At 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 at a sampling rate of 10 kHz, one must consider the relationship between sampling rate and Nyquist frequency. The Nyquist theorem states that in order to accurately sample a signal, the sampling frequency must be at least twice the highest frequency present in the signal.

At a sampling rate of 10 kHz, the highest frequency that can be accurately captured is 5 kHz (which is half of 10 kHz). The period of a wave is inversely related to its frequency, calculated using the formula:

[ \text{Period} = \frac{1}{\text{Frequency}} ]

For a 5 kHz signal:

[ \text{Period} = \frac{1}{5000 \text{ Hz}} = 0.0002 \text{ seconds} ]

[ \text{Period} = 0.2 \text{ msec} ]

This means that the shortest sine wave that can be resolved has a period of 0.2 msec. Any waveform shorter than this would not be accurately captured by the sampling frequency of 10 kHz, leading to potential distortion or incomplete representation of the waveform.

Therefore, the correct