Pendulum with Dirac Comb excitation There is a pendulum that is excited by a Dirac Comb.
$l \ddot\theta+b\dot \theta+g\theta=G\,\sum_{-\infty}^\infty\delta(t-nT)$
where $l, b, g, G$ are constants and $T=\dfrac{2\pi}{\omega}$.
Show that the resulting motion is given by
$\theta(t)=\dfrac{G}{Tl\omega^2}+\dfrac{2G\cos(\omega t-\frac{\pi}{2})}{Tb\omega}$+[terms with frequencies $\ge$ 2$\omega$]
and explain why the higher frequency terms are supressed.
My first take was to rearrange to
$ \ddot\theta+\dfrac{b}{l}\dot \theta+\omega^2\theta=\frac{G}{l}\,\sum_{-\infty}^\infty\delta(t-nT)$
where $\omega=\sqrt{\dfrac{g}{l}}$
Then, taking the Laplace transform of both sides I got
$\Theta(s)=\dfrac{G}{l\,\sqrt{\omega^2-\left(\frac{b}{2l}\right)^2}}\, \dfrac{\sqrt{\omega^2-\left(\frac{b}{2l}\right)^2}}{\left(s+\frac{b}{2l} \right)^2-\left(\left(\frac{b}{2l}\right)^2-\omega^2 \right)}\, \dfrac{1}{1-e^{-sT}}$
which, as far as I'm concerned transforms to
$\theta(t)=\dfrac{G}{l\,\sqrt{\omega^2-\left(\frac{b}{2l}\right)^2}}\sum_{n=0}^\infty\,H(t-nT)\,e^{-\frac{b}{2l}(t-nT)}\sin\left(\sqrt{\omega^2-\left(\frac{b}{2l}\right)^2}(t-nT) \right)$
And, assuming that this is a correct form of the solution, I can't see how that is equivalent with the function given in the question. I reckon it has something to do with using Fourier series/transform instead? If so, I'm not sure how to do that. Or, is there a way to convert my solution into the given one?
I've been struggling with this for a good few days now, so any help would be much appreciated.
 A: The distinction between apparent solutions via Laplace transform and via Fourier transform depends meaningfully on the function-space that the "target" function is in, and the desired solution is in.
Suppressing some of the constants to reduce clutter, consider the equation
$$
\theta''+a\theta'+\theta=\sum_{n\in\mathbb Z}\delta_n
$$
Under various external context assumptions, we might suppose that the solution $\theta$ is at worst a tempered distribution (the right-hand side is such). This excludes exponentially-growing (either forward or backward) solutions $\theta$. In particular, unless the constants $a,b$ on the left-hand side are such that solutions to the characteristic equation $\lambda^2+a\lambda+b=0$, the homogeneous equation has no tempered-distributions solutions (except $0$).
The Poisson summation formula is the assertion that $\sum_n\delta_n$ is its own Fourier transform. Thus, again up to essentially irrelevant constants,
$$
-x^2\widehat{\theta}+bix\widehat{\theta}+c\widehat{\theta}\;=\;\sum_n\delta_n
$$
which suggest that
$$
\widehat{\theta} \;=\; \sum_n {\delta_n\over -x^2+bix+c}
\;=\;
\sum_n {\delta_n \over -n^2+bin+c}
$$
Yes, there is a technical problem if the denominator should vanish... :)
Transforming back, again suppressing constants,
$$
\theta(t) \;=\; \sum_n {e^{2\pi int}\over -n^2+bin+c}
$$
This shows the mechanism...
