3

The Fourier transform of a function is given by $$f(ξ) = \int_{-∞}^{∞} f(x) e^{-2πixξ}dx$$

the paper I was reading from says that the test function $e^{2πixξ}$ is a periodic with period $2\pi/ξ$ and it continues by saying ("so integrating $f$ against this test function gives information about the extent to which this frequency occurs in $f$")

My question is how integrating our original function with this "test function" gives us information about the extent to which frequency occurs in $f$ ?

Abdelrahman
  • 522
  • 1
    That's the main observation behind entire Fourier analysis. You may want to start with wiki: https://en.m.wikipedia.org/wiki/Fourier_transform They have not bad description in "Background" section. – freakish Mar 26 '24 at 13:24
  • @freakish, the way Fourier transform describe the frequency extent of the function is derived from the Fourier series by taking that $c_{n}$ of the Fourier series and make the period goes to infinity, so the $c_{n}$ is equivalent to $f(ξ)$ but $c_{n}$ for discrete frequencies and $f(ξ)$ for continuous of frequencies, that what I Know and that what all the books that I've read from derive the Fourier transform, My question is :How this paper say just by integrating we show the frequency extent of the $f(x)$ ? – Abdelrahman Mar 26 '24 at 14:08
  • @freakish, another thing to note when I discuss with many users about this connection between Fourier series and Fourier transform , they just say this connection is not totally rigorous and the way Fourier transform is defined isn't by Limiting the Fourier series , I don't know what their point really is but I don't what's wrong with what I am saying all the books that I read from derive it the same way by the same procedure , if so what is the objection then ? – Abdelrahman Mar 26 '24 at 14:13
  • See my answer to your related question on "The Connection between Fourier transform and Fourier series (derivation and Intuition)". – Steven Clark Mar 26 '24 at 17:50

2 Answers2

1

The Fourier transform of a periodic function $f(x)$ is related to its frequency content since

$$\mathcal{F}_x\left[a_n \cos\left(2 \pi \frac{n}{P} x\right)\right](\omega )=\int_{-\infty}^{\infty} a_n \cos\left(2 \pi \frac{n}{P} x\right)\, e^{-2 i \pi \omega x} \, dx\\=\frac{a_n}{2}\, \delta\left(\omega-\frac{n}{P}\right)+\frac{a_n}{2}\, \delta\left(\frac{n}{P}+\omega\right)\tag{1}$$

and

$$\mathcal{F}_x\left[b_n \sin\left(2 \pi \frac{n}{P} x\right)\right](\omega)=\int\limits_{-\infty}^{\infty} b_n \sin\left(2 \pi \frac{n}{P} x\right)\, e^{-2 i \pi \omega x} \, dx\\=\frac{b_n}{2} i\, \delta\left(\frac{n}{P}+\omega\right)-\frac{b_n}{2} i\, \delta\left(\omega-\frac{n}{P}\right)\tag{2}$$

where $\delta(\omega)$ is the Dirac delta function, but the Fourier transform of a periodic function doesn't converge in the usual sense, rather it converges only in a distributional sense.

Its seems to me the notion that the Fourier transform of a non-periodic function $f(x)$ somehow gives the frequency content of the function $f(x)$ analogous to the coefficients of a Fourier series is perhaps a bit misguided since for a non-periodic function $f(x)$ I believe one has

$$c_n=\lim\limits_{P\to\infty}\left(\frac{1}{P} \int\limits_{-\frac{P}{2}}^{\frac{P}{2}} f(x)\, e^{-i 2 \pi \frac{n}{P} x} \, dx\right)=0\tag{3}.$$

For example consider the function

$$f(x)=e^{-\pi^2 x^2}\tag{4}$$

with Fourier transform

$$F(\omega)=\mathcal{F}_x[f(x)](\omega)=\int\limits_{-\infty}^{\infty} e^{-\pi^2 x^2}\, e^{-i 2 \pi \omega x} \, dx=\frac{e^{-\omega^2}}{\sqrt{\pi}}\tag{5}.$$

Now since $f(x)=e^{-\pi^2 x^2}$ is an even function of $x$ consider the Fourier $\cos$ series coefficient

$$a_n=\frac{2}{P} \int\limits_{-\frac{P}{2}}^{\frac{P}{2}} f(x) \cos\left(2 \pi \frac{n}{P} x\right) \, dx=\frac{2}{P} \int\limits_{-\frac{P}{2}}^{\frac{P}{2}} e^{-\pi^2 x^2} \cos\left(2 \pi \frac{n}{P} x\right) \, dx\\=\frac{i e^{-\frac{n^2}{P^2}}}{\sqrt{\pi} P} \left(\text{erfi}\left(\frac{n}{P}-\frac{i \pi P}{2}\right)-\text{erfi}\left(\frac{n}{P}+\frac{i \pi P}{2}\right)\right)\tag{6}$$

where the function $f(x)=e^{-\pi^2 x^2}$ can be approximated on the interval $-\frac{P}{2}<x<\frac{P}{2}$ by the Fourier $\cos$ series

$$f(x)\approx\frac{a_0}{2}+\lim\limits_{N->\infty}\left(\sum_{n=1}^N a_n \cos\left(2 \pi \frac{n}{P} x\right)\right)\tag{7}.$$

For the function $f(x)=e^{-\pi^2 x^2}$ note that

$$\underset{P\to\infty}{\text{lim}}a_n=\underset{P\to\infty}{\text{lim}}\left(\frac{i e^{-\frac{n^2}{P^2}}}{\sqrt{\pi} P} \left(\text{erfi}\left(\frac{n}{P}-\frac{i \pi P}{2}\right)-\text{erfi}\left(\frac{n}{P}+\frac{i \pi P}{2}\right)\right)\right)=0\tag{8}.$$

This is consistent with Parseval's theorem which can be stated as

$$\frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(x)^2 \, dx=\frac{1}{2} (a_0)^2+\sum\limits_{n=1}^\infty \left(a_n^2+b_n^2\right)\tag{9}$$

for a $2 \pi$-periodic representation of a function $f(x)$ and which I believe can be generalized as

$$\frac{2}{P} \int\limits_{-\frac{P}{2}}^{\frac{P}{2}} f(x)^2 \, dx=\frac{1}{2} (a_0)^2+\sum\limits_{n=1}^\infty \left(a_n^2+b_n^2\right)\tag{10}$$

for a $P$-periodic representation of a function $f(x)$ where $a_n$ and $b_n$ are the coefficients of the related $\cos$ and $\sin$ terms in the Fourier series representation of $f(x)$.

For the function $f(x)=e^{-\pi ^2 x^2}$ one has

$$\int_{-\infty }^{\infty } \left(e^{-\pi ^2 x^2}\right)^2 \, dx=\frac{1}{\sqrt{2 \pi }}\tag{11}$$

which implies for the function $f(x)=e^{-\pi ^2 x^2}$ one has

$$\lim_{P\to\infty}\left(\frac{2}{P} \int\limits_{-\frac{P}{2}}^{\frac{P}{2}} f(x)^2 \, dx\right)=\frac{1}{2} (a_0)^2+\sum\limits_{n=1}^\infty \left(a_n^2+b_n^2)\right)=0\tag{12}$$

which implies $a_n=b_n=0$.

Finally, the magnitude of the frequency $\omega_n=\frac{n}{P}$ in the Fourier $cos$ series of an even function $f(x)$ related to the term in formula (1) above is not the value of $F\left(\omega_n\right)$ where $F(\omega)$ is the Fourier transform of $f(x)$, but rather the "energy" at $F\left(\omega_n\right)$ which for an even function $f(x)$ can be calculated as

$$\lim\limits_{\epsilon\to 0^+}\left(2 \int_{\omega_n-\epsilon}^{\omega_n+\epsilon} F\left(\omega-\omega_n\right) \, d\omega\right)\tag{13}$$

which for $F(\omega)=\frac{e^{-\omega^2}}{\sqrt{\pi}}$ in formula (5) above evaluates to zero which is consistent with formulas (3), (8), and (12) above.

Steven Clark
  • 8,744
  • How then Fourier transform was derived ? And what was the main idea behind it ? – Abdelrahman Mar 30 '24 at 18:18
  • The Fourier series coefficient $c_n$ for the Fourier series

    $$f(x)\approx\sum\limits_n c_n, e^{i 2 \pi \frac{n}{P} x}$$

    can be evaluated as

    $$c_n=\frac{1}{P}, C_P\left(\frac{n}{P}\right)$$

    where

    $$C_P(\omega)=\int\limits_{-\frac{P}{2}}^{\frac{P}{2}} f(x) , e^{-i 2 \pi \omega x} , dx$$

    is a Fourier transform with a truncated integration range, and it is true that

    $$F(\omega)=\lim\limits_{P->\infty} C_P(\omega),$$

    but the observations in my answer above seem to be inconsistent with the notion that $F(\omega)$ represents the frequency content of $f(x)=e^{-\pi^2 x^2}$.

     – Steven Clark Mar 30 '24 at 19:28
  • @Mans Also note that $C_P(\omega)$ is not the same thing as $c_n$, and while $C_P(\omega)$ and $F(\omega)$ are both continuous functions of $\omega$ for the example function $f(x)=e^{-\pi^2 x^2}$ in my answer above, the "nested" Fourier series representation in formula (5) of my answer to your related question only evaluates $F(\omega)$ at rational values of $\omega$, but this is perhaps analogous to a Fourier series only evaluating $C_P(\omega)$ at a countably infinite number of values of $\omega$ even though $C_P(\omega)$ is continuous. – Steven Clark Mar 30 '24 at 19:43
  • @Mans It seems to me that saying $F(\omega)$ represents the frequency content of $f(x)$ is analogous to saying that $\frac{1}{P}, C_P(\omega)$ represents the frequency content of $f(x)$, but note there are an uncountably infinite number of values of $\omega$ for which $\frac{1}{P}, C_P(\omega)\ne 0$ which are not accounted for in the Fourier series for $f(x)$. – Steven Clark Mar 30 '24 at 20:23
0

Actually, the definition of the Fourier transform, namely $$ \hat{f}(\xi) = \int_\Bbb{R} f(x)e^{-2\pi ix\xi} \,\mathrm{d}x, $$ corresponds to a (hermitian) inner product $\langle f, e^{2\pi ix\xi} \rangle$, where $$ \langle \phi,\psi \rangle := \int_\Bbb{R} \phi(x)\overline{\psi(x)} \,\mathrm{d}x. $$ In the case of Fourier series, the trigonometric "monomials" $e^{2\pi ix\xi_n}$, with $\xi_n = \frac{n}{T}$, form an orthonormal basis of the Lebesgue space $L^2([-\frac{T}{2},\frac{T}{2}])$, which is a functional vector space for recall, in such a way that the expression $$ f(x) = \sum_n c_n e^{2\pi ix\xi_n} $$ is nothing else than the projection of $f$ onto the Fourier basis. In consequence, the coefficient $c_n = \langle f, e^{2\pi ix\xi_n} \rangle$ "measures" how much $f$ "contains" the sinusoid $e^{2\pi ix\xi_n}$ as its coordinate in that particular basis. It is to be noted that the Taylor series are actually constructed in the same manner with respect to a polynomial basis.

Also, note that this formalism was developped to study waves historically, that is why one speaks of the amplitude associated to the frequency $\xi_n$ when referring to $c_n$, even if those quantities have to be understood in an abstract sense and might carry other physical units.

Finally, mutatis mutandis, the Fourier transform can be interpreted in the same way, except for the fact that the trigonometric functions $e^{2\pi ix\xi}$ are meant to form the continuous basis of a larger space.

Abezhiko
  • 14,205
  • what about the derivation of Fourier transform from Fourier series , Is this was the original approach and the main one ,that transform limiting case of series, or the derivation of Fourier transform has another approach other than what I've mentioned, If so, How then Fourier transform decompose the function into a sum of harmonics . I understand it of the limiting case of Fourier series because the derivation was done by taking the limit of $c_{n}$ and it will approach the $f(ξ)$ , the $c_{n}$ tell the amplitude of each harmonics but as a discrete so the $f(ξ)$ but as a continuous version – Abdelrahman Mar 26 '24 at 17:28
  • If Fourier transform is an inner product, how then Fourier transform is used to analyze the function into its frequency components? – Abdelrahman Mar 26 '24 at 18:32
  • @Mans First question : note that $c_n = \hat{f}(\xi)$ in both the discrete and continuous cases (their definitions coincide, except for the domain of integration), but the original function is decomposed as $$f(x) = \sum_n c_n e^{2\pi ix\xi_n}$$ in the discrete case and $$f(x) = \int_\Bbb{R} \hat{f}(\xi) e^{2\pi ix\xi} ,\mathrm{d}\xi$$ in the continuous case. The limit is not considered inside $c_n$/$\hat{f}(\xi)$ but with respect to the summation, since the integral is nothing else than a continuous sum (in the sense that the summation index takes values in a continuous interval). – Abezhiko Mar 28 '24 at 15:22
  • @Mans Second question : the Fourier transform itself corresponds to the frequency component, because the inner product with a basis vector "selects" the associated component (when the basis is orthonormal). "Analyzing a function into its frequency components" is just an English sentence meaning precisely "computing its Fourier transform". – Abezhiko Mar 28 '24 at 15:26
  • I mean does the idea of the Fourier transform came by the limiting case of the Fourier series (as most of books derive it that way ) Or The main idea came by taking the inner product between our $f(x)$ and $e^{-2πixξ}$ because the inner product I see it as the projection of vector on the coordinate basis if we say that the inner product tells us the projection of $f(x)$ on some basis then accordingly that function could be reconstructed by that projection multiplied by the basis, So does Fourier transform came by the limit of Fourier series(as books do) or by the projection that I said – Abdelrahman Mar 28 '24 at 19:38
  • @Mans It was developped as a limit, both mathematically and historically, since Fourier series came first (when physicists were studying string vibrations and heat equation, if I'm not mistaken). However, it is to be noted that the Fourier series themselves are constructed as projections on the Fourier basis (of the Lebesgue space $L^2$). – Abezhiko Mar 29 '24 at 10:13