Why complex signal doesn't remove "beatings" while sampling?

Question

My motivation to this post has arisen from the accepted answer to Amplitude modulation vs sampling rate?

Sampling the signal with the frequency close to $f_s/2$ can lead to some kind of "beatings" of the sampled signal. It is clearly seen if I try to sample simple cosine signal:

Here FFT shows both positive and negative frequencies of the signal as it should be.

But if I try complex signal (here it is $e^{j2\pi f t}$) and left plot is real part of my complex signal:

In this case FFT shows only one (positive) frequency component, but beatings of the sampled signal still exist. My question is why?

The Python code to reproduce these plots:

fig, ax = plt.subplots(2,2, figsize=(15,5))
fs = 20
f = 8.8
t = np.arange(0, 2, 1/(fs*fs))
t_sampled = np.arange(0, 2, 1/fs)
signal_complex = np.exp(1j(2np.pift_sampled))
signal_real = np.cos(2np.pif*t_sampled) # cosine
ax[0,0].plot(t, np.cos(2np.pif*t))
ax[0,0].plot(t_sampled, signal_real, color='r', marker='o')
ax[0,0].set_title('Time domain (cosine signal)')
ax[1,0].plot(t, np.exp(1j(2np.pift)).real)
ax[1,0].plot(t_sampled, signal_complex, color='r', marker='o')
ax[1,0].set_title('Time domain (exponential signal)')
FFT_complex = np.fft.fft(signal_complex)
FFT_real = np.fft.fft(signal_real)
freqs_fft = np.fft.fftfreq(len(t_sampled), t_sampled[1])
ax[0,1].stem(freqs_fft, abs(FFT_real))
ax[0,1].set_title('FFT (cosine signal)')
ax[1,1].stem(freqs_fft, abs(FFT_complex))
ax[1,1].set_title('FFT (exponential signal)')
plt.tight_layout()

Answer 1

In this case FFT shows only one (positive) frequency component, but beatings of the sampled signal still exist. My question is why?

Why would it? Real vs. complex has nothing to do with the phenomenon we are looking at. The only difference between a real and complex sinusoid is that the former has two spectral lines and the latter only one.

You seem to be confusing two different effects here.

The so-called "beating" does NOT exist. It's simply a visual artifact of the way you draw the waveform. Most plotting routines simply connect samples with straight lines which is simply wrong. If you were to draw this using proper band-limited interpolation at each pixel (and with enough pixels) you get the correct waveform. This has no bearing on the frequency domain at all.

The spreading of the spectrum you see is spectral leakage. There are plenty of post on this forum about this. Spectral leakage occurs for sine waves whose frequency doesn't line up with the frequency grid of the FFT. As long as your frequency is an integer multiple of the bin resolution (sample rate divided by FFT length), you will only get a single spectral line, regardless whether there "visual beating" or not in your time domain graph.

In your case, the frequency resolution is 0.5Hz so if you choose 8.5Hz or 8Hz (instead of 8.8Hz) all the spectral artifacts will disappear.

EDIT: Derivation of the apparent beat at $(f_s/2-f)$

Let's say we have a sequence $x[n] = \cos(\omega n)$. If $\omega$ is close to Nyquist we can write this $\omega = \pi-\delta$ where $\delta$ is the frequency difference to Nyquist.

Hence we get $$x[n] = \cos(\pi n - \delta n) = \cos(\pi n )\cos(\delta n) + \sin(\pi n )\sin(\delta n)$$

$\sin(\pi n ) = 0$ and $\cos(\pi n ) = [+1,-1,+1,-1,...] = (-1)^n$ so the whole thing becomes

$$x[n] = (-1)^n\cos(\delta n) = (-1)^n\cos(2\pi(f_s/2-f) n) $$

Note that we haven't done any actual manipulation here, we've just written the samples of the sine wave in a different form.

You visually interpret this as something that's oscillating fast multiplied with a slow sine wave. That sort of looks like a beat, but it's not. It's still a perfectly good sine wave. There is no actual amplitude modulation whatsoever.

Answer 2

Through the commentary I now understand the crux of the OP's question which I will summarize as follows:

As detailed in the referenced post (https://dsp.stackexchange.com/a/70884/21048) I was able to explain how the underlying analog signal prior to sampling could have either been a single real tone OR a double-side-band-suppressed-carrier (DSB-SC) modulation with the suppressed carrier at $f_s/2$ (and there are infinitely many other solutions that would result in the same discrete samples that resulted). The OP has astutely observed that the same result (the beats in the time domain) can occur even if the input isn't real, which seems to counter the explanation provided.

The reason is the time domain plot shown is just the real component. In doing that the spectrum of the real component alone is identical to the first spectrum shown above which is the spectrum of the real signal NOT the complex signal. To observe the complex time domain signal we must observe two real signals (either magnitude and phase or real and imaginary). From a plot of magnitude and phase of the complex case, it will be clear that the magnitude is constant at all time samples while the same plot of magnitude and phase for the real case will show the beating as observed. If we choose to instead plot real and imaginary, then yes each of those will show beating but it is in such a way to maintain a constant magnitude (as one goes up the other goes down). Beating refers to an Amplitude Modulation, and with the complex waveform as $I+jQ$ we must consider both $I$ and $Q$ together to measure the amplitude, not individually. So plotting amplitude as $\sqrt{I^2+Q^2}$ will show that there is no beating (no AM), unless we take the real or take the imaginary only of the resulting waveform.

Consider the simple case for the spectrum of $e^{j \omega_o t}$ as plotted below without the added complexity of sampling, folding and spectral leakage:

$$e^{j \omega_o t} = \cos(\omega_o t) + j \sin (\omega_o t)$$

So if we were to only plot the real part of the above complex spectrum in the time domain, we would get $\cos(\omega_o t)$. However to be complete the spectrum of this would then appear as below:

So bottom line, the second time domain plot on the left showing beating together with a complex spectrum on the right is incomplete. The time domain plot is not the time domain for the spectrum given, but the time domain for the real portion of the spectrum given (which would then be identical to the spectrum in the upper plot).