1
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ 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. $\endgroup$
    – Jose27
    Commented Dec 4, 2016 at 1:01
  • $\begingroup$ @Jose27, thanks that was my approach. $\endgroup$ Commented Dec 4, 2016 at 8:44

1 Answer 1

1
$\begingroup$

$$f_T(t)=\sum_{m=-\infty}^\infty f (t-mT)$$ 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: $$\begin{align}\mathcal{F}(f_T(t))&=\mathcal{F}\left(\sum_{k=-\infty}^{\infty}c_ke^{i\frac{2k\pi}{T} t}\right)\\ &=\sum_{k=-\infty}^{\infty}c_k\mathcal{F}\left(e^{i\frac{2k\pi}{T} t}\right)\\ &=2\pi \sum_{k=-\infty}^{\infty}c_k\delta(\omega-\frac{2k\pi}{T}) \end{align}$$

$\endgroup$
2
  • $\begingroup$ You can't take $c_k$ out of the sum. $\endgroup$ Commented Dec 4, 2016 at 8:44
  • $\begingroup$ 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. $\endgroup$
    – msm
    Commented Dec 4, 2016 at 10:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .