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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?
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.
