How is the third 8:1 MUX used?

P and Q go to I0 and I1, S3 drives its S0 select and the other selects are tied to 0

P and Q go to I6 and I7 with S3 on all selects

All eight inputs carry P and Q alternately and S0 selects

2082 Shrawan · 4 marks

16:1 MUX using 8:1 MUXes

Short answer

Two 8:1 multiplexers select I0–I7 and I8–I15 using S2,S1,S0. Their outputs feed a third 8:1 mux used as a 2:1 mux: its I0 and I1 get the halves, S3 drives its select, and its other selects are tied to 0. Output Y = S3'P + S3Q.

Implement 16:1 multiplexer using only 8:1 multiplexers.

A 16:1 MUX has inputs $I_0$ – $I_{15}$ and four selects $S_3S_2S_1S_0$ . Using only 8:1 MUXes, three are needed:

MUX-1 (8:1): inputs $I_0$ – $I_7$ , selects $S_2S_1S_0$ , output $P$ .
MUX-2 (8:1): inputs $I_8$ – $I_{15}$ , selects $S_2S_1S_0$ , output $Q$ .
MUX-3 (8:1) used as a 2:1 MUX: $P$ to its $I_0$ , $Q$ to its $I_1$ ; its select $S_0$ is driven by $S_3$ and its other two selects are tied to 0.

         S2 S1 S0
        +--------+
 I0-I7->| MUX-1  |--P--+     +--------+
        +--------+     +---->| I0     |
        +--------+     +---->| I1 MUX3|---> Y
 I8-15->| MUX-2  |--Q--+     | I2-I7=0|
        +--------+           +--------+
         S2 S1 S0            0  0  S3
                             (S2 S1 S0 of MUX-3)

Y = S_3'P + S_3Q

$S_3$	Y
0	$P$ = one of $I_0$ – $I_7$
1	$Q$ = one of $I_8$ – $I_{15}$

Example: $S_3S_2S_1S_0 = 1101$ : $P = I_5$ , $Q = I_{13}$ ; MUX-3 select $= 001$ passes $Q$ , so $Y = I_{13}$ . Correct.

✓ Answer checked by Claude Opus 5.5 (high effort)

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