Question

The Fourier Transform of the sampled signal is given by

• A. The multiplication of shifted versions of the impulse signals
• B. The original signal's Fourier Transform
• C. A finite sum of unit step signal's Fourier Transform
• D. An infinite sum of shifted versions of the original signal's Fourier Transform

Hint:

Sampling in time domain produces periodic replicas in frequency domain separated by \(\omega_s\).

Answer 1

The Correct Option is D

Concept: Sampling in time corresponds to periodic repetition in the frequency domain.

Step 1: Represent the sampled signal.
\[ x_s(t) = x(t) \sum_{n=-\infty}^{\infty} \delta(t-nT). \]

Step 2: Apply Fourier Transform.
The transform becomes \[ X_s(j\omega) = \frac1T \sum_{k=-\infty}^{\infty} X\!\left( j(\omega-k\omega_s) \right). \]

Step 3: Interpret the expression.
The sampled spectrum consists of infinitely many shifted replicas of the original spectrum. Hence, \[ \boxed{ \text{An infinite sum of shifted versions of the original spectrum} } \]