Consider the 4-to-1 multiplexer with two select lines \(S_1\) and \(S_0\)…
2014
Consider the 4-to-1 multiplexer with two select lines \(S_1\) and \(S_0\) given below
The minimal sum-of-products form of the Boolean expression for the output \(F\)of the multiplexer is
- A.
\(\bar{P}Q + Q\bar{R} + P\bar{Q}R\) - B.
\(\bar{P}Q + \bar{P}Q\bar{R} + PQ\bar{R} + P\bar{Q}R\) - C.
\(\bar{P}QR + \bar{P}Q\bar{R} + Q\bar{R} + P\bar{Q}R\) - D.
\(PQ\bar{R}\)
Attempted by 116 students.
Show answer & explanation
Correct answer: A
Correct. With S1 = P and S0 = Q, the 4-to-1 MUX gives F = I0 P'Q' + I1 P'Q + I2 P Q' + I3 P Q. Substituting I0=0, I1=1, I2=R, I3=R' yields F = P'Q + P Q' R + P Q R'. By grouping prime implicants the minimal sum-of-products is F = P'Q + Q R' + P Q' R (this matches the given expression).
A video solution is available for this question — log in and enroll to watch it.