--- title: "K-map ΠM(0,2,4,6,7,8,10,13,15) and 8:1 MUX" subject: "Digital Logic" chapter: 3 key: "semester_3/digital_logic:kmap-pom-8to1-mux" source: https://www.bctengineer.com/q/semester_3/digital_logic/kmap-pom-8to1-mux site: BCT Engineering --- # K-map ΠM(0,2,4,6,7,8,10,13,15) and 8:1 MUX Digital Logic · Chapter 3: Combinational Logic Circuits - Asked in: 2081 Bhadra, 4+2 marks - Video: [K' Map using Max Terms (Part 2)](https://www.youtube.com/watch?v=lPT9Ltjij1s) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/kmap-pom-8to1-mux - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-3 ## Question Simplify the following Boolean logic function using K-map. Also realize the logic function using an 8:1 multiplexer and necessary logic gates. Y(A,B,C,D) = ΠM(0,2,4,6,7,8,10,13,15) ## Short answer Minimal SOP: Y = B'D + A'C'D + ABD'. Using an 8:1 multiplexer with A, B, C as select lines: I0=I1=I2=I4=I5 = D, I3 = 0, I6=I7 = D'. One inverter generates D'. ## Answer $Y(A,B,C,D) = \Pi M(0,2,4,6,7,8,10,13,15)$ means Y = 0 at these cells. So Y = 1 at the remaining cells: $$ Y = \Sigma m(1,3,5,9,11,12,14) $$ #### K-map simplification ```text CD AB 00 01 11 10 +---+---+---+---+ 00 | 0 | 1 | 1 | 0 | +---+---+---+---+ 01 | 0 | 1 | 0 | 0 | +---+---+---+---+ 11 | 1 | 0 | 0 | 1 | +---+---+---+---+ 10 | 0 | 1 | 1 | 0 | +---+---+---+---+ ``` Grouping the 1s: - Quad m(1,3,9,11) → $\overline{B}D$ - Pair m(1,5) → $\overline{A}\,\overline{C}D$ - Pair m(12,14) → $AB\overline{D}$ $$ Y = B'D + A'C'D + ABD' $$ (Grouping the 0s gives the minimal POS $Y = (A+D)(B+D)(\overline{A}+\overline{B}+\overline{D})(\overline{B}+\overline{C}+\overline{D})$, which has more terms, so the SOP form is used.) #### Realization with an 8:1 multiplexer Use A, B, C as the select lines $S_2 S_1 S_0$ and connect each data input $I_0$–$I_7$ to 0, 1, D or $\overline{D}$. For each value of ABC, compare Y for D = 0 and D = 1: | ABC (sel) | Minterms | Y at D=0 | Y at D=1 | Data input | |---|---|---|---|---| | 000 | 0, 1 | 0 | 1 | $I_0 = D$ | | 001 | 2, 3 | 0 | 1 | $I_1 = D$ | | 010 | 4, 5 | 0 | 1 | $I_2 = D$ | | 011 | 6, 7 | 0 | 0 | $I_3 = 0$ | | 100 | 8, 9 | 0 | 1 | $I_4 = D$ | | 101 | 10, 11 | 0 | 1 | $I_5 = D$ | | 110 | 12, 13 | 1 | 0 | $I_6 = \overline{D}$ | | 111 | 14, 15 | 1 | 0 | $I_7 = \overline{D}$ | ```text +-----------+ D ───── | I0 | D ───── | I1 | D ───── | I2 | 0 ───── | I3 8:1 |── Y D ───── | I4 MUX | D ───── | I5 | D' ───── | I6 | D' ───── | I7 | +-----------+ S2 S1 S0 A B C D ──[NOT]── D' ``` Only one NOT gate is needed besides the multiplexer. Check: for $ABC = 110$, Y = $\overline{D}$, so minterm 12 (D=0) gives 1 and minterm 13 gives 0, as required. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. PiM(0,2,4,6,7,8,10,13,15) means: A. Y=0 at these cells, 1 elsewhere B. Y=1 at these cells C. Y=0 elsewhere D. Don't cares Answer: A. Maxterms are zeros. (Key point: Minterms) 2. Minimal SOP: A. BD + AC B. B'D + A'C'D + ABD' C. A'B + CD D. B'D' + AC Answer: B. From the three groups. (Key point: SOP) 3. MUX select lines: A. A, B, D B. B, C, D C. A, B, C with data from D D. All four Answer: C. Data inputs are 0, 1, D or D'. (Key point: Selects) 4. I6 (ABC=110) equals: A. D B. 0 C. 1 D. D' Answer: D. Y is 1 at minterm 12 (D=0), 0 at 13. 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