Question

Consider a real-valued base-band signal \( x(t) \), band limited to 10 kHz. The Nyquist rate for the signal \( y(t) = x(t) \cdot \left( 1 + \frac{t}{2} \right) \) is

• A. 15 kHz
• B. 30 kHz
• C. 60 kHz
• D. 20 kHz

Hint:

When multiplying a signal by a linear function, the resulting signal's bandwidth increases, and the Nyquist rate is twice the new highest frequency component.

Answer 1

The Correct Option is B

The given signal \( x(t) \) is band-limited to 10 kHz, meaning that the highest frequency component in \( x(t) \) is 10 kHz. The Nyquist rate is the minimum sampling rate required to completely reconstruct the signal without aliasing, which is twice the highest frequency present in the signal. For the given signal \( y(t) = x(t) \cdot \left( 1 + \frac{t}{2} \right) \), this is a product of \( x(t) \) and another function. When multiplying two signals, the resulting signal's bandwidth is the sum of the individual bandwidths. Since \( x(t) \) has a bandwidth of 10 kHz, the bandwidth of the function \( \left( 1 + \frac{t}{2} \right) \) (which is a linear term) increases the bandwidth of the signal. The frequency components of \( y(t) \) are therefore doubled, leading to a Nyquist rate of 30 kHz, which is twice the highest frequency in the product. Thus, the Nyquist rate for \( y(t) \) is 30 kHz. Final Answer: 30 kHz