GATE Electrical 2015 Set 2 | Question: 38

Question

A Boolean function $f(A, B, C, D) = \prod  (1,5,12,15)$ is to be implemented using an $8 \times 1$ multiplexer ($A$ is $MSB$). The inputs $ABC$ are connected to the select inputs $S_{2} S_{1} S_{0}$ of the multiplexer respectively.

Which one of the following options gives the correct inputs to pins $0,1,2,3,4,5,6,7$ in order?

• A. $D, 0, D, 0, 0, 0, \overline{D}, D$ 
• B. $\overline{D}, 1, \overline{D}, 1, 1, 1, D, \overline{D}$
• C. $D, 1, D, 1, 1, 1, \overline{D}, D$ 
• D. $\overline{D}, 0, \overline{D}, 0, 0, 0,D, \overline{D}$

Comments

Detailed Video Solution: https://youtu.be/UaYjTM_tdC0 

Answer 1

We know that the general equation for the output($Z$) of an $8 \times 1 \ Mux$ is

\[ \large Z = \sum_{k=0}^{7} m_kI_k  \]

where $m_k$ is the minterm of the $3$ select variables (here $A, B, C$) and $I_k$ is the corresponding data input.

$\begin{align*}
f(A,B,C,D) &= \prod (1,5,12,15) \\
\\ &= \sum(0,2,3,4,6,7,8,9,10,11,13,14) = \sum_{k=0}^{7} m_kI_k \\
\\ &= \sum(\underbrace{\mathbf{000}0}_{\underset{\LARGE{I_0 = \bar D}}{\large{m_0\bar D}}},
\underbrace{\mathbf{001}0, \mathbf{001}1}_{\underset{\LARGE{I_1 = 1}}{\large{m_1(\bar D+D)}}},
\underbrace{\mathbf{010}0}_{\underset{\LARGE{I_2 = \bar D}}{\large{m_2\bar D}}},
\underbrace{\mathbf{011}0, \mathbf{011}1}_{\underset{\LARGE{I_3 = 1}}{\large{m_3(\bar D+D)}}},
\underbrace{\mathbf{100}0, \mathbf{100}1}_{\underset{\LARGE{I_4 = 1}}{\large{m_4(\bar D+D)}}},
\underbrace{\mathbf{101}0, \mathbf{101}1}_{\underset{\LARGE{I_5 = 1}}{\large{m_5(\bar D+D)}}},
\underbrace{\mathbf{110}1}_{\underset{\LARGE{I_6 = D}}{\large{m_6 D}}},
\underbrace{\mathbf{111}0}_{\underset{\LARGE{I_7 = \bar D}}{\large{m_7\bar D}}}) \\

\end{align*}$

$\therefore Ans\ - B.$

Here the shorter way would be to directly find out the four pins for which output of $f$ will depend on $D$ by the $CPOS$ form given, since there are only four inputs for which $f$ is $0$.