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The Fourier series of the cycloid can be expressed in terms of the Bessel functions of the first kind: $$J_n(x)=\frac{1}{\pi}\int_0^{\pi}\cos(nt-x\sin t)dt,\qquad n\in\mathbb{Z}\tag{1}$$ Using the cycloid parameterization $$y(t)=1-\cos t,\qquad x(t)=t-\sin t\tag{2}$$ which results in a period of $2\pi$ and a maximum value of $2$, the Fourier series of $y(t)$ ...
I haven't managed to completely evaluate the integral, but I've made some progress and perhaps someone can pick up where I leave off. The integral you gave is $$c_n = \int_0^{2\pi}\left(\frac{1-\cos\theta}{\pi}\right)^2 e^{-jn(\theta-\sin\theta)}d\theta.$$ The first thing I did is perform a Taylor expansion on that exponential term; $$c_n = \frac{1}{\pi^2}... 2 "Overlap add" or "overlap save" should work just fine here. See https://en.wikipedia.org/wiki/Overlap%E2%80%93add_method You'll have to truncate the Gaussian both in time and in frequency but since a Gaussian decays really fast, it's easy to find a length that's "good enough" for your application. You will also have to pick a ... 2 The Kaiser window of length L is:$$ w(t) = \begin{cases} \frac{1}{I_0(\beta)} \, I_0\left(\beta \sqrt{1 - \left(\frac{t}{L/2}\right)^2 } \right) \qquad & |t| \le L/2 \\ 0 & |t| > L/2 \\ \end{cases}$$where$$ I_0(u) \triangleq \sum\limits_{k=0}^{\infty} \frac{(-1)^k }{(k!)^2}\left( \frac{u}{2} \right)^{2k} $$is the zeroth-order Bessel ... 2 I'm a bit rusty with solving equations using the Laplace transform so please be indulgent if I made a mistake. your signal is$$ x = sin(\omega t + \theta)$$and you want to find$$ y = \int{sin(\omega t + \theta)dt} $$You can rewrite x as$$ x = a\sin(\omega t) + b\cos(\omega t)$$In Laplace$$ y(s) = \frac{1}{s} \{\frac{a\omega + bs}{s^2 + \omega^2}\} $$... 2 I recommend a standard dq-PLL, check this reference https://vbn.aau.dk/ws/portalfiles/portal/273236528/PLL_Review_RSER.pdf There are 2 required integrators. For algebraic loop reasons, the last integrator, the one that converts the angular frequency to the phase, should be a forward-Euler integrator. The integrator in the PI controller, can be a trapezoidal ... 1 Does such a procedure exist? Of course. All systems that you are describing are just FIR filters, i.e.$$y[n] = \sum_{k=0}^{N-1} h[k] \cdot x[n-k] $$and their Z-transform is$$H(z) = \sum_{k=0}^{N-1} h[k] \cdot z^{-k} $$The inverse becomes$$G(z) = H^{-1}(z) = \frac{1}{\displaystyle\sum_{k=0}^{N-1} h[k] \cdot z^{-k}}  This is an "all pole" ...