Question 2 of 8: Combinational Logic and a NOR-Only Exclusive-OR

Question 2: Combinational Logic and a NOR-Only Exclusive-OR (20 marks)

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 I: four two-input NAND gates wired as in Figure 2 — $A$ and $B$ drive the first gate whose output is $C$; $A$ and $C$ drive the gate whose output is $D$; $B$ and $C$ drive the gate whose output is $E$; $D$ and $E$ drive the output gate $F$. Part II: only two-input NOR gates are available.

Find. [a] $F(A,B)$; [b] the simplified form and the single equivalent gate; [c] the truth table at $C,D,E,F$; [d]–[g] a NOR-only exclusive-OR, with its truth table, expression, NOR-form algebra, and gate diagram.

[Figure not reproduced: Figure 2 — combinational network redrawn with the internal nodes $C$, $D$, $E$ labelled as in the question paper. See the official exam paper.]

Part I — analysis of the given network

Approach. Propagate the NAND function node by node, then apply DeMorgan to collapse the double inversions.

1. [a] Write each node. A two-input NAND realises $\overline{XY}$, so reading Figure 2 from left to right: $$C = \overline{AB}, \qquad D = \overline{AC}, \qquad E = \overline{BC}, \qquad F = \overline{DE}.$$ Substituting gives the general expression asked for in [a]: $$F = \overline{\overline{A\,\overline{AB}}\cdot\overline{B\,\overline{AB}}}.$$
2. [b] Apply DeMorgan. $\overline{DE} = \bar D + \bar E$, and each of $\bar D$, $\bar E$ is a double inversion that collapses: $$F = \bar D + \bar E = A\,\overline{AB} + B\,\overline{AB} = \overline{AB}\,(A + B).$$ Expanding $\overline{AB} = \bar A + \bar B$ and multiplying out: $$F = (\bar A + \bar B)(A+B) = \underbrace{\bar A A}_{0} + \bar A B + A\bar B + \underbrace{\bar B B}_{0},$$ so that $$\boxed{\;F = \bar A B + A \bar B = A \oplus B\;}$$ Yes — a single two-input exclusive-OR (EOR) gate replaces the whole four-NAND network.
3. [c] Tabulate the internal nodes. Evaluating $C$, then $D$ and $E$, then $F$ for each input combination gives the truth table below. $F$ is high only when the inputs differ, confirming the algebra.

Part II — exclusive-OR from NOR gates only

Approach. Write the EOR in sum-of-products form, then convert to a NOR-only form by double-inverting and applying DeMorgan, so that every product becomes a NOR of complemented literals. The complements themselves are generated by NOR gates.

1. [d] Truth table. The exclusive-OR is high when exactly one input is high.

2. [e] General expression. Directly from the two rows where the output is high, $$F = A \oplus B = \bar A B + A \bar B.$$
3. [f] Convert to NOR form. A NOR gate computes $\overline{X+Y}$, so the target is an expression built only from complemented sums. Start with the identity $$\overline{A + \overline{A+B}} = \bar A \,(A+B) = \bar A B,$$ which follows by DeMorgan on the outer bar and then distributing; symmetrically $$\overline{B + \overline{A+B}} = \bar B\,(A+B) = A\bar B.$$ Naming those two results $Y_2$ and $Y_3$ and NOR-ing them, $$Y_4 = \overline{Y_2 + Y_3} = \overline{\bar A B + A\bar B} = \overline{A \oplus B},$$ which is the exclusive-NOR. One final inversion — a NOR gate with both inputs tied together, since $\overline{X+X} = \bar X$ — gives the required function: $$\boxed{\;F = \overline{\overline{\;\overline{A+\overline{A+B}} \;+\;\overline{B+\overline{A+B}}\;}}\;=\;A\oplus B\;}$$ Five two-input NOR gates are needed.
4. [g] Gate array. $N_1$ forms $Y_1=\overline{A+B}$; $N_2$ and $N_3$ form $Y_2$ and $Y_3$; $N_4$ combines them into the exclusive-NOR $Y_4$; $N_5$, with both inputs tied to $Y_4$, acts as the inverter that produces $F$.