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

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?

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

mathman
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?
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.

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.

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.

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.
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.

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

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.

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.
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,
$$\frac{1}{T}\mathcal{F}\phi\left(\frac{k}{T}\right) = \frac{1}{T} \int_{-\infty}^{\infty} e^{-2\pi i (k/T)t}\phi(t) \, dt \\ = \frac{1}{T} \int_{0}^{T} e^{-2\pi i (k/T)t}\Phi(t) \, dt [/itex] But this is the k-th Fourier coefficient $c_k$ of $\Phi$. In other words, the Fourier series is a consequence of Fourier transform theory. [tex] \Phi(t) = \sum_{k = -\infty}^\infty c_k e^{2\pi ikt/T}, \qquad \text{where} \qquad c_k = \frac{1}{T} \int_{0}^{T} e^{-2\pi i (k/T)t}\Phi(t) \, dt$$

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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.

jasonRF