How to we do Boolean expreesion.

WBahn

Joined Mar 31, 2012
26,299
Can anyone explain how we do boolean expression. And how to do it.
One of the first steps should be to get the original circuit into nothing but one- and two-input gates. Can you change the 3-input AND gate into a subcircuit that uses only 2-input gates (doesn't matter what type).

Next is to translate the different one- and two-input gates into equivalent subcircuits that only use 2-input NAND gates. You can do those off to the side one at a time. At the most basic Boolean expression level, you only have three gates to worry about - NOT, OR, and AND. Can you make each of those out of just 2-input NAND gates? If not, that is the first thing you need to focus on.
 

bertus

Joined Apr 5, 2008
20,878
Hello,

I have taken your picture, cut out the essential part, adjusted brightness and contrast.

Ricky_sum_boolean.jpg

Please cut away any unwanted parts next time.

Bertus
 

WBahn

Joined Mar 31, 2012
26,299
Check your expression for Z that you have drawn on the diagram. Hint: The middle gate is not and OR gate.
 

thumb2

Joined Oct 4, 2015
122
I am not sure if you're expression for the NOR output is right.

You have a NOR with both its input inverted.
So it should be \((\overline{\bar A + \bar B})\).
Making a thruh table for the 2NOT -> NOR Gate I get:

\(\begin{array}{c c | c }
A & B & \overline Y\\
0 & 0 & 0\\
0 & 1 & 0\\
1 & 0 & 0\\
1 & 1 & 1
\end{array}
\)

where A and B are the non inverted inputs. So The final output can be reduced to

\(Y = (A \wedge B \wedge B)\)

For a & b ceck out NAND logic and NOR logic.
 
Last edited:

dl324

Joined Mar 30, 2015
11,885
\(Y = (A \wedge B \wedge B)\)
There is no Y in the original problem. If you meant Z, you dropped C.

This is the Homework Help forum and members are expected to provide guidance to the OP so they can learn how to do it themselves.
 
Last edited:
Top