What is happening when a signal is acquired at exactly Nyquist frequency?

Question

So I am doing some some simulations with 1V sinusoidal signals, and a 1 second acquisition at sampling frequency of 100 kHz.

Now I am slowly increasing the frequency of the sinusoidal signal. Trying 1 kHz, 10 kHz, 20 kHz, 40 kHz... Everything goes as planned and the FFT produces a spectrum with a mazimum equal to 1 V (aprox.) at the frequency of the signal. However at the exact nyquist frequency, 50 kHz, the amplitude is heavily reduced and there seems to be a DC component. Is there any reason why this occurs? What is failing in the theoretical Nyquist theorem.

Answer 1

Here's a waveform sampled at its zero crossing points: -

Image from here.

Based on the samples taken, you'd estimate that there was no signal and that ties in with this: -

at the exact Nyquist frequency, 50 kHz, the amplitude is heavily reduced

On the other hand, if the samples happened to be offset by 90° you'd conclude that the amplitude was correct.

there seems to be a DC component

Your image shows virtually nothing (1.6 mV) but, with some waveforms there might be a residual DC offset or, it might just be some small DC value from some signal processing chain.

What is failing in the theoretical Nyquist theorem.

Nothing as far as I can see. The theorem states that the sampling frequency has to be greater than the maximum signal frequency.

Answer 2

You get \$(-1)^n\sin(\phi)\$ where \$\phi\$ is the phase that the signal is at at the first sampling point. Essentially sampling a signal at exactly half the Nyquist frequency only recovers the cosine part of the cosine/sine quadrature pair. The sine part falls through the cracks because it is always zero at the sampling points.

So when sampling, you get an alternating signal with an amplitude between zero and the actual signal amplitude, depending on the phase you start sampling at.