How to Simplify Boolean Algebra Expressions

[Kbob08]

I just started Digital Systems coursework and it seems our professor felt like giving us a book that has no examples and progresses the class with no examples as well. So basically we are stuck trying to learn Boolean algebra reduction with no real guidence.

Great.

Anyways, I'm looking for some help:

Homework Statement

(x'y'+z')' +z +xy+ wz


2. The attempt at a solution

all I can see is this:

(x'y'+z')' +z +xy

 

[berkeman]

Have you learned about Karnaugh Maps yet? That's a handy visual tool for boolean algebra reduction:

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

 

[Kbob08]

Sorry, I'm having a very difficult day (IE trying to do all my homework and NONE of it, it seems, I can accomplish, either to my shoddy math skills or lack of guidence).

Anyways,

I really just need how to deal with that first expression:

(x'y'+z')'

There isn't an example in the book and the one example in the lecture notes doesn't include any sort of situaion. I have been trying to guess at it:


[(z+x')(z+y')]'+z +xy
(z+x') + (z+y')+z +xy
z +x' + y'+xy

The book says I should get x + y + z as my answer, but I don't see it. I assume we will learn those maps next since it's in the book, but I'm pretty sure this is the type of prof that wants it done his way and when he says it.

 

[berkeman]

To work on (x'y'+z')', use DeMorgan's Laws:

http://en.wikipedia.org/wiki/De_Morgan's_laws

first expand the x'y' term (express it as an OR of two terms), and then apply the appropriate DeMorgan's Law to take the NOT of the whole expression.

 

[chickens]

I just started Digital Systems coursework and it seems our professor felt like giving us a book that has no examples and progresses the class with no examples as well. So basically we are stuck trying to learn Boolean algebra reduction with no real guidence.

Great.

Anyways, I'm looking for some help:

Homework Statement

(x'y'+z')' +z +xy+ wz


2. The attempt at a solution

all I can see is this:

(x'y'+z')' +z +xy

(x'y'+z')' +z +xy+ wz
= use de morgan's law, (x.y)' = x' + y' | (x+y)' = x'.y'
= take out the common factor, anything +1 = 1
= z.1 + xy
= z + xy

hope this helps

 

[berkeman]

(x'y'+z')' +z +xy+ wz
= (x'y')'.z + z + xy + wz use de morgan's law, (x.y)' = x' + y' | (x+y)' = x'.y'
= z[(x'y')'+ 1 + w] + xy take out the common factor, anything +1 = 1
= z.1 + xy
= z + xy

hope this helps

I'm going to leave this solution for now because I think Bob has worked it out by now. I'm going to issue chickens a 0-point warning for posting a complete solution to a homework problem, however.

 

[chickens]

I'm going to leave this solution for now because I think Bob has worked it out by now. I'm going to issue chickens a 0-point warning for posting a complete solution to a homework problem, however.

didnt know i can't do that, really sorry, next time will post a guided solution rather than a complete one

 

[berkeman]

didnt know i can't do that, really sorry, next time will post a guided solution rather than a complete one

No worries. The PF is a special place. I have a feeling that you'll fit in well here.

 

[MathG]

hi you can use one very good programs for that .
http://www.phoenixbit.com/site/products.asp?productid=karnaughanalyzer

with that program you can get even the circuit.