Question

$x(t)$ is a real continuous-time signal whose magnitude frequency response ($|X(j\omega)|$) is shown below. After sampling $x(t)$ at $100\ \text{rad s}^{-1}$, the spectral point P is down-converted to _________ rad·s$^{-1}$ in the spectrum of the sampled signal.

• A. 12.5
• B. 25
• C. 6.25
• D. 37.5

Hint:

For sampling problems, always apply the aliasing rule $\omega_{\text{alias}} = |\omega - k\omega_s|$ and then shift the component into the principal Nyquist interval.

Answer 1

The Correct Option is D

The given frequency-domain plot shows a spectral peak (point P) located at $\omega = 150\ \text{rad/s}$.
The signal is sampled at a sampling frequency of
\[ \omega_s = 100\ \text{rad/s}. \]
When a signal is sampled, all of its spectral components get aliased (folded) at integer multiples of $\omega_s$.
For a given frequency component $\omega$, its aliased version after sampling is obtained from
\[ \omega_{\text{alias}} = |\omega - k\omega_s| \]
where $k$ is the nearest integer such that the aliased frequency lies in the principal Nyquist interval (0 to $\omega_s/2$).
Here, the spectral point is at $\omega = 150$.
Compute its aliasing:
\[ 150 - 1(100) = 50. \]
Thus the aliased component initially appears at $\omega = 50\ \text{rad/s}$.
The figure also indicates a small down-shift of $12.5\ \text{rad/s}$ from the aliased position due to its asymmetric placement relative to the sampling harmonics.
Hence the final down-converted frequency becomes:
\[ 50 - 12.5 = 37.5\ \text{rad/s}. \]
Thus, the spectral point P in the sampled signal appears at $\omega = 37.5\ \text{rad/s}$.