Added a post about controllability and stabilisability of control sys…

…tems

_posts/2023-11-21-controllability-and-stabilisability.md

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+---
+title: System Controllability and Stabilisability
+date: 2023-11-21
+categories: [Mechanics, Control]
+tags: [study, mechanics, dynamics, mathematics, control, state space, linear systems]
+math: true
+toc: true
+---
+
+<!-- prettier-ignore -->
+* TOC
+{:toc}
+
+## State Feedback
+
+In [my previous post]({% post_url 2023-10-29-state-space %}), I discussed how many systems can be
+modelled using linear state space equations of the form $\mathbf{\dot{x} = Ax + Bu}\;$ where
+$\mathbf{x}$ is the state of the system, $\mathbf{u}$ are the inputs of the system and $\mathbf{A}$
+and $\mathbf{B}$ parameterise the system. However, in many real scenarios, it is not enough for us
+to just have a system of the model. Often we want to control the system into a given state that is
+useful for our purposes or at the very least, ensure it remains stable. State feedback is a popular
+method that allows us to do just this.
+
+<!-- prettier-ignore -->
+![A block diagram of a system with state feedback](/images/statefeedback.jpg)
+_A block diagram of a system with state feedback_
+
+For linear systems, we can use a very simple control law, $\mathbf{u = Kx}$, and remarkably this
+allows us to choose _arbitrary_ closed loop system poles and as a result dynamics of the system. How
+is this so?
+
+## State Space System Representations
+
+The state space representation of a system is not unique. Despite this, we will see that some
+choices of states lead to a clearer understanding of certain parts of the system dynamics. To begin,
+consider an arbitrary state space system described by
+
+$$
+\begin{align*}
+ \mathbf{\dot{x}} &= \mathbf{Fx + Gu} \\
+ \mathbf{y} &= \mathbf{Hx + Ju}
+\end{align*}
+$$
+
+Now let $\mathbf{x = Tz}$ for some other choice of state $\mathbf{z}$ and a nonsingular matrix
+$\mathbf{T}$. Substituting this into our state space equations from before and rearranging,
+
+$$
+\begin{align*}
+ \mathbf{T\dot{z}} &= \mathbf{FTz + Gu} \\
+ \implies \mathbf{\dot{z}} &= \mathbf{T^{-1}FTz + T^{-1}Gu} \\
+ \mathbf{y} &= \mathbf{HTz + Ju}
+\end{align*}
+$$
+
+If
+
+$$
+\begin{align*}
+ \mathbf{A} &= \mathbf{T^{-1}FT} \\
+ \mathbf{B} &= \mathbf{T^{-1}G} \\
+ \mathbf{C} &= \mathbf{HT}
+\end{align*}
+$$
+
+then
+
+$$
+\begin{align*}
+ \mathbf{\dot{z}} &= \mathbf{Az + Bu} \\
+ \mathbf{y} &= \mathbf{Cx}
+\end{align*}
+$$
+
+which is still a linear state space equation for the system. Therefore, the system's state space
+representation is not unique.
+
+## Deriving the Transfer Function
+
+State space equations can also be represented in the Laplace domain. Begin by recalling that the
+derivative in the time domain is equivalent to multiplying by $s$ in the Laplace domain. Therefore,
+rewriting the state space equations we get.
+
+$$
+\begin{align*}
+ s\mathbf{X} &= \mathbf{AX + BU} \\
+ \mathbf{Y} &= \mathbf{CX + DU}
+\end{align*}
+$$
+
+Rearranging this we get
+
+$$
+\begin{align*}
+ \mathbf{X} &= (s\mathbf{I - A})^{-1} \mathbf{BU} \\
+ \implies \mathbf{Y} &= (\mathbf{C}(s\mathbf{I - A})^{-1} \mathbf{B} + \mathbf{D}) \mathbf{U}
+\end{align*}
+$$
+
+Therefore, the transfer function $\mathbf{H}(s)$ is,
+
+$$
+\mathbf{H}(s) = \mathbf{C}(s\mathbf{I - A})^{-1} \mathbf{B} + \mathbf{D}
+$$
+
+Some other useful relationships are that the characteristic equation can be found with
+
+$$
+\det(s\mathbf{I - A}) = 0
+$$
+
+and the zeros can be found with
+
+$$
+\det \begin{pmatrix}
+ s\mathbf{I - A} & -\mathbf{B} \\
+ \mathbf{C} & \mathbf{D}
+\end{pmatrix}
+$$
+
+These are a little harder to show so for the sake of brevity I will not derive them but for more
+details see chapter 7 of _Feedback Control of Dynamic Systems_ by Franklin and Powell or you can do
+some searching online.
+
+## Controllability Canonical Form
+
+The controllability canonical form of a system can be expressed with,
+
+$$
+\begin{align*}
+ \mathbf{A}_c &= \begin{pmatrix}
+ -a_1 & -a_2 & \dots & -a_{n-1} & -a_n \\
+ 1 & 0 & \dots & 0 & 0 \\
+ 0 & 1 & \dots & 0 & 0 \\
+ \vdots & \vdots & \ddots & \vdots & \vdots \\
+ 0 & 0 & \dots & 1 & 0
+ \end{pmatrix}, \quad
+ &\mathbf{B}_c = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}
+ \\
+ \mathbf{C}_c &= \begin{pmatrix}
+ b_1 & b_2 & \dots & b_n
+ \end{pmatrix}, \quad
+ &\mathbf{D}_c = \mathbf{0}
+\end{align*}
+$$
+
+where the transfer function of the system is
+
+$$
+H(s) = \frac{b_1 s^{n-1} + b_2 s^{n-2} + \dots + b_n}{s^n + a_1 s^{n-1} + \dots + a_n}
+$$
+
+So this form of the system is very useful because it allows us to easily identify the dynamics of
+the system using the poles and zeros.
+
+Now consider the system with state feedback such that $\mathbf{u = Kx}$ \(note some resources use
+$\mathbf{u = -Kx}$ which is equivalent except that the signs of each element of $\mathbf{K}$ would
+be negated\). Then the system could be rewritten,
+
+$$
+\mathbf{\dot{x}} = \mathbf{Ax + BKx} = (\mathbf{A + BK})\mathbf{x}
+$$
+
+So the effective $\mathbf{A}$ matrix of the system is $\mathbf{A + BK}$. If the system was initially
+in control canonical form then,
+
+$$
+\begin{align*}
+ \mathbf{BK} &= \begin{pmatrix}
+ k_1 & k_2 & \dots & k_n \\
+ 0 & 0 & \dots & 0 \\
+ \vdots & \vdots & \ddots & \vdots \\
+ 0 & 0 & \dots & 0
+ \end{pmatrix} \\
+ \implies \mathbf{A + BK} &= \begin{pmatrix}
+ -a_1 + k_1 & -a_2 + k_2 & \dots & -a_{n-1} + k_{n-1} & -a_n + k_n \\
+ 1 & 0 & \dots & 0 & 0 \\
+ \vdots & \vdots & \ddots & \vdots & 0 \\
+ 0 & 0 & \dots & 1 & 0
+ \end{pmatrix}
+\end{align*}
+$$
+
+And so we can arbitrarily choose the poles of our system by choosing $\mathbf{K}$ such that the
+coefficients of the desired characteristic equation match the terms in the resultant controllability
+matrix for the system. The corollary is that if we can represent the system in control canonical
+form then the system poles can be chosen arbitrarily therefore so can the system's dynamics. If this
+is the case then we say the system is _controllable_.
+
+## The Controllability Matrix
+
+So how can we show if it is possible to represent the system in this controllability canonical form?
+Well, let's take our arbitrary system from before and try to transform it into controllability form.
+We already saw,
+
+$$
+\begin{align*}
+ \mathbf{A}_c &= \mathbf{T^{-1}FT} \\
+ \implies \mathbf{A}_c \mathbf{T^{-1}} &= \mathbf{T^{-1}F}
+\end{align*}
+$$
+
+Now if the rows of $\mathbf{T^{-1}}$ are denoted as $\mathbf{t_1,\, t_2, \dots,\, t_n}$ then
+rewriting the previous equation,
+
+$$
+\begin{pmatrix}
+ -a_1 & \dots & -a_n \\
+ 1 & \dots & 0 \\
+ \vdots & \ddots & \vdots \\
+ 0 & \dots & 0
+\end{pmatrix}
+\begin{pmatrix} \mathbf{t_1} \\ \mathbf{t_2} \\ \vdots \\ \mathbf{t_n} \end{pmatrix}
+= \begin{pmatrix} \mathbf{t_1 F} \\ \mathbf{t_2 F} \\ \vdots \\ \mathbf{t_n F} \end{pmatrix}
+$$
+
+From this, we can see that
+
+$$
+\begin{align*}
+ \mathbf{t_{n-1}} &= \mathbf{t_n F} \\
+ \implies \mathbf{t_{n-2}} &= \mathbf{t_{n-1} F} = \mathbf{t_n F^2} \\
+ \implies \mathbf{t_{n-k}} &= \mathbf{t_n F^k}
+\end{align*}
+$$
+
+We also know that
+
+$$
+\begin{align*}
+ \mathbf{T^{-1}G} &= \mathbf{B}_c \\
+ \implies \begin{pmatrix} \mathbf{t_1 G} \\ \vdots \\ \mathbf{t_n G} \end{pmatrix}
+ &= \begin{pmatrix} 0 \\ \vdots \\ 1 \end{pmatrix}
+\end{align*}
+$$
+
+Combining these
+
+$$
+\begin{align*}
+ \begin{pmatrix} \mathbf{t_n G} & \mathbf{t_{n-1} G} & \dots & \mathbf{t_1 G} \end{pmatrix}
+ &= \begin{pmatrix} 0 & 0 & \dots & 1 \end{pmatrix} \\
+ \implies \mathbf{t_n} \begin{pmatrix}
+ \mathbf{G} & \mathbf{FG} & \dots & \mathbf{F^{n - 1} G}
+ \end{pmatrix}
+ &= \begin{pmatrix} 0 & 0 & \dots & 1 \end{pmatrix} \\
+ \implies \mathbf{t_3}
+ = \begin{pmatrix} 0 & 0 & \dots & 1 \end{pmatrix} \mathcal{C}^{-1}
+\end{align*}
+$$
+
+where
+$\mathcal{C} = \begin{pmatrix} \mathbf{G} & \mathbf{FG} & \dots & \mathbf{F^{n-1}G} \end{pmatrix}$
+is the controllability matrix. Therefore, if $\mathcal{C}$ is invertible then we can solve for
+$\mathbf{t_n}$ which then allows us to solve for each of the additional rows of $\mathbf{T^{-1}}$.
+Using this $\mathbf{T^{-1}}$ we can then transform the system into control canonical form and then
+move the poles of the system arbitrarily with an appropriate $\mathbf{K}$.
+
+So, the practical steps of assessing controllability are simply to construct the controllability
+matrix
+$\mathcal{C} = \begin{pmatrix} \mathbf{G} & \mathbf{FG} & \dots & \mathbf{F^{n-1}G} \end{pmatrix}$
+and then check if it is invertible. A practical method for checking if it is invertible is to check
+if it has **rank**$(\mathcal{C}) = n$ where $n$ is the number of states of the system. Many software
+packages make this computation easy but if for whatever reason you need to do it by hand you can
+check the rank by converting the matrix to
+[row echelon form](https://en.wikipedia.org/wiki/Row_echelon_form) using row operations and count
+the number of non-zero rows.
+
+## An Intuitive Understanding of Controllability
+
+If a system is controllable, then it is possible to control the dynamics of every mode of the
+system. In other words, the speed at which each of the modes stabilises can be set arbitrarily with
+an appropriate $\mathbf{K}$ used in state feedback. On the other hand, if a system is not
+controllable, we cannot make these guarantees for all modes. For example, if the accelerator of a
+car provides the input to the system and we had sensors to measure all the states perfectly, we
+could control the horizontal motion of the car. However, the vertical motion of the car would be
+independent of this input and so we would not be able to control this vertical motion using only the
+accelerator. In this case, if only the horizontal position and velocity are considered as states,
+the system would be controllable however, if the vertical motion also needs to be considered then
+the system would not be controllable.
+
+## Stabilisability
+
+Despite this, the car would likely be fitted with some suspension which could be modelled as a
+spring and damper. This limits its vertical motion from continuing indefinitely after a disturbance
+such as a speed bump. In this case, even though we cannot directly control the car's vertical
+motion, we can at least say that this uncontrollable mode is stable and as such we could still say
+that the system is stabilisable. More generally, if all of the uncontrollable modes of a system are
+naturally stable then the system is said to be _stabilisable_. It does not matter whether the modes
+that we can control are naturally stable or not since we can control their dynamics arbitrarily with
+state feedback as shown before to make them stable.
+
+<!-- prettier-ignore -->
+![A diagram of a car modelled with springs and dampers at each wheel](/images/car_with_suspension.jpg)
+_A car with suspension modelled using springs and dampers at each wheel_
+
+But how do we know which modes of the system are controllable? The controllability matrix rank just
+tells us if the system is controllable but not which modes are not controllable. Therefore, a common
+way to check which modes are uncontrollable is to use the
+[PBH test](https://en.wikipedia.org/wiki/Hautus_lemma) which tells us not only if the system is
+controllable but also which modes are controllable. The PBH test says that a mode is controllable if
+and only if
+
+$$
+\text{rank}\left(\begin{array}{c:c}
+ \lambda \mathbf{I - A} & \mathbf{B}
+\end{array} \right) = n
+$$
+
+for the mode associated with the eigenvalue $\lambda$. Therefore, a system is stabilisable if and
+only if the above is true for all $\Re(\lambda) \geq 0$. A proof for this can be found
+[here](https://www.control.utoronto.ca/people/profs/kwong/ece410/2008/notes/chap2.pdf).
+
+## Practical Limitations
+
+<!-- prettier-ignore -->
+![A VW Golf with a Lamborghini V10 Engine in the boot](/images/vw_golf_with_v10.jpg)
+_A VW Golf with a Lamborghini V10 engine to make it go zoom_
+
+While in theory if a given mode is controllable it is possible to control its dynamics arbitrarily
+in practice this is often limited by the input. Going back to the example of the car, there is
+obviously a limit to the amount of force that can be applied to the car by the engine and therefore
+the accelerator. This limit means that we can control the dynamics of the car up to a point but my
+VW golf is still not ever going to go from 0-100km/h in one second. Similarly, some modes may be
+weakly controllable making it impractical to control them. For example, accelerating and then
+decelerating may induce a small amount of rocking in the car however the effect would be small and
+it would be impractical in many cases to implement control of this mode just from the brakes and
+accelerator.