ece 331 – digital system design boolean algebra and standard forms of boolean expressions (lecture...

ECE 331 – Digital System Design

Boolean Algebraand

Standard Forms of Boolean Expressions

(Lecture #4)

The slides included herein were taken from the materials accompanying Fundamentals of Logic Design, 6th Edition, by Roth and Kinney,

and were used with permission from Cengage Learning.

Spring 2011 ECE 331 - Digital System Design 2

Basic Laws and TheoremsOperations with 0 and 1:1. X + 0 = X 1D. X • 1 = X2. X + 1 = 1 2D. X • 0 = 0 Idempotent laws:3. X + X = X 3D. X • X = X Involution law:4. (X')' = X Laws of complementarity:5. X + X' = 1 5D. X • X' = 0


Basic Laws and TheoremsCommutative laws:6. X + Y = Y + X 6D. XY = YX Associative laws:7. (X + Y) + Z = X + (Y + Z) 7D. (XY)Z = X(YZ) = XYZ = X + Y + Z Distributive laws:8. X(Y + Z) = XY + XZ 8D. X + YZ = (X + Y)(X + Z) Simplification theorems:9. XY + XY' = X 9D. (X + Y)(X + Y') = X10. X + XY = X 10D. X(X + Y) = X11. (X + Y')Y = XY 11D. XY' + Y = X + Y


Basic Laws and Theorems

DeMorgan's laws:12. (X + Y + Z +...)' = X'Y'Z'...12D. (XYZ...)' = X' + Y' + Z' +... Duality:13. (X + Y + Z +...)D = XYZ... 13D. (XYZ...)D = X + Y + Z +... Theorem for multiplying out and factoring:14. (X + Y)(X' + Z) = XZ + X'Y 14D. XY + X'Z = (X + Z)(X' + Y) Consensus theorem:15. XY + YZ + X'Z = XY + X'Z15D. (X + Y)(Y + Z)(X' + Z) = (X + Y)(X' + Z)


Duality (13)

The dual of a Boolean expression can be written by

Replacing AND with OR, and OR with AND Replacing 0 with 1, and 1 with 0 Leaving literals unchanged

See the Boolean laws and theorems, previously discussed, for examples of Boolean expressions and their duals.


Distributive Law: Example #1

Use the distributive law to multiply out the following Boolean expression:

F = (A+B).(C+D).(E+F)

Distributive law (8): X.(Y + Z) = X.Y + X.Z


Distributive Law: Example #2

Use the distributive law to factor the following Boolean expression:

F = A.B + C.D

Distributive law (8D): X + Y.Z = (X+Y).(X+Z)


Simplification Theorems: Example #1

Use the simplification theorems to simplify the following Boolean expression:

F = ABC' + AB'C' + A'BC'

Simplification Theorems (9 – 11): X.Y + X.Y' = X (X+Y).(X+Y') = XX + X.Y = X X.(X+Y) = X(X+Y').Y = X.Y X.Y' + Y = X+Y




F = (A'+B'+C').(A+B'+C').(B'+C)





F = AB'CD'E + ACD + ACF'GH' +ABCD'E +ACDE' + E'H'


(See Programmed Exercise 3.4 on page 75)


Consensus Theorem: Example #1

Use the consensus theorem to simplify the following Boolean expression:

F = ABC + BCD + A'CD + B'C'D'

Consensus Theorem: (15) X.Y + Y.Z + X'.Z = X.Y + X'.Z(15D) (X+Y).(Y+Z).(X'+Z) = (X+Y).(X'+Z)




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





F = AC' + AB'D + A'B'C + A'CD' + B'C'D'


(See Programmed Exercise 3.5 on page 77)


DeMorgan's Law

DeMorgan's Law: (12) (X + Y + Z + … )' = X'.Y'.Z'... (12D) (X.Y.Z… )' = X' +Y' + Z' …

X Y X + Y (X + Y)' X' Y' X'.Y'

0 0 0 1 1 1 1

0 1 1 0 1 0 0

1 0 1 0 0 1 0

1 1 1 0 0 0 0

Prove (using a truth table): (X+Y)' = X'.Y'


DeMorgan's Law

Graphical representation of DeMorgan's Law

xy

xy

x

y

(X.Y)' X' + Y'

xy

xy

x

y(X+Y)' X'.Y'


DeMorgan's Law: Example

DeMorgan's Law: (12) (X + Y + Z + … )' = X'.Y'.Z'... (12D) (X.Y.Z… )' = X' +Y' + Z' …

Find the complement of the following Boolean expression using DeMorgan's law:

F = (A + (BC)').((AD)' + C.(B' + D))


Simplifying Boolean Expressions

Boolean algebra can be used in several ways to simplify a Boolean expression:

Combine terms Eliminate redundant or consensus terms Eliminate redundant literals Add redundant terms to be combined with or

allow the elimination of other terms


Equivalency of Boolean Expressions

Two Boolean expressions are equivalent iff both expressions evaluate to the same value for all combinations of the variables in the expressions.

The equivalency can be proven using A Truth table Boolean algebra theorems to manipulate one

expression until it is identical to the other. Boolean algebra theorems to reduce both

expressions independently to the same expression.


Importance of Boolean Algebra Boolean algebra is used to simplify Boolean

expressions. Simpler expressions leads to simpler logic circuits.

Reduces cost Reduces area requirements Reduces power consumption

The objective of the digital circuit designer is to design and realize optimal digital circuits.

Thus, Boolean algebra is an important tool to the digital circuit designer.


Problem with Boolean Algebra

In general, there is no easy way to determine when a Boolean expression has been simplified to a minimum number of terms or a minimum number of literals.

Karnaugh Maps provide a better mechanism for the simplification of Boolean expressions.


Circuit Design: Example

For the following Boolean expression:

F(A,B,C) = A.B.C + A'.B.C + A.B'.C + A.B.C'

1. Draw the circuit diagram2. Simplify using Boolean algebra3. Draw the simplified circuit diagram


Standard Forms of Boolean Expressions


Standard Forms

There are two standard forms in which all Boolean expressions can be written:

1. Sum of Products (SOP)2. Product of Sums (POS)


Sum of Products (SOP)

Product Term Logical product = AND operation A product term is the ANDing of literals Examples: A.B, A'.B.C, A.C', B.C'.D', A.B.C.D

“Sum of” Logical sum = OR operation The sum of products is the ORing of product

terms.


Sum of Products (SOP)

The distributive laws are used to multiply out a general Boolean expression to obtain the sum of products (SOP) form.

The distributive laws are also used to convert a Boolean expression in POS form to one in SOP form.

A SOP expression is realized using a set of AND gates (one for each product term) driving a single OR gate (for the sum).


Product of Sums (POS)

Sum Term Logical sum = OR operation A sum term is the ORing of literals Examples: A+B, A'+B+C, A+C', B+C'+D'

“Product of” Logical product = AND operation The product of sums is the ANDing of sum

terms.


Product of Sums (POS)

The distributive laws are used to factor a general Boolean expression to obtain the product of sums (POS) form.

The distributive laws are also used to convert a Boolean expression in SOP form to one in POS form.

A POS expression is realized using a set of OR gates (one for each sum term) driving a single AND gate (for the product).


SOP and POS: Examples

For each of the following Boolean expressions, identify whether it is in SOP or POS form:

1. F(A,B,C) = (A+B).(A'+B'+C').(B+C')

2. F(A,B,C) = A.B.C + B'.C' + A.C' + A'.B.C'3. F(A,B,C) = A + B.C + B'.C' + A'.B'.C4. F(A,B,C) = (A'+B'+C).(B+C').(A+C').(B')5. F(A,B,C) = A.B.C + A'.(B+C) + (A+C').B6. F(A,B,C) = A + B + C


Questions?

ece 331 – digital system design boolean algebra and standard forms of boolean expressions (lecture...

Documents