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Can I apply the Fourier Transform to a Fourier series?

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What have you tried so far?: What have you tried so far? – draks ... Jul 1 '12 at 13:34
I have a Fourier series that belongs to a bipolar pulse (square wave). I have to filter that signal through an ideal low pass filter, h(f), with cutoff frequency at 4KHz. At the beginning I thought the Fourier series was the "same" as the Fourier Transform but then I realize that the Fourier series is x(t) not x(f). Then I started to think that maybe I should apply the Fourier transform but I am not sure if I should do that because as I said I thought the Fourier series was the same as Fourier transform. – marina Jul 1 '12 at 13:43
Now I am confuse. What I need to know if they are equally. @dranks – marina Jul 1 '12 at 13:47
@marina: Edit the question and put that comment in. – kennytm Jul 1 '12 at 15:18

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.

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