I have determined the complex Fourier series of the given periodic function $f(t) = e^{-t}$ over the interval $(-3,3)$ and $f(t+6)$ to be the following,
$$\sum_{n=-\infty}^\infty \left[ \frac{(-1)^n\sinh 3}{3+n\pi i}\right] e^{\pi i nt/3}$$
Note the period is length $6$, and we use one full period on the interval $(-3,3)$
However now I am trying to plot some points of the frequency spectrum and I am not quite sure how to proceed. I'm wondering if someone would be able to show me? or is there a method to use on wolfram alpha to create this frequency spectrum?
Here, the complex Fourier coefficients are $c_n(f) = \frac{1}{6} \int_{-3}^{3} \text{e}^{-t} \text{e}^{-n\text{i}\pi t / 3} \text{d}t$, i.e. $$ c_n(f) = \frac{(-1)^n\sinh(3)}{3 + \text{i}n\pi} \, . $$ The modulus $\chi_n$ of the $n$th harmonic --which sequence is commonly denoted by amplitude spectrum-- is ${\chi_n} = \sqrt{a_n(f)^2 + b_n(f)^2}$, where \begin{aligned} a_n(f) &= \phantom{\text{i}\, (} c_n(f) + c_{-n}(f) \phantom{)} = \phantom{-} 2\,\Re\, c_n(f)\\ b_n(f) &= \text{i} \left(c_n(f) - c_{-n}(f)\right) = -2\,\Im\, c_n(f)\, . \end{aligned} Therefore, $$ {\chi_n} = 2 \left|c_n(f)\right| = \frac{2\sinh(3)}{\sqrt{9 + n^2\pi^2}}\, . $$ The modulus $\chi_n$ of the $n$th harmonic can be displayed in Alpha, using
plot table 2 sinh(3)/sqrt(9 + n^2*pi^2) for n = 0..10:The phase spectrum $\varphi_n = \arctan(-b_n(f)/a_n(f))$ given by $$ \varphi_n = \arctan\left(\frac{\Im\, c_n(f)}{\Re\, c_n(f)}\right) = \arctan\left(-\frac{n\pi}{3}\right) $$ can be displayed in a similar manner using
plot table atan(-n pi / 3) for n=1..10: