GATE CSE 2019 | Question: 50

Question

What is the minimum number of $2$-input NOR gates required to implement a $4$ -variable function expressed in sum-of-minterms form as $f=\Sigma(0,2,5,7, 8, 10, 13, 15)?$ Assume that all the inputs and their complements are available. Answer: _______

Comments

@mrinmoyh 4 NOR gates are required when complemented inputs are not available.

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Here important statement given in question is:

Assume that all the inputs and their complements are available.

If complements are NOT available then minimum $4$ NOR gates($2$ – input) are required to implement Ex – NOR.

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Fascinated

Answer 1

K-Map for given minterms $\sum(0,2,5,7,8,10,13,15)$ is:

$f$= $bd+b'd'= (bd+b')(bd+d')=(b'+d)(b+d')$

It is given that inputs and their compliments are available, following is the implementation of $f$ using three $2$ input NOR gate:

Hence 3 is the correct answer

Comments

@Shashi Shekhar 1

Question given that sum of minterms(sum of product term )but they want to implement using only NOR GATE

Implement using maxterm or minterm answer should be same they only give minterm like $f(A,B,C,D)=\sum(0,2,5,7,8,10,13,15)$

If we can implement only NAND GATE then minterm is useful, but we can also represent like $f(A,B,C,D)=\Pi(1,3,4,6,9,11,12,14)$ if we implement using only NOR GATE then MAxterm will be useful.

So, the answer is $3$ see my answer below

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@Lakshman Patel RJIT

Correct answer is 3 not 4...

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@ akash.dinkar12

Yes, the answer should be ${\color{Orchid}3 }$, because they give both inputs (pure from and ${\color{Red} {complemented }}$  form) and wants ${\color{DarkOrange}{minimum} }$ ${\color{Magenta}{NOR} }$ GATES.

Answer 2

Answer is 3, just use de moragan' s law in this question. 

Answer 3

 

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