ECE 331 – Digital System Design

ECE 331 – Digital System Design

Boolean Algebra

(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.

Fall 2010 ECE 331 - Digital System Design 2

Converting POS to SOP

Given an expression in product-of-sums (POS) form, the corresponding sum-of-products (SOP) expression can be obtained by multiplying out, using the two distributive laws:

X(Y + Z) = XY + XZ (3-1)

(X + Y)(X + Z) = X + YZ (3-2)In addition, the following theorem is very useful for factoring and multiplying out:

(X + Y)(X′ + Z) = XZ + X′Y (3-3)


Boolean Algebra

Exercise:

Convert the following Boolean Expression in POS form to a Boolean Expression in SOP form.

(Hint: Multiply out the expression)

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


Boolean Algebra

Exercise:

Convert the following Boolean Expression in POS form to a Boolean Expression in SOP form.

(Hint: Multiply out the expression)

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


Boolean Algebra

If we were to multiply out by brute force, we would generate 162 terms, and 158 of these terms would then have to be eliminated to simplify the expression.

Instead, we will use the distributive laws to simplify the process.


Converting SOP to POS

The same theorems that are useful for multiplying out expressions are also useful for factoring.

By repeatedly applying (3-1), (3-2), and (3-3), an expression in sum-of-products (SOP) form can be converted to an expression in product-of-sums (POS) form.

X(Y + Z) = XY + XZ (3-1)

(X + Y)(X + Z) = X + YZ (3-2)

(X + Y)(X′ + Z) = XZ + X′Y (3-3)


Boolean Algebra

Exercise:

Convert the following Boolean Expression in SOP form to a Boolean Expression in POS form.

(Hint: Factor the expression)

F = ACE + BCE + ADE + BDE + ACF + BCF + ADF + BDF


Boolean Algebra

Exercise:

Convert the following Boolean Expression in SOP form to a Boolean Expression in POS form.

(Hint: Factor the expression)

F = AC + A'BD' + A'BE + A'C'DE


Boolean Algebra


Exclusive-OR and EquivalenceThe following theorems of Boolean Algebra apply to Exclusive-OR and Equivalence (Exclusive-NOR):


Boolean Algebra

Exercise:

Simplify the following Boolean Expression.


Boolean Algebra

Exercise:



Boolean Algebra

Exercise:


F(A,B,C) = (A xor B)' xor C


Consensus Theorem

The consensus theorem can be stated as follows:XY + X'Z + YZ = XY + X'Z (3-20)

Dual Form:(X + Y)(X’ + Z)(Y + Z) = (X + Y)(X’ + Z) (3-21)

XY + X'Z + YZ = XY + X'Z + (X + X')YZ = (XY + XYZ) + (X'Z + X'YZ)

= XY(1 + Z) + X'Z(1 + Y) = XY + X'Z

Consensus theorem proof:


Simplifying Boolean Expressions

1. Combining terms. Use the theorem XY + XY′ = X to combine two terms.

2. Eliminating terms. Use the theorem X + XY = X to eliminate redundant terms if possible; then try to apply the consensus theorem (XY + X′Z + YZ = XY + X′Z) to eliminate any consensus terms.

3. Eliminating literals. Use the theorem X + X’Y = X + Y to eliminate redundant literals. Simple factoring may be necessary before the theorem is applied.

4. Adding redundant terms. Redundant terms can be introduced in several ways such as adding xx′, multiplying by (x + x′), adding yz to xy + x′z, or adding xy to x. When possible, the added terms should be chosen so that they will combine with or eliminate other terms.


Boolean AlgebraExample:


Boolean Algebra

Simplify the following Boolean expression using Boolean algebra:

F(A,B,C,D) = A'BC'D' + A'B'C'D + A'BC'D + ABC'D + A'BCD + ABCD


Proving Validity of an EquationOften we will need to determine if an equation is valid for all combinations of values of the variables. Several methods can be used to determine if an equation is valid:

1. Construct a truth table and evaluate both sides of the equation for all combinations of values of the variables. (This method is rather tedious if the number of variables is large, and it certainly is not very elegant.)

2. Manipulate one side of the equation by applying various theorems until it is identical with the other side.

3. Reduce both sides of the equation independently to the same expression.

4. It is permissible to perform the same operation on both sides of the equation provided that the operation is reversible.


Boolean AlgebraExample: Show that

A'BD' + BCD + ABC' + AB'D = BC'D' + AD + A'BC

Solution: Starting with the left side,


Importance of Boolean Algebra

Boolean Algebra is used to simplify Boolean expressions.

– Through application of the Laws and Theorems discussed

Simpler expressions lead to simpler circuit realization, which, generally, reduces cost, area requirements, and power consumption.

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

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


Boolean Algebra

Exercise:

For the following Boolean expression

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

1. Using Boolean algebra, simplify the Boolean expression.2. Derive the Truth table for the simplified expression.3. From the Truth table, determine an equivalent Boolean expression.4. Draw the circuit diagram for the “cheapest” expression.


Questions?

ECE 331 – Digital System Design

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