1
$\begingroup$

Can I apply the Fourier Transform to a Fourier series?

  • 2
    @marina: [Edit](http://math.stackexchange.com/posts/165308/edit) the question and put that comment in.2012-07-01

1 Answers 1

1

Let's assume your square-wave Fourier series looks like this:

$ \begin{align} x_{\mathrm{square}}(t) & {} = \frac{4}{\pi} \sum_{k=1}^M{\sin{\left ((2k-1) 2\pi ft \right )}\over(2k-1)} \\ & {} = \frac{4}{\pi}\left (\sin(2\pi ft) + {1\over3}\sin(6\pi ft) + {1\over5}\sin(10\pi ft) + \cdots\right ) \end{align} \tag{1} $

When you apply the Fourier Transform to this, you use the property of linearity:

For any complex numbers $a_n$, if $h(x)=\sum_n a_n\cdot f_n(x)$,  then $ \hat{h}(\xi)=\sum_n a_n\cdot \hat{f_n}(\xi) $,

so the FT of ${\sin{\left ((2k-1) 2\pi ft \right )}\over(2k-1)}$ is

$ \mathcal{F}_t\left[{\sin{\left ((2k-1) 2\pi ft \right )}\over(2k-1)}\right](\omega)= i\frac{ \sqrt{\pi/2} \delta\left(\omega-f \pi (2k-1)\right)}{(2k-1)}-i\frac{ \sqrt{\pi/2} \delta(\omega+f \pi (2k-1))}{(2k-1)}, $ with $\delta(\cdot)$ being the Dirac $\delta$ function. Now you can cut the frequencies above your threshold, but you might have done this in $(1)$ already, so there is actual need to transform it.