I have a state space representation, system S1, in the form of:

$$ \frac{dx}{dt} = Ax + bu $$ $$y = c^Tx$$

This system is transformed with the state transform $$x=T z$$ into the system S2: $$ \frac{dz}{dt} = \begin{pmatrix} -1 & -2 \\-1 & -2 \end{pmatrix}z + \begin{pmatrix} 0 \\1 \end{pmatrix}u $$ $$y = \begin{pmatrix} 1 &-1 \end{pmatrix}z $$ T is the transformation matrix.
The only thing I know from system S1 is, that is in diagonal form.
So $A$ should look something like this: A = $\begin{pmatrix} a & 0 \\0 & d \end{pmatrix}$

I think I know how I could transform S1 into S2 but I don't know the other way.
Found some formulas like $$T^{-1}AT = \begin{pmatrix} -1 & -2 \\-1 & -2 \end{pmatrix}$$ and $$T^{-1}b = \begin{pmatrix} 0 \\1 \end{pmatrix}$$ I have a problem obtaining the transformation matrix.
I also thought about the eigenvalues of S2 (0, -3), but I don't know if I am right that these values are the diagonal points of the matrix A.


So with @Sasha 's help I got the system matrix A:
$$p_1 = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$$ $$p_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$ $$T = \begin{pmatrix} -2 & 1 \\1 & 1 \end{pmatrix}$$ $$T^{-1}\begin{pmatrix} -1 & -2 \\-1 & -2 \end{pmatrix}T = \begin{pmatrix} 0 & 0 \\0 & -3 \end{pmatrix}$$ $$b = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$ $$c = \begin{pmatrix} -\frac{2}{3} \\ -\frac{1}{3} \end{pmatrix}$$

Some additional infos:
Matrix A
Matrix A is the system matrix, and relates how the current state affects the state change x' . If the state change is not dependent on the current state, A will be the zero matrix. The exponential of the state matrix, eAt is called the state transition matrix.

Matrix B
Matrix B is the control matrix, and determines how the system input affects the state change. If the state change is not dependent on the system input, then B will be the zero matrix.

Matrix C
Matrix C is the output matrix, and determines the relationship between the system state and the system output.


Transformation formulas you wrote are correct. Because $A$ is diagonal, you can tell that $a$ and $d$ must eigenvalues. And the matrix $T$ can be constructed from eigenvectors of $\left( \begin{array}{cc} -1 & -2 \\ -1 & -2 \end{array} \right)$, so that columns of $T$ will be normalized eigenvectors.

  • $\begingroup$ Thank you, so in my case I only have to calculate the eigenvalues to get the A matrix. $\endgroup$ – madmax Oct 6 '11 at 13:14

Suppose an invertible transformation matrix is given such that the states of the transformed system becomes $q$ i.e. $q=Tx$. Then, we also know that $\dot q = T\dot x$. From these, we obtain $T^{-1}q=x$ and $T^{-1}\dot q=\dot x$. If you plug these in to the state space representation:

$$\begin{align} T^{-1}\dot q&= AT^{-1}q+Bu\\ y &= CT^{-1}q + Du \end{align} $$ Now multiply the first equation with $T$, $$\begin{align} \dot q&= TAT^{-1}q+TBu\\ y &= CT^{-1}q + Du \end{align} $$ This is the same system $G(s)$ in different state coordinates. In your particular case, things are slightly easier since you are looking for the transformation matrices that diagonalizes $A$ which is possible, for example, with eigenvalue decomposition.

  • $\begingroup$ Ok, but I think you just gave me the formulas I have already written in my post. $\endgroup$ – madmax Oct 6 '11 at 12:53
  • 1
    $\begingroup$ @madmax : And Sasha gave you the formulas of eigenvalue decomposition. I just tried to explain you where those formulas come from. $\endgroup$ – user13838 Oct 6 '11 at 13:14
  • $\begingroup$ Now you can use the matrix you have found and plug in to what I have given in my answer and you are done. $\endgroup$ – user13838 Oct 6 '11 at 13:16

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