Controllability and Observability

Controllability

Consider

$\frac{dx}{dt} = Ax(t) + Bu(t),$

where $A$ is $n \times n$ and $B$ is $n \times p$. However, for simplicity, in the derivations below throughout the chapter, we will assume that $p = 1$.

Consider the following:

$\frac{dx}{dt} = \begin{bmatrix} 0 & 1 \\ -4 & -5 \end{bmatrix} x + \begin{bmatrix} b_1 \\ b_2 \end{bmatrix} u$

In this case, if $b_1, b_2$ are both 0, it is evident that the system is not controllable: for every given $x(0)$, the future paths are uniquely determined.

Consider, now the more interesting case with $b_1 = 1 = -b_2$. In this case, if the initial condition $x(0) = \begin{bmatrix} x_1(0) \\ x_2(0) \end{bmatrix}$ takes values from the subspace determined by the line $x_1(0) + x_2(0) = 0$, then, for all $t > 0$, the state remains in this subspace. To see this, note that $\frac{d(x_1(t) + x_2(t))}{dt} = \frac{dx_1(t) + dx_2(t)}{dt} = 0$ so that the sum of the state components does not change and thus $x_1(t) + x_2(t)$ remains 0. Thus, this subspace, which is a strict subset of $\mathbb{R}^2$, is invariant no matter what control is applied: this system is not controllable.

The matrix $W_c$ above is called the controllability Grammian of $(A, B)$.

Reachable Set (from origin) and the Controllable Subspace. An implication of the proof of the equivalence between (ii) and (iii) is that the range space of $\mathcal{C}$ and the range space of the linear operator from the space of integrable control inputs $\mathcal{U} = \{u : \mathbb{R}_+ \to \mathbb{R}, \; \|u\|_1 < \infty\}$ to $\mathbb{R}^n$ defined with

$\bigcup_{t \geq 0} \left\{ \int_0^t e^{A(t-s)} Bu(s) ds, \quad u \in \mathcal{U} \right\}$

are equal. This set is called the reachable (from the origin) set. This set is also called the controllable subspace.

Exercise. The controllability property is invariant under an algebraically equivalent transformation of the coordinates: $\tilde{x} = Px$ for some invertible $P$.

Hint: Use the rank condition and show that with $\frac{d\tilde{x}}{dt} = \tilde{A}\tilde{x}(t) + \tilde{B}u(t)$ and $\tilde{A} = PAP^{-1}$, $\tilde{B} = PB$, and with $\mathcal{C} = \begin{bmatrix} B & AB & \cdots & A^{n-1}B \end{bmatrix}$, we have that the transformed controllability matrix writes as $\tilde{\mathcal{C}} = \begin{bmatrix} \tilde{B} & \tilde{A}\tilde{B} & \cdots & \tilde{A}^{n-1}\tilde{B} \end{bmatrix} = P\mathcal{C}$.

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Observability

In many problems a controller has access to only the inputs applied and outputs measured. A very important question is whether the controller can recover the state of the system through this information.

Consider

$\frac{dx}{dt} = Ax(t) + Bu(t), \qquad y(t) = Cx(t) + Du(t)$

In the above, we could consider without any loss that $u(t) = 0$ for all $t$, since the control terms appear in additive forms whose effects can be cancelled from the measurements.

Consider then

$\frac{dx}{dt} = Ax(t), \qquad y(t) = Cx(t)$

The measurement at time $t$ writes as:

$y(t) = Ce^{At}x(0)$

Taking the derivative:

$\frac{dy(t)}{dt} = CAe^{At}x(0)$

and taking the derivatives up to order $n - 1$, we obtain for $1 \leq k \leq n - 1$

$\frac{d^{k-1}y(t)}{dt^{k-1}} = CA^{k-1}e^{At}x(0)$

In matrix form, we can write the above as

$\begin{bmatrix} y(t) \\ \frac{dy(t)}{dt} \\ \vdots \\ \frac{d^{n-1}y(t)}{dt^{n-1}} \end{bmatrix} = \begin{bmatrix} C \\ CA \\ \vdots \\ CA^{n-1} \end{bmatrix} e^{At} x(0)$

Thus, the question of being able to recover $x(0)$ from the measurements becomes that of whether the observability matrix

$\mathcal{O} := \begin{bmatrix} C \\ CA \\ \vdots \\ CA^{n-1} \end{bmatrix}$

is full-rank or not. Note that adding further rows to this matrix does not increase the rank by the Cayley-Hamilton theorem. Thus, we can recover the initial state if the observability matrix is full-rank.

Furthermore, we have that $Ce^{At}$ is a linear combination of $\{CA^k, \; k = 0, 1, \cdots, n-1\}$. Therefore, if $x_0$ is orthogonal to $\{CA^k, \; k = 0, 1, \cdots, n-1\}$, then it is also orthogonal to $Ce^{At}$. In particular, if the observability matrix is not full-rank, then there exists a non-zero $x_0$ so that $Ce^{At}x_0 = 0$. Thus, we cannot distinguish between $x_0$ and the $0$ vector in $\mathbb{R}^n$ and thus the system is not observable.

The null-space of $\mathcal{O}$, that is, $\{v \in \mathbb{R}^n : \mathcal{O}v = 0\}$ is called the unobservable subspace.

The structure of the observability matrix $\mathcal{O}$ and the controllability matrix $\mathcal{C}$ leads to the following very important and useful duality result.

In view of Theorem (and in particular, now that we have related observability to a condition of the form given in Theorem(iii)), we have the following immediate result:

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Feedback and Pole Placement

Consider $u = -Kx$. Then,

$\frac{dx}{dt} = Ax(t) + Bu(t) = (A - BK)x(t)$

To see this result, first consider a system in the controllable canonical realization form (see the relevant section) with

$\frac{d}{dt}x(t) = Ax(t) + Bu(t), \qquad y(t) = Cx(t)$

$A_c = \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & & \ddots & 1 \\ -a_0 & -a_1 & -a_2 & \cdots & -a_{N-1} \end{bmatrix}$

$B_c = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{bmatrix}$

Note that, the eigenvalues of $A$ solve the characteristic polynomial whose coefficients are located in the bottom row of $A_c$ (see the proof of Theorem).

Now, apply $u = -Kx$ so that $u = \sum_{i=1}^{N} -k_i x_i$, leading to

$A - BK = \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & & \cdots & 0 \\ -(a_0 + k_1) & -(a_1 + k_2) & -(a_2 + k_3) & \cdots & -(a_{N-1} + k_N) \end{bmatrix}$

Once again, since the eigenvalues of this matrix solve the characteristic polynomial whose coefficients are located in the bottom row (by the proof of Theorem), and these coefficients can be placed by selecting the scalars $k_i$, we can arbitrarily place the eigenvalues of the closed-loop matrix by feedback.

Through a coordinate transformation $\tilde{x} = Px$, every controllable system $x' = Ax + Bu$ can be transformed to an algebraically equivalent linear system in the controllable canonical realization form $(A_c, B_c)$ above. As we saw, for a system in this form, a control can be found so that all the eigenvalues of the closed loop system are on the left-half plane. Finally, the system can be moved back to the original coordinates.

We now see how this (transformation into a controllable canonical realization form) is possible. With $\tilde{x} = Px$, we have that

$\frac{d\tilde{x}}{dt} = \tilde{A}\tilde{x}(t) + \tilde{B}u(t)$

with $\tilde{A} = PAP^{-1}$, $\tilde{B} = PB$. Now, if $(A, B)$ is controllable, we know that $\mathcal{C} = \begin{bmatrix} B & AB & \cdots & A^{n-1}B \end{bmatrix}$ is full-rank. The transformed controllability matrix writes as: $\tilde{\mathcal{C}} = \begin{bmatrix} \tilde{B} & \tilde{A}\tilde{B} & \cdots & \tilde{A}^{n-1}\tilde{B} \end{bmatrix} = P\mathcal{C}$. As a result,

$P = \tilde{\mathcal{C}} \mathcal{C}^{-1}$

whose validity follows from the fact that $\mathcal{C}$ is invertible. This leads us to the following conclusion.

The above then suggests a method to achieve stabilization through feedback: First transform into a controllable canonical realization form, place the eigenvalues through feedback, and transform the system back to the original coordinate.

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Observers and Observer Feedback

Consider

$\frac{dx}{dt} = Ax + Bu, \qquad y = Cx$

Suppose that the controller intends to track the state. A candidate for such a purpose is to write an observer system of the form

$\frac{d\hat{x}}{dt} = A\hat{x} + Bu + L(y - C\hat{x})$

We then obtain with $e = x - \hat{x}$, and subtracting the above two equations from one another

$\frac{de}{dt} = Ae - LCe = (A - LC)e$

Then, the question whether $e(t) \to 0$ is determined by whether the eigenvalues of $A - LC$ can be pushed to the left-half plane with some appropriate $L$. If the system is observable, then this is possible, with the same arguments applicable to the pole placement analysis presented in the previous section (note that controllability and observability are related to each other with a simple duality property that was presented in Theorem: that is $L^T$ can be selected so that $A^T - C^T L^T$ has all eigenvalue in the left-half plane, which will also imply that $A - LC$ will have the same property).

Now that under observability we have that the controller can track the state with asymptotically vanishing error, suppose that we consider

$\frac{dx}{dt} = Ax + Bu, \qquad y = Cx$

with the goal of stabilizing the actual system state $x(t)$.

Suppose that we run an observer, and that we consider the following feedback control policy

$u(t) = -K\hat{x}(t)$

where $K$ is what we used for pole placement, and $\hat{x}$ is what we used in our observer. In this case, we obtain the following relation:

$\begin{bmatrix} \frac{dx}{dt} \\ \frac{de}{dt} \end{bmatrix} = \begin{bmatrix} A - BK & BK \\ 0 & A - LC \end{bmatrix} \begin{bmatrix} x \\ e \end{bmatrix}$

Due to the upper triangular form, we conclude that $\begin{bmatrix} x(t) \\ e(t) \end{bmatrix} \to 0$ if both $A - BK$ and $A - LC$ are stable matrices; two conditions that we have already established under controllability and observability properties. Such a design leads to the separation principle for linear control systems: run an observer and apply the control as if the observer state is the actual state. This design is stabilizing.

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Canonical Forms

Controllable canonical form

If a model is not controllable, then we can construct a state transformation $\tilde{x} = Px$ with the form $\tilde{x} = \begin{bmatrix} \tilde{x}_c \\ \tilde{x}_{uc} \end{bmatrix}$ with

$\frac{d\tilde{x}}{dt} = \begin{bmatrix} A_c & A_{12} \\ 0 & A_{uc} \end{bmatrix} \begin{bmatrix} \tilde{x}_c \\ \tilde{x}_{uc} \end{bmatrix} + \begin{bmatrix} B_c \\ 0 \end{bmatrix} u$

$y = \begin{bmatrix} C_c & C_{uc} \end{bmatrix} \begin{bmatrix} \tilde{x}_c \\ \tilde{x}_{uc} \end{bmatrix}$

In the above $(A_c, B_c)$ is controllable. In the above, $A_{12}$ is some submatrix. The form above is called a controllable canonical form.

The matrix $P$ can be obtained with constructing $P^{-1}$ to consist of the following: Let $n_1$ be the rank of the controllability matrix $\mathcal{C}$. Then take the first $n_1$ columns of $P^{-1}$ to be $n_1$ linearly independent columns of $\mathcal{C}$, and the remaining $n - n_1$ columns are arbitrary vectors which make $P^{-1}$ invertible. If we write

$A_c = PAP^{-1}, \qquad B_c = PB,$

we have that

$P^{-1}A_c = AP^{-1}, \qquad P^{-1}B_c = B$

Using the fact that the controllable subspace is $A$ invariant, it follows that the structure of $A_c$ has to have the given structure.

An implication of the above analysis is that

$C(sI - A)^{-1}B + D = C_c(sI - A_c)^{-1}B_c + D$

Observable canonical form

A similar construction applies for observable canonical forms.

$\frac{d\tilde{x}}{dt} = \begin{bmatrix} A_o & 0 \\ A_{21} & A_{uo} \end{bmatrix} \begin{bmatrix} \tilde{x}_o \\ \tilde{x}_{uo} \end{bmatrix} + \begin{bmatrix} B_o \\ B_{uo} \end{bmatrix} u$

$y = \begin{bmatrix} C_o & 0 \end{bmatrix} \begin{bmatrix} \tilde{x}_o \\ \tilde{x}_{uo} \end{bmatrix}$

with the property that $(A_o, C_o)$ is observable.

An implication of the above analysis is that

$C(sI - A)^{-1}B + D = C_o(sI - A_o)^{-1}B_o + D$

Kalman decomposition

One can apply a joint construction, known as Kalman's decomposition. There exists a coordinate transformation so that

$z = Px$

with

$z = \begin{bmatrix} x_{co} \\ x_{c/uo} \\ x_{uc/o} \\ x_{uc/uo} \end{bmatrix}$

$\bar{A} = PAP^{-1}$

leads to

$\frac{dz}{dt} = \begin{bmatrix} A_{c/o} & 0 & A_{x/o} & 0 \\ A_{c/x} & A_{x/uo} & A_{x/x} & A_{x/uo} \\ 0 & 0 & A_{uc/o} & 0 \\ 0 & 0 & A_{uc/x} & A_{uc/uo} \end{bmatrix} z + \begin{bmatrix} B_{c/o} \\ B_{c/uo} \\ 0 \\ 0 \end{bmatrix} u$

$y = \begin{bmatrix} C_{c/o} & 0 & C_{uc/o} & 0 \end{bmatrix} z + Du,$

where $(A_{c/o}, B_{c/o}, C_{c/o})$ is both controllable and observable. Furthermore,

$C(sI - A)^{-1}B + D = C_{c/o}(sI - A_{c/o})^{-1}B_{c/o} + D$

A corollary of the above discussion is that the minimal realization; that is, the state-space realization with the smallest dimensions involving matrices, is attained when the system is both controllable and observable, as there are no redundant state variables.

Stabilizability and detectability. From the controllable canonical form, we can also establish the following result.

A linear system is stabilizable (in the sense of local or global asymptotic stability) by control if and only if $A_{uc}$, whenever exists, is a stable matrix (i.e., with eigenvalues strictly in the left half plane).

Define a control-free system to be detectable if whenever $y(t) \to 0$ then $x(t) \to 0$. A consequence of the observable canonical form is that a system is detectable if and only if $A_{uo}$, whenever exists, is a stable matrix.

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Using Riccati Equations to Find Stabilizing Linear Controllers [Optional]

While controllability and observability properties reveal what is possible or impossible with regard to stabilization, they don't directly present an easy-to-compute or constructive method for arriving at design.

One effective method is through Riccati equations. We will present the discussion for discrete-time, but the approach is essentially identical for continuous-time (with the stability conditions of linear systems, as noted earlier, being different).

Controller design via Riccati equations

Consider the following linear system

$x_{t+1} = Ax_t + Bu_t,$

where $x \in \mathbb{R}^n$, $u \in \mathbb{R}^m$.

Suppose that we would like to minimize the expression over all control laws:

$\sum_{t=0}^{\infty} x_t^T Q x_t + u_t^T R u_t$

with $R > 0$, $Q \geq 0$.

In the above, we established a method to find $K$ so that $A - BK$ is stable: Run the recursions, for any arbitrary initial condition, find the limit $P$ and select $u_t = -Kx_t$ with

$K = (B^T P B + R)^{-1} B^T P A$

This controller will be stabilizing.

Observer design via Riccati equations

A similar phenomenon as applies for observer design. In fact, with the above discussion, using the duality analysis presented earlier, we can directly design an observer so that the matrix $A - LC$ is stable. By writing the condition as the stability of $A^T - C^T L^T$, the question becomes that of finding $L^T$ for which $A^T - C^T L^T$ is a stable matrix.

Let $(A, C)$ be observable. In Theorem, if we replace $A$ with $A^T$, $B$ with $C^T$, and defining $W = BB^T$ for any $B$ with $(A, B)$ controllable, we obtain:

$S = W + ASA^T - ASC^T(CSC^T + R)^{-1}CSA^T.$

or the Riccati equations

$S_{t+1} = W + AS_tA^T - AS_tC^T(CS_tC^T + R)^{-1}CS_tA^T.$

whose limit as $t \to \infty$ for any initial $S_0$ will converge to a unique limit. Finally, taking

$L^T = (CSC^T + R)^{-1}CSA^T$

will lead to the conclusion that $A - LC$ is stable.

Putting controller and observer design together

Accordingly, all we need for the system:

$x_{t+1} = Ax_t + Bu_t, \qquad y_t = Cx_t$

is that $(A, B)$ be controllable and $(A, C)$ be observable. With the controller gain $K$ and observer gain $L$ from above, we can find $K$ and $L$ so that the system

$x_{k+1} = Ax_k - BK\hat{x}_k$

$\hat{x}_{k+1} = A\hat{x}_k + L(Cx_k - C\hat{x}_k)$

or, equivalently, with $e_k = x_k - \hat{x}_k$, the system defined with

$\begin{bmatrix} x_{k+1} \\ e_{k+1} \end{bmatrix} = \begin{bmatrix} A - BK & BK \\ 0 & A - LC \end{bmatrix} \begin{bmatrix} x_k \\ e_k \end{bmatrix}$

is stable.

In the above, the conditions on $(A, B)$ being controllable and $(A, C)$ being observable can be relaxed: controllability can be replaced with stabilizability and observability can be relaxed to detectability. While stability will be maintained, the only difference would be that $P$ or $S$ would not be guaranteed to be positive-definite.

Continuous-time case

A similar discussion as above applies for the continuous-time setup. We only discuss the control design, as the observer design follows from duality, as shown above.

Consider

$\frac{dx}{dt} = Ax + Bu$

Let $Q \geq 0$, $R > 0$. The only difference with the continuous-time is that the discrete-time Riccati equations above are replaced by a corresponding Riccati differential equation:

$-\frac{dP}{dt} = Q + A^T P + PA - PBR^{-1}B^T P.$

If $(A, B)$ is controllable and, with $Q = C^T C$, $(A, C)$ is observable, then there exists a unique positive-definite matrix $P$ such that the following algebraic Riccati equation is satisfied:

$Q + A^T P + PA - PBR^{-1}B^T P = 0$

With this $P$, the control given by

$u = -Kx = -R^{-1}B^T Px$

is so that $A - BK$ is stable.

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Applications and Exercises

Exercise. Recall that we had studied the controlled pendulum on a cart (see Figure 12.1). The non-linear mechanical/rotational dynamics equations were found to be

$M\frac{d^2 y}{dt^2} = u - m\frac{d^2}{dt^2}(y + l\sin(\theta)) = u - m\frac{d^2 y}{dt^2} + ml\frac{d^2\theta}{dt^2} - ml\left(\frac{d\theta}{dt}\right)^2 \sin(\theta)$

$m\frac{d^2\theta}{dt^2} = \frac{mg}{l}\sin(\theta) - \frac{m}{l}\frac{d^2 y}{dt^2}\cos(\theta)$

Around $\theta = 0$, $\frac{d\theta}{dt} = 0$, we apply the linear approximations $\sin(\theta) \approx \theta$ and $\cos(\theta) \approx 1$, and $\left(\frac{d\theta}{dt}\right)^2 \approx 0$ to arrive at

$M\frac{d^2 y}{dt^2} = u - \left(m\frac{d^2 y}{dt^2} + ml\frac{d^2\theta}{dt^2}\right)$

$l\frac{d^2\theta}{dt^2} = g\theta - \frac{d^2 y}{dt^2}$

Finally, writing $x_1 = y$, $x_2 = \frac{dy}{dt}$, $x_3 = \theta$, $x_4 = \frac{d\theta}{dt}$, we arrive at the linear model in state space form

$\frac{dx}{dt} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{-mg}{M} & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & \frac{(M+m)g}{Ml} & 0 \end{bmatrix} x + \begin{bmatrix} 0 \\ \frac{1}{M} \\ 0 \\ \frac{-1}{Ml} \end{bmatrix} u,$

where $x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$.

a) When is the linearized model controllable?

b) Does there exist a control policy with $u = -Kx$ that makes the closed loop linearized system stable? Select specific values for $M, m, l$ so that controllability holds, and accordingly find an explicit $K$.

c) With the controller in part b), can you conclude that through the arguments presented in the previous chapter (e.g. Theorem), that your (original non-linear) system is locally asymptotically stable?

Hint: a) With

$A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{-mg}{M} & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & \frac{(M+m)g}{Ml} & 0 \end{bmatrix}, \qquad B = \begin{bmatrix} 0 \\ \frac{1}{M} \\ 0 \\ \frac{-1}{Ml} \end{bmatrix}$

we have that

$\begin{bmatrix} B & AB & A^2B & A^3B \end{bmatrix} = \begin{bmatrix} 0 & \frac{1}{M} & 0 & \frac{mg}{M^2 l} \\ \frac{1}{M} & 0 & \frac{mg}{M^2 l} & 0 \\ 0 & \frac{-1}{Ml} & 0 & \frac{-(M+m)g}{M^2 l^2} \\ \frac{-1}{Ml} & 0 & \frac{-(M+m)g}{M^2 l^2} & 0 \end{bmatrix}$

You will be asked to find the condition for this system to be invertible in your homework assignment.

b) By controllability, we can place the eigenvalues of the matrix arbitrarily. Find an explicit $K$. You can use the method presented earlier in the chapter, or try to explicitly arrive at a stabilizing control matrix.

c) Then, by Theorem, the system is locally stable around the equilibrium point. Precisely explain why this is the case.

Exercise. Consider the linear system

$\frac{dx}{dt} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 1 & 1 \end{bmatrix} x + \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} u$

Is this system controllable? Does there exist a matrix $K$ so that with $u = Kx$, the eigenvalues of the closed-loop matrix: $\left(\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 1 & 1 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} K\right)$ can be arbitrarily assigned?

Exercise. Consider

$\frac{dx}{dt} = Ax + Bu, \qquad y = Cx$

with

$\frac{dx}{dt} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} x + \begin{bmatrix} 0 \\ 1 \end{bmatrix} u$

$y = \begin{bmatrix} 1 & 0 \end{bmatrix} x$

a) Is this system observable? b) Is this system controllable? c) Provide a stabilizing feedback control policy by running an observer.

Hint: a) and b) Yes. c) The system is both controllable and observable. If the system state were available, we could have $u = -Kx$ and select $K$ so that $A - BK$ is stable. Find such a $K$. Now, we can run an observer as explained in the relevant section:

$\frac{d\hat{x}}{dt} = A\hat{x} + Bu + L(y - C\hat{x})$

with the property that $A - LC$ is stable. Find such an $L$. Then, the control to be applied would be: $u_t = -K\hat{x}_t$. Find explicit values.

Exercise. a) Show that controllability is invariant under an algebraically equivalent transformation of the coordinates: $\tilde{x} = Px$ for some invertible $P$.

Hint: With $\frac{d\tilde{x}}{dt} = \tilde{A}\tilde{x}(t) + \tilde{B}u(t)$ and $\tilde{A} = PAP^{-1}$, $\tilde{B} = PB$, and with $\mathcal{C} = \begin{bmatrix} B & AB & \cdots & A^{n-1}B \end{bmatrix}$, we have that the transformed controllability matrix writes as $\tilde{\mathcal{C}} = \begin{bmatrix} \tilde{B} & \tilde{A}\tilde{B} & \cdots & \tilde{A}^{n-1}\tilde{B} \end{bmatrix} = P\mathcal{C}$. Since $P$ is invertible, $\text{rank}(\tilde{\mathcal{C}}) = \text{rank}(\mathcal{C})$.

b) Consider

$\frac{dx}{dt} = \begin{bmatrix} a & b \\ -b & a \end{bmatrix} x + \begin{bmatrix} 1 \\ 0 \end{bmatrix} u$

Express, through a transformation, this system in a controllable canonical realization form.

Hint: The characteristic polynomial of $A = \begin{bmatrix} a & b \\ -b & a \end{bmatrix}$ is $s^2 - 2as + (a^2 + b^2)$. The controllability matrix is $\mathcal{C} = \begin{bmatrix} 1 & a \\ 0 & -b \end{bmatrix}$, which has full rank when $b \neq 0$. The controllable canonical form has $A_c = \begin{bmatrix} 0 & 1 \\ -(a^2+b^2) & 2a \end{bmatrix}$ and $B_c = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$. The transformation $P = \tilde{\mathcal{C}}\mathcal{C}^{-1}$ where $\tilde{\mathcal{C}} = \begin{bmatrix} 0 & 1 \\ 1 & 2a \end{bmatrix}$.

Exercise. Consider

$\frac{dx}{dt} = Ax + Bu, \qquad y = Cx$

with

$\frac{dx}{dt} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} x + \begin{bmatrix} 0 \\ 1 \end{bmatrix} u$

$y = \begin{bmatrix} 1 & 0 \end{bmatrix} x$

a) Is this system observable? b) Is this system controllable? c) Provide a stabilizing feedback control policy by running an observer.

Note. The model here and the model for $P$ given in Exercise are related.

Solution. The system is both controllable and observable. If the system state were available, we could have $u = -Kx$ and select $K$ so that $A - BK$ is stable. Find such a $K$.

Now, we can run an observer as explain in the lecture notes:

$\frac{d\hat{x}}{dt} = A\hat{x} + Bu + L(y - C\hat{x})$

with the property that $A - LC$ is stable. Find such an $L$. Then, the control to be applied would be: $u_t = -K\hat{x}_t$.