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.

Answer 2

Instead of first finding the highest frequency in \(x(t)\) and then scaling it, the individual frequency components of \(y(t)=x(2t+5)\) can be written out directly and the highest one identified from that expression, then Nyquist's rule applied once at the end.

Since \(x(t) = \cos(6\pi t) + \sin(8\pi t)\), substituting \(t \to 2t+5\) gives \[ y(t) = \cos\big(6\pi(2t+5)\big) + \sin\big(8\pi(2t+5)\big) = \cos(12\pi t + 30\pi) + \sin(16\pi t + 40\pi) \] The constant phase offsets (\(30\pi\) and \(40\pi\), both integer multiples of \(2\pi\)) do not affect the frequency content, so the two angular frequencies present in \(y(t)\) are \(12\pi\) and \(16\pi\) rad/s, corresponding to \(6\) Hz and \(8\) Hz.

1. 8 samples/s: This equals the highest frequency itself (8 Hz), not twice it, so it falls short of the Nyquist rate required to avoid aliasing.
2. 12 samples/s: This equals twice the lower frequency component (6 Hz), not twice the highest one, so it under-samples the 8 Hz component.
3. 16 samples/s: This is exactly twice the highest frequency present in \(y(t)\), 8 Hz, which is the minimum rate needed to satisfy the sampling theorem for this signal.
4. 32 samples/s: This is four times the highest frequency, safely above the Nyquist rate but larger than the minimum rate the theorem actually requires.

Therefore, the correct answer is 16.