How do we know that the Fourier Transform gives us the correct Frequency Domain Representation?

Question

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 $\omega$?

Answer 1

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.

Answer 2

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.