Suppose that a continuous aperiodic signal $g(t)$ is Transformed by the Fourier Transform in $G(\omega)$, where $t$ is the time domain and $\omega$ the frequency domain. What would be the meaning of the value $G(\omega)$? Mathematically why we get that so close relationship between the domains?

  • $\begingroup$ I'm not quite sure what kind of detail you are looking for, but I think you might be looking for Fourier transforms in the context of operator theory. Those keywords might be sufficient to start down that long path. The fact that the Fourier transform is defined as an integral means that you have linearity (superposition) of signals. We also get a useful type of convolution when transforming the product of two signals. The fact that the Fourier transform may be considered a kind-of continuous Fourier series, gives the frequency interpretation of $w$. This is just one in a class of transforms. $\endgroup$
    – adfriedman
    Commented Aug 5, 2017 at 3:13
  • $\begingroup$ There have been a few days that i started to study the Fourier Transform in the Geophysics context.I know that the Fourier Transform have pretty good properieties such as linearity or the killing of derivates , what i would know is why to refer the domain of the Fourier Transform as the frequency $\endgroup$ Commented Aug 5, 2017 at 3:26
  • $\begingroup$ Joan, have you used the Laplace Transform in the context of linear differential equations? There is a relationship between the two. But your question sounds as if you're wondering what is gained in transforming to the frequency domain and manipulating the frequency-domain functions. $\endgroup$ Commented Aug 5, 2017 at 3:43
  • $\begingroup$ I just have seen a few video classes about Laplace Transform in the context of linear differential equations. They could call the Fourier Transform of g(t) is a function G(z) where z is a just a variable without any specific meaning, but why that is "frequency domain"? $\endgroup$ Commented Aug 5, 2017 at 3:51
  • $\begingroup$ i think they would use the letter "$s$" in $G(s)$ for the Laplace Transform. when it's "$z$", then i suspect they mean the so-called Z-Transform. $\endgroup$ Commented Aug 5, 2017 at 4:57

1 Answer 1


What would be the meaning of the value G($\omega$)?

Using the notation you gave, a function $f(t)$ can be expressed by its Fourier transform (I excluded the constant parameters) as: $$f(t) = \int_{-\infty}^{\infty} G(\omega)e^{i\omega t}d\omega$$ In this form, the original signal is expressed as a 'sum' over all frequencies $\omega$ of the functions $e^{i\omega t}$. The function $G(\omega)$ gives the amplitude and the phase (this is a complex value) for the specific function $e^{i\omega t}$ that corresponds to $\omega$.

Mathematically why we get that so close relationship between the domains?

Can you rephrase this one? What I guess you ask here is why there is a change from time to frequency. If that is the case, I'll try cover it.

Your signal is a function that depends of $t$. By applying a Fourier transform on it, you basically break that function in a 'sum' of $e^{i\omega t}$ functions that depend on the frequency $\omega$, each function having a given amplitude and a phase. As mentioned above, $G(\omega)$ gives you just that, for each frequency needed, in order to reconstruct your signal. That is why the Fourier Transform of the signal aka $G(\omega)$ is a function of frequency, and not time.

I would suggest, you look at this video: https://www.youtube.com/watch?v=r18Gi8lSkfM

It should help with the idea behind Fourier transform, and what it does, although the video focuses on periodic functions. Once you get that part, going to your specific signals should be OK.


From what I understand from your comment, you asked if one can use the FT to 'encode' some data, such that whenever is needed it can be decoded back and analysed. Yes, this can be done, not sure if it is worth it in terms of data storage, but it is possible.

I'll also add this because it might help get a qualitative understanding of the FT and why it is used most of the time. You can think of this frequency in the musical sense. Say you have a sound generator, that can make one perfect sine wave sound of a given frequency. Now, you wish to generate a 'sound' that results from combining more sine wave frequencies generated individually using one generator each. So at the end, let's assume that you have 1000 generators, each generating a different a sound with a given frequency, and for each generator/frequency you have the ability to control the volume. If you record this using a microphone, you'll get a signal. If you take the FT of that signal, surprise, surprise, what you get are the frequencies and the volume settings you used to generate it in the first place (also with some phase, but again, it isn't important here). You will see if you plot $|G(\omega)|$, that for each frequency you used, say 1kHz, 2kHz, 3kHz and 4kHz, you'll get a very narrow peak around the values $\omega =$ 1kHz, 2kHz, 3kHz and 4kHz and that the value $|G(\omega)|$ at those peaks will be proportional with the volume that you used.

From this, consider that you have a generator for every real number and that you can adjust the volume on all of them as before. Start all of them after they are set up, record the signal with a microphone, do the FT of the signal and you will be in your scenario with functions that are not periodic.

Now, the reason I mentioned all of that is to link it with what I assume the geophysics people use it for and that is to check what frequencies make up an earthquake (or any other earth vibration) and what amplitude (aka volume) does it have. It is the same reasoning as above, with the only difference that now you don't know the frequencies and the volumes on the generators, you only have the resulting signal. What do you do in this case if you want the set up parameters? You take the FT. Why does it matter? Because at some frequencies, different structures might resonate and they can collapse even for small intensity of vibrations. This would be an example.

  • $\begingroup$ That was really useful for me thank you. So computationally when you get a signal you do the Fourier transform of this and collect a countable number of points to reconstruct a signal that look likes your original signal, is that right? $\endgroup$ Commented Aug 5, 2017 at 4:22
  • $\begingroup$ I'll add to the answer because of the character limit. $\endgroup$ Commented Aug 5, 2017 at 12:32

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