simplifying boolean expression

A

ant9985

New Member

• Apr 26, 2005

• #1

hi, is there anyone that can simplify this boolean expression algebraically?

PG'CK + PGCK' + PGCK

i know how to simplify some expressions but this on im real stuck on :?
Changing it to PC(G'K+GK'+GK) is about as far as i can get really, any help would be appreciated, thnx

 

evandude

New Member

• Apr 26, 2005

• #2

well think about it... the term in parentheses is true if GK=01, 10, or 11... so in other words it is true if EITHER of G and K are true. so it should reduce to just (G+K)

So the whole thing is PC(G+K)

 

mechie

New Member

• Apr 26, 2005

• #3

Boolean simplification

PC(G'K+GK'+GK) ...

if I add another set of brackets and swap the order of the last two terms around ...
PC(G'K+GK'+GK)
PC((G'K+GK)+GK')
is that implying that G is a 'don't care' ?

 

evandude

New Member

• Apr 26, 2005

• #4

you can do the same thing with K. simply put the GK term in there twice (doesn't change anything)... then you end up with ((G'K+GK)+(GK+GK')) which makes G a don't care for the first expression, and K a don't care for the second... in other words the whole thing is (K+G) like i said before.

 

Phasor

Member

• Apr 26, 2005

• #5

evandude said:

So the whole thing is PC(G+K)

Click to expand...

I concur with evandude - It is proven quite simply with a Karnaugh map.

 

E

eeyouth

New Member

• Apr 26, 2005

• #6

Here is another solution

PG'CK + PGCK' + PGCK
PG’CK + PGC(K’ + K)
PG’CK + PGC
PC(G’K + G) ; G’K + G = G + K
PC(G + K)

 

A

ant9985

New Member

• Apr 27, 2005

• #7

oh cool, thnx a lot for the help all