I have this system here: enter image description here

In this example the state space representation $ \frac{dx}{dt} = Ax + bu $ and the corresponding transition matrix $\Phi(t)$ is asked for.

So to get the state space, I think this method is the fastest one. For that my solution is: $$\frac{dx}{dt} = \begin{pmatrix} 0&1&0 \\0&0&1 \\ -6 & -11 & -6\end{pmatrix}x + \begin{pmatrix} 0 \\0 \\1 \end{pmatrix}u$$

$$y = \begin{pmatrix} -2 &-6 &-2 \end{pmatrix}x + u $$

But in this case, getting $\Phi(s) = (sI-A)^{-1}$, is hard work. ;-) And at my exam this is wasting too much time.

So there is another method getting the state space.
There you use the eigenvectors and put them in the diagonal of the A matrix like that:

$$A = \begin{pmatrix} -1&0&0 \\0&-2&0 \\ 0 & 0 & -3\end{pmatrix}$$ Then you use this equation: $$G(s) = c^T(sI-A)^{-1}b+d$$ (d = 1 I think)
and use coefficient comparison.

But here is my problem. I don't know how I shall solve this equation.

  • $\begingroup$ Skype advises me to call to the Australian telephone number in your matrix. $\endgroup$ – Ilya Oct 11 '11 at 15:48
  • $\begingroup$ That answer is helping a lot! $\endgroup$ – madmax Oct 11 '11 at 21:21

There is no free lunch when it comes to these type of problems. Either you do the $(sI-A)^{-1}$ or you diagonalize the $A$ matrix by finding the eigenvalues and eigenvectors i.e. $V^{-1}AV = \Lambda$, then, using the state space transformation $$ \left[\begin{array}{c|c}A &B\\ \hline C&D \end{array}\right] \longrightarrow \left[\begin{array}{c|c} V^{-1}AV &V^{-1}B\\ \hline CV&D \end{array}\right] $$ This would also take time as much as the Laplace method would. The fastest way is to exploit the structure of the problems for which the instructors usually spend a lot of time to come up with.

The question actually asks you to identify the modes of the system just by inspection and to come up with a state space representation (you have already found the A matrix): Let the $B$ matrix be all-ones vector such that $$ (sI-A)^{-1}B = \begin{pmatrix} s+1&0&0 \\0&s+2&0 \\ 0 & 0 & s+3\end{pmatrix}^{-1}\pmatrix{1\\1\\1} = \pmatrix{\frac{1}{s+1}\\\frac{1}{s+2}\\\frac{1}{s+3}} $$ Then, use $C$ and $D$ matrices to match the output given in the question: $$ C(sI-A)^{-1}B + D = \pmatrix{1 &-2&-1}\pmatrix{\frac{1}{s+1}\\\frac{1}{s+2}\\\frac{1}{s+3}} + 1 $$ which gives the desired system representation $$ G(s) = \left[\begin{array}{c|c}A &B\\ \hline C&D \end{array}\right] = \left[\begin{array}{ccc|c}-1 & & &1\\&-2&&1\\&&-3&1\\\hline 1&-2&-1&1 \end{array}\right] $$ Transition matrix $\Phi(t)$ is simply the scalar exponentiation of the diagonal entries i.e. $$ \Phi(t) = e^{At } = \begin{pmatrix} e^{-t}&0&0 \\0&e^{-2t}&0 \\ 0 & 0 & e^{-3t}\end{pmatrix} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.