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Below solid black curve is the bandwidth of a signal(a 4kHz RC low pass filter). And when we sample the signal at 8kHz rate due to folding the blue dashed line shows the aliasing:

enter image description here

1-) It looks very symmetric but how can we calculate the overall shape of the original bandwidth plus the aliasing? Should we add or multiply these curves?

2-) And from here I found the above curve which the author is able to quantify 50:1 signal energy to alias ratio. What is meant by that and how can that be calculated?

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  • \$\begingroup\$ For the above vurve, the author would be wrong. \$\endgroup\$ Commented Apr 20, 2020 at 11:39
  • \$\begingroup\$ The author says this: Can you fix this problem by increasing the filter order? Not really. Figure 7 on page 10 shows the result of using a 6th-order Butterworth low pass filter instead of a 1st-order filter. We’ve improved the situation So he says we can't improve the situation then he says the situation has improved. Gobbledy gook. \$\endgroup\$
    – Andy aka
    Commented Apr 20, 2020 at 11:41
  • \$\begingroup\$ hey, @TimWescott, you might want to weigh in here :) To be honest, your phrasing isn't that good. \$\endgroup\$ Commented Apr 20, 2020 at 11:49
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    \$\begingroup\$ Prepare for the ensuing bun-fight LOL \$\endgroup\$
    – Andy aka
    Commented Apr 20, 2020 at 11:52
  • \$\begingroup\$ @Andyaka a weapon from a more civilized time *grin* \$\endgroup\$ Commented Apr 20, 2020 at 12:09

2 Answers 2

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You add the energies in the signal and aliased curves.

Now if you assume the aliasing is completely de-correlated with the signal, you can treat them as two independent sources (like a "random walk" with 2 steps in statistics) you can use an RMS summation of the amplitudes.

Since the crossover at 4 kHz has both amplitudes as 0.7, or -3dB, the result is simply a flat line at 1 independent of frequency.

But it's a dangerous assumption and probably wrong for certain signals, so it's worth considering a worst case scenario, where the aliasing is perfectly correlated and in phase with the signal : in this case you can add amplitudes, giving a peak of 1.4 (+3dB) at 4 kHz.

With regard to:

Can you fix this problem by increasing the filter order? Not really. Figure 7 on page 10 shows the result of using a 6th-order Butterworth low pass filter instead of a 1st-order filter. We’ve improved the situation

Well he's right : you can't fix the problem simply by increasing filter order. You can improve it (decreasing the "alias" energy below 4 kHz and narrowing the +3dB peak in the worst case, thus decreasing its area). Thus you have improved the situation somewhat, but not fixed it.

A fix : if you need to reduce the aliasing energy below some level : would involve reducing the filter cutoff frequency - as well as optionally increasing the order - such that the attenuation at Fs/2 (4 kHz here) met your target.

And regarding comments to the other answer : aliasing below 4 kHz is the system response (due to the sampling process) to energy after the filter above 4 kHz. (which btw sounds nasty).

Aliasing at 7 kHz is huge because the filter has no attenuation at 1 kHz. It would also sound nasty but will be removed by the DAC's reconstruction filter.

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  • \$\begingroup\$ Thanks for this nice answer. But I have a question. In other answer the user in the comments says "There's no "aliasing at 7 kHz"." On the other hand you have written "Aliasing at 7 kHz is huge because... " So Im confused about where the aliasing starts. Is there aliasing at 7kHz or not? \$\endgroup\$
    – GNZ
    Commented Apr 20, 2020 at 17:21
  • \$\begingroup\$ Yes of course. A 1 kHz input signal will be aliased to 7 kHz by the sampling process, as well as appearing as a 1 kHz signal. Normally that 7kHz component would be removed by he DAC reconstruction filter (which is now generally a digital filter before an oversampled DAC) \$\endgroup\$
    – user16324
    Commented Apr 20, 2020 at 17:33
  • \$\begingroup\$ I never thought a lower frequency would be aliased to higher frequency. I thought aliasing only happens from higher freq to lower freq. I started to feel like I don't know some fundamental concept here.,, So just to be clear I wan to check if I understand what you mean; is this correct? --> 1kHz input signal will be aliased to 7kHz but 7kHz input will not be aliased to 1kHz. \$\endgroup\$
    – GNZ
    Commented Apr 20, 2020 at 19:35
  • \$\begingroup\$ Almost. With the given anti-alias filter, 1 kHz will alias to 7 kHz (and 9 kHz, 15 kHz, 17 kHz, and every n*Fs +/- 1 kHz). 7 kHz WILL alias to 1 kHz but with amplitude < 0.05 (looks like 0.03 from my reading of your graph). \$\endgroup\$
    – user16324
    Commented Apr 20, 2020 at 19:52
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It looks very symmetric but how can we calculate the overall shape of the original bandwidth plus the aliasing? Should we add or multiply these curves?

Neither!

This is just an amplitude spectrum plot. There's nothing said about the phase of the signals at the individual frequencies.

For example, if the signal at originally 12 kHz was highly correlated, in that it was the same, but with a negative sign, as the signal at 4 kHz, the sum of nyquist zone + alias would cancel out. If it was in-phase, it would constructively add.

If we assume the signal to be uncorrelated, e.g. by being white noise, OK; the powers would add. Powers adding up means you get the square root of the resulting power as amplitude; so not even in that case could you just add the amplitude graphs.

(That's one of the practical reasons why we use power spectral densities and not amplitude spectral densities to describe stochastic signals, by the way.)

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  • \$\begingroup\$ Will the signal be contaminated by aliasing only for inside the first Nyquist zone? I mean in my question above 4kHz the blue aliasing line is almost one very high. Will this cause any aliasing above 4kHz? (sampling rate is 8kHz) \$\endgroup\$
    – GNZ
    Commented Apr 20, 2020 at 12:23
  • \$\begingroup\$ @GNZ I think you need to revisit the very basics of sampling: if you sample at 8 kHz, there's no "signal" above 4 kHz that you can unambiguously represent. (or I am misunderstanding your comment. \$\endgroup\$ Commented Apr 20, 2020 at 12:25
  • \$\begingroup\$ Oh I think you got my confusion now. Under 4kHz there is no aliasing anyway in this case then. What confused me the "entire" blue line is associated with aliasing but in fact the aliasing occurs only below 4kHz in this case. Did I get it correct? \$\endgroup\$
    – GNZ
    Commented Apr 20, 2020 at 12:27
  • \$\begingroup\$ the blue line is the alias \$\endgroup\$ Commented Apr 20, 2020 at 12:34
  • \$\begingroup\$ Sorry I meant the blue line for above 4kHz there is no alias for the signal correct "for 8kHz" sampling rate? \$\endgroup\$
    – GNZ
    Commented Apr 20, 2020 at 12:36

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