CODEWITHSUNDEEP
boolean-expressionboolean-operationscircuit
John Alberto
John Alberto

Reputation: 437

simplifying a boolean expression?

I have these 2 boolean expressions for 2 circuits that I have tried simplifying for a long time and I just cant figure it out.

a) AB'C' + ABD + AB'CD'

b) (A+C+D) . (A+B'+C) . (A+B'+C'+D)

I know for a the answer is AB'C' + AB'D' + ABD and then for b the answer is (A+C+D) . (A+B'+C) . (A+B'+D) but I have no idea how to arrive at the solutions for either two. I have tried using the fact that A+A' is 1 as well as the distributive and combining properties but its just not working out. I've been stuck on these last 2 for hours now. I would really appreciate any help anyone can give.

Upvotes: 0

Views: 1093

Answers (2)

jonasn
jonasn

Reputation: 91

I'll do the first one:

AB'C' + ABD + AB'CD'

The last term is the "most specified" term. Try to find another term that is less specified (fewer variables) but has many common factors.

In this case, the first term is similar, with two variables in common. It lacks the variable D. (D+D') = 1, so you can expand this term with (D+D') without changing the expression. Expand this term with (D+D'):

AB'C'(D+D') + ABD + AB'CD'

Now distribute AB'C' over the recently added parenthesis:

AB'C'D + AB'C'D' + ABD + AB'CD'

Duplicate the second term:

AB'C'D + AB'C'D' + AB'C'D' + ABD + AB'CD'

Use one of the copies to collapse the recently expanded term back to what is was from the beginning:

AB'C'D + AB'C'D' + AB'C'D' + ABD + AB'CD'

AB'C'(D + D') + AB'C'D' + ABD + AB'CD'

AB'C' + AB'C'D' + ABD + AB'CD'

Rearrange the terms and collapse the two four-factor terms in a similar way:

AB'C' + AB'C'D' + AB'CD' + ABD

AB'C' + AB'(C'+C)D' + ABD

AB'C' + AB'D' + ABD

Presto!

The second example can be done the same way. Identify the "most specified" term and expand another term that is similar with the variable that's missing.

Upvotes: 2

Paul R
Paul R

Reputation: 212979

Use a truth table, e.g. for (a):

   A  B  C  D  X
   0  0  0  0  0
   0  0  0  1  0
   0  0  1  0  0
   0  0  1  1  0
   0  1  0  0  0
   0  1  0  1  0
   0  1  1  0  0
   0  1  1  1  0
   1  0  0  0  1
   1  0  0  1  1
   1  0  1  0  1
   1  0  1  1  0
   1  1  0  0  0
   1  1  0  1  1
   1  1  1  0  0
   1  1  1  1  1

Then by inspection this reduces to:

  AB'C' + ABD + AB'CD'

= A(B'C' + BD + B'CD')

= A(B'(C' + CD') + BD)

(although this is not the only possible solution of course)

Upvotes: 0

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