--- title: "Function Σm(0,2,3,5,7,8,10,13,15) using 1:8 DEMUX" subject: "Digital Logic" chapter: 4 key: "semester_3/digital_logic:demux-1to8-realize-function" source: https://www.bctengineer.com/q/semester_3/digital_logic/demux-1to8-realize-function site: BCT Engineering --- # Function Σm(0,2,3,5,7,8,10,13,15) using 1:8 DEMUX Digital Logic · Chapter 4: Data Processing Circuits - Asked in: 2080 Bhadra, 6 marks - Video: [Boolean Expressions Implementation Using Demultiplexers: Designing and Solved Examples](https://www.youtube.com/watch?v=EGMChjUxauE) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/demux-1to8-realize-function - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-4 ## Question Realize the following logic function using a single 1:8 demultiplexer and necessary logic gates. Y(A,B,C,D) = Σm(0,2,3,5,7,8,10,13,15) ## Short answer Use a 1:8 DEMUX with data=1 and selects B,C,D. Outputs Y0,Y2,Y5,Y7 cover all terms except those requiring A'Y3. Final Y = Y0 + Y2 + Y5 + Y7 + A'Y3. Needs a NOT, a 2-input AND, and a 5-input OR. ## Answer A 1:8 DEMUX with data input tied to 1 produces the 8 minterms of its three select variables. The 4-variable function is realized by choosing three variables as selects and handling the fourth with a gate. #### Choice of select lines Use $B, C, D$ as selects ($S_2 S_1 S_0$) and the data input $= 1$. Then output $Y_j$ is 1 when $BCD = j$. Each $Y_j$ covers two minterms of the function: $j$ (with $A=0$) and $j+8$ (with $A=1$). | BCD = j | Minterm j (A=0) | Minterm j+8 (A=1) | Contribution | |---|---|---|---| | 0 | 0 ✓ | 8 ✓ | $Y_0$ | | 1 | 1 ✗ | 9 ✗ | – | | 2 | 2 ✓ | 10 ✓ | $Y_2$ | | 3 | 3 ✓ | 11 ✗ | $A' Y_3$ | | 4 | 4 ✗ | 12 ✗ | – | | 5 | 5 ✓ | 13 ✓ | $Y_5$ | | 6 | 6 ✗ | 14 ✗ | – | | 7 | 7 ✓ | 15 ✓ | $Y_7$ | Hence $$ Y = Y_0 + Y_2 + Y_5 + Y_7 + A'Y_3 $$ #### Circuit ```text +---------+ 1 ------>| D Y0 |-------> Y0 | Y1 | (not used) B --->S2 | Y2 |-------> Y2 C --->S1 | 1:8 Y3 |-------> Y3 D --->S0 | Y4 | (not used) | Y5 |-------> Y5 | Y6 | (not used) | Y7 |-------> Y7 +---------+ A ---[NOT]--- A' A', Y3 ---[AND]--- A'Y3 Y0, Y2, Y5, Y7, A'Y3 ---[OR 5]---> Y ``` Gates: one NOT (for $A'$), one 2-input AND ($A'Y_3$) and one 5-input OR. **Check:** minterm 11 ($A=1, BCD=011$): only $Y_3 = 1$, but $A' = 0$, so $Y = 0$ (correct, 11 is not in the list). Minterm 3 ($A=0$): $A'Y_3 = 1$, so $Y = 1$. All nine minterms 0, 2, 3, 5, 7, 8, 10, 13, 15 give $Y = 1$. (For reference, the minimal SOP is $Y = B'D' + BD + A'B'C$, which matches the result above.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the 1:8 DEMUX used? A. Selects B, C, D with data input 1 to produce minterms of B, C, D B. Selects A, B, C with data input D C. All four variables as selects D. Data input tied to 0 Answer: A. Tying data to 1 turns the DEMUX into a 3-to-8 decoder; variable A is handled with a gate. (Key point: Selects and data input) 2. What is the resulting expression? A. Y0 + Y2 + Y5 + Y7 + A'Y3 B. Y0 + Y2 + Y3 + Y5 + Y7 C. Y0 + Y2 + Y5 + Y7 + A Y3 D. Y1 + Y4 + Y6 Answer: A. For BCD = 0, 2, 5, 7 both A = 0 and A = 1 minterms are in the function; for 3 only minterm 3 is. (Key point: Final function) 3. Why is Y3 ANDed with A'? A. Only minterm 3 (A = 0) is in the function, not 11 B. Minterm 11 is a don't-care C. Y3 is inverted D. To reduce delay Answer: A. Y3 is active for both 3 and 11, but only 3 belongs to the function, so A' blocks 11. (Key point: Why A' with Y3) 4. Which gates are needed besides the DEMUX? A. One NOT, one AND and one 5-input OR B. Three AND and one OR C. Two XOR D. One NAND only Answer: A. A NOT for A', an AND for A'Y3 and an OR for the five terms. 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