I have read a lot about the Fourier transform up until now, and I now seem to understand why the formula for the Fourier transform gives us the frequency domain representation of the signal.

However, looking at a derivation from my textbook, I cannot get my head around exactly how we can deduce from looking at the following relation:

$$f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}d\omega e^{i\omega t} \int_{-\infty}^{\infty}du f(u) e^{-i\omega u}$$

that the second set of operations (multiplying by $e^{-i\omega u}$ and then integrating with respect to $t$) would transfer to the frequency domain. In the proof, the variable $u$ was taken to be equal to $t$ at one point, presumably so as not to confuse it with the $t$ outside of the integral.

I can see that the above relation says that if you apply the various operators and multiplications to $f(u)$ then you will end up back where you started. I can also see that one of the sets of operators is an integral in $\omega$ which for me suggests, but does not directly show how we transfer from the time to the frequency domain and back using the relation. Even though I can show on paper that we get a function of $\omega$ from the integral.

How do we know that this equation, $F(\omega)$, is indeed a function that gives the relative amplitudes of the various frequencies (of the time domain), and not simply some other arbitrary function of the variable $\omega$?

  • 1
    $\begingroup$ Note the outer integration is summing up oscillators with angular frequency $\omega $ weighted according to F. Try to intuitively think of integration as continuous summation. $\endgroup$ – Ian Apr 5 '17 at 14:14
  • $\begingroup$ Thanks - I can sort of see it. I can see why it works, but I don't know what motivated someone to derive it this way if that makes sense? Sort of the same way I might see that the handle of a window does indeed open and shut it, but not why. $\endgroup$ – Resquiens Apr 5 '17 at 14:25
  • 1
    $\begingroup$ Well, it depends what you already know. If you know about Fourier series already, consider what happens in Fourier series when the domain gets bigger. The frequencies get closer together! So you might hope that on the whole line, you would have a frequency representation that involves all frequencies. You might also hope that the "coefficients" in this representation could be understood as some kind of limit of the Fourier coefficients. The Fourier transform is exactly the realization of these hopes. $\endgroup$ – Ian Apr 5 '17 at 20:14
  • $\begingroup$ Thanks. I have been studying it and I think I understand exactly why it gives the correct frequency domain representation now :) Do you happen to know, also, why they have chosen such strange ordering of the differentials $d\omega$ and $du$ in the above. It seems to me like in usual notation that would indicate that they were empty integrals, but I'm not too certain whether the differentials are exactly the same as those in your usual $\frac{dy}{dx}$ (are they Leibnitz infinitesimals or limit differentials?) math.stackexchange.com/questions/21199/… $\endgroup$ – Resquiens Apr 6 '17 at 20:42
  • 1
    $\begingroup$ They're the same differentials as you always see in integration. You could put them at the end if you wanted, it is just popular to put them on the left so that the whole object is a kind of operator. This is especially common in physics and just in general when integrands get complicated. $\endgroup$ – Ian Apr 6 '17 at 23:24

If you already understand Fourier series, then the Fourier transform can be understood as a realization of the limit of the Fourier series of $f$ on $[-L,L]$ as $L \to \infty$. Since we are seeking intuition, we will ignore convergence issues, and write:

$$f(x)=\sum_{n=-\infty}^\infty \frac{1}{2L} e^{i \frac{n \pi}{L} x} \int_{-L}^L e^{-i \frac{n \pi}{L} y} f(y) dy$$

for $x \in [-L,L]$. Now look at the numbers $\frac{n \pi}{L}$. These are evenly spaced, and as $L \to \infty$ the spacing between them goes to zero. And we are dividing by something proportional to this spacing. So the sum is, roughly speaking, a Riemann sum for an improper integral. Again, we're looking for intuition, so I'll leave it at that instead of trying to explain what that means.

So we can send $L \to \infty$ on the right side, and the left side does not depend on $L$, so it will converge by default. Since we want to identify $\omega=n \pi/L$, we should have the right $\Delta \omega$ factor in the sum, which is $\pi/L$. The way to make that happen is to write:

$$f(x)=\frac{1}{2\pi} \sum_{n=-\infty}^\infty \frac{\pi}{L} e^{i \frac{n\pi}{L} x} \int_{-\infty}^\infty e^{-i \frac{n \pi}{L} y} f(y) dy.$$

Then we send $L \to \infty$ and from what we know about Riemann integration we should find

$$f(x)=\frac{1}{2\pi} \int_{-\infty}^\infty e^{i \omega x} \int_{-\infty}^\infty e^{-i \omega y} f(y) dy d \omega$$

which is exactly the Fourier transform inversion formula. The version you wrote just moves the differentials to the front, which has no semantic significance at all, it is just notational preference.

Everything we did here works exactly as written if $f$ is Schwartz class. There is more work to be done for less nice $f$, and this work turns out to be easier when starting directly from the definition, instead of trying to pass through Fourier series.

  • $\begingroup$ Thanks. What really made this click for me was when I realised, as you show in your answer through deriving the units of $F(\omega)$, that the $d\omega$ that we "take out" out of the Fourier transform integral, means that instead of the continuous 'weighting' function (which could be thought of as the continuous equivalent of $c_n$ for the Fourier series) being a function of the scalar multiple of $\omega_0$, it is a function of $n\omega_0$ where $\omega_0->d\omega$, therefore it is a function of the continuous $\omega$, according to the definition of the limit/Riemann integral. $\endgroup$ – Resquiens Apr 7 '17 at 10:13

What do you mean with "give the amplitude of the various frequencies" ? What is your model for the signal ?

  • In the case of the Fourier transform, the model is $$f(t) = \int_{-\infty}^\infty g(\omega) e^{i\omega t}d\omega$$ where $g(\omega)$ is some function, or some distribution (as the Dirac delta). The Fourier transform is the operator letting us recover $g(\omega)$ from $f$.

  • In the case of the Fourier series, the model is $$f(t) = \sum_{n=-\infty}^\infty c_n e^{i \omega n}$$ and the operator $f \mapsto \frac{1}{2\pi}\int_0^{2\pi} f(t) e^{-i n t}dt$ let us recover the coefficients $c_n$

  • In the case of the discrete Fourier transform, the model is $$f(n) = \sum_{k=0}^{N-1} C_k e^{2 i\pi nk/N}$$ and the operator $f \mapsto \frac{1}{N}\sum_{k=0}^{N-1} f(n) e^{-2 i\pi nk/N}$ let us recover the coefficients $C_k$

  • There are other frequency and time-frequency representations : the short-time Fourier transform, the wavelet and constant-Q transform, the filter-bank, the Wigner distribution function and the reassigned spectrogram, the Laplace and Z transform, all coming with their own model for the signal.

  • $\begingroup$ Thanks. Just out of interest, is the discrete Fourier Transform derived from the continuous Fourier Transform or the Fourier Series? $\endgroup$ – Resquiens Apr 7 '17 at 10:02
  • 1
    $\begingroup$ @Resquiens Unfortunately for unifying all the Fourier theory you need the Fourier transform of distributions. In some sense everything is derived from the discrete Fourier transform $\endgroup$ – reuns Apr 7 '17 at 10:05
  • $\begingroup$ @Resquiens The DFT is, in a way, its own matter, because it is just about interpolating a function known at finitely many points into a particular form. $\endgroup$ – Ian Apr 7 '17 at 11:14
  • 1
    $\begingroup$ @Resquiens As for the Fourier series/transform distinction, distribution theory actually shows us that the Fourier series is a special case of the Fourier transform. Specifically, if you take a function $f$ on $[-L,L]$ and periodically extend it to $\mathbb{R}$ (which can always be done, even if $f$ is not periodic), the Fourier transform of $f$ is a linear combination of Dirac deltas at the frequencies $n \pi/L$ for $n \in \mathbb{Z}$, and the coefficients are the corresponding Fourier coefficients (or perhaps some universal multiple of them, depending on your Fourier transform convention). $\endgroup$ – Ian Apr 7 '17 at 11:14
  • $\begingroup$ @Resquiens One of the great things about the distributional Fourier transform is that technically, you only need to handle the Schwartz class case (the one rigorously handled by analysis like my answer) to get off the ground. After that the extension follows from duality and the unitary property of the FT. $\endgroup$ – Ian Apr 7 '17 at 11:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.