Given $x_1(t),X_1(j\omega), x_2(t)=\sum_{k=-\infty}^{\infty}x_1(t-6k)$, find Fourier series coefficient of $x_2(t)$.

Looking up the FT table, I got $X_2(j\omega)=\sum_{k=-\infty}^{\infty}e^{-j\omega 6k}X_1(j\omega)$. Fourier transform can be represented as a summation of FS terms, so FS coefficient of $x_2(t)$, $a_k$, is $X_1(j\omega)$, is this correct?

  • $\begingroup$ Have you heard of Poisson's sum formula? $\endgroup$ – Matt L. Apr 9 '20 at 10:08
  • $\begingroup$ @MattL. No until I looked it up just now, and I don't know how to apply it, since $x_2(t)$ here is shifted. $\endgroup$ – keanehui Apr 9 '20 at 10:20
  • $\begingroup$ Eq. (4) in this answer is what you need. $\endgroup$ – Matt L. Apr 9 '20 at 16:10

Figure out that $x_2(t)$ is basically, sum of shifted copies of $x_1(t)$ which can be written as follows : $$x_2(t) = x_1(t) * \sum^{\infty}_{k=-\infty} \delta(t-6k)$$, where, $*$ represents convolution operation. therefore, the Fourier Representation of $x_2(t)$ will be product of the fourier representation of $x_1(t)$ and the pulse train $\sum^{\infty}_{k=-\infty} \delta(t-6k)$.

Fourier Transform of a pulse train $\sum^{\infty}_{k=-\infty} \delta(t-6k)$ is given by the following :

$$\mathscr F \Big\{ \sum^{\infty}_{k=-\infty} \delta(t-6k) \Big\} = \frac{1}{6} \sum^{\infty}_{k=-\infty} \delta(f-\tfrac{k}{6})$$

Which basically means that Fourier Representation of $x_2(t)$ becomes : $$X_2(f) = X_1(f) \cdot \frac{1}{6} \sum^{\infty}_{k=-\infty} \delta(f-\tfrac{k}{6})$$

So, $f = \frac{1}{6}$ is the fundamental frequency, since you have periodized $x_1(t)$ by $T = 6$. And, therefore, at all multiples of $\frac{1}{6}$, you will get a Fourier coeff, which will be equal to $\frac{1}{6} X_1(\frac{k}{6})$.

What this means is, $X_2(f)$ is nothing but sampled version of $X_1(f)$ at $f = \frac{k}{6}$ and hence, $X_2(f)$ is discrete. Which is evident from the fact that $x_2(t)$ was periodized in the first place.

  • $\begingroup$ Thank you. And is it possible to graph $X_2(j\omega)$? If so, how? Since it's continuous infinitely sum of shifted $\endgroup$ – keanehui Apr 13 '20 at 13:32
  • 1
    $\begingroup$ Hi, $X(j\omega)$ is DTFT representation which is used to analyze discrete time infinite support sequences. $x_2(t)$ is still continuous time signal, it is just periodised version of $x_1(t)$, so we can only transform if using Fourier Series which I have explained above. $X_2(\omega)$ will just be sampled version of $X_1(\omega)$ at $\omega = 2\pi k/6$. $\endgroup$ – DSP Rookie Apr 13 '20 at 13:40
  • $\begingroup$ Is scaling needed, because of the $\frac{1}{6}$? $\endgroup$ – keanehui Apr 13 '20 at 13:58
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    $\begingroup$ Yes, amplitude scaling of $\frac{1}{6}$ is required as well. $\endgroup$ – DSP Rookie Apr 13 '20 at 14:00

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