# System feedback: why does not output feedback change observability of a (linear time-invariant) system?

Consider teh following system written on state-space form:

\begin{align} \dot{x}(t) &= Ax(t) + Bu(t) \\ y(t) &= Cx(t) \end{align}

where $$x\in\mathbb{R}^{n}$$, $$A\in\mathbb{R}^{n\times n}$$, $$B\in\mathbb{R}^{n\times k}$$, $$u\in\mathbb{R}^{k}$$, $$C\in\mathbb{R}^{m\times n}$$ and $$y\in\mathbb{R}^{m}$$. When we want to stabilize such a system we may use some feedback. If we have information about the whole state $$x$$, then we call it state feedback. If only the output $$y$$ is measurable, then we may use this information to create some feedback, and we call it output feedback. Mathematically we have that

\begin{align} \text{state feedback}:\quad u(t) = Kx(t) + v(t), \\ \text{output feedback}:\quad u(t) = Ky(t) + v(t). \end{align}

Here $$K$$ is some matrix that we design. As I see it, state feedback is just a special case of output feedback since $$Ky = KCx$$. Now to my question, it apperently holds that under any such feedback the reachability, or controllability, of the system remains unchanged. Is it also the case that observablity for the system is the same, after a feedback has been applied? This claim is made in my book, at least that output feedback leaves system observability the same, but should not this then be also the case for state feedback? How do we see this, that observability for the new system with output feedback is the same as for the original system without feedback?

In order to show this one could use the Hautus test to check for observability

$$\text{rank}\! \begin{bmatrix} C \\ A - \lambda\,I \end{bmatrix} = n, \forall\ \lambda \in \mathbb{C}. \tag{1}$$

When using output feedback one can substitute $$A$$ with $$A - B\,K\,C$$, such that the system is observable if

$$\text{rank}\! \begin{bmatrix} C \\ A - B\,K\,C - \lambda\,I \end{bmatrix} = n, \forall\ \lambda \in \mathbb{C}. \tag{2}$$

However, the matrix in $$(2)$$, from which you need to calculate the rank, can also be written as

$$\begin{bmatrix} C \\ A - B\,K\,C - \lambda\,I \end{bmatrix} = \begin{bmatrix} I & 0 \\ -B\,K & I \end{bmatrix} \begin{bmatrix} C \\ A - \lambda\,I \end{bmatrix}, \tag{3}$$

whose most right matrix is the same as in $$(1)$$ and the matrix multiplied by it is always full rank, due to its lower block triangular structure and the fact that the matrices on the diagonal are full rank. Therefore, the condition for observability of $$(3)$$ is identical to that of $$(2)$$.

This same factorization can not be done in general when using state feedback, since then you would need to use $$B\,K$$ instead of $$B\,K\,C$$.

• I see, thanks! One question before I green check mark your asnwer: for state feedback, is not $C$ just the identity $n$-by-$n$ matrix? Why could we not argue in the same way for that case, or am I missing something? Jan 3, 2021 at 15:53
• @SimpleProgrammer the state feedback gain, for example obtained from pole placement or LQR, does assume that the full state is "known". However, due to certainty equivalence it is also possible to design the state feedback gain and state observer separately such that if both are designed properly the closedloop would be stable, where $u(t)=K\,\hat{x}(t)$ with $\hat{x}(t)$ the estimate of $x(t)$. The combination of this gain the observer would technically not be called state feedback, but output feedback instead. But yes, in order to have true state feedback it would be required that $C=I$. Jan 3, 2021 at 17:24
• @SimpleProgrammer However, from a mathematical perspective it would be possible. For example if some other operator can measure the entire state and uses that to apply a control signal. In that case another operator which only measures $y(t)=C\,x(t)$, and who wants to find an estimate of the full state, could get observability issues depending on the choice for full state feedback gain $K$. Jan 3, 2021 at 17:28