The propagation delay of the \(2 \times 1\) MUX shown in the circuit is \(10 \, {ns}\). Consider the propagation delay of the inverter as \(0 \, {ns}\). If \(S\) is set to 1, then the output \(Y\) is \(\_\_\_\_\).
Show Hint
In MUX analysis, focus on the select line's behavior to determine the output. Propagation delay affects timing but not the frequency of the signal.
Step 1: Understand the \(2 \times 1\) MUX operation.
The \(2 \times 1\) MUX has two inputs, \(I_0\) and \(I_1\), and a select line \(S\). The output \(Y\) is determined by the value of \(S\):
\[
Y =
\begin{cases}
I_0, & {if } S = 0,
I_1, & {if } S = 1.
\end{cases}
\]
When \(S = 1\), the output \(Y\) directly reflects \(I_1\), which is a square wave of frequency \(50 \, {MHz}\).
Step 2: Consider the propagation delay.
The propagation delay of the MUX is given as \(10 \, {ns}\). However, this delay does not affect the frequency of the output square wave; it only shifts the signal in time.
Step 3: Confirm the output.
When \(S = 1\), the output \(Y\) is a square wave with a frequency identical to \(I_1\), which is \(50 \, {MHz}\).
Final Answer:
\[
\boxed{{(2) } {A square wave of frequency } 50 \, {MHz}}
\]
