What is the point of Fourier Series if you can do the Fourier Transform?

1. Jul 22, 2009

Rib5

Hey,

I was wondering. Since the Fourier Series coefficients can just be represented in the form of a Fourier Transform, what is the point of ever finding the Fourier coefficients and not doing the transform?

2. Jul 23, 2009

trambolin

For example, to reconstruct a bandlimited signal from fourier coefficients (happen to be the samples in time domain)

3. Jul 23, 2009

mathman

Fourier series are defined for periodic functions (finite interval) while Fourier transforms are defined for functions integrable over the real line. They just have different contexts.

4. Jul 23, 2009

maverick_starstrider

Yes, but the OP's point is that all functions that can be expanded in fourier series can also be fourier transformed so why bother with fourier series at all. And I can't say I really have a reason to disagree.

5. Jul 23, 2009

Rib5

I asked my teacher about it in class today, and he pretty much said the same thing. He said that the main reason we learn about fourier series is that it helps students understand the fourier transform more intuitively, and also for historical reasons about how the fourier transform was developed.

6. Jul 24, 2009

John Creighto

How about, you do the Fourier series because it is easier.

7. Jul 24, 2009

mathman

An example: the density function for the normal distribution is Ke-x2/2. It has a Fourier transform, but you cannot expand it with a Fourier series.

8. Jul 26, 2009

matonski

The OP's question concerns the other way around. Any function that has a Fourier series also has a Fourier transform. How are they related? The following argument has been taken from Brad Osgood's excellent course that's available online.

Let $\Phi$ be a periodic function of period T. Then $\Phi$ can be written as
$$\Phi(t) = \phi(t) \ast \sum_{k = -\infty}^\infty \delta(t-kT)$$
where $\phi$ is one period of $\Phi$ and $\ast$ denotes convolution. Now, by the convolution theorem (the Fourier tranform of a convolution is the product of the Fourier transforms),
\begin{align*} \mathcal{F}\Phi(s) &= \Bigl(\mathcal{F}\phi(s) \Bigr) \frac{1}{T}\sum_{k = -\infty}^\infty \delta\left(s-\frac{k}{T}\right) \\ &= \frac{1}{T}\sum_{k = -\infty}^\infty \mathcal{F}\phi\left(\frac{k}{T}\right)\delta\left(s-\frac{k}{T}\right) \end{align*}

Taking the inverse Fourier transform gives us back $\Phi$. Since $\mathcal{F}\phi\left(k/T\right)$ are constants and the inverse Fourier transform of a shifted delta function is a complex exponential,
$$\Phi(t) = \sum_{k = -\infty}^\infty \frac{1}{T}\mathcal{F}\phi\left(\frac{k}{T}\right) e^{2\pi i k t/T}$$
However,
$$\frac{1}{T}\mathcal{F}\phi\left(\frac{k}{T}\right) = \frac{1}{T} \int_{-\infty}^{\infty} e^{-2\pi i (k/T)t}\phi(t) \, dt \\ = \frac{1}{T} \int_{0}^{T} e^{-2\pi i (k/T)t}\Phi(t) \, dt [/itex] But this is the k-th Fourier coefficient $c_k$ of $\Phi$. In other words, the Fourier series is a consequence of Fourier transform theory. [tex] \Phi(t) = \sum_{k = -\infty}^\infty c_k e^{2\pi ikt/T}, \qquad \text{where} \qquad c_k = \frac{1}{T} \int_{0}^{T} e^{-2\pi i (k/T)t}\Phi(t) \, dt$$

Last edited: Jul 27, 2009
9. Jul 28, 2009

pellman

One reason for using Fourier series is simply practical. When analyzing something numerically you have to stop somewhere. You can't calculate it to the infinite decimal place. In calculating an integral this amounts to (1) using a finite domain, and (2) approximating the integral with a summation. With the Fourier transform integral the most convenient and obvious way to do this is to use a Fourier series. Even then, it is not a true Fourier series since you have to stop the summation after a finite number of terms.

Indeed, computational physicists use the terms "Fourier transform" and "Fourier series" interchangeably in casual conversation.

10. Jul 28, 2009

jasonRF

A lot of times Fourier series and transforms show up in applications like solving linear eigenvalue problems. For some problems sinusoids are the eigenfunctions, and if the operator has a discrete set of eigenvalues it is natural to represent the solution to the eigenvalue problem in the form of a Fourier series - it kind of "falls out" when solving the problem and throwing an integral transform at the problem (especially without an understanding of Fourier series!) can be more cumbersome. If the operator has a continuous set of eigenvalues then it is natural to represent the solution as an integral transform. Some problems even have both kinds of eigenvalues, so the solution can be nicely written as a the combination of a sum and an integral. Usually there are clear physical interpretations for the various terms, so it is nice to keep them separate. Facility dealing with both the series and the transform allows you to use whatever is most convenient for the problem at hand, and to easily make the transition to other sets of basis functions that many applications require.

Also, as noted by pellman, Fourier series show up in numerics all the time. And in the era of digitized data, the DFT (usually implemented as an FFT whenever possible) is used all the time to look at spectral content or to quickly do convolutions. What is the DFT? It effectively assumes your data are periodic and that you only have one period sampled, so it does the periodic extension of the data (like you learn about with simple Fourier series problems!) and finds the discrete Fourier series.

So of course you are right, and it is probably good for you to lump them together in your head (along with Laplace transforms, etc.), but I would claim that making yourself work through some of the Fourier series results in a class is still a useful exercise.

jason