Question

Consider the signal $x(t) = \cos(6\pi t) + \sin(8\pi t)$, where $t$ is in seconds. The Nyquist sampling rate (in samples/second) for the signal $y(t) = x(2t + 5)$ is

• A. 8
• B. 12
• C. 16
• D. 32

Hint:

Time compression increases frequency, while time expansion decreases frequency in signals.

Answer 1

The Correct Option is C

Step 1: Identify frequency components of $x(t)$.
Given:
\[ x(t) = \cos(6\pi t) + \sin(8\pi t) \]
Angular frequencies are $6\pi$ and $8\pi$ rad/s.
Step 2: Convert angular frequency to frequency in Hz.
\[ f_1 = \frac{6\pi}{2\pi} = 3\text{ Hz}, \quad f_2 = \frac{8\pi}{2\pi} = 4\text{ Hz} \]
Maximum frequency in $x(t)$ is $4$ Hz.
Step 3: Effect of time scaling in $y(t) = x(2t+5)$.
Time scaling by a factor of $2$ increases all frequencies by a factor of $2$.
\[ f_{\max,y} = 2 \times 4 = 8\text{ Hz} \]
Step 4: Apply Nyquist sampling theorem.
\[ f_s = 2 f_{\max,y} = 2 \times 8 = 16 \text{ samples/s} \]
Step 5: Final conclusion.
The Nyquist sampling rate is 16 samples per second.