0
$\begingroup$

XOR by NAND and NOR implementation How do we demonstrate using Boolean algebra that these NAND and NOR gate combinations are XOR gates?

$\endgroup$
  • $\begingroup$ By boolean algebra, do you mean something like $\overline{\overline{A\overline{AB}}\,\overline{\overline{AB}B}} ={A\overline{AB}+\overline{AB}B} =A(\overline{A} + \overline{B})+(\overline{A} + \overline{B})B =A\overline B+\overline{A}B $? $\endgroup$ – Apass.Jack Apr 16 at 12:12
  • $\begingroup$ Yes, thankyou @Apass.Jack $\endgroup$ – Shane Apr 16 at 13:20
  • $\begingroup$ $\overline{\overline{A + \overline{A+B}}+\overline{\overline{A+B}+B}} =(A+\overline{A+B})(\overline{A+B}+B) =(A+\overline{A}\,\overline{B})(\overline{A}\,\overline{B}+B) =(AB+\overline{A}\,\overline{B})$. Then, $\overline{(AB+\overline{A}\,\overline{B}) + (AB+\overline{A}\,\overline{B})} =\overline{(AB+\overline{A}\,\overline{B})} =\overline{AB}\,\overline{\overline{A}\,\overline{B}} =\overline{AB}(A+B) =(\overline{A} + \overline{B})(A+B) =\overline{A}B + \overline{B}A =A\overline{B} + B\overline{A} $ $\endgroup$ – Apass.Jack Apr 23 at 9:44
1
$\begingroup$

The nice thing about small Boolean circuits like this, is there are only a few possible inputs.

With two input wires, there are $2^2=4$ possible inputs to your circuit: $00, 01, 10, 11$. You can test each one and write down its output. Then compare that to the XOR gate's outputs.

If the two line up, then the circuits are equivalent! And if they're not, something has to be adjusted.

$\endgroup$
0
$\begingroup$

Just write down what the output of every circuit is. In the first example, the left circuit is X = NAND (a, b), top middle is Y = NAND (a, X), bottom middle is Z = NAND (X, b), right = NAND (Y, Z).

Substitute everything into the eqation for the right circuit, then use Boolean algebra.

$\endgroup$

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.