What I'm trying to understand for a few days is how Fourier transform is derived from Fourier series.
I know what is Fourier series. Using Fourier series, I could approximate a shape and draw the shape using JavaScript. (Please see my work if you are interested: https://logic-finder.github.io/memory/FourierSeriesExploration/chapter4/chapter4-en.html)
So this time I challenge to conquer Fourier transform. I've searched about this a little bit but in vain. It was tough to understand. As far as I understand, since Fourier series requires a period $T$ therefore it can't be applied to non-periodic waves, which have no period. But if we consider non-periodic waves as waves having an infinite period, now we can think of a Fourier series of them. … But here is a point where my first question arises: Let's suppose we have a complicated wave for ten seconds. Surely we don't know its period, since we do not know the future. But why can't we just say it's period is ten seconds? What's wrong with Fourier series? What is a meaning of Fourier transform? Fourier series still can give us the wave's components: $a0$, $an$, $bn$.
Anyways, where I'm having a trouble is this part:
Let
$$C_n=\frac{1}{T}\int^\frac{T}{2}_{-\frac{T}{2}}g(t)e^{-i2\pi nft}dt$$
$$g(t)=\sum_{n=-\infty}^\infty C_ne^{i2\pi nft}$$
where $T$ is period, $f$ is frequency, $C_n$ is Fourier coefficients (complex number form), and $g(t)$ is Fourier series (complex number form).
Since $g(t)$ has $C_n$ and $C_n$ is defined in that way, we can write $g(t)$ as follows:
$$g(t)=\sum_{n=-\infty}^\infty \left[\frac{1}{T}\int^\frac{T}{2}_{-\frac{T}{2}}g(t)e^{-i2\pi nft}dt\right]e^{i2\pi nft}$$
By definition, it is obvious that
$$\frac{1}{T}=f$$
Therefore, $g(t)$ can be rewritten as follows:
$$g(t)=\sum_{n=-\infty}^\infty \left[\int^\frac{T}{2}_{-\frac{T}{2}}g(t)e^{-i2\pi nft}dt\right]e^{i2\pi nft}\cdot f$$
So far, so good! I can explain every parts of the equation. (e.g. the reason that sigma starts from $-\infty$ and ends at $\infty$). But from here is where I lost.
If $T\to\infty$, $g(t)$ becomes:
$$\lim_{T\to \infty}g(t)=\lim_{T\to \infty}\sum_{n=-\infty}^\infty \left[\int^\frac{T}{2}_{-\frac{T}{2}}g(t)e^{-i2\pi nft}dt\right]e^{i2\pi nft}\cdot f$$
But at here somehow the equation changes magically, as resulting in the below:
$$\lim_{T\to \infty}g(t)=\int^\infty_{-\infty}\left[\int^\infty_{-\infty} g(t)e^{-i2\pi ft}dt\right]e^{i2\pi ft}\cdot df$$
It is observed that $\lim$, $\sum$, $n$, $f$ disappeared and $\int$ appeared newly. As far as I understand, As $T$ approaches to $\infty$, by definition, $f$ approaches to $0$. Thus intervals between $n$th multiples of fundamental frequency $f$ become dense, which results in the spectrum becoming continuous finally (The spectrum is no longer discrete like the existing Fourier coefficients–$a0$, $an$, $bn$).
But I don't understand what creates that newly added integral and why it starts from $-\infty$ and ends at $\infty$ (is it due to sigma?). So… what the equation means is that if I integrate continuous spectrum, then I will get the original wave??? It… it is hard to imagine/accept…
Also, I wonder whether it is possible to derive Fourier transform from Fourier series (trigonometric form)
Thank you for reading my long question.
p.s. My math level is at AP Calculus AB… I'll do my best if you help me, but if possible, please explain in an easy way… Thank you very much!
Best Answer
When you say that $f=1/T$ is "the frequency" you are actually referring to the [fundamental frequency][2]. There are more frequencies in the Fourier series: $f_n=n/T$ for $n\in\mathbb{Z}$ (yes, we also talk about negative frequencies). Then $1/T$ is the distance between consecutive frequencies, $\Delta f = f_{n+1}-f_{n} = 1/T.$ Therefore you get $$ g(t) =\sum_{n=-\infty}^\infty \left[\int^\frac{T}{2}_{-\frac{T}{2}}g(t)e^{-i2\pi nft}dt\right]e^{i2\pi f_n t} \Delta f. $$ Perhaps it's more natural now that the sum becomes an integral as $T\to\infty$?