GATE CSE 1997 | Question: 71

Question

Let $f=(\bar{w} + y)(\bar{x} +y)(w+\bar{x}+z)(\bar{w}+z)(\bar{x}+z)$

• A. Express $f$ as the minimal sum of products. Write only the answer.

• B. If the output line is stuck at $0$, for how many input combinations will the value of $f$ be correct?

Comments

@bikram sir what is the meaning of second point .m not getting this k map

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@ set2018 

second point want to say , 

first find the f as the minimal sum of products , now consider that the output is 0 , now find for how many input combinations you are getting output as 0.

output line is stuck at 0

means output is 0 , i.e f = 0.

Answer 1

Answer of question A: $w'x'+yz$

Answer of question B:

Stuck at $0$, means output is fixed at $0$ (No matter what the input is). We got $0$ for $9$ input combinations (Check K-Map). So, answer is 9.

Comments

take two times complement and reduced one complete using demorgans law then u will get sum of product(min term) after that take complements of min terms

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In case anyone wondering how to solve without K-map

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

$f'=wy'+xy'+w'xz'+wz'+xz'$

      $=wy'+xy'+wz'+xz'(w'+1)$

      $=wy'+xy'+wz'+xz'$

      $=y'(w+x)+z'(w+x)$

      $=(w+x)(y'+z')$

$(f')'=x'w'+yz$

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Our SOP is: $w’x’ + yz$. To get output 0, both the first term and second term should be zero

$w’x’ $ can be zero in three ways. Similarily, $yz$ can be zero in three ways.

Total:   $3 * 3 = 9   $ ways.