The Gateway to Computer Science Excellence
+3 votes

Q,R,S are true for sure but how p is true ???

in Digital Logic by Boss (20.1k points)
retagged by | 228 views
I am not sure about the associativity of XNOR ...for Three input.

$\begin{align*} \text{Let Q} &= A \odot B \odot C \\ &= \underbrace{\color{red}{A \odot B}} \; \odot \; C \\ &= {\color{red}{X}} \odot C \\ &=\overline{{\color{red}{X}}}.\;\overline{C} \;\; + {\color{red}{X}}.\;C \\ &= {{\bf \color{red}{\left ( A \oplus B \right )}}}.\;\overline{C} \;\; + \overline{{\bf \color{red}{\left ( A \oplus B \right )}}}.\;C \\ &={\color{blue}{\bf P}}.\;\overline{C} \; + \; \overline{{\color{blue}{\bf P}}}.\;C \\ &= {\color{blue}{\bf P}} \oplus C \\ &= A \oplus B \oplus C \\ \end{align*}$

but this does not resolve the above QS.
Both XOR and XNOR are associative.
XOR AND XNOR Both are Associative and Commutative

2 Answers

+2 votes
Best answer

Yes, for odd number of variable XOR gate behaves similar to XNOR gate.

  • Output of $\oplus$ is $1$ for odd number of $1's$

Ex: $0\oplus1\oplus1\oplus1\oplus0 = 1$ (odd number of 1's)

  • Output of $\odot$ is $1$ for even number of $0's$

Ex: $1\odot1 = 1$ (0 number of $0's$) and $1\odot0\odot0\odot1 = 1$(even $0's$)

Input to a gate can either be $0$ or $1$. So, If the number of input variables is odd. then either $0's$ as input are odd and $1's$ as input are even (or)  $0's$ are even and $1's$ are odd.

I mean to say if a XOR gate takes odd number of inputs, then either 0's are odd (or) $1's$ are odd.

If number of input variables is odd and $1's$ in it are odd, then output of XOR gate is $1$ (remember XOR gives $1$ for odd number of $1's$ in input), and if $1's$ are odd, then it contains even number of $0's$, which implies that output of XNOR is also $1$(XNOR gives $1$ for even number of zeros)

If number of input variables is odd and $0's$ are odd, then output of XNOR is $0$ (XNOR is $1$ for even number of $0's$), thus number of $1's$ in input is even, and for even number of $1's$ in input XOR gives $0$.

Although it has become bit complicated but It can be seen from above two cases that when number of input variables to a XOR and XNOR gate are odd, they behave exactly same.

So, for even number of inputs $\color{red}{\oplus = \overline{\odot}}$ and odd number of inputs $\color{red}{\oplus = \odot}$

by Boss (28.8k points)
selected by

Hello, I have another similar question, where they are considering exnor to be complement of exor for any number of inputs even or odd. And hence answer for this question is given as D.

This link too says exnor is defined to be negation of exor.

Kindly help as to what is definition of exnor which is valid for any number of inputs.

0 votes

all option are correct.

by Boss (36.7k points)
edited by

Related questions

0 votes
2 answers
asked Jan 16, 2017 in Digital Logic by firki lama Junior (681 points) | 277 views
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,737 questions
57,384 answers
105,340 users