Implement the following boolean expression using 8:1 mux \[f\left(A,B,C,D\right)=\bar{A}\ \bar{B}\bar{C}+ABC+\bar{B}CD+\bar{A}CD\]

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written 2.9 years ago by

The given boolean expression is not in standard SOP form.

First convert this into standard SOP form.

$\begin{align*} F(A,B,C,D) &=\bar A\bar B\bar D(C+\bar C)+ABC(D+\bar D)+\bar BCD(A+\bar A)+\bar ACD(B+\bar B)\\[2ex] &=\bar A\bar BC\bar D+\bar A\bar B\bar C\bar D+ABCD+ABC\bar D\\& +A\bar BCD+\bar A\bar BCD+\bar A BCD+\bar A\bar BCD\\[2ex] &=\bar A\bar BC\bar D+\bar A\bar B\bar C\bar D+ABCD+ABC\bar D\\& +A\bar B CD+\bar A\bar BCD+\bar A BCD \end{align*}$

The truth table for this standard SOP form,

No	Minterm	A	B	C	D	Y
0	$\bar A\bar B\bar C\bar D$	0	0	0	0	1
1		0	0	0	1	0
2	$\bar A\bar BC\bar D$	0	0	1	0	1
3	$\bar A\bar BCD$	0	0	1	1	1
4		0	1	0	0	0
5		0	1	0	1	0
6		0	1	1	0	0
7	$\bar ABCD$	0	1	1	1	1
8		1	0	0	0	0
9		1	0	0	1	0
10		1	0	1	0	0
11	$A\bar BCD$	1	0	1	1	1
12		1	1	0	0	0
13		1	1	0	1	0
14	$ABC\bar D$	1	1	1	0	1
15	$ABCD$	1	1	1	1	1

Implementation table,

	$D_0$	$D_1$	$D_2$	$D_3$	$D_4$	$D_5$	$D_6$	$D_7$
$\bar A$	0	1	2	3	4	5	6	7
$A$ \| 8 \| 9 \| 10 \| 11 \| 12 \| 13 \| 14 \| 15 \| \| \| $\bar A$ \| 0 \| $\bar A$ \| 1 \| 0 \| 0 \| $A$	1

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