Boolean algebra

Hi guys.. Well ive looked at my lecture notes and well they have examples and they are veryyyyyyyyy basic basically i get the concept but the examples are too simple to understand Something difficult...
So i got this question for Tutorial and I basically dont understand of it... I was wondering if u guys could help me out

Simplify the following Boolean expressions. In each case, express the result in Sum of Products form.

(a) PQ’ + Q’P + RST + R’ST + TR’S
(b) AB’ ( C + D’ ) + AB’C’D

Any feedback i'll be very greatful for.... even maybe notes you may wanna direct me too which are good would be greatly appreciated.. thanks in advance... I'm not sure if this is the right section so excuse me if it is in the wrong section :)

vitabrits

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Are P, Q, A and B individual data elements? If they are then the first

a. (P*Q) + (Q*P) + (R*S*T) + (R*S*T) + (T*R*S)

I don't know the significance of the ' character, so the above may not be completly correct.

Ancient Dragon

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I'm kind of confused as to what the ' character is too... Are you sure it isn't one of the logical AND or NOT characters? Did you copy the problem down right?

http://en.wikipedia.org/wiki/Boolean_algebra

-Fredric

Daishi

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Yup thats exactly how its worded in the Work sheet! well nvm that Website link you gave me was extremely helpful! i was able to do it through that many thanks :)

vitabrits

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(a) PQ’ + Q’P + RST + R’ST + TR’S
P'Q+ST(R+R'+R')
PQ' and Q'P are equivalent expressions, and redunancy is discarded.
R+R' (R or not R) is always true, so:
PQ'+ST+R'ST
Additionally,
PQ' + ST(1+R') == PQ' + ST

(b) AB’ ( C + D’ ) + AB’C’D
AB'C+AB'D'+AB'C'D==AB'(C'+D'+C'D)==AB'(C'(1+D)+D')==AB'(C'+D')==AB'C'+AB'D'

I freakin' love logic. It's really handy to get some philosophy credits and it's practical in application. But just wait until you get to K-maps. They're made of pure Boolean awesome.

Drowzee

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Nerd ;)

prog-bman

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