Consider the circuit below and determine the output function f(x, y, z).

Concept

A 2-to-1 multiplexer passes exactly one of its two data inputs through to the output: it forwards I0 while the select line s is 0 and forwards I1 while s is 1, which in Boolean form is \(I_0\bar{s}+I_1s\). When multiplexers are cascaded, handle one stage at a time — write the earlier stage's output as an intermediate expression, substitute that expression into the data input of the next stage, and then simplify.

Application

1. Let a be the output of the first multiplexer. With I0 = x, I1 = \(\bar{y}\), and s = z, \(a=x\bar{z}+\bar{y}z\).

2. For the second multiplexer, I0 = a, I1 = x, and s = y. Therefore \(f=a\bar{y}+xy\).

3. Substituting a gives \(f=(x\bar{z}+\bar{y}z)\bar{y}+xy=x\bar{y}\bar{z}+\bar{y}z+xy\).

4. Using \(y+\bar{y}\bar{z}=y+\bar{z}\), simplify to \(f=x\bar{z}+xy+\bar{y}z\).

Cross-check

When y = 1, the second multiplexer selects x, and the simplified expression becomes x. When y = 0, the circuit gives \(x\bar{z}+z=x+z\), which is also obtained from the simplified expression.

Hence \(f(x,y,z)=x\bar{z}+xy+\bar{y}z\).