# Logic Circuits for Boolean Expressions

1. Using Boolean algebra, reduce the following Boolean expression to its simplest form and implement it using your method of choice.

F = A'B'D + ABC' + AB'CD + BC + A'CD + BCD

2. Using a Karnaugh map (K-map), reduce the following Boolean to its simplest form and implement it using SOP (Sum of Products)

F = B'C'D + B'D + ABC + BC'D' +ABD' + A'B'D

3. Using a Karnaugh map (K-map), reduce the following Boolean to its simplest form and implement it using POS (Products of Sum)

F(A, B, C, D) = âˆ‘(0,2,3,5,7,8,9,10,11,13,15)

Show karnaugh map steps and details.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please see attached for the full solution to the given problems.

1. Using Boolean algebra, reduce the following Boolean expression to its simplest form and implement it using your method of choice.

F = A'B'D + ABC' + AB'CD + BC + A'CD + BCD

Solution: We first group the terms with common logic:

F = A'B'D + B(AC'+C) + CD(AB'+A') + BCD

Then we apply Boolean algebra laws to simplify the highlighted terms:

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

Then we distribute the terms:

F = A'B'D + AB + AC + ACD + B'CD + BCD

Then we group terms common logic:

F = A'B'D + AB + AC + ACD + CD(B'+ B)

Then we apply Boolean algebra laws to simplify the highlighted terms:

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

Then we group terms common logic:

F = A'B'D + AB + AC + CD(A+1)

Then we apply Boolean algebra laws to simplify the highlighted terms to get the final answer of:

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

Using K-map and SOP, we also get the same answer:

ABCD 00 01 11

10

00 0 1

1 0

01 0

0 1

1

11 1 1 1 1

10 0 0 1 0

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

2. Using a Karnaugh map (K-map), reduce the following Boolean to its simplest form and implement it using SOP (Sum of Products)

F = B'C'D + B'D + ABC + BC'D' +ABD' + A'B'D

Solution: Using K-map and SOP, we have:

A'B' A'B

AB AB'

C'D' 0

1 1 0

C'D 1 0 0 1

CD 1 0 1

1

CD' 0 0 1 0

Grouping the 1's, the final expression in SOP is:

F = B'D + BC'D' + ABC

3. Using a Karnaugh map (K-map), reduce the following Boolean to its simplest form and implement it using POS (Products of Sum)

F(A, B, C, D) = âˆ‘(0,2,3,5,7,8,9,10,11,13,15)

Solution: Using K-map and POS, we have:

ABCD 00

01 11

10

00 0 1 0 0

01 1 0

0 1

11 1

0 0 1

10 0 0 0 0

The product of sum terms are placed in the K-map as 0's. Grouping the 0's, the final expression in POS is:

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

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