--- title: "K-map Π(0,2,5,8,10)+d, SOP and NOR" subject: "Digital Logic" chapter: 3 key: "semester_3/digital_logic:kmap-maxterm-sop-nor-025" source: https://www.bctengineer.com/q/semester_3/digital_logic/kmap-maxterm-sop-nor-025 site: BCT Engineering --- # K-map Π(0,2,5,8,10)+d, SOP and NOR Digital Logic · Chapter 3: Combinational Logic Circuits - Asked in: 2069 Chaitra, 4+4 marks - Video: [Minimize the following expression using K map and realize using NOR gate](https://www.youtube.com/watch?v=2HpvaRRDPK8) — Maharshi Sanand Yadav T - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/kmap-maxterm-sop-nor-025 - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-3 ## Question Simplify F(A,B,C,D) = Π(0,2,5,8,10) + d(7,15). Write its standard SOP and implement the simplified circuit using NOR gates only. ## Short answer The standard SOP from the maxterm list is Σm(1,3,4,6,9,11,12,13,14) + d(7,15). Simplified SOP is F = B' D + B D' + A D; the POS form is F = (B+D)(A+B'+D'). NOR implementation uses the POS: generate B' and D' with NOR inverters, use a NOR for (B+D) and a NOR for (A+B'+D'), then combine with a final NOR. ## Answer Given $F(A,B,C,D) = \Pi M(0,2,5,8,10) + d(7,15)$. The listed maxterms are the 0s; the other cells (except don't-cares 7, 15) are 1s. #### Standard SOP The standard (canonical) SOP is the sum of the minterms not in the maxterm list: $$ F = \Sigma m(1,3,4,6,9,11,12,13,14) + d(7,15) $$ $$ \begin{aligned} F = {} & \overline{A}\,\overline{B}\,\overline{C}D + \overline{A}\,\overline{B}CD + \overline{A}B\overline{C}\,\overline{D} + \overline{A}BC\overline{D} + A\overline{B}\,\overline{C}D \\ & + A\overline{B}CD + AB\overline{C}\,\overline{D} + AB\overline{C}D + ABC\overline{D} \end{aligned} $$ #### K-map ```text CD AB 00 01 11 10 +---+---+---+---+ 00 | 0 | 1 | 1 | 0 | +---+---+---+---+ 01 | 1 | 0 | X | 1 | +---+---+---+---+ 11 | 1 | 1 | X | 1 | +---+---+---+---+ 10 | 0 | 1 | 1 | 0 | +---+---+---+---+ ``` **Grouping 1s (minimal SOP):** - Quad m(1,3,9,11) → $\overline{B}D$ - Quad m(4,6,12,14) → $B\overline{D}$ - Quad m(9,11,13,15) → $AD$ $$ F = \overline{B}D + B\overline{D} + AD $$ (The last term may also be taken as $AB$, quad 12, 13, 14, 15; both are minimal.) **Grouping 0s (minimal POS):** - Quad of 0s M(0,2,8,10) → $(B+D)$ - Pair of 0s M(5,7) → $(A+\overline{B}+\overline{D})$ $$ F = (B+D)(A+\overline{B}+\overline{D}) $$ #### Implementation using NOR gates only The POS form has fewer terms, and POS maps directly to a two-level NOR-NOR circuit: $$ F = \overline{\overline{(B+D)} + \overline{(A+\overline{B}+\overline{D})}} $$ ```text B ──[NOR]── B' (inputs tied) D ──[NOR]── D' (inputs tied) B, D ──[NOR]── P1 ─┐ ├──[NOR]── F A, B', D' ──[NOR]── P2 ─┘ ``` Gates: 2 NOR inverters, one 2-input NOR, one 3-input NOR and one 2-input output NOR: 5 NOR gates. Check with $ABCD = 0101$ (m5): $B+D = 1$, $A+\overline{B}+\overline{D} = 0$, so F = 0, correct. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In F = ΠM(0,2,5,8,10) + d(7,15), what do the listed numbers represent? A. Cells where F = 1 B. Cells where F = 0 C. Don't-care cells D. Cells that are unused Answer: B. The maxterms in the list are the 0s of F; all other cells except don't-cares are 1. (Key point: Maxterm list meaning) 2. Which is the standard SOP minterm list? A. Σm(0,2,5,8,10) B. Σm(1,3,4,6,9,11,12,13,14) + d(7,15) C. Σm(1,3,4,6,7,9,11,15) D. Σm(4,6,12,14) Answer: B. The minterms are all numbers not in the maxterm list and not don't-cares. (Key point: Standard SOP) 3. What is the minimal POS found by grouping 0s? A. (B + D)(A + B' + D') B. (B' + D')(A + B + D) C. (A + B)(C + D) D. B'D + BD' + AD Answer: A. Quad M(0,2,8,10) = B + D and pair M(5,7) = A + B' + D'. (Key point: Minimal POS) 4. Why is the POS form well suited to NOR-only implementation? A. POS maps to a two-level NOR-NOR circuit B. POS needs only AND gates C. POS has no complemented inputs D. POS maps to NAND-NAND Answer: A. Each sum term is a NOR gate and the output NOR combines them: F = ((B+D)' + (A+B'+D')')'. This takes 5 NOR gates. 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