Logic functions and Minimization



 
  
        
    
   
  

  
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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'
GATE CS 2013    Digital Logic & Number representation    Logic functions and Minimization    
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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.

gatecs2012Kmap
A
b'd'
B
b'd' + b'c' 
C
b'd' + a'b'c'd' 
D
 b'd' + b'c' + c'd'
Digital Logic & Number representation    Logic functions and Minimization    
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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? 

GATECS201113
A
A
B
B
C
C
D
D
GATE CS 2011    Digital Logic & Number representation    Logic functions and Minimization    
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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: 
 CSE_2010_06_ans
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
GATE-CS-2009    Digital Logic & Number representation    Logic functions and Minimization    
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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? 
7

A) 8  


B) 9

C) 10

D) 11 
A
A
B
B
C
C
D
D
Digital Logic & Number representation    GATE CS 2008    Logic functions and Minimization    
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Question 7 Explanation: 
 107
Question 8
Given f1, f3 and f in canonical sum of products form (in decimal) for the circuit
14

15

A) sigma_Em(4, 6)

B) sigma_Em(4, 8)

C) sigma_Em(6, 8)

D) sigma_Em(4, 6, 8)
A
A
B
B
C
C
D
D
Digital Logic & Number representation    GATE CS 2008    Logic functions and Minimization    
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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 (D) m(4,5,6,7,8)(6,8)+(1,6,15)

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


s is  ( C) part.
Question 9
If P, Q, R are Boolean variables, then (P + Q')(PQ' + PR)(P'R' + Q') simplifies

(P+overline{Q})(P.overline{Q} + P.R)(overline{P}.overline{R}+overline{Q})

= (P+Q)(P.Q + P.R.overline{Q})

= (P+Q)(P.overline{Q})[because (1+R)=1]

= Poverline{Q}
A
PQ'
B
PR'
C
PQ' + R
D
PR'' + Q
Digital Logic & Number representation    GATE CS 2008    Logic functions and Minimization    
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Question 9 Explanation: 
  (P+\overline{Q})(P.\overline{Q} + P.R)(\overline{P}.\overline{R}+\overline{Q})  = (P+Q)(P.Q + P.R.\overline{Q})  = (P+Q)(P.\overline{Q})[\because (1+R)=1]  = P\overline{Q}  
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.
Digital Logic & Number representation    GATE-CS-2007    Logic functions and Minimization    
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Question 10 Explanation: 
 19

On solving K-MAP we get ZX’+XZ’

so  it is independent of w,y

Ans (B) part.
                
    
  

  	
    
   
  
    

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