Question

Let \(x_1(t)\) and \(x_2(t)\) be two band-limited signals having bandwidth \(B = 4\pi \times 10^{3}\) rad/s each. In the figure below, the Nyquist sampling frequency, in rad/s, required to sample \(y(t)\), is _____________

• A. $20\pi\times 10^{3}$
• B. $40\pi\times 10^{3}$
• C. $8\pi\times 10^{3}$
• D. $32\pi\times 10^{3}$

Hint:

After mixing with $\cos(\omega_c t)$, the new highest frequency is $\omega_c + B$. For sums of branches, take the largest among them, then double it for the Nyquist angular rate.

Answer 1

The Correct Option is D

Step 1: Modulation spectra
Multiplication by $\cos(\omega_c t)$ shifts the spectrum: $x(t)\cos(\omega_c t)\ \Rightarrow\ \tfrac{1}{2}\big[X(\omega-\omega_c)+X(\omega+\omega_c)\big]$.
Each $x_k(t)$ is band-limited to $|\omega|\le B=4\pi\times 10^{3}$.
Step 2: Highest frequency in $y(t)$
Top branch: $\omega_{c1}=4\pi\times 10^{3}$ gives bands on $[\,\omega_{c1}-B,\ \omega_{c1}+B\,]=[\,0,\ 8\pi\times 10^{3}\,]$ (and symmetric negative).
Bottom branch: $\omega_{c2}=12\pi\times 10^{3}$ gives bands on $[\,8\pi\times 10^{3},\ 16\pi\times 10^{3}\,]$ (and symmetric negative).
Hence $y(t)$ contains frequencies up to $\omega_{\max}=16\pi\times 10^{3}\ \text{rad/s}$.
Step 3: Nyquist rate
Nyquist angular sampling frequency $\omega_s \ge 2\omega_{\max}=2. 16\pi\times 10^{3}=32\pi\times 10^{3}\ \text{rad/s}$.
\[ \boxed{32\pi\times 10^{3}\ \text{rad/s}} \]