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suppose I have the following LTI system:

$$ \begin{split} \dot{\mathbf{x}} &= \mathbf{A}\mathbf{x} + \mathbf{B}\mathbf{u} \\ \mathbf{y} &= \mathbf{C}\mathbf{x} + \mathbf{D}\mathbf{u} \end{split} $$

with

$$ \mathbf{A} = \begin{bmatrix} 0 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}, \, \mathbf{B} = \begin{bmatrix} 0 & 0 \\ 0.1 & 0 \\ 0 & 0 \\ 0 & 0.1 \\ 0 & 0 \\ -0.08 & 0 \end{bmatrix}, \, \mathbf{C}^T = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}, \, \mathbf{D} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\,. $$

Since the controllability matrix of the pair $(\mathbf{A}, \mathbf{B})$ has rank 4, the system has two uncontrollable modes.

I have two general questions now (actually three), for which I chose the above system as an example:

  • Can the uncontrollable modes of the system be connected to the states of the system? I.e. can I say which states are uncontrollable? Which would be the uncontrollable states here?
  • According to Matlab, the eigenvalues of the uncontrollable subsystem are located at $-8.76\times10^{-17} \pm i8.83\times 10^{-9}$... so is the sub-system stable or unstable?
  • If unstable, is there nothing one can do about that in this specific case?

EDIT: Thanks for the answers, however those have created further questions for me:

If I change the $\mathbf{C}$ matrix of the system to identity matrix $\mathbf{I}$ (to output all states) and then use the Matlab command minreal, I get these reduced matrices:

$$ \mathbf{A}_{red} = \begin{bmatrix} 0.3942 & 0 & -0.1924 & 0 \\ 0 & 0 & 0 & 0 \\ 0.8076 & 0 & -0.3942 & 0 \\ 0 & -1 & 0 & 0 \end{bmatrix}\, , \mathbf{B}_{red} = \begin{bmatrix} -0.1151 & 0 \\ 0 & 0.1 \\ 0.0562 & 0 \\ 0 & 0 \\ \end{bmatrix} $$

My questions are now:

  • Should I use the full or the reduced system for controller design (e.g. when using the lqr method)?
  • If I use the reduced system for controller design, how would I implement this controller into a feedback loop in the full system (the dimension of my feedback matrix $\mathbf{K}$ wouldn't match)?
  • Are the states of the reduced system still physically meaningfull? Or are those kind of "virtual" states?
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For this you can use Hautus lemma. This states that $(A, B)$ is controllable if,

$$ \text{rank}\left[A-\lambda\,I,\, B\right]=n\quad \forall\, \lambda \in\mathbb{C} \tag{1}, $$

where $n$ is the size of $A$ (which is square). It can be noted that all $\lambda$ unequal to eigenvalues of $A$ already yield full rank, so you only need to try eigenvalues of $A$ in $(1)$.

If for an eigenvalue it is not full rank, then that eigenvalue is not controllable and can not be changed with any feedback law. However if all uncontrollable modes are stable (eigenvalues located in the open left hand plane) then the system is still not controllable but is stabilizable. And if you are only interested in stabilizability, then you only have to check rank for all eigenvalues with a nonnegative real part (whether you can control unstable modes).


As for your followup questions, the LQR algorithm should be able to handle uncontrollable states as long as they are stable (eigenvalues with negative real part). This is because it just tries to minimize some quadratic function, but since the feedback gain does not influence the rate at which the uncontrollable states go to zero, then the optimal feedback gain also can't be influenced by it. So you could use both systems, however it might be less intuitive for the reduced system to choose a $Q$ matrix, because the entire state of the complete system might be based on physically meaningful quantities, which might be lost/be more complicated when reducing it. And to answer your last question, the reduced states will still be some what meaningful, because they will just be some linear combination of the previous states. However if the physical units of those states are not the same, then that linear transformation would also have to include units conversion. The reduction technique does not show those, so in that sense the reduced states could be less meaningful.

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  • $\begingroup$ Thanks for the answer, I have updated my post because I have further questions. $\endgroup$ – SampleTime Apr 5 '17 at 19:11
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Can the uncontrollable modes of the system be connected to the states of the system? I.e. can I say which states are uncontrollable? Which would be the uncontrollable states here?

In principle, the answer is yes. However, you will usually not see this directly in a given representation of the system. What you can do is performing a Kalman decomposition of the system into controllable and uncontrollable (and even observable/unobservable modes). See here for the transformation matrices. There is even a Matlab-command for computing it.

According to Matlab, the eigenvalues of the uncontrollable subsystem are located at −8.76×10−17±i8.83×10−9−8.76×10−17±i8.83×10−9... so is the sub-system stable or unstable?

I don't know how you got to these eigenvalues (seems to be a numerical issue because they're practically zero). Actually, the eigenvalues of the given matrix are all located at 0 which can be seen easily due to the triangular struture of the matrix. This means that the whole system (and thus also the uncontrollable subsystem) is (marginally) stable but not unstable.

If unstable, is there nothing one can do about that in this specific case?

If a given system has unstable uncontrollable modes (i.e. it is not stabilizable) there is, indeed, nothing you can do about it. That's why people usually like to have at least stabilizable systems.

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  • $\begingroup$ Thank you too for the answer, I have updated my question with additional questions. $\endgroup$ – SampleTime Apr 5 '17 at 19:12

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