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Suppose that there are two periodic signals with particular frequencies. Two signals are then combined into one signal.

Suppose that we take finite samples of these signals. (so, finite time of transform.)

Then, is it possible to figure out the frequency content of the two original signals using any kind of transform?

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  • \$\begingroup\$ What is a time-limited periodic signal? Not that it's really relevant to your question. \$\endgroup\$
    – The Photon
    Jun 14, 2012 at 5:08
  • \$\begingroup\$ @ThePhoton I edited the question. Hope this clarifies:) \$\endgroup\$ Jun 14, 2012 at 5:12
  • \$\begingroup\$ "unique frequencies" means f1/f2 is irrational. Is that what you mean? \$\endgroup\$
    – stevenvh
    Jun 14, 2012 at 7:20
  • \$\begingroup\$ @stevenvh not really. I never knew unique frequencies mean f1/f2 being irrational.. \$\endgroup\$ Jun 14, 2012 at 7:36
  • \$\begingroup\$ If f1/f2 is rational with reduced fraction a/b, then the b-th harmonic of f1 and the a-th harmonic of f2 coincide. It's the "unique" that confused me. If f1/f2 is irrational they will never have coinciding harmonics. \$\endgroup\$
    – stevenvh
    Jun 14, 2012 at 7:42

5 Answers 5

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What you are asking about falls into the realm of spectral estimation, of which frequency estimation is a particular case.

In general, with no prior knowledge about the frequencies in your original signals, you cannot do what you ask. One fundamental reason is aliasing. If you sample at some frequency \$f_s\$, you cannot distinguish input frequencies \$nf_s - f_0\$, \$nf_s + f_0\$ for different n from 0 to infinity. For example, if you sample at 100 Hz, you can't tell the difference between a 20 Hz input, an 80 Hz input, 120 Hz, 180, 220, etc.

Another limitation, if you know nothing about your input frequencies, is that the precision with which you can estimate your input frequencies is limited by the length of time you sample for. For example, if you sample at say 100 Hz for 1 s, you might (very roughly speaking) just barely be able to distinguish between a 20 Hz input and a 21 Hz input. If you sample for 100 s, you might barely be able to distinguish between a 20 Hz input and a 20.01 Hz input, etc.

These limitations apply whether you have an input that is formed by combining two independent sources or if you just have a single, pure, sinewave input and you want to estimate its frequency. Of course the second limit has some relevance if you have two inputs at closely spaced frequencies and you want to be able to separate them.

You may also be able to get some help with this at dsp.stackexchange.com, though the typical answer over there requires a substantial amount of mathematical background that may make them difficult to understand.

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  • \$\begingroup\$ Well, if you sample at 100 Hz you cannot reconstruct a 120 Hz signal anyway \$\endgroup\$
    – clabacchio
    Jun 15, 2012 at 5:43
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    \$\begingroup\$ @clabacchio, you can if you have a priori knowledge that the signal is bandlimited in a 50 Hz band. For example, if you know all content of the signal is between 100 and 150 Hz. \$\endgroup\$
    – The Photon
    Jun 19, 2012 at 21:45
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If you know that a signal is periodic, you need only one period to calculate the Fourier transform. The problem is determining what will be the period of the two signals together, once you sum them.

You say that the signals have unique frequencies, I guess you mean sinewaves. Having unique frequencies, they will have a peak in the spectrum, and nothing else (ideally).

Being the Fourier transform linear, the sum of the signals will have a spectrum whih is the sum of the spectra of the single sinewaves, so it will have two peaks in the exact same position of the other two signals.

If the sum signal has a finite period, it will be the minimum necessary interval on which you have to integrate to transform it, otherwise you will introduce an error and you won't be able to reconstruct the original sinewaves.

Note that finite samples has two implications: you sample on a finite interval in time, and you sample with a finite frequency. Both are constraints for the signals you want to transform: the first works only if the transformed signal has a finite period (as we said), the second is given by the Nyquist-Shannon theorem, and means that you need to sample at least at 2x the maximum frequency of the signal. You can observe that the first constraints puts a lower limit in the frequency of the signal, and the second puts an upper limit.

As a side note: you won't be able to reconstruct the signals if their spectra overlap, so if they are sinewaves, if their frequencies are equal.

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If the two frequencies are unique, ie) f1/f2 = irrational, then the product signal will not be periodic (I have a proof for this if you care)

You cannot do a Fourier series on a non-periodic signal. The best you could do is an approximate Fourier analysis by assuming the period is some acceptable value. This will lead to an approximate reconstruction.

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Quite often we "combine" several signals, send them over a single wire or fiber or through the air, as a single combined signal, and at the far end separate them back into more-or-less the original signals.

There are dozens of different ways we "combine" signals.

Then, is it possible to figure out the frequency content ... using any kind of transform?

We almost always use a fast Fourier transform to figure out the frequency content of a signal. (A few people are experimenting with other techniques for finding the frequency content of a signals -- the Hartley transform, the fast Walsh–Hadamard transform, the chirplet transform, wavelet transforms, etc.)

Suppose that we take finite samples of these signals. (so, finite time of transform.)

In practice, whenever we sample a signal, we always take a finite number of samples of a signal, over some extremely limited amount of time.

We must already know (or guess) at the frequency range of interest in order to select the proper anti-aliasing filter and sample rate.

Then, is it possible to figure out the frequency content of the two original signals using any kind of transform?

You must know something about how the signals were combined, and something about the original signals, in order to get any useful information out of the combined signal.

For example, say you know that the original signals were audio in the human hearing range, and the first signal was AM modulated to 540 kHz, while the second signal was AM modulated to 1610 kHz, and then the two modulated signals were added together. In that case it's pretty easy to figure out the frequency content of each of the two original audio signals, looking at the display of a RF spectrum analyzer (which is often implemented by taking a finite number of samples and then applying a fast Fourier transform). It's also pretty easy to demodulate and recover signal A and signal B that are more-or-less the same as the original first signal and second signal, respectively.

For another example, say you know that the original signals were audio in the human hearing range, and they were AM modulated and then the modulated signals were added together, but you don't know ahead of time what particular modulation frequency was used. In that case it's pretty easy to look at the display of a RF spectrum analyzer and figure out what modulation frequencies were used. Then you can tell if the modulation frequencies are far enough apart, and if so, you can demodulate and recover signal A and signal B that are more-or-less the same as the original first signal and second signal; and figure out the frequency content of signal A and signal B -- but it's impossible to tell if the first signal goes with signal A and the second signal goes with signal B, or vice-versa.

On the other hand, if all you know is that the original signals were audio in the human hearing range, and then they were simply added together, (or if each one was AM modulated to approximately the same frequency and then added), it's pretty much impossible to separate them. You can figure out the frequency content of the combined signal, by looking at the spectrum -- the Fourier transform of the combined signal. If you are lucky, and the signals were not synchronized with each other, you can figure out some information about the frequency content of the two original signals. When you see "quiet" frequencies in the combined signal, both the first and the second signal must be quiet at that frequency. When you see "loud" frequencies in the combined signal, either the first or the second signal must be producing that frequency. But without more information, it's impossible to tell if that "loud" frequency is only being produced by the first signal, or only by the second signal, or by both the first and second signals.

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If you run a fourier transform over a finite sampling period which is not an integral number of signal periods, you will experience spectral distortion as what you are actually doing is taking the transform of the input signal convolved with your off/on-for-a-while/off-again sampling window. This will give you the transform of the input signal multiplied by the transform of the rectangular sampling window, which is a sync function ((sin x)/x).

To avoid such frequency domain distortion, it is common to multiply the input sample data by a window function, which has the effect of "fading" the signal in and out gradually to avoid the distortions caused by the abrupt start and end of the sampling. There are various window functions with advantages and disadvantages; a simple one is (cos x) + 1 scaled to the period of sampling.

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  • \$\begingroup\$ Chris, your main point is correct, but the windowing effect is multiplication in the time domain and convolution in the transform domain, not the other way around like you have it. \$\endgroup\$
    – The Photon
    Aug 14, 2012 at 17:45

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