- The Fourier series is a decomposition of a periodic function into sums of sinusoids with periods less than it. Representing a function with longer period requires sinusoids with longer periods, which is the same as requiring a denser range of frequencies.
- A non-periodic function can be understood as a function with an infinitely long period, which requires a frequency range of all real numbers.
- So the Fourier transform can be understood as a generalization of the Fourier series that represents all functions as sums of sinusoids. Interestingly, the inverse Fourier series (i.e. the expression for the coefficients) was already an integral, albeit with finite domain, while the Fourier series already had an infinite domain (all integers), albeit not an integral. In the limit where you get the Fourier transform, though, they both magically become the same.
- Anyway, it's then straightforward to see the Fourier relationship between the sinusoid and the Dirac delta function, etc.
- In general, the Fourier transform can be seen as a "change of basis" for functions. Looking at the sinusoids as a basis in this sense, one can immediately infer e.g. Parseval's theorem and its generalisation the Rayleigh energy theorem as the "Pythagoras theorem" for this basis.
- Other changes of basis or linear transformations lead to other integral transforms.
Consider a function with period 1 -- computing its Fourier series, you write it as:
\[f(x) = \sum\limits_{n = - \infty }^\infty {{a_n}{e^{i2\pi \,\,nx}}} \]
Where
\[{a_n} = \int_{-1}^1 {f(x){e^{ - 2\pi inx}}dx} \]
That's all standard and trivial. But suppose you wanted to study a function with a higher period (we will tend this period to infinity) -- what would that look like? Well, consider $g(x)=f(x/L)$, which is this function we're looking for -- then we can rewrite the above identities as:
\[g(xL) = \sum\limits_{n = - \infty }^\infty {{a_n}{e^{i2\pi {\kern 1pt} {\kern 1pt} nx}}} \Rightarrow g(x) = \sum\limits_{n = - \infty }^\infty {{a_n}{e^{i2\pi {\kern 1pt} {\kern 1pt} nx/L}}} \]
\[{a_n} = \int_{-1}^1 {g(xL){e^{ - 2\pi inx}}dx} \Rightarrow {a_n} = \int_{-L}^L {g(x){e^{ - 2\pi inx/L}}dx/L} \]
Where we transformed $x\to x/L$.
This seems all too trivial and useless, and maybe you're looking for a little trick to turn this into something interesting. But tricks must typically also arise from some sort of insight. Let's assume for a moment that we didn't know anything about variable substitutions or transformations like the kind we did above (and indeed, the idea behind variable substitutions also comes from a geometric understanding of the corresponding transformation) and think about how we may re-think the Fourier transform in its context.
Well, if the function's period is $P$, in other words it is stretched out by $P$, the same logic must be used to derive the Fourier series for the new function as for the function with period 1 -- specifically, sines and cosines with longer periods than $P$ don't matter (their coefficient must be zero, because otherwise you've introduced an element into the function that doesn't repeat with that period), but those with shorter, divisible periods matter, because they influence the value of the function within the period, perturbing it by little bits to get to the right function.
So when dealing with our new period $L$, one would expect periods that are fractions of $L$, i.e. $L/n$, as opposed to just $1/n$. So $n/L$ is "more important" than $n$, and indeed it seems very easy to transform the summation into one in terms of this new variable, which we will still call $n$ (i.e. transform $n/L\to n$):
\[g(x) = \sum\limits_n^{} {{a_n}{e^{i2\pi nx}}} \]
\[{a_n} = \frac{1}{L}\int_{-L}^L {g(x){e^{ - 2\pi inx}}dx} \]
Where we labeled $a_{nL}$ as just $a_n$, because that's just a subscript, the labeling doesn't matter. Just remember that $n$ is no longer just an integer/multiple of 1, but a multiple of any fraction $1/L$.
Now note how a non-periodic function is just a function with infinite period, i.e. $L\to\infty$. So $n$ stops being a discrete integer and starts approaching a continuous variable, which we'll call $s$, writing $a_n$ as $a(s)ds$ (why the $ds$? because the increment in $n$ is just $1/L$, which appears in the expression for $a_n$).
\[g(x) = \int_{ - \infty }^\infty {ds\,\,a(s){e^{i2\pi sx}}} \]
\[a(s) = \int_{ - \infty }^\infty {dx\,\,g(x){e^{ - i2\pi sx}}} \]
Which is just a pretty satisfying result.
Recall again the expressions we got for the Fourier transform and its inverse:
\[f(t) = \int_{ - \infty }^\infty {ds\,\,\hat f(s){e^{i2\pi ts}}} \]
\[\hat f(s) = \int_{ - \infty }^\infty {dt{\kern 1pt} {\kern 1pt} f(t){e^{ - i2\pi st}}} \]
(We typically say the Fourier transform maps time-domain functions to frequency-domain ones, so we consider the latter to be the Fourier transform and the first equation to be its inverse.) Note how you can easily turn the first one into an actual Fourier transform, by transforming $s\to -s$:
\[f(t) = \int_{ - \infty }^\infty {ds\,\,\hat f( - s){e^{ - i2\pi ts}}} \]
In other words:
\[{\mathcal{F}^{ - 1}}\left\{ {f(s)} \right\} = \mathcal{F}\left\{ {f( - s)} \right\}\]
And of course that means ${\mathcal{F}^4} = I$, the identity operator (kind of like the derivative on complex exponentials/sine and cosine, is it not?).
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