# Boolean expression simplification

Discussion in 'Homework Help' started by KTD108, Sep 14, 2008.

1. ### KTD108 Thread Starter New Member

Sep 14, 2008
4
0
Boolean expression simplification confuses me beyond belief. Please help...

1) F=(AC+(B'D)')'(E+D')'+C
2) F=((BD)'(E XNOR C)'+CA)'
3) A XNOR (A+B)
4) A XNOR B XNOR (AB)

My hw is due tomorrow at 10, so please help!! Thanks!!

Jan 28, 2005
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3. ### KTD108 Thread Starter New Member

Sep 14, 2008
4
0
Ok, so I figured out 3 and 4...the first 2 are still confusing me tho. This is what I have for the first one...

(AC+(B'D)')'(E+D')'+C

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

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

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

= (A'B'DE' + B'C'DE' + C

= B'DE'(A' + C') + C

But I don't know how to get it so there is only 1 C term...any advice?

4. ### silvrstring Active Member

Mar 27, 2008
159
0
KTD108,

Take your (A' + C'), and transform it to (AC)'.
You are then left with F=(AC)'B'DE' + C.

From there, you can apply the inverse of F to manipulate the equation so you can get C where you want it (i.e., F'=((AC)'B'DE' + C)'

When you get it to a form that only has one of each term, you can take F'' = F to find your final simplified equation. If I'm right, you will end up with only 4 terms.

Let us know how your progress goes.

5. ### KTD108 Thread Starter New Member

Sep 14, 2008
4
0
Hmm..ok I must be doing something wrong then. I'm supposed to draw the CMOS circuit for:

(AC+(B'D)')'(E+D')'+C

And the only way I know/understand how to draw the CMOS circuit is to simplify the equation and then draw it. I think all 5 terms should still be used though. Is there an easier way to draw the circuit without simplifying?

6. ### silvrstring Active Member

Mar 27, 2008
159
0
KTD108,

I can't help you with the CMOS drawing, but I'll show you how I simplified the Boolean eqn. You can then look it over to see if it all adds up.

If F = B'DE'(A' + C') + C then F = (AC)'B'DE' + C
F' = ((AC)'B'DE' + C)' = ((AC)'B'DE')' C'
= ((AC)'' + B'' + D' +E'') C'
= (AC + B + D' + E) C'
= A0 + BC' + D'C' + EC'
= BC' + D'C' + EC'

Now F'' = F = (BC' + D'C' + EC')' = (BC')' (D'C')' (EC')'
= (B' + C)(D + C)(E' + C)
= (B'D + B'C + CD + C)(E' + C)
= B'DE' + B'DC + B'CE' + B'C + CDE' + CD + CE' + C
= B'DE' + C(B'D +B'E' + B' +DE' + D + E' +1)
= B'DE' + C (because Any + 1 = 1)

The algebra makes sense, but you'll want to look over it, and make a truth table to compare to the one you started with.

Hope this helped. Good luck on number 2.