Question

Let an analog x(t) signal be sampled. It is assumed that the highest frequency component of this signal is at 2000 Hz. Circle the correct answers and justify below the chosen answers:

I. We can sample with fe=2000 Hz.
II. We can sample with fe=4010 Hz.
III. We can sample with fe=8000 Hz.
IV. We can sample with Te=0.002 sec.
V. We can sample with Te=0.0002 sec.

Answer 1

The correct sampling rates for an analog signal with a highest frequency component of 2000 Hz are 4010 Hz, 8000 Hz, and when the sampling period is 0.0002 sec. Sampling at 2000 Hz is not sufficient as it does not meet the Nyquist criterion.

Step-by-step explanation:

The question deals with the sampling of an analog signal with the highest frequency component of 2000 Hz. According to the Nyquist theorem, to accurately sample and reconstruct a signal, the sampling frequency, fe, must be at least twice the highest frequency component of the signal, known as the Nyquist rate. This means that the signal must be sampled at a frequency greater than or equal to 4000 Hz.

• We cannot sample with fe=2000 Hz because it is equal to the highest frequency and not twice the highest frequency.
• We can sample with fe=4010 Hz because it exceeds the Nyquist rate of 4000 Hz.
• We can sample with fe=8000 Hz because it far exceeds the Nyquist rate.
• We cannot sample with Te=0.002 sec because this translates to a sampling frequency of 500 Hz, which is not sufficient.
• We can sample with Te=0.0002 sec, which equates to a sampling frequency of 5000 Hz, exceeding the Nyquist rate.

Therefore, the correct answers are II, III, and V.