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Σ(0,2,3,6,7,8,12,13,15) using 8×1 MUX

Short answer

Use A,B,C as select lines and D as data input. Determine each MUX input Ii from the minterms: I0=D', I1=1, I2=0, I3=1, I4=D', I5=0, I6=1, I7=D. Connect accordingly with one inverter for D'.

Implement the following function using 8×1 MUX. F(A,B,C,D) = Σ(0,2,3,6,7,8,12,13,15)

Use A,B,CA, B, C as select lines (S2S1S0S_2S_1S_0) and DD (LSB) as the data variable. Input IiI_i corresponds to minterms 2i2i (with D=0D=0) and 2i+12i+1 (with D=1D=1).

Implementation table for F=Σ(0,2,3,6,7,8,12,13,15)F = \Sigma(0,2,3,6,7,8,12,13,15) (bold = minterm in F):

I0I_0I1I_1I2I_2I3I_3I4I_4I5I_5I6I_6I7I_7
D′D' row02468101214
DD row13579111315
InputD′D'101D′D'01DD

Rule: both entries bold gives 1, neither gives 0, only the upper (D′D') row gives D′D', and only the lower (DD) row gives DD.

Circuit:

D' --> I0 +--------+
1  --> I1 |        |
0  --> I2 |  8x1   |
1  --> I3 |  MUX   |--> F
D' --> I4 |        |
0  --> I5 |        |
1  --> I6 |        |
D  --> I7 +--------+
           S2 S1 S0
           A  B  C

Logic 1 is tied to VCCV_{CC}, logic 0 to ground, and a NOT gate gives D′D'.

Verification:

ABCDMintermSelected inputF
00000I0=D′=1I_0 = D' = 11
00011I0=D′=0I_0 = D' = 00
110113I6=1I_6 = 11
111014I7=D=0I_7 = D = 00
111115I7=D=1I_7 = D = 11

All agree with the given minterm list, so the function is realized with one 8×1 MUX and one inverter.

✓ Answer checked by Claude Opus 5.5 (high effort)

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