# Boolean Alg Help

Discussion in 'Homework Help' started by 20FinalKonekt12, Feb 3, 2009.

1. ### 20FinalKonekt12 Thread Starter New Member

Feb 3, 2009
4
0
Its been a while since I did Boolean simplification... I know tisk tisk

Good thing I came across this sick forum.

OK, so here's the expression I got from a simple circuit:

X = ((A'B')*(C'D'))+((C'D')*(B+D))'

If someone could show me a step by step that would be great, I'm mostly getting stuck on ((C'D')*(B+D))' half of the expression.
Not sure how the simplification rules handle this? DeMorgans? Idk, but any help is greatly appreciated

Also, if someone could show me how to fill this into a kmap that would be nice too. I know how kmaps work once its all filled out, just not sure how to apply this expression to the map.

I appreciate any help, thank you

2. ### vvkannan Active Member

Aug 9, 2008
138
11
Hello 20Finalkonekt12,

you can use Demorgan's theorems for the second part[(AB)' = A'+B']
((C'D')*(B+D))' = ((C'D')' + (B+D)')

=(C'' + D'') + (B'D') = C + D + B'D'

For karnaugh mapping refer this page
http://www.allaboutcircuits.com/vol_4/chpt_8/6.html

3. ### 20FinalKonekt12 Thread Starter New Member

Feb 3, 2009
4
0
Ok cool,

so does the first half, ((A'B')*(C'D')), stop in this form:
(A'+B')(C'+D') ?

4. ### Alexei Smirnov Active Member

Jan 7, 2009
43
1
(A'B')(C'D') = (A+B)'(C+D)' ?

5. ### 20FinalKonekt12 Thread Starter New Member

Feb 3, 2009
4
0
...yeh that would be the question

6. ### Alexei Smirnov Active Member

Jan 7, 2009
43
1
It's not a question, it's true: A'B' = (A+B)'

The question was why
(A'B')*(C'D') stop in this form (A'+B')(C'+D')

7. ### Ratch New Member

Mar 20, 2007
1,068
3
Alexei Smirnov,

You do not have to use the asterisk in your logic expressions.

(A'B')*(C'D') = (A'B')(C'D') = A'B'C'D' = (A +B + C + D)'

Ratch

8. ### Alexei Smirnov Active Member

Jan 7, 2009
43
1
Shame on me... I've missed the last step:
(A'B')(C'D') = (A+B)'(C+D)' = ((A+B)+(C+D))' = (A+B+C+D)'

Still not (A'+B')(C'+D') though.

9. ### 20FinalKonekt12 Thread Starter New Member

Feb 3, 2009
4
0
Awesome, thanks for all of your help so far

So what im getting from combined posts is the the simplified function is:
(A+B+C+D)'+C+D+(BD)'

And then to applying this to a kmap:

Does this look right?