Question

A signal contains frequencies up to 4$\pi$ rad/sec. Minimum sampling frequency is:

• A. 4$\pi$
• B. 8$\pi$
• C. 2$\pi$
• D. 16$\pi$

Hint:

Always pay close attention to the units:
If the maximum frequency is in Hz ($f_m$), the minimum sampling frequency is $f_s = 2f_m$ Hz.
If it is in rad/s ($\omega_m$), the minimum sampling frequency is $\omega_s = 2\omega_m$ rad/s. The scaling factor of 2 remains identical.

Answer 1

The Correct Option is B

Step 1: Understanding the Problem:
This question asks for the minimum sampling rate (Nyquist rate) required to represent a continuous-time signal in the discrete-time domain without losing any information.

Step 2: Key Formula or Approach:
According to the Nyquist-Shannon Sampling Theorem, a band-limited continuous-time signal can be fully reconstructed from its samples if the sampling frequency ($\omega_s$) is at least twice the maximum frequency component ($\omega_m$) present in the signal:
\[ \omega_s \geq 2\omega_m \]

Step 3: Detailed Explanation:

• The maximum frequency component contained in the given signal is:
\[ \omega_m = 4\pi \text{ rad/sec} \]
• To prevent aliasing, the sampling rate must satisfy the Nyquist criterion.

• The minimum boundary for the sampling rate (called the Nyquist rate) is:
\[ \omega_s = 2 \omega_m \]
• Substituting the value of $\omega_m$:
\[ \omega_s = 2 \times (4\pi) = 8\pi \text{ rad/sec} \]
• If we sample at any frequency lower than $8\pi$ rad/sec, high-frequency components will overlap with low-frequency components, causing irreversible distortion (aliasing).

Step 4: Final Answer
Thus, the minimum sampling frequency is $8\pi$ rad/sec, which corresponds to option (B).