Determining the solution of a closed loop system

Consider the system

\ddot{q}(t)=u(t)

and

y=\begin{pmatrix} q(t) \\ \dot{q}(t) \end{pmatrix}

Take

\begin{pmatrix} \epsilon\sin{(\omega t)} \\ \epsilon\sin{(\omega t)} \end{pmatrix}

as the noise, with

\epsilon

small and

\omega

large.

So the output we have at our disposal for controlling the system is

\begin{pmatrix} q(t) \\ \dot{q}(t) \end{pmatrix} +\begin{pmatrix} \epsilon\sin{(\omega t)} \\ \epsilon\sin{(\omega t)} \end{pmatrix}

.

Choose

u(t)=\begin{pmatrix} -1 & -2 \end{pmatrix}

and

y(t)=-2q'(t)-q(t)

,

Then applying this to our system, we get

u(t)=-2(q'(t)+\epsilon\sin{(\omega t)})-(q(t)+\epsilon\sin{(\omega t)})

I want to determine the solution of this closed loop system and that the perturbation on the output

y

remains bounded...Or more precisely, the perturbation on the output

y

is bounded by

C\frac{\epsilon}{\omega}

, where

C>0

is independent of

\epsilon >0

and

\omega\ge 1

u(t)=-2q'(t)-q(t)-3\epsilon\sin{(\omega t)}

Thus

0=\ddot{q}(t)+2\dot{q}(t)+q(t)+3\epsilon\sin{(\omega t)}

Any tips on proceeding from here?

I get the inhomogeneous second order ODE,

\ddot{q}(t)+2\dot{q}(t)+q(t)=-3\epsilon\sin{(\omega t)}

.

How should one proceed with this?

The homogeneous part of the equation is,

q_{h}(t)=c_{1}e^{-t}+c_{2}te^{-t}

(edited 9 years ago)

No one?

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