Question 4 of 5: Boolean Simplification and K-map Minimization

Question 4: Boolean Simplification and K-map Minimization (24 marks, sub-mark split uncertain — see Verify note)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Part (a): a 2-input circuit — input $A$ feeds one AND input directly, input $B$ feeds the other AND input through an inverter; the AND output feeds a final inverter producing $X$. Part (b): $Y(A,B,C,D)=\sum M(1,2,4,5,6,7,10,11,15)$ (9 minterms out of 16, remaining 7 are 0-terms — no don't-cares).

Find. Part (a) the simplified expression for $X$. Part (b)(i) the K-map. (ii) minimal POS for $Y$. (iii) a NOR-gate-only realization of that POS.

1. Part (a). Reading the gates directly: $X = \overline{A\cdot \bar B}$. Applying De Morgan's theorem, $$X = \overline{A\cdot\bar B} = \bar A + B$$ — a single 2-input OR gate (with $A$ inverted) replaces the original AND-then-NOT pair.
2. Part (b)(i) — K-map. Plot all 16 minterms on a 4-variable map (rows $AB$, columns $CD$, both in Gray-code order) and mark the 9 given minterms as 1, the remaining 7 (0,3,8,9,12,13,14) as 0.

3. Part (b)(ii) — POS via the complement. Minimizing the 0-cells (i.e. $\bar Y$) directly is exactly as valid as minimizing the 1-cells, and each resulting product term of $\bar Y$ becomes one OR-sum-term of $Y$ under De Morgan. The four essential groups above (each covers at least one 0-cell no other group reaches, so all four are required) give $$\bar Y = B'C'D' + A'B'CD + ABD' + AC'$$ Complementing term-by-term: $$\boxed{Y = (B+C+D)(A+B+C'+D')(A'+B'+D)(A'+C)}$$
4. Part (b)(iii) — NOR-only realization. A NOR-NOR two-level network realizes a POS expression exactly as a NAND-NAND network realizes an SOP: feed each sum-term's literals into its own first-level NOR gate (a NOR gate outputs the complement of the OR of its inputs, i.e. the complement of that one sum-term); then a single second-level NOR gate combines the four first-level outputs, since $\text{NOR}(s_1',s_2',s_3',s_4') = (s_1'+s_2'+s_3'+s_4')' = s_1 s_2 s_3 s_4 = Y$. The complemented literals $A',B',C',D'$ needed inside some sum-terms are each generated by a 2-input NOR gate with its two inputs tied together (NOR(x,x)$=\bar x$), so the whole network uses NOR gates exclusively.