP and POS

° Easiest way to understand: Do examples!

° Next time: More logic gates!