Question

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 \(\_\_\_\_\). 

• A. A square wave of frequency \(100 \, {MHz}\)
• B. A square wave of frequency \(50 \, {MHz}\)
• C. Constant at 0
• D. Constant at 1

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.

Answer 1

The Correct Option is B

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: \[ \boxed{{(2) } {A square wave of frequency } 50 \, {MHz}} \]