Ques 16: The function AB’C + A’BC + ABC’ + A’B’C + AB’C’ is equivalent to
GATE 2004
(A) AC’ + AB + A’C
(B) AB’ + AC’ + A’C
(C) A’B + AC’ + AB’
(D) A’B + AC + AB’
Ans: (B) AB’ + AC’ + A’C
Solution:
Ques 17: Which are the essential prime implicants of the following Boolean function?
GATE 2004
f(a,b,c)=a′c+ac′+b′c
(A) a’c and ac’
(B) a’c and b’c
(C) a’c only
(D) ac’ and bc’
Ans: (A) a’c and ac’
Solution: Prime implicant of f is an implicant that is minimal – that is, the removal of any literal from product term results in a non-implicant for f.
Essential prime implicant is an prime implicant that cover an output of the function that no combination of other prime implicants is able to cover.
f(a,b,c)=a′c+ac′+b′c
We can write these product of sum terms into canonical product of sum form.
f(a,b,c)=a’c(b+b’) + ac'(b+b’) + b’c(a+a’)
f(a,b,c)=∑(1,3,4,5,6)
Now draw the k map
Prime implicants are:a′c,b′c,ab′,ac′
Essential prime implicants are:a′c,ac′
Ques 18: Which of the following expressions is equivalent to (A⊕B)⊕C
GATE 2005
(A) (A+B+C)(A’+B’+C’)
(B) (A+B+C)(A’+B’+C)
(C) ABC+A'(B⊕C)+B'(A⊕C)
(D) None of these
Ans: (C) ABC+A'(B⊕C)+B'(A⊕C)
Solution: (A⊕B)⊕C
=(A’B + AB’) ⊕ C
=(A’B + AB’) C’ + (A’B + AB’)’ C
=A’BC’ + AB’C’ + (A+B’)(A’+B)C
=A’BC’ + AB’C’ + (AB + A’B’)C
=A’BC’ + AB’C’ + ABC +A’B’C
=A’BC’ + AB’C’ + ABC +A’B’C + A’B’C
=ABC + A’ (BC’ + B’C) + B'(AC’ + A’C)
=ABC + A'(B ⊕ C) + B'(A ⊕ C)
Ques 19: A two-way switch has three terminals a, b and c. In ON position (logic value 1), a is connected to b, and in OFF position, a is connected to c. Two of these two-way switches S1 and S2 are connected to a bulb as shown below.
GATE IT 2005
Which of the following expressions, if true, will always result in the lighting of the bulb ?
(A) S1. S2′
(B) S1 + S2
(C) (S1 ⊕ S2)’
(D) S1 ⊕ S2
Ans: (C) (S1 ⊕ S2)’
Solution: Bulb will glow when both switch are either ON or OFF
which is nothing but the operation of X-NOR
Ques 20: Consider the following circuit.
GATE 2005
Which of the following is True?
(A) f is independent of x
(B) f is independent of y
(C) f is independent of z
(D) None of X, Y, Z is redundant
Ans :(A) f is independent of x
Solution: The expression will be
f=[(x.y′)′.(y.z)]′
=[(x′+y).(y.z)]′
=[x′.y.z+y.z]′
=[(x′+1).(y.z)]′
=[1.(y.z)]′
=[y.z]′
=y′+z′
The final expression only contains y and z,
Therefore, answer will be (a) f is Independent of x