GATE CSE 1987 | Question: 1-IV

Question

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

• A. $AB+B\bar{C}+\bar{C}A+\bar{B}\bar{C}$
• B. $A\oplus B\oplus C$
• C. $A \oplus B$
• D. $\bar{A} \bar{B} C+ \bar{A} B \bar{C}+A \bar{B} \bar{C}$

Comments

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.

Answer 1

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$

Comments

Hey, can you explain your answer, please? Shouldn't it be A'B'C + A'BC' + AB'C' + ABC?

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yes, that is

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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.

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Can you explain what you mean by un-prime form here?

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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,

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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

 

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@srestha how to know if A is lsb or B is lsb for the select lines?

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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.

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yes, just general assumption

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@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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how have you assumed msb and lsb?

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@Abhineet Singh If nothing mentioned then by default rightmost is LSB and leftmost is MSB.

Answer 2

Option B

Put c=1

Comments

cool

thanks

Answer 3

(B)

You wont find any explanation like this