i have a homework to simplify
F(x,y,z)=xyz'+yzy'+yz(x+y'+(yz)')+z(x+z')+x'(y+z)' to a point of two gates of two entries...

so far i have done:

xyz'+yz(x+y'+z')+xz+x'y'z'= xyz'+xyz+xz+x'y'z'=xy+xz+x'y'z'=x(y+z)+x'y'z'

any help would be appreciated

 

Are XOR gates allowed?

 

i guess the question sais two gates. i dont think there is a restriction

 

I also should have asked if complemented inputs are allowed, but that's a bit of a moot point.

If XOR gates are allowed, then it can be done using two two-input gates a couple different ways. Either with a NOR and an XOR are with an OR and an XNOR.

Working with XOR gates in Boolean logic is tricky, but doable. Look at your result thus far and see if you can work the last term into a form that is compatable with the first term as far as being of the form of an XOR logic function (i.e., a'b+ab').

It might help to look at the truth table for the function you are implementing and then keep in mind that an XOR gate can be thought of as a programmable inverter.

z = x XOR y

Is the same as

IF x=0 THEN z=y ELSE z = y'

 

ive reached to x(y+z)+x'(y+z)'

 

It would be helpful if you showed your work in getting to that point. That way, we can determine the path you are taking and provide appropriate guidance.

 

ive reached to x(y+z)+x'(y+z)'

That's correct.

You really need to show your work because, if it hadn't been correct, it would have been all but impossible to give you any useful feedback to help you find and fix your error.

Now, what two fuctions, f and g, is the above expression the XOR of?

 

finally found the xor a'b+ab'
a=x
b=(y+z)'
thnx for the help guys

 

Very good!

Now, we could have said

a = x'
b = (y+z)

But then we would have needed three gates -- an XOR, an OR, and a NOT. But for how you've set it up, you just need an XOR and a NOR.

The other way we could have done it would have been to use the XNOR, whose defining expression is

ab + a'b'

In which case you would have had

F = x XNOR (y OR z)

It may not seem like it, but forcing you to find it on your own has almost certainly taught you more that you would have learned had we just given you the answer and explained it.

Good job on accepting the challenge and putting forth the effort to slog through it.

 

thank you very much for the help and guidance guys

 

Hey, you did all the work. Thank YOU for being willing to do so!

 

i have a homework to simplify
F(x,y,z)=xyz'+yzy'+yz(x+y'+(yz)')+z(x+z')+x'(y+z)' to a point of two gates of two entries...

so far i have done:

xyz'+yz(x+y'+z')+xz+x'y'z'= xyz'+xyz+xz+x'y'z'=xy+xz+x'y'z'=x(y+z)+x'y'z'

any help would be appreciated

Although LampriX got it solved, I thought it might be useful to others that run across this thread to see the solution method.

F(x,y,z)=xyz'+yzy'+yz(x+y'+(yz)')+z(x+z')+x'(y+z)'

The first thing to note as that anytime we have something of the form xx' we can set any factor in which it is a part equal to 0. Hence the second term, yzy', goes away immediately. If we then distribute yz across the three term sum that it is ANDed with, we get

F(x,y,z)=xyz'+yzx+yzy'+(yz)(yz)'+z(x+z')+x'(y+z)'

Here we can see that the third term and the fourth term both go away for the same reason.

F(x,y,z)=xyz'+yzx+z(x+z')+x'(y+z)'

We can then factor out xy from the first two terms to get

F(x,y,z)=xy(z+z')+z(x+z')+x'(y+z)'

And since (z+z') is 1, this reduces by

If we distribute the z we get xz + zz', which becomes just xz, leaving us with

F(x,y,z)=xy+xz+x'(y+z)'

We can now factor out the x from the first two terms leaving us with

F(x,y,z)=x(y+z)+x'(y+z)'

At this point, we might be able to recognize this as A EQU B, which is also known as A XNOR B. If we do, this becomes

F(x,y,z)=x XNOR (y OR z)

And this is an acceptable solution. If we don't spot this, then hopefully we can spot that we basically have an XOR gate. We can force it into the form we need by double inverting the (y+z) factor in the first term.

F(x,y,z)=x[(y+z)']'+x'[(y+z)']

And now we have A XOR B were A=x and B=(y+z)', which becomes

F(x,y,z) = x XOR (y NOR z)