Using Fourier transform to compute Fourier series. 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). 
 A: 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 $c_k$.
A: 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 $n$.
