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GATE | GATE CS 2010 | Question 65

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The Boolean expression for the output \’f\’ of the multiplexer shown below is:

 


(A)

(P(XOR)Q(XOR)R)\’

(B)

P(XOR)Q(XOR)R

(C)

(P+Q+R)\’

(D)

P+Q+R



Answer: (B)

Explanation:

For 4 to 1 mux truth table SEL                  INPUT                 O/P

Q P R R’ R’ R F
0 0 X X X 1 1
0 1 X X 1 X 1
1 0 X 1 X X 1
1 1 1 X X X 1

 

In the 1st column there are 4 NOR Gates, number them as 1 to 4 ( top to down).

In the 2nd column there are 2 NOR Gates, number them as 5 and 6 ( top to down).

In the 3rd column there is only 1 NOR Gate, number it as 7.

1st numbered Gate gives output as : ( P + Q )’
2nd numbered Gate gives output as : ( Q + R )’
3rd numbered Gate gives output as : ( P + R )’
4th numbered Gate gives output as : ( R + Q )’

5th numbered Gate gives output as :
(( P + Q )’ + ( Q + R )’)’
= ((P + Q)” . ( Q + R )”) ( De Morgan’s law)
= (P + Q ) . ( Q + R ) ( Idempotent Law, A” = A)
= (PQ + PR + Q + QR )
= (Q(1 + P + R) + PR) 

=> Q + PR ( as, 1 + ” any boolean expression” = 1 )

Similarly 6th numbered Gate gives output as : R + PQ   (as this time R is common here)

Now 7th numbered Gate gives output as :
((Q + PR) + (R + PQ))’
= (Q( 1+P) + R(1+P))’
= (Q+R)’ 


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Last Updated : 14 Feb, 2018
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