# Given a signal and its Fourier transform, find FS coefficient of the shifted sum of the signal

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?

• Have you heard of Poisson's sum formula? – Matt L. Apr 9 '20 at 10:08
• @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. – keanehui Apr 9 '20 at 10:20
• Eq. (4) in this answer is what you need. – 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.

• Thank you. And is it possible to graph $X_2(j\omega)$? If so, how? Since it's continuous infinitely sum of shifted – keanehui Apr 13 '20 at 13:32
• 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 $\omega = 2\pi k/6$. – DSP Rookie Apr 13 '20 at 13:40
• Is scaling needed, because of the $\frac{1}{6}$? – keanehui Apr 13 '20 at 13:58
• Yes, amplitude scaling of $\frac{1}{6}$ is required as well. – DSP Rookie Apr 13 '20 at 14:00