What would happen if I am using the maximum frequency as the sample rate for sampling a pure tone sine wave?

For example, a $10\ \rm kHz$ sampling frequency for a $10\ \rm kHz$ monotone sine wave. What effect would aliasing produce at the output?

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    $\begingroup$ Could you elaborate on what exactly it is that you don't understand? You sample a sine wave once per period, so what will the result be? $\endgroup$ – Matt L. Dec 3 '20 at 12:58
  • $\begingroup$ yeah, what will be the result be? $\endgroup$ – Mark Steven Dec 4 '20 at 10:48

In contrast to common misunderstanding, aliasing does not affect the process of sampling, rather it affects the process of reconstruction. i.e., the samples themselves do not contain error, but their interpretation at the reconstruction (or anything related with it) will be in error.

So, if you sample a sine wave $$x(t) = A \sin( 2 \pi f_0 t + \theta) $$ at a sampling frequency of $F_s = f_0$, (which is undersampled according to Nyquist-Shannon sampling theorem for bandlimited signals), then your samples will be $$ x[n] = A \sin(\theta) $$ a constant signal! When reconstructed back into continuous waveform at the interpolator output, you will get a DC signal, $x(t) = A \sin(\theta)$, as opposed to the original sine wave.

The error is due to the false assumption at the interpolation stage that those samples $x[n]$ came from an original continuous-time signal which was band-limited to $f_0/2$, Nyquist frequency of sampling frequency $f_0$, which is not the case in reality.

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    $\begingroup$ But if you can make sure that no DC can enter the sampling process, that the only input signal trigging that constant sampled waveform is that sine of f0=fs, then you can safely reconstruct a sine at >= fs/2. Which means that Nyquist works for a bandwidth limited to <fs/2, not necessarily the baseband range of [0...fs/2> $\endgroup$ – Knut Inge Dec 3 '20 at 23:03

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