In general, if A has distinct eigenvalues, it can be diagonalized using similarity transformation. Consider a square matrix A which has distinct eigenvalues $\lambda_1,\;\lambda_2,\;\ldots\;\lambda_n$ . It is required to find a transformation matrix P which will convert A into a diagonal form

$\displaystyle \Lambda=\begin{bmatrix}\lambda_1&0&\ldots&0\\ 0& \lambda_2&\ldots&0\\ 0&0&\ldots& \lambda_n\end{bmatrix}$

through similarity transformation AP = P∧ . If $v_1,\;v_2,\;\ldots,\;v_n$ are the eigenvectors of matrix A corresponding to eigenvalues $\lambda_1,\;\lambda_2,\;\ldots\;\lambda_n$ , then we know $Av_i=\lambda v_i$ . This gives

$\displaystyle A\left[v_1\;v_2\;\ldots\;v_n\right]=[v_1\;v_2\;\ldots\;v_n]\begin... ...bda_1&0&\ldots&0\\ 0& \lambda_2&\ldots&0\\ 0&0&\ldots& \lambda_n\end{bmatrix}$

Thus $P=\left[v_1\;v_2\;\ldots\;v_n\right]$ . Consider the following state model.

$\displaystyle \boldsymbol{x}(k+1) = A \boldsymbol{x}(k) + B u(k)$

If P transforms the state vector $\boldsymbol{x}(k)$ to $\boldsymbol{z}(k)$ through the relation

$\displaystyle \boldsymbol{x}(k) = P \boldsymbol{z}(k),$ or, $\displaystyle \;\;\; \boldsymbol{z}(k) = P^{-1} \boldsymbol{x}(k)$
then the modified state space model becomes
$\displaystyle \boldsymbol{z}(k+1) = P^{-1}AP \boldsymbol{z}(k) + P^{-1}B u(k)$
where $P^{-1}AP =\Lambda$ .

3. Computation of $\Phi (t)$

We have seen that to derive the state space model of a sampled data system, we need to know the continuous time state transition matrix $\Phi (t)=e^{At}$ .

3.1 Using Inverse Laplace Transform

For the system $\dot{\mathbf{x}}(t)=A\mathbf{x}(t)+Bu(t)$ , the state transition matrix e^At can be computed as,

$\displaystyle e^{At}$ $\displaystyle =\mathscr{L}^{-1}\big \{ (sI-A)^{-1} \big \}$