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GATE | Gate IT 2008 | Question 9

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What Boolean function does the circuit below realize ?


(A) xz+x’z’
(B) xz’+x’z
(C) x’y’+yz
(D) xy+y’z’


Answer: (B)

Explanation: There 0, 2, 5, 7 are enable input in given decoder.

Therefore, given K-Map should be as following:

So, output of above K-Map is,

= xz + x'z' 
= (x⊙z)
= x (ex-nor) z 

But, there given gate is NOR instead of OR, therefore above output will be negatated.
So, output function f is,

f = (xz + x'z')'
= x'z + xz'
= (x⊕z)
= x (xor) z 

So, option (B) is correct.

This explanation is contributed by Deep Shah.


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Last Updated : 20 Aug, 2019
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