Fourier Transform on a compact support

Question

Is there a Fourier Transform of a function that exists only on an interval like [0,1]? A solution might be to use a window function, that’s what others told me. But that’s not what I want. My concrete problem is: I use a fourier-series because my function is 2$\pi$ -periodic. In the next step, I just have subset of 2$\pi$, like [0,$\pi$]. I don’t know how or whether I can make a transformation of a function, that only exists on that angular range. It’s like "is there any Fourier transform on a compact carrier"?
Best regards
Loki

Answer 1

You can use a Fourier series on any interval $[a,b]$ by mapping the interval linearly to $[0,2\pi]$. For example, if you have a function $f$ defined on $[0,\pi]$, you can define a function $\tilde{f}$ on $[0,2\pi]$ by $\tilde{f}(x) = f(2x)$, expand it as a Fourier series (let us assume it converges) to obtain

$$ \tilde{f}(x) = \sum_{n \in \mathbb{Z}} a_n e^{i n x} $$

but then

$$ f(x) = \tilde{f} \left( \frac{x}{2} \right) = \sum_{n \in \mathbb{Z}} a_n e^{\frac{i n x}{2}}. $$

You can also write a formula for the coefficients $a_n$ in terms of $f$ alone by performing a variable change and this generalizes to arbitrary intervals.