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The output $F$ of the below multiplexer circuit can be represented by

  1. $AB+B\bar{C}+\bar{C}A+\bar{B}\bar{C}$
  2. $A\oplus B\oplus C$
  3. $A \oplus B$
  4. $\bar{A} \bar{B} C+ \bar{A} B \bar{C}+A \bar{B} \bar{C}$
in Digital Logic by Boss (30.2k points)
edited by | 830 views
0
An easy would be to convert the expression obtained to 0s and 1s. And then, you can notice the fact that there are all possible combinations of odd  $\#1$s in each of the terms. Thus it would be the XOR of the inputs.

1 Answer

+13 votes
Best answer
Answer is B)

$A'B'C+AB'C'+A'BC'+ABC$

$=(A+B')(A'+B)C+A'BC'+AB'C'$

$=$$A\oplus B\oplus C$
by Veteran (117k points)
edited by
0
Hey, can you explain your answer, please? Shouldn't it be A'B'C + A'BC' + AB'C' + ABC?
0
yes, that is
0
yes, it's A⊕B⊕C,  ex-or is an odd function, it means number of boolean variables in un-prime form will be odd.
0
Can you explain what you mean by un-prime form here?
+1
Thehobo03
A variable has two forms p(non-prime) p'(prime (or) complemented). in exor function there will always be odd number of non-prime variables in each minterm,
0

can you please explain how you got

=(A+B')(A'+B)C+A'BC'+AB'C'

by using

A'B'C+AB'C'+A'BC'+ABC

I am unable to understand this part can uh  please explain

@srestha

 

0
@srestha how to know if A is lsb or B is lsb for the select lines?
0
It is like that if not given than you have to take MSB RHS side i.e. B here and LSB LHS i.e. A.
+1
yes, just general assumption
+2

@air1ankit

it is simple

$A'B'C+ABC=\left ( A\odot B \right )C$

Now $\left ( A\odot B \right )=\left ( A '+B\right )\left ( A+B' \right )$

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