The minimized expression of the Boolean function Y(P, Q, R) implemented by the multiplexer (MUX) circuit shown in the figure is:
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For multiplexer circuits, systematically determine the output based on the select inputs and the corresponding data inputs. Remember the standard form of a multiplexer output:
$$Output = \sum_{i=0}^{2^n - 1} ({minterm}_i \cdot {DataInput}_i)$$
where $n$ is the number of select inputs.
Step 1: Analyze the first multiplexer. The first multiplexer has select inputs $S_1 = P$ and $S_0 = Q$. The data inputs are connected as follows: $S_1S_0 = 00$: Input = 0 $S_1S_0 = 01$: Input = 1 $S_1S_0 = 10$: Input = 1 $S_1S_0 = 11$: Input = 0 The output of the first multiplexer, let's call it Z, is given by: $$Z = \overline{P}\overline{Q}(0) + \overline{P}Q(1) + P\overline{Q}(1) + PQ(0)$$ $$Z = \overline{P}Q + P\overline{Q}$$ This is the expression for the XOR gate: $$Z = P \oplus Q$$ Step 2: Analyze the second multiplexer. The second multiplexer has select input $S = R$. The data inputs are connected as follows: $S = 0$: Input = Z $S = 1$: Input = 1 The output of the second multiplexer is Y, given by: $$Y = \overline{R}(Z) + R(1)$$ $$Y = \overline{R}(P \oplus Q) + R$$ Step 3: Simplify the expression for Y. We can use the Boolean algebra identity $x + \overline{x}y = x + y$. Let $x = R$ and $y = P \oplus Q$. $$Y = R + \overline{R}(P \oplus Q) = R + (P \oplus Q)$$
