Virtual Gate Test Series: Digital Logic - NAND Gates

Question

The minimum number of $2$-input $NAND$ gates required to implement the function $F = (x' + y')(z + w)$ is ______

Comments

Answer is $4$. This question is already asked in Gateoverflow (search)

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plz give a linK??

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See this

Answer 1

Total 4 gates require as the expression can be minimized to( ((xy)'z)'((xy)'w)')'

Answer 2

NAND means we require AND-OR combination but the given expression is in OR-AND so complement the expression, hence one NAND gate for this,

now after complementing the expression becomes

f ' = (x ' + y ' ) ' + ( z + w ) '

=(xy)+(z'w')

this is of the form AND-OR realization and requires 3 gates.

Hence in all we require 4 NAND gates.

Correct if wrong.

Comments

Answer is correct, but your solution is wrong. (Try to realize it)

Answer 3

Ans is 4

F = (x' + y')(z + w)

=(xy)^c(z+w)       take (xy)^c=P

=Pz + Pw

and Nand -nand reliazation is (And  Or ) reliazation

Answer 4

understood. First we have to convert it in suitable form for NAND gate and applying De Morgan law we can convert it .total 4 nos NAND gates will be required.