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Digital Logic Worked Example: <boolean equations>

Module by: Cathy Huang. E-mail the authorEdited By: Elec and Comp 326

Summary: Boolean equation simplification practice examples

Module: Digital Logic Worked Example: Boolean Equations

Simplify the following Boolean equations.
Sketch combinational circuit implementation for each equation.

Exercise 1

Y=B⁢¬C+B⁢C+A⁢B Y B C B C A B

Solution

Y=B⁢¬C+B⁢C+A⁢B Y B C B C A B
apply Distributivity Thm:
Y=B⁢(¬C+C)+A⁢B Y B C C A B
apply Complements Thm:
Y=B+A⁢B Y B A B
apply Distributivity Thm:
Y=B⁢(1+A) Y B 1 A
apply Null Element Thm:
Y=B Y B

Figure 1
schematics

Exercise 2

Y=B⁢A+B⁢C⁢(B+C) Y B A B C B C

Solution


 Y=B⁢A+B⁢C⁢(B+C) Y B A B C B C
apply Distributivity:
Y=A⁢B+B⁢B⁢C+B⁢C⁢C Y A B B B C B C C
apply Idempotency
Y=A⁢B+B⁢C+B⁢C Y A B B C B C
apply apply Idempotency
Y=A⁢B+B⁢C Y A B B C
apply Distributivity
Y=B⁢(A+C) Y B A C

Figure 2
schematics2

Exercise 3

Y=¬A⁢(C⁢(C+B)+¬B⁢C) Y A C C B B C

Solution

Y=¬A⁢(C⁢(C+B)+¬B⁢C) Y A C C B B C
apply Distributivity
Y=¬A⁢(C⁢C+C⁢B+¬B⁢C) Y A C C C B B C
apply Idempotency
Y=¬A⁢(C+C⁢B+C⁢¬B) Y A C C B C B
apply Distributivity
Y=¬A⁢(C+C⁢(B+¬B)) Y A C C B B
apply Complements
Y=¬A⁢(C+C) Y A C C
apply Idempotency
Y=¬A⁢C Y A C

Figure 3
schematics3

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