1
$\begingroup$

I want to derive the Fourier transform of the impulse train. So far I have gotten up to this point. $$p(t) = \sum_{n = -\infty}^{\infty}\delta(t - nT_s)$$ $$P(\omega) = \int_{-\infty}^{\infty}p(t).e^{-j\omega t}dt$$ $$P(\omega) = \int_{-\infty}^{\infty}\sum_{n = -\infty}^{\infty}\delta(t - nT_s).e^{-j\omega t}dt$$ Interchanging summation and integration, $$P(\omega) = \sum_{n = -\infty}^{\infty}\int_{-\infty}^{\infty}\delta(t - nT_s).e^{-j\omega t}dt$$ This yields $$P(\omega) = \sum_{n = -\infty}^{\infty}e^{-j\omega nT_s}$$

But I know the transform function is $$P(\omega) = \frac{2\pi}{T_s}\sum_{n = -\infty}^{\infty}\delta(\omega - n\omega_s)$$

Can anyone point me in a direction where I can get the exponential term into a delta function? :)

$\endgroup$
1
$\begingroup$

You should start by writing the Fourier series of representation of $p(t)$, i.e.:

$$ p(t) = \sum_k a_k e^{ik2\pi t/T} $$

where

$$ a_k = \frac{1}{T} \int_{-T/2}^{T/2} p(t)e^{-ik2\pi t/T} dt $$

From the delta definition of $p(t)$, it follows that $a_k=1$ for all $k$. The Fourier transform, $P(\omega)$, follows from the Fourier transform of an exponential, yielding:

$$ P(\omega) = 2\pi \sum_k \delta\left(\omega-\frac{2\pi}{T}k\right) $$

I appear to have lost a factor of $T$, but this gives you the general idea.

$\endgroup$
  • $\begingroup$ in my opinion $\sum_n \delta(t-n) = \sum_k e^{2i \pi k t}$ is exactly the solution to the problem, thus the problem is understanding the Fourier transform itself. in fact, if you assume the Fourier series inversion theorem for functions L1 on one period (and for distributions = limits of such functions) then the OP question is trivial. $\endgroup$ – reuns Dec 31 '15 at 1:20
  • 1
    $\begingroup$ Is it $a_k = \frac{1}{T}$ for all k? So there comes the missing factor of T :) So the Fourier transform of $e^{j\omega_0t}$ is $2\pi \delta (\omega - \omega_0)$ and that gives us the $P(\omega)$ $\endgroup$ – Blogger Dec 31 '15 at 2:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.