boolean algebra help

[ant9985]

hi, wasnt sure if this was right place to post this but here goes:
im having trouble minimizing this logic expression and its giving me a bit of grief as its the only thing left on the assignment to do. iv look in books and on the net to try and teach myself again but i just cant seem to understand it, if anyone could give me a step by step minimization proccess for this equation i would appreciate it very much:
the ' before the letter is meant to be a NOT sign above it

Y = 'A'B'C'DE+'A'B'CD'E+'AB'C'DE+AB'C'DE+AB'CD'E

thnx

 

[aibelectronics]

reply

It would help if you know these laws:
1) A + 'A = 1
2) A .AND. 'A = 0
3) A .OR. A = A
4) A. AND. A = A
5) 1 .AND. A = A
6) 1. OR. A = 1
7) AB .OR. 'B = A .OR. 'B
8 A .OR. A'B = A

I hope my notations are clear.
These plus a little algebra I believe would enable you minimise the equation.

 

[Russlk]

Also, look up DeMorgan's theorum.

 

[spindrah]

Y = 'A'B'C'DE+'A'B'CD'E+'AB'C'DE+AB'C'DE+AB'CD'E

pull out the NOT C and use DeMorgan's theorum
you will get something like Y='C('DE+D'E(AB+'A'B))

 

[Dean Huster]

A quick plug into a Karnaugh map will at least combine the first and third terms into 'A'C'DE, but that's all that method will get you.

Dean

 

[jimbob jones]

Well first things first. It seems that if you think this this is the right site to be posting such a question then you are obviously retarted. Secondly you sound like a complete ****** anyway. Stay away from this forum!!

Moderator.

 

[spindrah]

Well first things first. It seems that if you think this this is the right site to be posting such a question then you are obviously retarted. Secondly you sound like a complete [bleep] anyway. Stay away from this forum!!

Moderator.

why is he "retarted"??

boolean algebra is part of electronics!

 

[ant9985]

thnx for help all

@jimbob jones, i think u need to take some kind of medication

 

[Dean Huster]

"retarted" -- is that when you put the astringency back into a worn-out persimmon? Or does it have to do with a lady of the evening who is getting back into the profession after a short hiatus?

Yes, considering the the simplification of Boolean expressions is a major part of the software for implementing a lot of custom programmed ICs, I'd think it a major part of the profession/hobby. I had it as an integral part of my digital logic fundamentals course, and getting an expression into the "sum-of-the-products" form for playing with a Karnaugh map was commonplace. Unfortunately, although a Karnaugh map can work miracles on simplification, it does have its limitations, e.g., XOR expressions always hang them up into a checkerboard pattern that cannot be simplified.

Dean

 

[spindrah]

Well if you want sum of products OF
Y='C('DE+D'E(AB+'A'B))

then you get
y = 'C'DE+'CD'EAB+'CD'E'A'B