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We sample each other sample (01010101). Now say we have a signal - that is inside the original sample which has the pattern of (010-1). Decimation gives us 000 for this sub-signal. How to prove this is an example of aliasing? Or why else do we have lost information?

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  • $\begingroup$ I'm missing something in the explanation...you want to retain every other sample (which you show by 01010101, meaning we skip a sample, keep a sample, skip...). I don't understand what the "-" means in the pattern "010-1" (which is part of the original signal), and why decimating it would yield "000". $\endgroup$ – Nigel Redmon May 9 '20 at 18:16
  • $\begingroup$ Hi Nigel, thanks. ‘-‘ It’s minus one. Like in sinus. Also, what are the distances between the samples? $\endgroup$ – Vitali Pom May 9 '20 at 18:18
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    $\begingroup$ I'm not 100% sure I understand everything you're implying. I gave a more detailed answer, feel free to discuss it there and I can discuss it mroe if needed. $\endgroup$ – Nigel Redmon May 9 '20 at 18:53
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You really need to look at the problem in the frequency domain. There are too many bit patterns that will create an obvious dilemma with respect to decimation, but these will invariably be in violation of the rules of the sampling theorem—something obvious in looking at the frequency spectrum, but perhaps not obvious looking at a list of sample values.

For instance, the simple case of decimating the signal 10101010... by a factor of two can yield 000... or 111..., depending on where you start. But the original signal is at half the sample rate (for instance, a cosine at half the sample rate, amplitude 0.5, with a DC offset of 0.5). This does not satisfy the requirements of the sample rate being greater than twice the highest frequency component.

You'll find that other patterns that seem to produce results that make no sense will have similar problems with the initial condition. And while some cases, like the example I gave, are fairly easy to reason out, in general you won't be able to tell if the result of decimation is aliased without comparing it in some manner to the original.

That is, there is no way of known whether a series of samples is aliased without knowing more. For instance, you might know that the samples are a recording of flute solo—in that case, it might be very easy to answer whether it is aliased, even eithout the original signal to compare. Otherwise, if you can assume nothing, even bad sounding audio may be as intended.

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  • $\begingroup$ Checking now the sampling theorem you’ve mentioned. Thanks @Nigel. $\endgroup$ – Vitali Pom May 9 '20 at 18:57
  • $\begingroup$ Hi Nigel (again :) ). Thanks for being ready to help. Could you explain using the sampling theorem how many times at least a sine function should be sampled? $\endgroup$ – Vitali Pom May 9 '20 at 19:16
  • $\begingroup$ Saw something now. So sinus shall be sampled at least two times? It has 3 zeroes and it’s asymmetric , so it makes sense. And so in my original answer there might be no aliasing, but after decimation the sampling theorem is not held. Uhm, hurray? $\endgroup$ – Vitali Pom May 9 '20 at 19:22
  • $\begingroup$ Hi Vitali—The sample rate must be more than twice the frequency of the highest component frequency in the signal. So that means there must be something more than two samples per cycle of the sine wave. The hard part to wrap your head around, I'm sure, will be "how much more?" In theory, any more at all. In practice we need a realizable lowpass filter just below half the sample rate, so you'll never really be so close to the edge case that you're sampling a 23.999999 kHz sine at 48 kHz—it would be wiped out by the filter anyway. $\endgroup$ – Nigel Redmon May 9 '20 at 19:26
  • $\begingroup$ But if sine wave goes in full cycle once per second. Then its frequency is 1/second. Hence twice is twice a second. I think sampling twice shall be enough. Think about it’s shape. No? $\endgroup$ – Vitali Pom May 9 '20 at 19:29
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The whole point of aliasing and the sampling theoreme is that you cannot (generally) know what an aliased signal was, as you cannot represent infinite bandwidth using finite infornation.

If you have extra knowledge of the input signal (eg that DC is not possible, then you might deduce what a string of 1-1-1 was originally.

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  • $\begingroup$ Does the wide most signal participating in the original signal ‘set’ its width? $\endgroup$ – Vitali Pom May 9 '20 at 17:51
  • $\begingroup$ Answering myself: highest frequency is the one that sets the sample rate. This is according to Sampling Theorem. $\endgroup$ – Vitali Pom May 9 '20 at 19:31
  • $\begingroup$ In simple terms, sample-rate > 2*highest frequency. By how much depends on your filter design/delay requirements. More accurately, it depends on the bandwidth of your continous domain signal. If you are sampling a signal of bandwidth 1kHz that is modulated to 1Mhz, it is (in principle) possible to sample that signal using >2kHz sampling rate. If you can enforce rigid assumptions about your signal you might even go lower than that (compressed sensing stuff that I do not know). $\endgroup$ – Knut Inge May 10 '20 at 15:00
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Sampling theorem says that we shall ALWAYS sample at least twice as the sampling rate. So perhaps in my 1st example there is no aliasing. But Sampling Theorem is not being held after decimation. (Thanks Nigel for helping me to succeed. You rock!)

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