What is the difference between Fourier series and Fourier transformation?

What's the difference between Fourier transformations and Fourier Series?

Are they the same, where a transformation is just used when its applied (i.e. not used in pure mathematics)?

• Fourier series is an discrete analogue of Fourier transformation Commented Oct 25, 2012 at 21:22
• @Dean Here is something that I found extremely useful for Fourier transforms. Link Commented Oct 25, 2012 at 22:36
• @M.Strochyk: why would one even needed the discrete analogue if the continuous one seem to cover all types of functions? Whether I deal with periodic and non periodic function I should be able to apply the continuous FT? Commented Jun 28, 2016 at 21:33

The Fourier series is used to represent a periodic function by a discrete sum of complex exponentials, while the Fourier transform is then used to represent a general, nonperiodic function by a continuous superposition or integral of complex exponentials. The Fourier transform can be viewed as the limit of the Fourier series of a function with the period approaches to infinity, so the limits of integration change from one period to $(-\infty,\infty)$.

In a classical approach it would not be possible to use the Fourier transform for a periodic function which cannot be in $\mathbb{L}_1(-\infty,\infty)$. The use of generalized functions, however, frees us of that restriction and makes it possible to look at the Fourier transform of a periodic function. It can be shown that the Fourier series coefficients of a periodic function are sampled values of the Fourier transform of one period of the function.

• +1. Thanks for pointing out the relations between FS and FT! In the last sentence, "the Fourier transform of one period of the function." Do you mean the function is truncated to be zero outside one period and still defined over $(-\infty, \infty)$, and then the FT applies to the truncated function over $(-\infty, \infty)$?
– Tim
Commented Jan 1, 2013 at 16:10
• @Tim you are right. Commented Jan 1, 2013 at 17:20
• Why would that be necessary, e.g. does that matter? Is it equivalent to say that our function is periodic with infinite period, so that we still integrate over a single period? Commented May 31, 2017 at 9:28
• Quick question: you write both the Fourier Series and Fourier Transform use complex exponentials, but they use imaginary exponentials - why do they fix the real component to 0? Commented Jul 23, 2021 at 15:53
• Another quick question: what are generalized functions? You use the term in your sentence "The use of generalized functions, however, frees us of that restriction and makes it possible to look at the Fourier transform of a periodic function." Commented Jul 23, 2021 at 15:55

Fourier transform and Fourier series are two manifestations of a similar idea, namely, to write general functions as "superpositions" (whether integrals or sums) of some special class of functions. Exponentials $x\rightarrow e^{itx}$ (or, equivalently, expressing the same thing in sines and cosines via Euler's identity $e^{iy}=\cos y+i\sin y$) have the virtue that they are eigenfunctions for differentiation, that is, differentiation just multiplies them: ${d\over dx}e^{itx}=it\cdot e^{itx}$. This makes exponentials very convenient for solving differential equations, for example.

A periodic function provably can be expressed as a "discrete" superposition of exponentials, that is, a sum. A non-periodic, but decaying, function does not admit an expression as discrete superposition of exponentials, but only a continuous superposition, namely, the integral that shows up in Fourier inversion for Fourier transforms.

In both case, there are several technical points that must be addressed, somewhat different in the two situations, but the issues are very similar in spirit.

• +1! Thanks! This reply confirms my recent understanding. "A periodic function provably can be expressed as a "discrete" superposition of exponentials," should "A periodic function" be "A periodic function which is $L^1$ in a single period" instead?
– Tim
Commented Jan 1, 2013 at 16:03
• @Tim... Oh, yes, certainly, "locally $L^1$" is necessary for the relevant integrals to be literal. But, also, suitable "generalized functions" (distributions) have similar expressions, convergent in a suitable topology. Commented Jan 1, 2013 at 17:44
• do you know any resource on Fourier transform with mathematics focused? Like how does it connect with complex analysis, topology, group theory? Commented Sep 25, 2017 at 7:07
• @RylanSchaeffer, a "generalized function", also called a "distribution", is a larger class of "functions" that includes things like Dirac's "delta function" $\delta$, which has the property that $\int f(x)\,\delta(x)\;dx=f(0)$... although no function with pointwise values can have that property. L. Schwartz and S. Sobolev developed these ideas. Commented Jul 24, 2021 at 16:19
• @RylanSchaeffer, the two uses are mostly unrelated... Commented Jul 24, 2021 at 18:56

Fourier transform is used to transform periodic and non-periodic signals from time domain to frequency domain. It can also transform Fourier series into the frequency domain, as Fourier series is nothing but a simplified form of time domain periodic function.

Fourier series

1. Periodic function => converts into a discrete exponential or sine and cosine function.

2. Non-periodic function => not applicable

Fourier transform

1. Periodic function => converts its Fourier series in the frequency domain.

2. non-Periodic function => converts it into continuous frequency domain.

If you have a locally compact Abelian group $$G$$ you can define a group called the Pontryagin dual group - $$\widehat{G}$$. You can define a Haar measure on $$G$$, $$\mu$$. We can define the Fourier transform of a function $$f\in L^1(G)$$:

$$\widehat f(\chi)=\int_Gf(x)\overline{\chi(x)}d\mu(x)$$

$$\widehat f(\chi)$$ is a bounded continuous function that vanishes at infinity on $$\widehat{G}$$.

If $$G=\Bbb R$$ then $$\widehat{G}=\Bbb R$$ and we have the regular Fourier transform.

If $$G=S^1$$ then $$\widehat{G}=\Bbb Z$$ and we have the Fourier series (an example of a Fourier transform).

• $\widehat{G}$ is a representation of G, its elements are called the "characters" of elements of G. Commented Apr 14, 2019 at 16:05
• Suggestion: write it as $\widehat{f}(\chi)$, not $\widehat{f(\chi)}$. The latter makes it look like $f$ takes $\chi$ as an argument, which it does not. Commented Apr 17, 2020 at 15:33

I see you’re asking probably about the most “meaningful two ones” that are seemingly differ, but yet comparable if you compare them one to one and these are discrete time signal’s fourier transform and discrete time signal’s fourier series, while both signals are periodic.

I’ll explain it in the simplest way possible.

Imagine you have a drawing in a map (don’t imagine an nth dimensional map, but it might be). Both of them represent a point on this drawing on this map. (Map is a globe map, don’t go too far.) Now it’s simple, each representation solves a different problem, but both are representing the same point. It’s just one representation represent a point as a point as we know it (alas series, with e and i or j and w0) and one representation (the one with the deltas) represents the same point as + <0,delta(y)>. We know the first representation. I won’t explain. But in the second representation if we sum up all e-s with i-s or j-s we’ll get what? A superposition. Meaning that we’ll know the final superposition, but won’t know what the projection is. We won’t know what x,y is because in signal x,y are not trivial to see as a point on the map while they are in superposition, so we need a coordinate system instead. The Fourier Transform shows us the same point, but like in the map - with coordinates. Alas we see the coefficients. And we see the coefficients thanks to the deltas. Otherwise we’d had a sum of e’s which is sort of “summable” sometimes and in order to find each coefficient we’d had to project and calculate the projection for each coefficient. Here we SEE the coefficients WITHOUT calculating each of them thanks to the deltas.

All this is because the inner product in signal’s dimension is “less straight forward” than the inner product in a globe map. But yet it is straight forward for a computer if you do it repeatedly and ineffectively. :)

That’s it, hope it helps.

I think one of the best intuitive explanation of the relation between Fourier series and Fourier transform was given in [Alan V. Oppenheim, Alan S. Willsky, Hamid Nawab. "Signals and Systems", 2ne edition, Section 4.1.1, pages 286-289]

I think the statement

"The Fourier transform can be viewed as the limit of the Fourier series of a function with the period approaches to infinity."

is misleading as it is the Fourier series coefficients which are derived from a truncated Fourier transform. The Fourier series for $$f(x)$$ converges to $$f(x)$$ over some period, whereas the Fourier transform does not generally converge to $$f(x)$$ (except in the rare case when $$f(x)$$ is its own Fourier transform).

Consider the "nested" Fourier series representation

$$f(x)\approx \underset{N, f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \frac{\mu(2 n-1) }{2 n-1} \left(\frac{3 F(0)}{8}\\+\frac{1}{2} \sum\limits_{k=1}^{2 f (2 n-1)} (-1)^k\, \cos\left(\frac{\pi k}{2 n-1}\right) \left(F\left(\frac{k}{4 n-2}\right) e^{\frac{i \pi k x}{2 n-1}}+F\left(-\frac{k}{4 n-2}\right) e^{-\frac{i \pi k x}{2 n-1}}\right)\\-\frac{1}{8} \sum\limits_{k=1}^{4 f (2 n-1)} (-1)^k\, \left(F\left(\frac{k}{8 n-4}\right) e^{\frac{i \pi k x}{4 n-2}}+F\left(-\frac{k}{8 n-4}\right) e^{-\frac{i \pi k x}{4 n-2}}\right)\right)\right)\tag{1}$$

where

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

is the Fourier transform of $$f(x)$$.

My conjectured Fourier inversion theorem can be stated as follows:

A function $$f(x)$$ is recoverable from its Fourier transform $$F(\omega)$$ if-and-only-if the "nested" Fourier series representation of $$f(x)$$ converges to $$f(x)$$ for all $$x\in\mathbb{R}$$.

When $$f(x)$$ is an even function formula (1) above simplifies to

$$f(x)\approx \underset{N, f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \frac{\mu(2 n-1)}{2 n-1} \left(\frac{3 F(0)}{8}\\+\sum _{k=1}^{2 f (2 n-1)} (-1)^k\, \cos\left(\frac{\pi k}{2 n-1}\right)\, F\left(\frac{k}{4 n-2}\right)\, \cos\left(\frac{\pi k x}{2 n-1}\right)\\-\frac{1}{4} \sum\limits_{k=1}^{4 f (2 n-1)} (-1)^k\, F\left(\frac{k}{8 n-4}\right)\, \cos\left(\frac{\pi k x}{4 n-2}\right)\right)\right),\quad f(-x)=f(x)\tag{3}.$$

When $$f(x)$$ is an odd function formula (1) above simplifies to

$$f(x)\approx \underset{N, f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \frac{\mu(2 n-1)}{2 n-1} \left(\frac{3 F(0)}{8}\\+i \sum\limits_{k=1}^{2 f (2 n-1)} (-1)^k\, \cos\left(\frac{\pi k}{2 n-1}\right)\, F\left(\frac{k}{4 n-2}\right)\, \sin\left(\frac{\pi k x}{2 n-1}\right)\\-\frac{1}{4}i \sum\limits_{k=1}^{4 f (2 n-1)} (-1)^k\, F\left(\frac{k}{8 n-4}\right)\, \sin\left(\frac{\pi k x}{4 n-2}\right)\right)\right),\quad f(-x)=-f(x)\tag{4}.$$

See my related MSE question or MO question for additional information.