# Need help simplifying boolean equation to SOP format

Discussion in 'Homework Help' started by redbuckeye84, May 12, 2008.

1. ### redbuckeye84 Thread Starter New Member

May 12, 2008
2
0
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.

2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
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?

3. ### redbuckeye84 Thread Starter New Member

May 12, 2008
2
0
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???

4. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Take a look at the material in the contained in the AAC ebook on the topic of boolean algebra.

hgmjr

5. ### mik3 Senior Member

Feb 4, 2008
4,846
63
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

Last edited: May 13, 2008
6. ### Ratch New Member

Mar 20, 2007
1,068
3
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

Last edited: May 13, 2008
7. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Ohh yes you are right

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