# Boolean algebra problem...

Discussion in 'Homework Help' started by mystero, Jan 21, 2011.

1. ### mystero Thread Starter New Member

Jan 21, 2011
5
0
x'y'z' + wx'z' + wx'y + w'x'z

I'm not sure what to do after i factor x' out... or even if thats the right thing to do in the first place

x'(y'z' + wz' + wy + w'z)

2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Err... what did you say it was that you where supposed to do again?
Oh, wait! You didn't say!

3. ### mystero Thread Starter New Member

Jan 21, 2011
5
0
"Simplify the following expression"

4. ### Wendy Moderator

Mar 24, 2008
20,993
2,731
Simplification of Boolean equation. It is a start, but you can do it again within the parenthesis. This probably should be in the homework section.

Are you supposed to show the gate arrangement after you are finished?

5. ### mystero Thread Starter New Member

Jan 21, 2011
5
0
no i just need to simplify that expression thats it but i dont know how to simplify the rest in the parenthesis...

6. ### Wendy Moderator

Mar 24, 2008
20,993
2,731
Basically same step. z' is a common denominator in two of the groups. So is w. Pick on and do the same thing.

I'm better at simplifying gates than Boolean algebra. I'm pretty rusty at the math.

7. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
I suggest making a Karnaugh map and simplifying it. It is a more algorithmic and visual approach.

8. ### mystero Thread Starter New Member

Jan 21, 2011
5
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we havent done those yet

9. ### Wendy Moderator

Mar 24, 2008
20,993
2,731
He is saying the same thing I was. You will study it, and soon. Basically it is a way of picking gates from within a truth table.

I have a favorite illustration on how to simplify gates once drawn. It may be too soon for you, but it will eventually come in handy.

I also have a set of templates, including gates, I use to draw logic for online posting (like above). You can download PaintCAD from my blog, which is hosted by AAC. Blogs and albums are part of the AAC site, as well as the book. I really like this site.

Bill's Index

10. ### mystero Thread Starter New Member

Jan 21, 2011
5
0
can anyone help with simplifying that expression though? what would be the next step

11. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
There is not much you can do after that. All I see is:
x'(y'z'+z'w(y+y')+yw+zw')=
x'(y'z'+yz'w+y'z'w+yw+zw')=
x'(y'z'(1+w)+yw(1+z')+zw')=
x'(y'z'+yw+zw').

As I said, without a Karnaugh map, this doesn't make much sense.

12. ### Wendy Moderator

Mar 24, 2008
20,993
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It is enough to make using gates though.

13. ### Buga New Member

Jan 22, 2011
4
0
x'y'z'+wx'z'+wx'y+w'x'z
x'y'z'+wx'y+w'x'z (wx'z' is consensus of x'y'z' and wx'y)

And that's it. It does not get any simpler. I verified it with Quine-McCluskey which you will undoubtedly learn as the primary form of boolean simplification used by computers, and it is guaranteed to get the minimal expression.

14. ### Wendy Moderator

Mar 24, 2008
20,993
2,731
Not necessarily. The solution showed in post #11 used qty 4 2 input gates, qty 1 3 input gates, and 4 inverters.

Yours used qty 4 3 input gates and 4 inverters. You are going to find digital electronics is not so cut and dried as you think. Your number of gates are fewer, but what if all you have is 2 input gates?

15. ### Buga New Member

Jan 22, 2011
4
0
His homework is probably just looking for boolean simplification, not necessarily simpler gate things. Introductory coursework usually concentrates on simplifying boolean expressions before gate simplification. Obviously other solutions are better for getting a minimal cost circuit design. And I totally would not implement such a function unless there was something like a PLA or PAL available.

16. ### Anamaria New Member

Feb 4, 2011
1
0
Hey everyone...

I have to find the dual function for
f=x1x2'+x1'x2x3'
I don't understand what a dual function means.

Can you help me?

Thanks

17. ### Papabravo Expert

Feb 24, 2006
11,083
2,159
The original expression is in a standard form called "sum of products". Each of the product terms has three of four boolean variables. In order to "simplify" the expression you need to reduce either the number of terms or the number of variables in some terms. If you write out a truth table you should be able to spot a pattern that will suggest simplification if it is possible. Otherwise, transforming the expression may not count as a "simplification".

The other standard form is called the "product of sums". It may result in a simplification or not depending on your point of view.

Because two of the terms overlap I see the simplification being:

w'x'z + x'y'z' + wx'y

which covers the input set.

Last edited: Feb 4, 2011