If a system has a sampling rate of 10 kHz, what is the shortest sine wave that can be resolved?

To determine the shortest sine wave that can be resolved by a system with a sampling rate of 10 kHz, one must understand the Nyquist theorem, which states that to accurately sample a signal without aliasing, the sampling frequency must be at least twice that of the highest frequency component of the signal.

In this case, a sampling rate of 10 kHz means that the system can take 10,000 samples per second. Consequently, the time interval between each sample is calculated by taking the reciprocal of the sampling rate:

1 / 10,000 Hz = 0.0001 seconds = 0.1 ms.

The shortest sine wave that can be resolved by this system is determined by how many samples are needed to accurately describe one complete cycle of the sine wave. A sine wave typically requires at least two samples per cycle to be resolved — one for the positive half and one for the negative half. Therefore, the minimum period of the sine wave that can be accurately captured is twice the sampling interval:

0.1 ms (sampling interval) multiplied by 2 equals 0.2 ms.

This means that the shortest sine wave that can be resolved at a 10 kHz sampling rate is 0.2 milliseconds. This