- #1

Expirobo

- 2

- 0

LTI System:

[itex]\dot{x}=\left[ \begin{array}{cc}

0& 1 \\

0 & 0 \\

\end{array} \right]x+\left[ \begin{array}{c}

0 \\

1 \\

\end{array} \right]u [/itex]

The performance measure is given by:

[itex]V=\int_0^\infty \left( x'\left[ \begin{array}{cc}

1 & 0 \\

0 & 2 \\

\end{array} \right]x+u^2\right)\mathrm{d}t.

[/itex]

So the initial Q is :

[itex]

Q=\left[ \begin{array}{cc}

1 & 0 \\

0 & 2 \\

\end{array} \right]

[/itex]

and R = 1

However, we want to choose a new matrix for Q and R so that we have stable eigenvalues:

[itex]

\lambda=-1 \\

\lambda=-2 \\

[/itex]

I have gone round and round with this, but I cannot see how to work back through the Hamiltonian Matrix to achieve a desired value for Q & R.

The Hamiltonian is:

[itex]

H=\left[

\begin{array}{cc}

A & -BR^{-1}B' \\

-Q & -A' \\

\end{array} \right]

[/itex]

and the stable Eigenvalues of H (ones with negative real parts) are exactly the eigenvalues of the closed loop system A-BK.

Please help if you can. Even something to point me in the right direction will help.

BTW, just iterating by hand the matrix Q, I came up with eig(H) = -1,-2 using:

[itex]

Q=\left[ \begin{array}{cc}

4 & 0 \\

0 & 5 \\

\end{array} \right]

[/itex]

Yet, I must show how to achieve something like this mathematically.

Thanks!