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# How can I find $B$ and $D$ matrix from lower triangular Toeplitz matrix? - MOESP

I'm doing some system identification with the MOESP-method. Right know I have learn how get $C$ and $A$ matrices from the extended observability matrix $\Gamma$

Example:

Assume that we have our observability matrix $\Gamma$ as:

$$\Gamma = \begin{bmatrix}

C\\

CA\\

CA^2\\

CA^3\\

CA^4\\

CA^5\\

CA^6\\

\vdots\\

CA^{n-1}

\end{bmatrix}$$

We only know the state dimension $x(k) \in \Re^n$ and output vector $y(k) \in \Re^{p}$

Then we can find $C$ very easy by doing:

$$C = \Gamma(1:p, 1:n)$$

But for $A$ matrix it seems to be more difficult, but it isn't.

We assume this:

$$\bar{\Gamma} = \begin{bmatrix}

CA\\

CA^2\\

CA^3\\

CA^4\\

CA^5\\

CA^6\\

\vdots\\

\end{bmatrix}$$

And

$$\underline{\Gamma} = \begin{bmatrix}

CA^2\\

CA^3\\

CA^4\\

CA^5\\

CA^6\\

\vdots\\

CA^{n-1}

\end{bmatrix}$$

Then we assume that

$$\bar{\Gamma}A = \underline{\Gamma}$$

And I want to find $A$. So the solution is to use Moore-Penrose inverse because we are dealing with non-square matrices:

$$A = \b