problem with some gate

Thread Starter

roeyhaim

Joined Nov 22, 2010
6
hello guys.
I'm stuck with a problem and need help...

This is the Q:
Introducing new operator :) ("smile")
with the following logic: A:)B = A'B
Express function F(x,y,z) =∑(4,6,7) using only :)
operator and construct a circuit diagram for this function using ONLY "smile" gate.
I tried for all night and didn't get it....
any suggestion?
 

thatoneguy

Joined Feb 19, 2009
6,359
Where are you stuck?

\(A\smile B= A\overline{B}\)

Apply that to the function, have you had previous examples of applying logic to a function?
 

Georacer

Joined Nov 25, 2009
5,182
You did well simplifying the function. Your task from now on is to convert any single operation to a smile operation. Start with a factorization.

Here are some hints:
a+b'=(a'b)'
x'=(x'1)

Can you figure it out?
 

renotenz

Joined Oct 25, 2010
11
well i tried to simplify the function.
I got F(x,y,z) = xy + xz'

but i dont know how to continue...
You can still simplify that function actually... there's one common letter there

After that you shall see a pattern which Georacer already wrote.

Also remember that : A'B = BA'

Cheers :) <= Smile function for you :D
 

Thread Starter

roeyhaim

Joined Nov 22, 2010
6
I'm sorry, but i still don't get it...

I give up...

My brain is gonna explode.

Thanks for trying :)

Now I scared from the yellow smile :D
 

Georacer

Joined Nov 25, 2009
5,182
Don't give up yet. Try a bit more. You have problems doing the factorization? Finding the part to be substituted?

Post any scribbles you have done, even if they are completely wrong. Don't let a good thread end without a closure.
 

Thread Starter

roeyhaim

Joined Nov 22, 2010
6
OK...
but please dont laugh at me...

i thought maybe to put the x out and then to play a little with the y

maybe to do some demorgan.

but the Q is what to do with the +

any hint / advise?
 

renotenz

Joined Oct 25, 2010
11
There it is.

Put X out, then you'll get X(Y + Z')

See any resembelance with what Georacer has typed?

Don't forget that your :) is A'B, same as BA', same as X'Y, YX', etc...
 

Thread Starter

roeyhaim

Joined Nov 22, 2010
6
so, tell me if im right

now i got X(y'z)'

and put this in circuit :

X:)1 = X'
Y:)Z = Y'Z
Y'Z:)1 = (Y'Z)'
X':)Y'Z = X''(Y'Z)'

X''(Y'Z)' = X(Y+Z')

Am I right?
This is the solution?
 

Georacer

Joined Nov 25, 2009
5,182
Not quite.
Y'Z:)1 = (Y'Z)' is correct only with parentheses: (Y'Z):)1 = (Y'Z)' and
X':)Y'Z = X''(Y'Z)' is wrong.
You have also written down some unneded equations. So let's sum it up:

xy+xz'=
x(y+z')=
x(y'z)'=
(y'z)'x=
...

Can you take it from here and substitute for the :) operation? Make one step at a time and start from inside out.
 

thatoneguy

Joined Feb 19, 2009
6,359
From the original question, the solution should have the :) operator only, so go through the steps to substitute it, and leave it (and ONLY the :)) in for the final solution.
 

Thread Starter

roeyhaim

Joined Nov 22, 2010
6
sorry but im lust again...
I feel so :eek:

What i need is to figure out how to get
X(Y'Z)' by input to a :) gate only two variables?

with no other gates?
just ONE AND ONLY :) gate?
 
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