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I have found an exercise on a signal processing book that asks to compute the Fourier series of a function by using its Fourier Transform, let:

$$ x(t) = \sum_{n=-\infty}^{\infty} \Lambda \left( \dfrac{t-nT}{T/2} \right) $$

I did compute the FT of this function, I used that a trianglePulse can be decomposed into two convoluted rectangularPulse. It led to:

$$ x(t) = \Pi \left( \dfrac{t}{T/2} \right) \star \Pi \left( \dfrac{t}{T/2} \right) \star \sum_{n=-\infty}^{\infty} \delta (t-nT) $$

Thus, the Fourier transform is:

$$X(\omega)= \dfrac{4}{\omega^2} sin^2 \left( \dfrac{\omega T}{4} \right)\cdot \dfrac{2\pi}{T} \sum_{n=-\infty}^{\infty} \delta \left( \omega - n\dfrac{2\pi}{T} \right) $$

Where: $$FT\left[\sum_{n=-\infty}^{\infty} \delta (t-nT) \right] = \sum_{n=-\infty}^{\infty} e^{-jnT\omega} = \dfrac{2\pi}{T} \sum_{n=-\infty}^{\infty} \delta \left( \omega - n\dfrac{2\pi}{T} \right)$$

I know that FT is a generalization of the Fourier Series. But I don't know how to continue in order to find the Fourier series. What should I try to do to solve the exercise? (I just need to know how to continue, it's not necessary that you do it for me).

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  • $\begingroup$ There should be some complex exponential terms in $X(\omega)$ coming from the fourier transform. Real and imaginary parts combine to give cosine and sine terms. $\endgroup$ – Paul May 6 '15 at 15:32
  • $\begingroup$ Please, check my edit. I obtained a complex exponential. Should I decompose it into $\sin$ and $\cos$ ? Thank you $\endgroup$ – Airman01 May 6 '15 at 15:40
  • $\begingroup$ This seems correct :) $\endgroup$ – krvolok May 6 '15 at 15:43
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Once you have obtained the Fourier Transform:

\begin{equation} \mathcal{F} \big\{ x(t)\big\} = X(\omega) = \dfrac{4}{\omega^2} \sin^2 \left( \dfrac{\omega T}{4} \right)\cdot \dfrac{2\pi}{T} \sum_{n=-\infty}^{\infty} \delta \left( \omega - n\dfrac{2\pi}{T} \right) \end{equation}

You can take into consideration that for any function $b(\omega)$ and a certain pulse delay $\omega_0$ you have:

\begin{equation} b(\omega) · \delta(\omega-\omega_0) = b(\omega_0)\delta(\omega-\omega_0) \tag{1} \end{equation}

Substituting $\omega_n = n\dfrac{2\pi}{T}$ we can rewrite you Fourier transform as following:

\begin{equation} X(\omega) = \dfrac{4}{\omega^2} \sin^2 \left( \dfrac{\omega T}{4} \right)\cdot \sum_{n=-\infty}^{\infty} \dfrac{\omega_n}{n} · \delta \left( \omega - \omega_n \right) \end{equation}

Now considering (1):

\begin{equation} X(\omega) = \sum_{n=-\infty}^{\infty} \dfrac{4}{\omega_n^2} ·\sin^2\Big(\dfrac{\omega_nT}{4}\Big)·\dfrac{\omega_n}{n}·\delta \left( \omega - \omega_n \right) \end{equation}

Working it out we finally obtain:

\begin{equation} X(\omega) = \sum_{n=-\infty}^{\infty} \dfrac{2T}{n^2\pi} ·\sin^2\Big(\dfrac{n\pi}{2}\Big)·\delta \left( \omega - n\dfrac{2\pi}{T} \right) = \sum_{n=-\infty}^{\infty} \alpha_n·\delta \left( \omega - n\dfrac{2\pi}{T} \right) \end{equation}

Where $\alpha_n = \dfrac{2T}{n^2\pi} ·\sin^2\Big(\dfrac{n\pi}{2}\Big)$ denotes the amplitude for each sample $n$.

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  • $\begingroup$ Thanks @luikn! It was helpful and now I could check that the exercise result was correct. I would vote your answer, but I can't yet. Thanks $\endgroup$ – Airman01 May 6 '15 at 18:59
  • $\begingroup$ Don't worry, we are here to help each other :) @Airman01 $\endgroup$ – lucasrodesg May 6 '15 at 21:25
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If you have a $T$-periodic function $x(t)$, its Fourier series is given by

$$x(t)=\sum_{k=-\infty}^{\infty}c_ke^{j2\pi kt/T}\tag{1}$$

where $c_k$ are the complex Fourier coefficients. Taking the Fourier transform of (1) gives

$$X(j\omega)=2\pi \sum_{k=-\infty}^{\infty}c_k\delta\left(\omega-\frac{2\pi k}{T}\right)\tag{2}$$

So, given the Fourier transform (2), you can directly read off the Fourier coefficients $c_k$.

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  • $\begingroup$ Wow alright. So does the problem reduces to saying: $ c_k = \dfrac{4}{T\omega^2} \sin^{2} \left( \dfrac{\omega T}{4} \right) $ ? (Being that the coefficient in the Fourier Series) $\endgroup$ – Airman01 May 6 '15 at 16:22
  • $\begingroup$ @Airman01: Yes, with $\omega=2\pi k/T$ because $f(\omega)\delta(\omega-\omega_0)=f(\omega_0)\delta(\omega-\omega_0)$. $\endgroup$ – Matt L. May 6 '15 at 20:18

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