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Sep 19th, 2005

Boolean algebra

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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

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vitabrits

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Sep 19th, 2005

Re: Boolean algebra

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.

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Re: Boolean algebra

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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Re: Boolean algebra

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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Re: Boolean algebra

(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.

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Re: Boolean algebra

Nerd

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Re: Boolean algebra

You say that now, but it's really cool how you can grid out a logic equation and make it really efficient, or discover if it's not going to work...

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