- #1
X89codered89X
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Hi all,
I have a linear algebra question relating actually to control systems (applied differential equations)
for the linear system
[itex]
{\dot{\vec{{x}}} = {\bf{A}}{\vec{{x}}} + {\bf{B}}}{\vec{{u}}}\\
\\
A \in \mathbb{R}^{ nxn }\\
B \in \mathbb{R}^{ nx1 }\\
[/itex]
In class, we formed a transformation matrix P using the controllability matrix [itex] M_c [/itex] as a basis (assuming it is full rank).
[itex]
M_c = [ {\bf{B \;AB \;A^2B\;...\;A^{n-1}B}}]
[/itex]
and there is a second matrix with a less established name. Given that the characteristic equation of the system is [itex] |I\lambda -A| = \lambda^n + \alpha_1 \lambda^{n-1} +... + \alpha_{n-1}\lambda + \alpha_n= 0 [/itex], we then construct a second matrix, call it M_2, which is given below.
[itex]
{\bf{M}}_2 =
\begin{bmatrix}
\alpha_{n-1} & \alpha_{n-2} & \cdots & \alpha_1 & 1 \\
\alpha_{n-2} & \cdots & \alpha_1 & 1 & 0 \\
\vdots & \alpha_1 & 1 & 0 & 0\\
\alpha_1 & 1 & 0 & \cdots & 0\\
1 & 0 & 0& \cdots & 0 \\
\end{bmatrix}
[/itex]
then the transformation matrix is then given by
[itex]
P^{-1} = M_c M_2
[/itex]and then applying the transformation always gives.. and this is what I don't understand...
[itex]
{\overline{\bf{A}}} = {\bf{PAP}}^{-1} =
\begin{bmatrix}
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & \vdots \\
\vdots & \vdots & 0 & 1 & 0\\
0 & 0 & \cdots &0& 1\\
-\alpha_{1} & -\alpha_{2} & \cdots & -\alpha_{n-1}& -\alpha_{n}\\
\end{bmatrix}
[/itex]
Now I'm just looking for intuition is to why this is true. I know that this only works if the controllability matrix is full rank, which can the be used as a basis for the new transformation, but I don't get how exactly the M_2 matrix is using it to transform into the canonical form... Can someone explain this to me? thanks...All Right! I think I'm done editin LATEX ...
I have a linear algebra question relating actually to control systems (applied differential equations)
for the linear system
[itex]
{\dot{\vec{{x}}} = {\bf{A}}{\vec{{x}}} + {\bf{B}}}{\vec{{u}}}\\
\\
A \in \mathbb{R}^{ nxn }\\
B \in \mathbb{R}^{ nx1 }\\
[/itex]
In class, we formed a transformation matrix P using the controllability matrix [itex] M_c [/itex] as a basis (assuming it is full rank).
[itex]
M_c = [ {\bf{B \;AB \;A^2B\;...\;A^{n-1}B}}]
[/itex]
and there is a second matrix with a less established name. Given that the characteristic equation of the system is [itex] |I\lambda -A| = \lambda^n + \alpha_1 \lambda^{n-1} +... + \alpha_{n-1}\lambda + \alpha_n= 0 [/itex], we then construct a second matrix, call it M_2, which is given below.
[itex]
{\bf{M}}_2 =
\begin{bmatrix}
\alpha_{n-1} & \alpha_{n-2} & \cdots & \alpha_1 & 1 \\
\alpha_{n-2} & \cdots & \alpha_1 & 1 & 0 \\
\vdots & \alpha_1 & 1 & 0 & 0\\
\alpha_1 & 1 & 0 & \cdots & 0\\
1 & 0 & 0& \cdots & 0 \\
\end{bmatrix}
[/itex]
then the transformation matrix is then given by
[itex]
P^{-1} = M_c M_2
[/itex]and then applying the transformation always gives.. and this is what I don't understand...
[itex]
{\overline{\bf{A}}} = {\bf{PAP}}^{-1} =
\begin{bmatrix}
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & \vdots \\
\vdots & \vdots & 0 & 1 & 0\\
0 & 0 & \cdots &0& 1\\
-\alpha_{1} & -\alpha_{2} & \cdots & -\alpha_{n-1}& -\alpha_{n}\\
\end{bmatrix}
[/itex]
Now I'm just looking for intuition is to why this is true. I know that this only works if the controllability matrix is full rank, which can the be used as a basis for the new transformation, but I don't get how exactly the M_2 matrix is using it to transform into the canonical form... Can someone explain this to me? thanks...All Right! I think I'm done editin LATEX ...
Last edited: