x= AB'(A'BC+DB)(BC+CD')+ABC+BC'D'+D(B'+C')

' = not......
Any help would be appreciated. I've read over the proofs but am having trouble getting started.

 

x= AB'(A'BC+DB)(BC+CD')+ABC+BC'D'+D(B'+C')

' = not......
Any help would be appreciated. I've read over the proofs but am having trouble getting started.

A good way to do it is to make a truth table according to your expression and then draw a Karnaugh map and simplify it.

Is that ok or you need to do it by using boolean algebra?

 

Unfortunately I have to simplify it down by using boolean algebra.....
How could i make a truth table by using that expression? I could take the results from the truth table then puts those in a k-map to pull the expression out. Which way is easier???

 

Take a look at the material in the contained in the AAC ebook on the topic of boolean algebra.

hgmjr

 

If you make the multiplications you get:

(AB'A'BC+AB'DB)(AB'BC+AB'CD')+ABC+BC'D+DB'+DC'

then AB'A'BC=0 , AB'DB=0 , AB'BC=0 because they have AA' and BB' in them

and BC'D+DC'=D[BC'+C']=D[C'(B+1)]=DC'

thus you get:

AB'CD'+ABC+DB'+DC'

i dont know if this is the simplest SOP but it is simplified

if you want the simplest do it with Karnaugh map

 

redbuckeye84,

The first expression AB'(A'BC+DB)(BC+CD')=0 so all that is left is ABC+BC'D'+D(B'+C') mik3 made an error in reduction.

Map it out on a 4 variable K-map and find out how the terms combine. Then chose the Boolean theorem that fits the groupings. The K-map also will find the correct answer or answers if there is more than one. Ask if you need more help. Ratch

 

redbuckeye84,

The first expression AB'(A'BC+DB)(BC+CD')=0 so all that is left is ABC+BC'D'+D(B'+C') mik3 made an error in reduction.

Map it out on a 4 variable K-map and find out how the terms combine. Then chose the Boolean theorem that fits the groupings. The K-map also will find the correct answer or answers if there is more than one. Ask if you need more help. Ratch

Ohh yes you are right

But because he wants SOP the answer is ABC+DB'+DC'