0
\$\begingroup\$

I'm looking for ways to simplify a configuration in which the output of an AND gate flows into an XNOR gate along with a third variable.

The function of such a configuration is as follows;

f(a,b,c) = (a*b) XNOR c

Using boolean algebra, how would you write that expression out so that it could be easily simplified?

A schematic is pasted in below;

enter image description here

Cheers

\$\endgroup\$
  • \$\begingroup\$ Can you write a truth table for the whole expression? \$\endgroup\$ – The Photon Oct 8 '16 at 1:43
  • 4
    \$\begingroup\$ Define cheaper. Define simpler. What's the price of an AND gate, a NOT gate, a 2-in MUX, and an XNOR? And does your logic include Q and Q' outputs? The expression is pretty easy, though: F = A' C' + B' C' + A B C = (A B)' C' + A B C \$\endgroup\$ – jonk Oct 8 '16 at 2:08
  • \$\begingroup\$ @jonk - I erased the comment where I said you did a mistake because you didn't \$\endgroup\$ – Claudio Avi Chami Oct 8 '16 at 7:02
3
\$\begingroup\$

This is the truth table: enter image description here

As you can see, it shows you how will the output react to different combinations of inputs. The "BOOLEAN" section shows the conditions for which that particular combination yields an output of 1 or TRUE. In other words, that is the algebraic expression for a TRUE output.

Now if any of these BOOLEAN expressions is 1, the output is also 1, so when you write f, you will put a + which signifies OR between them and obtain the rule for the circuit's behaviour.

Next you will simplify the equation. Take a look at this page. Find the appropriate rule and then use it to get your result. Don't worry, it should be headache free.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thank you voters, I appreciate that you considered sending feedback to me \$\endgroup\$ – Daniel Tork Apr 18 '17 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.