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

 

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

 

Ok cool,

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

 

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

 

...yeh that would be the question

 

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')

 

Alexei Smirnov,

A'B')*(C'D') stop in this form (A'+B')(C'+D')

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

 

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.

 

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?