- #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
<br /> <br /> {\dot{\vec{{x}}} = {\bf{A}}{\vec{{x}}} + {\bf{B}}}{\vec{{u}}}\\<br /> \\<br /> <br /> A \in \mathbb{R}^{ nxn }\\<br /> B \in \mathbb{R}^{ nx1 }\\<br />
In class, we formed a transformation matrix P using the controllability matrix M_c as a basis (assuming it is full rank).
<br /> <br /> M_c = [ {\bf{B \;AB \;A^2B\;...\;A^{n-1}B}}]<br />
and there is a second matrix with a less established name. Given that the characteristic equation of the system is |I\lambda -A| = \lambda^n + \alpha_1 \lambda^{n-1} +... + \alpha_{n-1}\lambda + \alpha_n= 0, we then construct a second matrix, call it M_2, which is given below.
<br /> {\bf{M}}_2 = <br /> \begin{bmatrix}<br /> \alpha_{n-1} & \alpha_{n-2} & \cdots & \alpha_1 & 1 \\<br /> \alpha_{n-2} & \cdots & \alpha_1 & 1 & 0 \\<br /> \vdots & \alpha_1 & 1 & 0 & 0\\<br /> \alpha_1 & 1 & 0 & \cdots & 0\\<br /> 1 & 0 & 0& \cdots & 0 \\<br /> \end{bmatrix}<br /> <br />
then the transformation matrix is then given by
<br /> <br /> P^{-1} = M_c M_2<br /> <br />and then applying the transformation always gives.. and this is what I don't understand...
<br /> {\overline{\bf{A}}} = {\bf{PAP}}^{-1} = <br /> <br /> \begin{bmatrix}<br /> 0 & 1 & 0 & \cdots & 0 \\<br /> 0 & 0 & 1 & \cdots & \vdots \\<br /> \vdots & \vdots & 0 & 1 & 0\\<br /> 0 & 0 & \cdots &0& 1\\<br /> -\alpha_{1} & -\alpha_{2} & \cdots & -\alpha_{n-1}& -\alpha_{n}\\<br /> \end{bmatrix}<br /> <br />
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
<br /> <br /> {\dot{\vec{{x}}} = {\bf{A}}{\vec{{x}}} + {\bf{B}}}{\vec{{u}}}\\<br /> \\<br /> <br /> A \in \mathbb{R}^{ nxn }\\<br /> B \in \mathbb{R}^{ nx1 }\\<br />
In class, we formed a transformation matrix P using the controllability matrix M_c as a basis (assuming it is full rank).
<br /> <br /> M_c = [ {\bf{B \;AB \;A^2B\;...\;A^{n-1}B}}]<br />
and there is a second matrix with a less established name. Given that the characteristic equation of the system is |I\lambda -A| = \lambda^n + \alpha_1 \lambda^{n-1} +... + \alpha_{n-1}\lambda + \alpha_n= 0, we then construct a second matrix, call it M_2, which is given below.
<br /> {\bf{M}}_2 = <br /> \begin{bmatrix}<br /> \alpha_{n-1} & \alpha_{n-2} & \cdots & \alpha_1 & 1 \\<br /> \alpha_{n-2} & \cdots & \alpha_1 & 1 & 0 \\<br /> \vdots & \alpha_1 & 1 & 0 & 0\\<br /> \alpha_1 & 1 & 0 & \cdots & 0\\<br /> 1 & 0 & 0& \cdots & 0 \\<br /> \end{bmatrix}<br /> <br />
then the transformation matrix is then given by
<br /> <br /> P^{-1} = M_c M_2<br /> <br />and then applying the transformation always gives.. and this is what I don't understand...
<br /> {\overline{\bf{A}}} = {\bf{PAP}}^{-1} = <br /> <br /> \begin{bmatrix}<br /> 0 & 1 & 0 & \cdots & 0 \\<br /> 0 & 0 & 1 & \cdots & \vdots \\<br /> \vdots & \vdots & 0 & 1 & 0\\<br /> 0 & 0 & \cdots &0& 1\\<br /> -\alpha_{1} & -\alpha_{2} & \cdots & -\alpha_{n-1}& -\alpha_{n}\\<br /> \end{bmatrix}<br /> <br />
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 ...
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