Fourier transform of $\sum_{n=-\infty }^\infty e^{-|t-2n|}$ I need to find Fourier transform of

$$\sum_{n=-\infty }^\infty  e^{-\lvert t-2n\rvert }$$

I found the solution online. But I think the choice of limits in equation (1) of the solution is wrong. It should be probably from $-\infty$ to $2n$ for the first integral and from $2n$ to $\infty$ for the second integral. But I am confused about what the limits would be as the both integration and summation is involved. and I am not sure whose limits to change (probably limits of integration must be changed only and that of summation must be kept as it is)

Given Solution:



Please tell me what the limits will be and the explanation. Also if possible please solve the problem.

Original answer:

$$ \frac{(1-e^{-2(1+jw)})}{(1-e^{-2})(1+jw)}-\frac{(e^{-2})(1-e^{-2(1+jw)})}{(1-e^{-2})(1-jw)}$$
First of all, the signal is periodic, so its spectrum must be discrete. Hence, your final solution ("Original answer") cannot be correct.

I would go the following way:


Calculate $X(f)=\mathcal{F}\left\{\exp(-\lvert t\rvert)\right\}$
Apply the time-shifting property of the Fourier Transform to transform all time-shifts of the function.
See that you can factor out $X(f)$ from the resulting sum and then see that the terms in the sum correspond to a comb function where the diracs are $\frac{1}{2}$ apart from each other.


Edit: Below is some code to plot the function:

Fs = 100  # some sampling frequency for discretizing the signal to plot it
T = 10    # the time range
t = np.arange(-T, T, 1/Fs)  # the time samples

# define the basis function
f = lambda t: np.exp(-abs(t))

# plot the basis function and the periodic extension
plt.plot(t, f(t))
plt.plot(t, sum(f(t-2*n) for n in range(-10, 10)))




Edit: Regarding your comment:

$$
\begin{align}
\mathcal{F}\left\{e^{−\lvert t\rvert}\right\}&amp;=X_1(j\omega)=\frac{2}{1+\omega^2}\\
\mathcal{F}\left\{\sum_{n}e^{-\lvert t-2n\rvert}\right\}&amp;=\sum_nX(j\omega)\exp\left(j2n \omega\right)\\
&amp;=X(j\omega)\sum_n\exp(j2n \omega)
\end{align}
$$

Here's the trick: 
$$\sum_n\exp(j2n \omega)=\frac{1}{2}\sum_n\delta\left(\omega-\frac{2\pi}{2}n\right)$$

Have a look at the wikipedia article and also this article for example on why this holds.