To implement a 4-variable function with a single 8:1 MUX, what is connected to the select lines?

A, B, C with D (or its complement, 0, 1) on the data side

B, C, D with A on the data side

Why is I2 connected to logic 1?

Neither minterm 4 nor 5 is in F

2079 Bhadra · 5 marks

Σm(2,4,5,7,10,14) using a single 8:1 MUX

Short answer

Connect A, B, C to the 8:1 MUX select lines. Set data inputs: I0=0, I1=D', I2=1, I3=D, I4=0, I5=D', I6=0, I7=D'. The output F implements Σm(2,4,5,7,10,14) with an inverter for D'.

Implement the following Boolean function using a single 8:1 multiplexer. F(A,B,C,D) = Σm(2,4,5,7,10,14).

Use $A, B, C$ as the select lines $S_2S_1S_0$ of the 8:1 MUX and feed each data input with $0$ , $1$ , $D$ or $D'$ .

Implementation table

Each data input $I_j$ corresponds to $ABC = j$ and covers minterms $2j$ ( $D=0$ ) and $2j+1$ ( $D=1$ ). Required minterms: 2, 4, 5, 7, 10, 14.

$I_j$	ABC	Minterms (D=0, D=1)	In F?	$I_j$
$I_0$	000	0, 1	no, no	0
$I_1$	001	2, 3	yes, no	$D'$
$I_2$	010	4, 5	yes, yes	1
$I_3$	011	6, 7	no, yes	$D$
$I_4$	100	8, 9	no, no	0
$I_5$	101	10, 11	yes, no	$D'$
$I_6$	110	12, 13	no, no	0
$I_7$	111	14, 15	yes, no	$D'$

The same result in the usual two-row form:

	$I_0$	$I_1$	$I_2$	$I_3$	$I_4$	$I_5$	$I_6$	$I_7$
$D'$	0	2	4	6	8	10	12	14
$D$	1	3	5	7	9	11	13	15
Input	0	$D'$	1	$D$	0	$D'$	0	$D'$

Circuit

             +--------+
  0 -------->| I0     |
  D --[NOT]->| I1     |
  1 -------->| I2     |
  D -------->| I3 8:1 |----> F
  0 -------->| I4 MUX |
  D' ------->| I5     |
  0 -------->| I6     |
  D' ------->| I7     |
             +--------+
               S2 S1 S0
               A  B  C

Only one inverter is needed besides the MUX.

Check: $ABCD = 1010$ (minterm 10): select $101$ picks $I_5 = D' = 1$ , so $F = 1$ . $ABCD = 1011$ (minterm 11): $I_5 = D' = 0$ , so $F = 0$ . Correct.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Σm(2,4,5,7,10,14) using a single 8:1 MUX

Implementation table

Circuit

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