Design a Multiplexer Circuit for A and B Outputs Based on C Input

[STEMucator]

Homework Statement

Design a multiplexer circuit given the data such that:

When $C = 0$, the output $X = A$.
When $C = 1$, the output $X = B$.

The truth table is displayed below:

Homework Equations

The Attempt at a Solution

I wanted to make sure I understood this and I wasn't overlooking anything.

Looking at the $A$, $C$ and $P$ columns, I deduced I needed a term of the form $A \cdot \bar C$.

Looking at the $B$, $C$ and $Q$ columns, I deduced I needed a term of the form $B \cdot C$.

Looking at the $P$, $Q$ and $X$ columns, I see I need a gate that satisfies $\bar P \cdot Q + P \cdot \bar Q + P \cdot Q$.

That last expression can be simplified to:

$\bar P \cdot Q + P \cdot \bar Q + P \cdot Q = \bar P \cdot Q + P(\bar Q + Q) = \bar P \cdot Q + P = Q + P$

So from what I can gather, $Q + P = B \cdot C + A \cdot \bar C$.

Hence $X = Q + P = B \cdot C + A \cdot \bar C$.

Thank you.

 

[gneill]

Looks okay. You might have considered using a Karnaugh Map approach which would yield the two terms by inspection.

 

[NascentOxygen]

How did you fill in the P and Q columns?

 

[STEMucator]

How did you fill in the P and Q columns?

When $C = 0$, the value of $Q$ does not matter and the value of $X$ depends only on $A$.

Similarly when $C = 1$, the value of $P$ does not matter and the value of $X$ depends only on $B$.

This gives half a column of $Q$s and $P$s that do not matter. I could have filled those in with a $d$ or $-$, but that was the problem statement provided and they seemed to have used a $0$.

 

[NascentOxygen]

This question is a bit puzzling. It's as though that table belongs to some other question and accidently got mixed in here.

The task is wholly described in just these lines:

Design a multiplexer circuit given the data such that:

When C=0, the output X=A.
When C=1, the output X=B.

The table with signals P and Q is either a mistake or a red herring, IMO.