Is Fourier series a sampled version of Fourier transform?

Question

I recently learned about dtft and how dft/dfs is the sampled version of dtft. I was wondering if Fourier series is also obtainable by sampling Fourier transform? I am a noob in the subject so sorry if this is stupid

Answer 1

There are 4 versions of Fourier transforms that are all close cousins.

It's all due to the basic property that "sampling in one domain corresponds to periodicity in the other domain". If a time signal is periodic, the frequency domain signal is discrete. If a frequency domain signal is periodic, the time domain signal is discrete.

So you have four variants

> Time                       Frequency                 Transform 
Continuous & aperiodic       Continuous & aperiodic    Fourier Transform
Discrete   & aperiodic       Continuous & periodic     Discrete Time Fourier Transform
Continuous &  periodic       Discrete   & aperiodic    Fourier Series
Discrete   &  periodic       Discrete   & periodic     Discrete Fourier Transform

The flavors are different since a continuous signal requires integration and a discrete signal needs summation.

The naming conventions are non-intuitive: it would be easier if the DFT would have been called Discrete Fourier Series and the DTFT called DFT, but it is the way it is.

Which flavor you need to use depends on the nature of your signals and depending on what assumptions around the signals you can make, you can use one flavor to calculate a different one.

Answer 2

Yes, for time-limited functions it is possible to obtain the Fourier series coefficients by sampling the Fourier transform. This is the dual case of the more common form of the sampling theorem, stating that a band-limited function is fully characterized by its (time-domain) samples.

The same is true in the discrete domain. Note that only for finite length signals is the DFT a sampled version of the DTFT.

Let $g(t)$ be a function that is zero outside the interval $[0,T]$. Its Fourier transform is

$$\begin{align}G(j\omega)&=\int_{-\infty}^{\infty}g(t)e^{-j\omega t}dt\\&=\int_{0}^{T}g(t)e^{-j\omega t}dt\tag{1}\end{align}$$

In the interval $t\in[0,T]$, the function $g(t)$ can be described by its Fourier coefficients:

$$c_k=\frac{1}{T}\int_{0}^Tg(t)e^{-j\frac{2\pi k}{T}t}dt\tag{2}$$

Comparing $(1)$ and $(2)$ shows that the Fourier coefficients $c_k$ are (scaled) samples of the Fourier transform $G(j\omega)$:

$$c_k=\frac{1}{T}G\left(j\frac{2\pi k}{T}\right)\tag{3}$$

In general, for functions that are not necessarily time-limited, the samples of the Fourier transform given by $(3)$ are the Fourier series coefficients of an aliased version of the original function. This is shown by Poisson's sum formula:

$$\sum_{n=-\infty}^{\infty}g(t-nT)=\frac{1}{T}\sum_{k=-\infty}^{\infty}G\left(j\frac{2\pi k}{T}\right)e^{j\frac{2\pi k}{T}t}\tag{4}$$

Of course, if $g(t)$ is zero outside the interval $[0,T]$ we have

$$g(t)=\frac{1}{T}\sum_{k=-\infty}^{\infty}G\left(j\frac{2\pi k}{T}\right)e^{j\frac{2\pi k}{T}t},\qquad t\in[0,T]\tag{5}$$

Answer 3

Not stupid.

For a real valued (band limited) periodic signal, the coefficients for the Fourier series can be found directly from the Discrete Fourier Series (DFT/FFT) done on one period of the cycle.

See my answer here for details:

Using fourier coefficients to reconstruct data in matlab

As for naming, DFT is the term used for both the transform being done and the results of that transform, which can lead to confusion. Macleod uses the term Discrete Fourier Spectrum (DFS) for the results of a DFT. I agree that that would be a better convention, but it is not very well known.

Answer 4

Not exactly,

With continuous periodic functions, one is free to select any fundamental frequency. The Fourier series doesn’t suffer spectral leakage.

In a discretely samples sequence, exact periods are a rational multiple of the sample period. A continuous time periodic function isn’t necessarily periodic when sampled.