Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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

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

share|cite|improve this question

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.

enter image description here

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:

enter image description here

share|cite|improve this answer
Nice and complete answer. – Dmoreno May 14 '14 at 8:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.