Logic functions and Minimization

Last Updated : 19 Nov, 2018

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Question 1

Which one of the following expressions does NOT represent exclusive NOR of x and y?

A	xy+x'y'
B	x⊕y'
C	x'⊕y
D	x'⊕y'

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Question 1 Explanation:

By Definition of XNOR, $x\odot y = x' y' + xy$ So Option-A is correct. Also by Definition of XOR, $x\oplus y = x' y + xy'$ Option-B is $x\oplus y' = x' y' + x(y')' = x' y' + xy = x\odot y$ So Option-B is also correct. Option-C is $x' \oplus y = (x')' y + x' y' = x' y' + xy = x\odot y$ Option-C is also correct. Option-D x'⊕y' = x''y' + x'y'' = xy' + x'y = x⊕y ≠ x⊙y Therefore option (D) is false. This explanation is provided by Chirag Manwani.

Question 2

What is the minimal form of the Karnaugh map shown below? Assume that X denotes a don’t care term.

A	b'd'
B	b'd' + b'c'
C	b'd' + a'b'c'd'
D	b'd' + b'c' + c'd'

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Question 2 Explanation:

There are two prime implicants in the following K-Map-

Prime Implicant higlighted in Green = $b'c'$ Prime Implicant higlighted in Orange = $b'd'$ So the Boolean expression is- $b'c' + b'd'$ Therefore option (B) is correct. This explanation is provided by Chirag Manwani.

Question 3

Which one of the following circuits is NOT equivalent to a 2-input XNOR (exclusive NOR) gate?

A	A
B	B
C	C
D	D

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Question 3 Explanation:

All options except D produce XOR as described below : $NOT({a \bigoplus b}) = XNOR \\ NOT(NOT(a)\bigoplus NOT(b)) = NOT(aNOT(b) + NOT(a)b) = XNOR \\ NOT(a) \bigoplus b = ab + NOT(ab)= XNOR \\ NOT(NOT(aNOT(b))(NOT(b)+a)) = NOT(NOT(ab)+ba) = NOT(XNOR)$

Question 4

The simplified SOP (Sum Of Product) form of the boolean expression (P + Q' + R') . (P + Q' + R) . (P + Q + R') is

A	(P'.Q + R')
B	(P + Q'.R')
C	(P'.Q + R)
D	(P.Q + R)

GATE CS 2011 Digital Logic & Number representation Logic functions and Minimization
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Question 4 Explanation:

See following : (P+Q'+R').(P+Q'+R).(P+Q+R') = $\prod(3, 2, 1) = \sum(0, 4, 5, 6, 7)$ gate2011A24

From the K-map, POS form is : P + Q'.R'

Question 5

The minterm expansion of f(P, Q, R) = PQ + QR' + PR' is

A	m2 + m4 + m6 + m7
B	m0 + m1 + m3 + m5
C	m0 + m1 + m6 + m7
D	m2 + m3 + m4 + m5

GATE CS 2010 Digital Logic & Number representation Logic functions and Minimization
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Question 5 Explanation:

K-map,

= PQ + QR’ + PR’ 
= PQ(R+R’) + (P+P’)QR’ + P(Q+Q’)R’
= PQR + PQR’ +PQR’ +P’QR’ + PQR’ + PQ’R’ 
= PQR(m7) + PQR'(m6)+P’QR'(m2) +PQ’R'(m4) 
= m2 + m4 + m6 + m7

Option (A) is correct.

Question 6

What is the minimum number of gates required to implement the Boolean function (AB+C)if we have to use only 2-input NOR gates?

A	2
B	3
C	4
D	5

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Question 6 Explanation:

AB+C = (A+C)(B+C) = ((A+C)' + (B+C)')' So, '3' 2-input NOR gates are required.

Question 7

In the Karnaugh map shown below, X denotes a don't care term. What is the minimal form of the function represented by the Karnaugh map?



A)   


B) 

C) 

D)

A	A
B	B
C	C
D	D

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Question 7 Explanation:

One group consists of (0000, 0010, 1000, 1010) which gives b’d’ Other group is (0000, 0001, 1000, 1001) which gives a’d’ So, solution is b’d’ + a’d’

Question 8

Given f₁, f₃ and f in canonical sum of products form (in decimal) for the circuit

A) m(4, 6)

B) m(4, 8)

C) m(6, 8)

D) m(4, 6, 8)

A	A
B	B
C	C
D	D

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Question 8 Explanation:

From logic diagram we have f=f1.f2+f3 f=m(4,5,6,7,8).f2+m(1,6,15)----(1) from eq(1) we need to find such f2 so that we can get f=m(1,6,8,15) eq(1) says we can get m(1,6,15) from f3 ,so only 8 left now from option (a,b,d) we get (4,6), (4,8) ,(4,6,8) respectively for m(4,5,6,7,8)f2 which is not required as m4 is undesired. But option (C) m(4,5,6,7,8)(6,8)+(1,6,15)

m(6,8)+m(1,6,15)
m(1,6,8,15)an

Option (C) is correct.

Question 9

If P, Q, R are Boolean variables, then (P + Q')(PQ' + PR)(P'R' + Q') simplifies

A	PQ'
B	PR'
C	PQ' + R
D	PR'' + Q

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Question 9 Explanation:

Step by step explanation :

= (P + Q’)(PQ’ + PR)(P’R’ + Q’)
= (PPQ’ + PPR + PQ’Q’ + PQ’R) (P’R’ + Q’) 
= (PQ’ + PR + PQ’ + PQ’R) (P’R’ + Q’)
= (PP’Q’R’ + PP’R’R + PP’Q’R’ + PP’Q’RR’ + PQ’Q’ + PQ’R + PQ’Q’ + PQ’Q’R)
= (0 + 0 + 0 + 0 + PQ’ + PQ’R + PQ’ + PQ’R)
= PQ’ + PQ’R
= PQ'(1 + R)
= PQ’

So, option (A) is correct.

Question 10

Consider the following Boolean function of four variables: f(w,x,y,z) = ∑(1,3,4,6,9,11,12,14) The function is:

A	independent of one variables.
B	independent of two variables.
C	independent of three variables.
D	dependent on all the variables.

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Question 10 Explanation:

On solving K-MAP we get ZX’+XZ’ so it is independent of w,y Ans (B) part.

There are 104 questions to complete.

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Logic functions and Minimization

Logic functions and Minimization

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