K-Map

Question 1

Consider the minterm list form of a Boolean function F given below.

    F(P, Q, R, S) = Σm(0, 2, 5, 7, 9, 11) + d(3, 8, 10, 12, 14)

Here, m denotes a minterm and d denotes a don’t care term. The number of essential prime implicants of the function F is _______ .

A
3
B
4
C
5
D
6
       Digital-Logic-Design       K-Map       Gate 2018
Question 1 Explanation: 
f = Σ(0, 2, 5, 7, 9, 11) + d(3, 8, 10, 12, 14)

There are 3 prime implicant i.e., P’QS, Q’S’ and PQ’ and all are essential.
Because 0 and 2 are correct by only Q’S’, 5 and 7 are covered by only P’QS and 8 and 9 are covered by only PQ’.
Question 2
 

Consider the Karnaugh map given below, where X represents “don’t care” and blank represents 0.

Assume for all inputs , the respective complements  are also available. The above logic is implemented using 2-input NOR gates only. The minimum number of gates required is _________.

 
A
1
B
2
C
3
D
4
       Digital-Logic-Design       K-Map       Gate 2017 set-01
Question 2 Explanation: 
Given K-Map represents the function f(a, b, c, d) = a' c = a'(c' )' = (a + c')'
As all variables and their complements are available we can implement the function with only one NOR Gate.
Question 3

Given f(w,x,y,z) = Σm(0,1,2,3,7,8,10) + Σd(5,6,11,15), where d represents the don’t-care condition in Karnaugh maps. Which of the following is a minimum product-of-sums (POS) form of f(w,x,y,z)?

A
B
C
D
       Digital-Logic-Design       K-Map       GATE 2017(set-02)
Question 3 Explanation: 
f(w,x,y,z) = Σm(0,1,2,3,7,8,10) + Σd(5,6,11,15)
K-Map for the function f is

Consider maxterms in K-map to represent function in product-of-sums (POS) form
f(w,x,y,z) = (w' + z')(x' + z)
Question 4
The total number of prime implicants of the function f(w,x,y,z= Σ(0, 2, 4, 5, 6, 10) is ______.
A
3
B
4
C
2
D
1
       Digital-Logic-Design       K-Map       GATE 2015(Set-03)
Question 4 Explanation: 

Total 3 prime implicants are there.
Question 5
A
B
C
D
       Digital-Logic-Design       K-Map       Gate 2012
Question 5 Explanation: 
Question 6
       
A
B
C
D
       Digital-Logic-Design       K-Map       Gate-2008
Question 6 Explanation: 

Question 7
 
A
B
C
D

Hence, option (A) matches.
Question 8
 
A
c’d’+ ad’ + abc’ + (ac)’d
B
(ac)’ + c’d’ + ad’ + abc’d
C
(ac)’ + ad’ + abc’ + c’d
D
b’c’d’ + acd’ + (ac)’ + abc’
       Digital-Logic-Design       K-Map       Gate 2007-IT
Question 8 Explanation: 
Let's check for option (C):
a'c' + ad' + abc' + c'd

Not equivalent to the K-map, we get in previous question.
Question 9
Consider a Boolean function f (w, x, y, z). Suppose that exactly one of its inputs is allowed to change at a time. If the function happens to be true for two input vectors i1 = w1x1y1z1 and i2 w2x2y2z2〉, we would like the function to remain true as the input changes from vectors i1 to i2 (i1 and i2 differ in exactly one bit position), without becoming false momentarily. Let f(w, x, y, z) = ∑(5, 7, 11, 12, 13, 15). Which of the following cube covers of f will ensure that the required property is satisfied?
A
B
C
D
       Digital-Logic-Design       K-Map       Gate-2006
Question 9 Explanation: 
Static hazard is the situation where, when one input variable changes, the output changes momentarily before stabilizing to the correct value. The most commonly used method to eliminate static hazards is to add redundant logic (consensus terms in the logic expression).
f = X1 * X2 + X1' * X3
If (X1,X2,X3) = (1,1,1) then f=1 because X1 * X2 =1 X1' * X3 = 0.
Let the input is changed from 111 to 011 , then f = 1 because X1 * X2 = 0 X1' * X3 =1.
The output f will be momentarily 0 if AND gate X1 * X2 is faster than the AND gate X1' * X3.
This Hazard can be avoided by adding the term X2 * X3 (because X1 is in true form in first term and in complement form in the second term . So pick the fixed terms X2 and X3 from both terms) to f i.e f = X1 * X2 + X1' * X3 + X2 * X3
Option D is equivalent to f(w, x, y, z) = ∑(5,7,11,12,13,15)
Question 10
 
A
aȼc and acȼ
B
aȼc and bȼc
C
aȼc only
D
acȼ and bcȼ
       Digital-Logic-Design       K-Map       Gate-2004
Question 10 Explanation: 
From given function 'f' we can draw,

There are two EPI,
A'C and AC'.
Question 11
   
A
(11, 9)
B
(9, 13)
C
(9, 10)
D
(11, 11)
       Digital-Logic-Design       K-Map       Gate-2003
Question 11 Explanation: 
For SOP,

⇒ w'y' + z'wx' + xyz'
Total 8 literals are there.
For POS,

⇒ (z' + w')(z' + y')(w' + x')(x + z + w)
Total 9 literals are there.
Question 12
 
A
xz+y'z
B
xz'+zx'
C
x'y+zx'
D
None of the above
       Digital-Logic-Design       K-Map       Gate-2002
Question 12 Explanation: 

⇒ xz' + zx'
Question 13
   
A
B
C
D
None of the above
       Digital-Logic-Design       K-Map       Gate-2000
Question 13 Explanation: 
Given k-map gives xy + xy + wz

⇒ wy + wz + xy
Question 14
 
A
B
C
D
       Digital-Logic-Design       K-Map       Gate-1999
Question 14 Explanation: 

⇒ CD+AD = D(A+C)
Question 15
   
A
A⋅B
B
AB+BC+CA
C
D
None of the above
       Digital-Logic-Design       K-Map       Gate-1998
Question 15 Explanation: 
Question 16
 
A
B
xz is a minterm of f
C
xz is an implicant of f
D
y is a prime implicant of f
       Digital-Logic-Design       K-Map       Gate-1997
Question 16 Explanation: 
In sum of terms,any term is an implicant because it implies the function. So xz is an implicant and hence 'C' is the answer.
Question 17
   
A
B
C
D
E
None of the above
       Digital-Logic-Design       K-Map       Gate-1996
Question 17 Explanation: 
Correct option is

Question 18
 
A
ABC + B'C' + A'C'
       Digital-Logic-Design       K-Map       Gate-1992
Question 18 Explanation: 
We can write this as

⇒ ABC + B'C' + A'C'
Question 19
 
A
xy+y'z
B
wx'y'+xy+xz
C
w'x+y'z+xy
D
xz+y
       Digital-Logic-Design       K-Map       Gate-2001
Question 19 Explanation: 

⇒ y'z + xy
There are 19 questions to complete.