Karnaugh Maps and Minimization Questions and Answers

Answer: A

The minimized expression is \(\bar{b}_{1}\bar{b}_{0} + \bar{b}_{2}\bar{b}_{0} + b_{1}\bar{b}_{2}b_{3}\).

Group the minterms 0, 2, 4, 8, 10, 11, 12 on a Karnaugh map:

• \(\bar{b}_{1}\bar{b}_{0}\) covers 0, 4, 8, 12 (the whole column with \(b_{1}b_{0} = 00\)).
• \(\bar{b}_{2}\bar{b}_{0}\) covers 0, 2, 8, 10 (the four corners).
• \(b_{3}\bar{b}_{2}b_{1}\) covers 10 and 11, and is the only way to include 11.

All three are essential prime implicants, and together they cover every minterm exactly as required. Option B leaves out minterm 11. Option C's \(b_{1}b_{2}b_{3}\) covers 14 and 15, which are not in F, and misses 4, 11 and 12. Option D's \(\bar{b}_{3}\) would make F true for 1, 3, 5, 6 and 7, which are not minterms.

Answer: B

The minimal sum of products is B′.

With A as the most significant bit, the minterms 0-3 are 0000-0011 and 8-11 are 1000-1011. In all eight, B = 0, and A, C and D take every combination of values. So F = 1 exactly when B = 0:

\[ F = \Sigma m(0,1,2,3,8,9,10,11) = B' \]

On a Karnaugh map these eight cells form a single group of eight covering the two B′ rows, which reduces to the single literal B′.

A′ would cover minterms 0-7, including 4-7 where B = 1. A′.B′ + A.B covers 0-3 and 12-15, the wrong second half. A′ + B′ + C′ + D′ is 1 for every minterm except 15.

Answer: A

The expression simplifies to \(A\) XOR \(C\).

Group the terms in pairs that differ only in \(B\):

\[ \begin{aligned} \bar{A}C(B + \bar{B}) + A\bar{C}(B + \bar{B}) &= \bar{A}C + A\bar{C} \\ &= A \oplus C \end{aligned} \]

The output is 1 exactly when \(A\) and \(C\) differ, whatever \(B\) is. \(A\) AND \(\bar{C}\) keeps only the second pair of terms. The output is not constant: with \(A = C = 0\) every term is 0, and with \(A = 0, C = 1\) the first term is 1, so neither 0 nor 1 is right.