Hi, I'm having trouble simplifying the following boolean functions:

1) wxy + wx'z' + wx'y' + w'xz' + w'xy' + w'y'z'
2) x' + wz' + wy' + y'z' + w'yz
3) x'z' + w'yz' + w'x'y + wxz + xy'z + wy'z'
4) wx + wy' + xy' + w'yz' + w'x'y

I will be drawing a logic circuit with them. I can use any gate, but I want to use the least amount of gates as possible.

Cheers.

 

What is the problem? Can't figure out the schematic (or how to draw these)?

 

Sorry, I didn't go into much detail.

So I'm using those four expressions to draw a logic circuit. I can draw the circuit with the expressions as they are now without any problem (just using and/or gates) but that uses a lot of gates. I want to have the least amount of gates as possible. I know I can draw the same expression but using nor, nand, xor, xnor etc gates, but I'm not sure how.

So basically, how can I draw a logic circuit of each function, using the least amount of gates as possible?

 

For example, I had the expression w'y + xz + wy' + x'z' and used an xor gate, an xnor gate, and an or gate to draw it. ((w xor y) + (x xnor z))

 

No. You cannot use XOR, XNOR. These are all AND.

I would use Karnaugh maps.

 

First, make the network, then simplify. Some people use Boolian Algebra, I do it graphically. The end result should be the same.

 

No. You cannot use XOR, XNOR. These are all AND.

I would use Karnaugh maps.

I've already simplified using karnuagh maps, but I know I can draw it using a combination of gates (not restricted to AND/OR)

 

So draw the 1st problem,

wxy + wx'z' + wx'y' + w'xz' + w'xy' + w'y'z'

We'll see what we can do.

 

So draw the 1st problem,

wxy + wx'z' + wx'y' + w'xz' + w'xy' + w'y'z'

We'll see what we can do.

Diagram for 1)

 

The K-map almost always gives you the least amount of logic gates and you can use the same type of gates (NOR, NAND) to build it every time, which is very cost effective. You should post a gate schematic based on you minimized by K-map function.

The use of the XOR and NXOR gate isn't algorithmically defined. It requires intuition and experience to know when to use it, but when successfully done it really saves you a lot of gates. I suggest you leave it for now.

In your case, you have four Boolean functions. Do you want to build them in the same board? If so, you can combine common factors from different functions to save gates, but this way you have to find them by manipulating the function using Boolean algebra; the K-map won't help you there.
For example, the factor wy' is common in all four functions and can be produced once and then used by all of them.
However, this method isn't algorithmic either. It resembles more to a brain puzzle and isn't advised for quick circuit design.

 

True enough. You are also ignoring the inverts on the inputs. They have to be accounted for, and count as a gate in and of themselves. Truth table may be the better way to go. It has been 3<mumble> years for me.

Redrawing what you have...

I hate to say it, but I think Boolean math may be the way to go.

wxy + wx'z' + wx'y' + w'xz' + w'xy' + w'y'z'

wxy + wx'(y+z) + w'(y'(x + z'))

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

I hate giving bad advice. Sorry about that.

 

I think I have solved the first one, but you need to post yours first. If you are stuck I'll put it up in a day. I'll also show the dead end I went down.

Of course, there is nothing that says I got it right.

 

As I said, if you want to build only one function, use the truth table and the K-map.
If you want to introduce XOR gates that might benefit your result, you have to sweat a bit more.
If you want to combine several Boolean functions in the same circuit, then you 'd better know what you 're doing.

Tell us what it is you want exactly and show some effort in that direction and we 'll propose some solutions too.

 

Edit: I did this one 1) wxy + wx'z' + wx'y' + w'xz' + w'xy' + w'y'z'

I got answer using k-map (been about 5 years since I did one) not sure it is right.

My answer has 5 gates AND, OR, XOR, AND, and NAND can anyone confirm the gates are correct?

Note: You can wait till the OP gets it.

Tim S.

 

@shahta01

Which function did you simplify? For each K-map generally there isn't one unique solution.

 

Which function did you simplify?

I added which one to the post; sorry about not doing it at first.

Tim S.