I am working on LTI system study and I am somewhat confused about what a system being observable actually means from the perspective of the application.

For LTI systems observability proofs usually involve taking the output vector $y(t)$ and $n$ derivatives and solving a system of linear equations using the fact the observability matrix is full rank.

My question is, how are the derivatives of the output vector supposed to be found? It seems like this theorem is purely of theoretical significance since I don't see a way for someone to compute the output vectors derivatives from measurements.


Usually, the derivatives of the output vector are not supposed to be found; instead, an observer is constructed. An observer is an auxiliary system that provides an estimate of the state of a given system.

Consider, for example, a LTI system \begin{equation}\tag{1} \left\{ \begin{array}{l} \dot x= Ax\\ y=Cx, \end{array}\right. \end{equation} where $x\in\mathbb R^n$ is a state vector, $y\in\mathbb R$ is a scalar output, the system (1) is observable, i.e. the observability matrix $$ V=\left( \begin{array}{c} C\\AC\\A^2C\\\vdots\\A^{n-1}C \end{array}\right) $$ is invertible. Also consider the observer system $$ \dot{\hat x}= A\hat x+F(y-C\hat x),\tag{2} $$ where $F$ is some matrix of size $n\times 1$. The input $y$ of the observer is the output of the real system (1). Denote the error vector $e(t)=x-\hat x$; $$ \dot e= Ax-A\hat x-F(y-C\hat x)= A(x-\hat x)-F(Cx-C\hat x)= (A-FC)e.\tag{3} $$ Since $V$ is invertible, it is always possible to choose (using, for example, the Ackermann's formula: $F^T= (0\; 0\; \ldots\; 1) (V^T)^{-1} P(A^T)$) the column $F$ to make the system (3) asymptotically stable, i.e. $$ \lim_{t\to\infty} e(t)= \lim_{t\to\infty} (x(t)-\hat x(t))=0. $$ Hence, $\hat x(t)$ is an estimation of the (unknown) state $x(t)$ of the system (1).


I am assuming that you are talking about taking measurement samples at a fixed interval (sample rate) from a continuous system. Assuming that the sample rate is sufficiently high to capture all the dynamics of the system and that aliasing wouldn't be a problem. In this case you could just filter the signal with different filters, which are equal to the different order derivatives from zero/low up to sufficiently high frequencies. Examples for this would be finite differencing methods. However this might be limited by the noise present in the output and finite differencing methods in general are not great for this. So you might want to look at general IIR filters instead of just those FIR filters.

But this approach of checking observability already assumes that you know the state space model of the system

$$ \dot{x} = A\,x + B\,u \tag{1} $$

$$ y = C\,x + D\,u \tag{2} $$

with $x\in\mathbb{R}^n$. Observability implies that it would be possible to reconstruct the full state information, $x$, when you are only given $y$ and $u$, and their first $n-1$ derivatives with respect to time (potentially less if the system has multiple outputs).

For example when $n = 3$ and you know $y$, $\dot{y}$, $\ddot{y}$, $u$, $\dot{u}$ and $\ddot{u}$ at a given time, then it can be shown when using the chain rule and equations $(1)$ and $(2)$ that they can be related to each other and the full state $x$ according to

$$ \dot{y} = C\,A\,x + C\,B\,u + D\,\dot{u} \tag{3} $$

$$ \ddot{y} = C\,A^2\,x + C\,A\,B\,u + C\,B\,\dot{u} + D\,\ddot{u} \tag{4} $$

Combining this with equation $(2)$ allows these to be written as

$$ \begin{bmatrix} y - D\,u \\ \dot{y} - C\,B\,u - D\,\dot{u} \\ \ddot{y} - C\,A\,B\,u - C\,B\,\dot{u} - D\,\ddot{u} \end{bmatrix} = \begin{bmatrix} C \\ A\,C \\ A^2\,C \end{bmatrix} x $$

with the matrix multiplying by $x$ the observability matrix (for $n=3$). So if that matrix is full rank you can simply solve for $x$ by inverting that matrix (potentially when it is not square using a pseudo inverse).

However this explanation of observability is merely to give an example that the full state $x$ could be recovered given these derivatives of $y$ and $u$ in an ideal case without unknown disturbances and perfect knowledge of the model. However in practice you would use different methods of reconstructing the full state if you want to do state feedback, such as Kalman filters.

However often for control purposes you would identify a model from input/output data, which should already result by definition in an observable model. You could also derive a model from first principles, so when deriving the differential equations from Newton's laws or Euler–Lagrange (at least for mechanical systems). This might result into unobservable models. One example of such system would be two masses with potentially spring, dampers and other forces acting on them, but the output only shows the difference in position of the two masses. So even if the output shows a constant value, it could still be that the masses are moving together, whose dynamics can not be observed.

  • 1
    $\begingroup$ Great answer. A small note you can also use the Euler-Lagrange formalism for electrical systems. $\endgroup$ – MrYouMath Nov 9 '17 at 9:03

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.