- 2075 Chaitra · 5 marks
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 as select lines () and (LSB) as the data variable. Input corresponds to minterms (with ) and (with ).
Implementation table for (bold = minterm in F):
| row | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 |
| row | 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 |
| Input | 1 | 0 | 1 | 0 | 1 |
Rule: both entries bold gives 1, neither gives 0, only the upper () row gives , and only the lower () row gives .
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 , logic 0 to ground, and a NOT gate gives .
Verification:
| ABCD | Minterm | Selected input | F |
|---|---|---|---|
| 0000 | 0 | 1 | |
| 0001 | 1 | 0 | |
| 1101 | 13 | 1 | |
| 1110 | 14 | 0 | |
| 1111 | 15 | 1 |
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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