Im trying to solve this ODE and find a simplified expression for $x(t)$.
$$\ddot x+4x=-2\sum_{n=1}^{4} e^{in\pi}u(t-n\pi);\space x(0)=0=\dot x(0),i=\sqrt{-1}$$
First i found the the laplace transform of the left hand side:
$$\mathcal{L}\{\ddot x+4x\}=s^2X(s)+4X(s)$$
But im not sure where to even start, to find the laplace transform of the right hand side? Cant find any similar examples, only general forms like this: $(t-\tau)^n e^{-\alpha (t-\tau)} \cdot u(t-\tau)$, that i cant tell whether or not they apply here or not?
Update #1 @ 6/3/14, 20:54) I now have obtained the simplified form:
$$x(t)=-\frac{1}{2}\sum_{n=1}^{4} (-1)^{n}u(t-n\pi)(1-\cos(2t)$$
But im really stumped on how to graph this? I would know how to graph it, if it was just $x(t)=u(t-n\pi)(1-\cos(2t)$ , but i really dont know what to do with this summation thing, in terms of plotting?
Note that in a system with step delays at different times, the solution of the system at the time that a new step function "activates" is equivalent to a new set of initial conditions at that time.
So if you have a system with some solution trajectory, and at time $T$ you have an input multiplied by a step, then when $t < T_1$, that input has zero effect. When $t = T_1$, you use the solution up to that point as your initial conditions. Do the same at $T_2, T_3,$ etc.