Which one of the following expressions does NOT represent exclusive NOR of x and y?
Question 1 Explanation:
By Definition of XNOR, So Option-A is correct. Also by Definition of XOR, Option-B is So Option-B is also correct. Option-C is 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.
b'd' + b'c'
b'd' + a'b'c'd'
b'd' + b'c' + c'd'
Question 2 Explanation:
There are two prime implicants in the following K-Map- Prime Implicant higlighted in Green = Prime Implicant higlighted in Orange = So the Boolean expression is- Therefore option (B) is correct. This explanation is provided by Chirag Manwani.
Question 3 Explanation:
All options except D produce XOR as described below :
The simplified SOP (Sum Of Product) form of the boolean expression (P + Q' + R') . (P + Q' + R) . (P + Q + R') is
(P'.Q + R')
(P + Q'.R')
(P'.Q + R)
(P.Q + R)
Question 4 Explanation:
See following : (P+Q'+R').(P+Q'+R).(P+Q+R') = From the K-map, POS form is : P + Q'.R'
The minterm expansion of f(P, Q, R) = PQ + QR' + PR' is
m2 + m4 + m6 + m7
m0 + m1 + m3 + m5
m0 + m1 + m6 + m7
m2 + m3 + m4 + m5
Question 5 Explanation:
= 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 + m7Option (A) is correct.
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?
Question 6 Explanation:
AB+C = (A+C)(B+C) = ((A+C)' + (B+C)')' So, '3' 2-input NOR gates are required.
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?
Given f1, f3 and f in canonical sum of products form (in decimal) for the circuit
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)
If P, Q, R are Boolean variables, then (P + Q')(PQ' + PR)(P'R' + Q') simplifies
PQ' + R
PR'' + Q
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.
Consider the following Boolean function of four variables: f(w,x,y,z) = ∑(1,3,4,6,9,11,12,14) The function is:
independent of one variables.
independent of two variables.
independent of three variables.
dependent on all the variables.
There are 104 questions to complete.
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