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I am somewhat confused on how the standard form for an uncontrollable system can be used to implement state-feedback on the controllable subspace.

  • I understand how to use feedback to achieve pole-placement in completely controllable systems.

  • I don't understand how this can then be applied to controllable subspaces of uncontrollable systems.

For example, suppose I have an LTI system with eigenvalues {-1, ,1, 2} where eigenvalues {1, 2} are part of the controllable submatrix while -1 is uncontrollable. How do change the pole-locations for the controllable eigenvalues?

Thank you.

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Since you only mentioned state feedback I assume that you are dealing with a system of the following form

$$ \dot{x} = A\,x + B\,u $$

with $x \in \mathbb{R}^n$, $u \in \mathbb{R}^m$ and you want to find a state feedback $u = -K\,x$ such that the dynamics of the controllable parts of the system can be chosen. There entire state $x$ is considered to be known and thus observability does not have to be taken into consideration.

By looking at the controllable and uncontrollable states it is possible to find a similarity transformation which decomposes these states, similar to the Kalman decomposition but without also considering the observability. In order to do this you need the controllability matrix $\mathcal{C}$ defined as

$$ \mathcal{C} = \begin{bmatrix} B & A\,B & \cdots & A^{n-1}B \end{bmatrix}. $$

Now a controllable decomposition can be found using the similarity transformation $T = \begin{bmatrix}T_1 & T_2\end{bmatrix}$, with the columns of $T_1$ having the same span as the controllability matrix and $T_2$ chosen such that $T$ is invertible. In the new coordinates, denoted by $x = T\,\hat{x}$, the dynamics are as follows

$$ \dot{\hat{x}} = \underbrace{T^{-1} A\,T}_{\hat{A}}\,\hat{x} + \underbrace{T^{-1} B}_{\hat{B}}\,u $$

such that

$$ \hat{A} = \begin{bmatrix} A_{11} & A_{12} \\ 0 & A_{22} \end{bmatrix}, \quad \hat{B} = \begin{bmatrix} B_1 \\ 0 \end{bmatrix} $$

where $A_{11}$ has the same size as the rank of $\mathcal{C}$ and the pair $(A_{11}, B_1)$ is controllable (their associated controllability matrix is full rank). So for this pair you can do pole placement, such that the poles of $A_{11}-B_1\,K_1$ can be chosen arbitrarily. Therefore $u = -\begin{bmatrix}K_1 & 0\end{bmatrix} \hat{x}$ will place the controllable poles of $\hat{A}$ arbitrarily as well.

By using the definition of $\hat{x}$ it is possible to express this state feedback in terms of $x$ instead, namely $\hat{x} = T^{-1} x$. Or in other words your feedback gain can be found with $K = \begin{bmatrix}K_1 & 0\end{bmatrix}\,T^{-1}$.


For example given the system

$$ A = \begin{bmatrix} 4 & 0 & 0 \\ -4 & 0 & -2 \\ -2 & -2 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}, $$

which has the eigenvalues 2, -2 and 4. This pair has the following controllability matrix

$$ \mathcal{C} = \begin{bmatrix} 1 & 4 & 16 \\ 0 & -2 & -12 \\ -1 & -2 & -4 \end{bmatrix}, $$

with has rank 2. It can be shown that its first two columns are linearly independent of each other, so an option for the similarity transformation would be

$$ T = \begin{bmatrix} 1 & 2 & 0 \\ 0 & -1 & 0 \\ -1 & -1 & 1 \end{bmatrix}. $$

For the added last column it can be shown that it makes $T$ invertible (full rank). Performing the similarity transformation gives

$$ \hat{A} = \begin{bmatrix} 0 & -4 & -4 \\ 2 & 6 & 2 \\ 0 & 0 & -2 \end{bmatrix}, \quad \hat{B} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, $$

such that

$$ A_{11} = \begin{bmatrix} 0 & -4 \\ 2 & 6 \end{bmatrix}, \quad B_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}. $$

Finding a feedback gain for this pair, which places the poles at -1 and -2 gives $K_1 = \begin{bmatrix}9 & 24\end{bmatrix}$. So the actual state feedback gain can be calculated with

$$ K = \begin{bmatrix}K_1 & 0\end{bmatrix}\,T^{-1} = \begin{bmatrix}9 & -6 & 0\end{bmatrix}, $$

which also places the two controllable poles of $A$, which are initially located at 2 and 4, at -1 and -2 respectively.

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  • $\begingroup$ Great answer. The combination of general procedure and an example is very instructive. $\endgroup$ – MrYouMath Dec 1 '17 at 20:19

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