# ODE Laplace Transforms: what impulse brings an oscillating system to rest?

$2y''+y'+2y=\delta(t-5)$

$y(0)=0, y'(0)=0$

Consider the system given by ODE above in which an oscillation is excited by a unit impulse at $t=5$. Suppose that it is desired to bring the system to rest again after one cycle.

Determine the impulse $k\delta(t-t_{0})$ that should be applied to system in order to accomplish this objective (bring sys to rest).

Now how can I calculate the extra impulse to applied at time $T$? I have no idea. I added an extra impulse to the system and tried to set $y(t)=0$ at that time $T$ but no luck. How can I kill a sinusoidal oscillation by applying an extra impulse?

Books answer: $-\exp(-T/4)\delta(t-5-T)$

We want to find $k\,\delta(t-(5+T))$ which will bring to rest the solution of $$2y''+y'+2y=\delta(t-5)\color{blue}{+k\,\delta(t-(5+T))},\quad y(0)=0,\ y'(0)=0,\tag{1}$$ at time $t_*=5+T$, where $t_*$ is the time that the original problem (in black above) completes one full cycle. And by "bring to rest at $t=t_*$", we mean that $y(t)\equiv 0$ for $t\ge t_*$.

Taking the Laplace transform of $(1)$, we see $$Y(s)=e^{-5s}\cdot {1\over 2s^2+s+2}+k\,e^{-(5+T)s}\cdot {1\over 2s^2+s+2}.$$ By standard techniques, $$\mathscr{L}^{-1}\left[{1\over 2s^2+s+2}\right]= {2\over \sqrt{15}}e^{-t/4}\sin\left({\sqrt{15}\over 4}t\right)=:g(t),$$ so the Convolution Theorem yields \begin{align} y(t)&=\delta(t-5)*g(t)+k\,\delta(t-(5+T))*g(t)\\ &=\int_0^t g(s)\delta(t-s-5)\,ds+k\int_0^t g(s)\delta(t-s-(5+T))\,ds\\ &=g(t-5)u(t-5)+kg(t-(5+T))u(t-(5+T)),\tag{2} \end{align} where $u(t)$ denotes the unit step function.

Let $t=t_*$ be the value which completes the first full cycle of the (exponentially decaying) sinusoid, denoted by the black dot. To find $t_*$, use the first term in $(2)$: $$g(t-5)=0\implies \sin\left({\sqrt{15}\over 4}(t-5)\right)=0\implies t=5+{4n\pi\over \sqrt{15}},$$ and $n=2$ produces the zero we seek: $$t_*=5+{8\pi\over \sqrt{15}} \implies \boxed{T=t_*-5={8\pi\over \sqrt{15}}},$$ from the way we set up the original problem (so that $T$ is the time from $t=5$ until we apply the second impulse).

Since we already have $y(t_*)=0$, we just need to find $k$ such that $y'(t_*)=0$, then we can conclude $y(t)\equiv 0$ for $t>t_*$. To accomplish this, \begin{align*} \lim_{t\to t_*^+}y'(t)&=g'(t_*^+-5)+k\,g'(t_*^+-(5+T))\\ &=g'\left({8\pi\over \sqrt{15}}^+\right)+k\,g'(0^+)\\ &={1\over 2}\exp(-2\pi/\sqrt{15})+k\cdot{1\over2}, \end{align*} which vanishes when $$\boxed{ k=-\exp(-2\pi/\sqrt{15})=-\exp\left(-{8\pi\over \sqrt{15}}/4\right)=-\exp(-T/4).}$$

Here's a graph of the solution $(2)$ with the values of $T$ and $k$ found above: • Nice and complete answer. – Dmoreno May 14 '14 at 8:01