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Question

What is the minimum sampling frequency (Nyquist sampling rate) needed to sample any signal having maximum frequency component $\omega_m$ ?

The correct answer is
$$2\omega_m$$

Nyquist Sampling Rate Determination

The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency must be at least twice the highest frequency component present in the signal. This minimum sampling frequency is known as the Nyquist rate.

Applying the Nyquist Theorem

Let the maximum frequency component of the signal be denoted by $f_{max}$. According to the theorem, the minimum sampling frequency, $f_s$, required is:

$f_s \ge 2 f_{max}$

The question provides the maximum frequency component in terms of angular frequency, $\omega_m$. The relationship between angular frequency ($\omega$) and frequency ($f$) is $\omega = 2\pi f$. Therefore, the maximum angular frequency is $\omega_m = 2\pi f_{max}$.

The sampling frequency in terms of angular frequency, $\omega_s$, must satisfy:

$\omega_s \ge 2 \omega_{max}$

Substituting $\omega_{max}$ with the given $\omega_m$, the minimum sampling angular frequency ($\omega_s$) is:

$\omega_s \ge 2 \omega_m$

Thus, the minimum sampling frequency (Nyquist sampling rate) needed is $2\omega_m$.

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