Aliasing beating frequency

Question

As I understand aliasing is because you sample a high frequency with a low sample rate, and the sample points don't have enough information about the high frequency component. Here is an example of aliasing:

w = 5.0;(*signal frequency*) 
dt = 0.66125;(*sampling rate*)
f[x_] := Sin[w x]
ls = Table[f[x], {x, dt, 200 dt, dt}];(*sample points*)

ListPlot[ls, PlotRange -> All, Joined -> True, 
 DataRange -> {dt, 200 dt}]

we can see in the plot that there is beating. And the Fourier transform gives only one frequency(except the symmetric negative frequency one).

DFT[A_, ht_] := 
 RotateRight[
  ht/Sqrt[2 \[Pi]]*
   Fourier[RotateLeft[A, Length[A]/2 - 1], 
    FourierParameters -> {1, 1}], Length[A]/2 - 1]

ListPlot[Abs[DFT[ls, dt]]^2, PlotRange -> All, Joined -> True, 
 DataRange -> {-1/dt/2, 1/dt/2}, Axes -> False, Frame -> True]

question:

1. Why there is the beating but only one peak in the Fourier transform? Since there is only one frequency in the sample, what is it beating against?
2. How to calculate the beating frequency and the aliasing frequency?

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Update 1

As chirlu points out there is not beating, if it is the beating because of adjacent frequencies, we would expect to see two separated peaks if we increase the Fourier transform resolution. Here I increase the frequency resolution by including more sample points, and we can see it is a single peak, not adjacent peaks.

ls = Table[f[x], {x, dt, 800 dt, dt}];(*sample points*)ListPlot[ls, 
 PlotRange -> All, Joined -> True, DataRange -> {dt, 200 dt}]

---

Update 2

Here is an example using sound:

Play[Sin[3000 t], {t, 0, 2}, SampleRate -> 6000]

Play[Sin[3000 t], {t, 0, 2}, SampleRate -> 950]

So use a lower sample rate, we can actually hear the beating!

Answer 1

Let’s look at an example. I’m going to use a sample rate of 50 throughout this post (i.e. the functions are sampled at 0, 0.02, 0.04, 0.06, …) and simple sine waves of the form $f_a(t) = \sin (2\pi\cdot a\cdot t)$, where $a$ is the frequency of the sine wave. For example, this is a plot of $f_3$:

The red impulses indicate the sample values. The green line, on the other hand, is the continuous function that is being sampled. The sample values all lie on this curve, and it goes through three complete cycles, as one would expect.

Aliasing

Because the sample rate is 50, anything above the Nyquist frequency of 25 is going to alias. As an example, here is a plot of $f_{53}$; mind the axes, I zoomed in to one tenth of the above, because otherwise the image would have become quite crowded.

There is slightly less than one sample per cycle, so the sampled values don’t follow the continuous waveform very closely. Actually, if all you saw were the sample values without the green curve, like so …

… you might be tempted to draw a much slower wave. Here is what happens when we plot not only $f_{53}$, but $f_3$ at the same time:

While the two sine waves are quite dissimilar in appearance, they happen to coincide at each and every sample point. There is absolutely no way to tell them apart given only the sample values. That’s aliasing. In fact, not only the samples of $f_{53}$ are exactly the same as those of $f_3$, but $f_{103}$, $f_{153}$, $f_{-47}$ etc. alias in the same way.

Put differently, due to sampling at a frequency of 50, the ranges 0–25, 50–75, 100–125, etc. collapse. What about the “upper half”, from the Nyquist frequency to the sampling frequency? It is mirrored; $f_{47}$ is the same as $-f_3$, and in general the frequency range 25–50 aliases to 25–0, with the samples inverted (or, equivalently, a phase shift by $\pi$).

This means, for your question, that the effect you are observing is not due to you using signals that exceed the Nyquist frequency; you can replace any higher-than-Nyquist-frequency signal with its alias in the Nyquist range and will get exactly the same results.

The “beating”

So what causes the apparent “beating”? Here is the very first plot of $f_3$ again, but this time it shows the sample values and linear interpolation between them (i.e. straight line segments from each sample point to the next), not the actual continuous waveform that was sampled:

The interpolated waveform is a bit edgy, but overall a good approximation of the original sine wave. Unfortunately, this is only because the frequency is very low relative to the sample rate. Here is $f_7$:

Urgh. It worsens as we get to higher frequencies, closer to the Nyquist frequency; this is $f_{23}$:

Unfortunately, even if we don’t draw those straight lines, our eye will try to connect the dots all on its own, and the impression is similar: It looks like a slow oscillation modulated onto a fast one. And that impression actually isn’t completely beside the point:

The same sample values, from $f_{23}$, can equally well explained by a ring modulation of $f_2$ and $f_{25}$ (with some phase adjustment), namely, $g(t) = -f_2(t)\cdot f_{25}(t+0.01)$. Thus, $g$ and $f_{23}$ are aliases, when sampled at a rate of 50. Here is a close-up again, showing that $g$ and $f_{23}$ in fact meet at 0, 0.02, 0.04 etc.:

A different way to write $g$ is $g(t) = 0.5 f_{23}(t) - 0.5 f_{27}(t)$, a sum of two sine waves of similar frequency, exactly the case where beating could be expected to happen. So it is beating after all, isn’t it? Not really, because $f_{27}$ is beyond the Nyquist frequency; as outlined above, it is an alias of $-f_{50-27}=-f_{23}$, and so $g(t)$ becomes $0.5 f_{23}(t) - 0.5 f_{27}(t) = 0.5 f_{23}(t) + 0.5 f_{23}(t) = f_{23}(t)$.

This is, within the baseband from 0 to 25, the only possible interpretation of those sample values.

Conclusion

Watch out for improper interpolation! When reconstructing a continuous signal from samples, the limited bandwidth must be taken into account (band-limited interpolation). Unfortunately, our eyes are not good at this; and many software waveform editors and resamplers aren’t, either.

Answer 2

If there is apparent beating in the time domain, and you zoom into your apparent one peak in the frequency domain, you will find the main lobe or skirt around that peak is slightly fatter than if there is no beating. This is because there is really a second peak (and potentially more peaks) hidden inside and/or under that wider main lobe or it's skirts. It's not always one pure sinusoid frequency causing a peak in a zoomed-out FFT result. And two adjacent frequency sinusoids can cause an amplitude modulation in the envelope of the linear sum (beating).

Answer 3

Sudarsun is correct. You aren't seeing the additional frequencies because your sample rate is only 1.5Hz. You are limited due to the periodicity of the fourrier transform to a band of .0.75Hz to 0.75Hz. If you do it on paper you will see that sure enough by repeating the spectrum of the 5Hz sinewave by n*1.5Hz in both the + and - direction you will see exactly what you show in your graphics for a frequency span of -0.75Hz to 0.75Hz. You were mistaken to increase the resolution of the FFT, no need for that that. what you needed to do was increase the sampling rate without changing your original signal. To do that just pack 7 zeros between sample points. This will give you a sampling rate of 1.5*8 = 12Hz. I'll bet you see all the frequencies you think you should see...

Answer 4

If your signal is of frequency $f_0$ and you are sampling at $F_s$, then you can find all aliasing frequencies by $\sin(2\pi(f_0-mF_s)t)$ where $m$ is an integer such that $|f_0 - mF_s|< F_s/2$.

So if $f_0 = 6$ and $F_s = 10$, you have for $m = 1$, an alias at frequency of 4 Hz. However there are other aliases at $m = 2, 3$ etc. with frequencies of 14 Hz, 24 Hz etc.

Charan Langton

www.complextoreal.com You can see my book "Intuitive Guide to Fourier Analysis ...."