Can you help me with some information about how to derive Fourier transform from Fourier series? Is there any connection between these two mathematical notions?

The amplitude spectrum can be determined from coefficients of Fourier series or from Fourier transform?


  • 2
    $\begingroup$ There is indeed some connection between these two concepts. For example have a look at the Poisson summation formula: "In mathematics, the Poisson summation formula is an equation that relates the Fourier series coefficients of the periodic summation of a function to values of the function's continuous Fourier transform." en.wikipedia.org/wiki/Poisson_summation_formula $\endgroup$ – Fritz Sep 27 '16 at 18:45
  • $\begingroup$ do you know a proof of the Fourier series for say $f \in C^1$ ? and of the Fourier transform of $e^{-x^2}$ ? from this and a limit $T \to \infty$ you can show the Fourier inversion theorem for any $L^1 \cap C^1$ function $\endgroup$ – reuns Sep 27 '16 at 20:50

Here is some intuition which may help (for a complete and rigorous treatment, see references at the end):

Assume a function $f$ is an even function and can be written as:

$$ f(x) = \sum_{n=0}^\infty a_n \cos(2\pi n x) $$

now those $\cos(2\pi nx)$ are orthogonal in that $\langle \cos(2\pi n\cdot),\cos(2\pi m\cdot)\rangle = \kappa\delta_{nm}$ for some appropriate constant $\kappa$ and with $\langle g,h\rangle=\int_{-\infty}^\infty f(x)\overline{g(x)}\mathrm{d} x$.

Now, brushing under the rug some technical details, if you take $\langle f, \cos(2\pi mx)\rangle$ intuitively all terms but one will disappear thanks to the orthogonality of the basis:

$$ a_m \propto \langle f, \cos(2\pi m\cdot)\rangle $$

You can generalize the idea to functions that are not even but following the same idea, the basis functions are then $\exp(-2i\pi k x)$ (there may be a factor in there depending on the convention but it doesn't change the reasoning) and you have in general

$$ a_k \propto \langle f, \exp(-2i\pi k \cdot)\rangle$$

and now it's only one step to define a function out of this by simply taking continuous $k$ say $\omega$:

$\hat f(\omega) \propto \langle f, \exp(-2i\pi \omega \cdot)\rangle$

which is the FT. Now depending on your church (math, physics, engineering, compsci), you'll have $\pi$ here and there and the proportionality constant will change but the rough idea stays the same.

Putting some more beautiful math under the aforementioned rug (functional analyst if you read this, please accept my apologies) you can go from the sum above to an integral and have a "continuous sum" for $f$:

$$ f \propto \langle \hat f, \exp(-2i\pi x \cdot) \rangle = \int_{-\infty}^\infty \hat f(\omega)\exp(2i\pi\omega x)\mathrm{d}\omega $$

the right hand side is the inverse FT and the part in the middle makes explicit the fact that the FT is a dual representation.

(I am aware this is a rough development, I'm just trying to give some intuition, for a complete and more rigorous development, you could look at something like http://lpsa.swarthmore.edu/Fourier/Xforms/FXformIntro.html)

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We can't use Fourier series if the signal we want to analyse is non-periodic. So the the idea was to say a non-periodic signal is a periodic signal with an infinite period. As you will see, this trick transforms the fourier serie equation into the fourier transform.

First we have to rewrite the fourier series basis in term of frequency with $\omega_k = \frac{2k\pi}{T}$ :

$B = \{\cos{(\omega_0 x)}, \cos{(\omega_1 x)}, \sin{(\omega_1 x)}, \cos{(\omega_2 x)}, \sin{(\omega_2 x)}, ... \cos{(\omega_n x)}, \sin{(\omega_n x)}\}$

So let's see what happens to the frequencies if we grow up the period T:


So as T is growing, the gaps between the differents $\omega_k$ become more and more smaller, and when $T\to +\infty$, the discrete variable $\omega_k$ becomes a continuous variable $\omega$.

Ok, so now we have to transform the discrete fourier series to the continuous fourier transform.

Recall the fourier series complex notation:

$$f(x) = \sum_{k=-\infty}^{+\infty}c_ke^{i\omega_kx}$$ $$c_k = \frac{1}{T} \int_{-T/2}^{T/2}f(x)e^{-i\omega_kx}dx$$

So as we said, when $T\to +\infty$, $\omega_k \to \omega$. So the $\sum$ becomes a $\int$, and $\Delta \omega $ becomes an integration variable $d\omega$.

We need a $\Delta \omega$:

$\Delta \omega = \omega_{k+1} - \omega_k = \frac{2(k+1)\pi}{T} - \frac{2k\pi}{T} = \frac{2\pi}{T}$

Let's refactor again the fourier series in order to reveal $\Delta \omega$:

$$f(x) = \frac{1}{T} \sum_{k=-\infty}^{+\infty}c_ke^{i\omega_kx}$$ $$c_k = \int_{-T/2}^{T/2}f(x)e^{-i\omega_kx}dx$$

and then:

$$f(x) = \frac{1}{2\pi} \sum_{k=-\infty}^{+\infty}c_ke^{i\omega_kx}\cdot \Delta \omega$$ $$c_k = \int_{-T/2}^{T/2}f(x)e^{-i\omega_kx}dx$$

Now let's $T\to +\infty$ which lead us to the Fourier transform:

$$f(x) = \frac{1}{2\pi} \int_{-\infty}^{+\infty}F(\omega)e^{i\omega}\cdot d\omega$$ $$F(\omega) = \int_{-\infty}^{+\infty}f(x)e^{-i\omega x}dx$$

You will find the complete approach here:



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