# Fourier transform of an infinite sum

Find the fourier transform of

$$\sum_{m=-\infty}^\infty f (t-mT)$$

Where $f (t) = 1$ for $0 <t < T$ and 0 otherwise. I am not sure how to tackle it ? Also I have problems if it's correct to interchange summation with integration.

• Your sum is equal to 1, except at integer multiples of $T$. Therefore the Fourier transform is a constant times the Dirac mass at 0. Commented Dec 4, 2016 at 1:01
• @Jose27, thanks that was my approach. Commented Dec 4, 2016 at 8:44

$$f_T(t)=\sum_{m=-\infty}^\infty f (t-mT)$$ is a periodic function with period $T$, and within each period it is equal to $f(t)$.
It has a Fourier series representation which is simple (it is just like a square wave). Assume the FS representation is $$f_T(t)=\sum_{k=-\infty}^{\infty}c_ke^{i\frac{2k\pi }{T} t}$$ Take the term-by-term Fourier transform of the series, using the linearity of Fourier transform: \begin{align}\mathcal{F}(f_T(t))&=\mathcal{F}\left(\sum_{k=-\infty}^{\infty}c_ke^{i\frac{2k\pi}{T} t}\right)\\ &=\sum_{k=-\infty}^{\infty}c_k\mathcal{F}\left(e^{i\frac{2k\pi}{T} t}\right)\\ &=2\pi \sum_{k=-\infty}^{\infty}c_k\delta(\omega-\frac{2k\pi}{T}) \end{align}
• You can't take $c_k$ out of the sum. Commented Dec 4, 2016 at 8:44
• Yes, it was a mistake in writing the sum. Note that the FT of a constant $\frac{k}{2\pi}$ is $k\delta(\omega)$ and your function is not a constant.