# ECE 486 Control Systems Lecture 20

## State-Space Design (Continued)

Last time we detailed our discussion of state-space notions. We learned the conversion between given state-space model and transfer functions. Given state-space model of a system, we can compute its transfer function; conversely, for a given transfer function, there are many realizations associated with it.

We also saw three different canonical forms, i.e., controllable canonical form, observable canonical form and modal form. To continue in this lecture, we will explore the effect of pole-zero cancellations on internal stability. Further we study the effect of coordinate transformations on the properties of a given state-space model, transfer function, open-loop poles and controllability etc. We shall derive how to convert the state-space model of any completely controllable system to CCF.

### OCF with Arbitrary Zeros

Recall at the end of last lecture we noticed that a system model in Controllable Canonical Form for a transfer function with arbitrary zero is always completely controllable. This is because the zero does not enter the controllability matrix. Does this hold for realization in Observable Canonical Form too?

Start with a state-space model in Controllable Canonical Form

\begin{align*} \left(\begin{matrix} \dot{x}_1 \\ \dot{x}_2 \end{matrix}\right) &= \underbrace{\left(\begin{matrix} 0 & 1 \\ -{6} &-{5}\end{matrix}\right)}_{A}\left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right) + \underbrace{\left(\begin{matrix} 0 \\ 1\end{matrix}\right)}_{B} u \\ y &= \underbrace{\left(\begin{matrix} -z & 1 \end{matrix}\right)}_{C} \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right). \end{align*}

We can convert it to Observable Canonical Form by $A \mapsto A^T, B \mapsto C^T, C \mapsto B^T$,

\begin{align*} \left(\begin{matrix} \dot{x}_1 \\ \dot{x}_2 \end{matrix}\right) &= \underbrace{\left(\begin{matrix} 0 & -6 \\ 1 &-{5}\end{matrix}\right)}_{\bar{A}\, =\, A^T}\left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right) + \underbrace{\left(\begin{matrix} -z \\ 1\end{matrix}\right)}_{\bar{B}\, =\, C^T} u, \\ y &= \underbrace{\left(\begin{matrix} 0 & 1 \end{matrix}\right)}_{\bar{C}\, =\, B^T} \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right). \end{align*}

We already know that this system realizes the same transfer function as the original system we started with.

But is it also completely controllable? We can compute its controllability matrix

\begin{align*} {\cal C}(\bar{A},\bar{B}) &= \left[ \begin{matrix} \bar{B} \, | \, \bar{A}\bar{B} \end{matrix}\right], \\ \text{ where }\, \bar{A} \bar{B} &= \left(\begin{matrix} 0 & -6 \\ 1 &-5\end{matrix}\right)\left(\begin{matrix} -z \\ 1\end{matrix}\right) \\ &= \left( \begin{matrix} -6 \\ -z-5\end{matrix}\right) \\ \implies \, {\cal C}(\bar{A},\bar{B}) &= \left( \begin{matrix} -z & -6 \\ 1 & -z-5 \end{matrix}\right). \\ \det {\cal C} &= z(z+5)+6 \\ &= z^2 + 5z + 6 \\ &= 0 \text{ for $z=-2$ or $z=-3$.} \end{align*}

Therefore the OCF realization of the transfer function $G(s) = \dfrac{s-z}{s^2 + 5s + 6}$ is not completely controllable when $z=-2$ or $-3$, even though the CCF is always controllable.

#### Pole-Zero Cancellations

Let’s examine what actually happened to $G(s)$ when $z=-2$ made OCF not completely controllable.

\begin{align*}\require{cancel} G(s) &= \left. \frac{s-z}{s^2 + 5s+6}\right|_{z=-2} \\ &= \frac{\cancel{s+2}}{(\cancel{s+2})(s+3)} \\ &= \frac{1}{s+3}. \end{align*}

Here pole-zero cancellation happened. When $z=-2$, $G(s)$ is a first order transfer function, which can always be realized by a first order controllable model,

$\dot{x}_1 = -3x_1 + u, \, y = x_1 \quad \implies \quad G(s) = \frac{1}{s+3}.$

On the other hand, if we can look at this from another point of view, consider the transfer function

$G(s) = \frac{1}{s+3}.$

We can realize it using a one-dimensional controllable state-space model

$\dot{x}_1 = -3x_1 + u, \, y = x_1,$

or a noncontrollable two-dimensional state-space model

\begin{align*} \left(\begin{matrix} \dot{x}_1 \\ \dot{x}_2 \end{matrix}\right) &= \left(\begin{matrix} 0 & -6 \\ 1 &-{5}\end{matrix}\right)\left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right) + \left(\begin{matrix} 2 \\ 1\end{matrix}\right) u, \\ y &= \left(\begin{matrix} 1 & 0 \end{matrix}\right) \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right). \end{align*}

Obviously, it is not the best way to realize a first order transfer function with more than one state. (This is therefore a nonminimum realization.)

Thus, even the state dimension of a realization of a given transfer function is not unique.

Out of all the nonminimum realizations of a given transfer function, one type of realization is really bad. Consider the realization of

$G(s) = \frac{1}{s+3}$

using a two-state state-space model,

\begin{align*} \dot{x}_1 &= - 3x_1 + u \\ \dot{x}_2 &= 100x_2 \tag{1} \label{d20_eq1} \\ y &= x_1. \end{align*}

We know using one state $x_1$ is enough for realization of $G(s) = \frac{1}{s+3}$ so the second state $x_2$ in Equation \eqref{d20_eq1} can be treated as redundancy. Similarly we can add arbitrary equations of $x_3, x_4, \ldots$, all of which are independent of $x_1$ and $u$ but the resulting high dimension state-space models are all corresponding to the same transfer functions $G(s)$.

We’d like to make several remarks here.

• In the above nonminimum realization, $x_2$ is not affected by the input $u$, i.e., it is an autonomous mode, not controllable. And $x_2$ is not visible from the output either.
• $x_2$ also does not change the transfer function.
• And yet, the dynamics of $x_2$ is horrible — $x_2(t) \propto e^{100 t}$.

A (lower order) transfer function can mask undesirable internal state behavior (of higher dimensional state-space model).

#### Pole-Zero Cancellations and Stability

In case of a pole-zero cancellation, the transfer function contains much less information than the state-space model because some dynamics are hidden.

These dynamics can be either good (stable) or bad (unstable), but we cannot tell from the transfer function.

Therefore, our original definition of stability saying no RHP poles means stability is flawed. Because there can be RHP eigenvalues of the system matrix $A$ that are canceled by zeros, yet they still have dynamics associated with them.

So we shall modify the definition of Internal Stability in state-space parlance. A state-space model with matrices $(A,B,C,D)$ is said to be internally stable if all eigenvalues of the $A$ matrix are in the Left Half Plane.

Regarding Equation \eqref{d20_eq1}, rewrite it in matrix form

\begin{align*} \left(\begin{matrix} \dot{x}_1 \\ \dot{x}_2 \end{matrix}\right) &= \left(\begin{matrix} -3 & 0 \\ 0 & 100 \end{matrix}\right)\left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right) + \left(\begin{matrix} 1 \\ 0\end{matrix}\right) u, \\ y &= \left(\begin{matrix} 1 & 0 \end{matrix}\right) \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right). \end{align*}

As a state-space realization, it is associated with transfer function

\begin{align*}\require{cancel} G(s) &= C(sI - A)^{-1}B \\ &= \left(\begin{matrix} 1 & 0 \end{matrix}\right) \left( sI -\left(\begin{matrix} -3 & 0 \\ 0 & 100 \end{matrix}\right) \right)^{-1} \left(\begin{matrix} 1 \\ 0\end{matrix}\right) \\ &= \frac{\cancel{s - 100}}{(s+3)(\cancel{s-100})}\\ &= \frac{1}{s+3}. \end{align*}

We see that the unstable mode associated with RHP open-loop (unstable) pole $s = 100$ is not visible through the transfer function $G(s) = \frac{1}{s+3}$ even if $G(s)$ itself is closed-loop stable.

Hence the internal stability is a stronger condition for stability and it is equivalent to having no RHP open-loop poles and no pole-zero cancellations in RHP.

### Coordinate Transformations

Now that we have seen that a given transfer function can have many different state-space realizations, we would like a systematic procedure of generating such realizations, preferably with favorable properties such as controllability. One such procedure is by means of coordinate transformations.

Consider the map

\begin{align*} \tau : x \mapsto \bar{x} &= Tx, \text{ where }\, T \in \RR^{n \times n} \text{ is nonsingular } \\ \iff x &= T^{-1}\bar{x}, \end{align*}

meaning we can go back and forth between the coordinate systems.

For example,

\begin{align*} \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right) \longmapsto \left(\begin{matrix} \bar{x}_1 \\ \bar{x}_2 \end{matrix}\right) = \left(\begin{matrix} x_1 + x_2 \\ x_1 - x_2 \end{matrix}\right). \end{align*}

This can be represented as linear transformation

\begin{align*} \bar{x} = Tx, \text{ where }\, T = \left( \begin{matrix} 1 & 1 \\ 1 & -1 \end{matrix}\right). \end{align*}

The transformation is invertible since it has nonzero determinant $\det T = -2$, and

\begin{align*} T^{-1} &= \frac{1}{\det T}\left( \begin{matrix} -1 & -1 \\ -1 & 1\end{matrix}\right) = \left( \begin{matrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & - \frac{1}{2} \end{matrix}\right) \end{align*}

Or we can see $\bar{x}$ in terms of $x$ directly

\begin{align*} \bar{x}_1 + \bar{x}_2 &= 2 x_1, \\ \bar{x}_1 - \bar{x}_2 &= 2 x_2. \end{align*}

#### Coordinate Transformations and State-Space Models

Consider a state-space model

\begin{align*} \dot{x} &= Ax + Bu, \\ y &= Cx \end{align*}

and a change of coordinates $\bar{x} = Tx$ where $T$ invertible.

What does the system look like in the new coordinates?

\begin{align*} \dot{\bar{x}} &= \dot{Tx} &\text{(by definition)}\\ &= T\dot{x} &\text{(by linearity of derivative)} \\ &= T(Ax + Bu) \\ &= T(AT^{-1}\bar{x} + Bu) &\text{(by $x = T^{-1}\bar{x}$)} \\ &= \underbrace{TAT^{-1}}_{\bar{A}}\bar{x} + \underbrace{TB}_{\bar{B}}u, \\ y &= Cx \\ &= \underbrace{CT^{-1}}_{\bar{C}}\bar{x}. \end{align*}

Therefore, under the linear transform $\tau : x \mapsto Tx$, we have

\begin{align*} \begin{array}{l} \dot{x} = Ax + Bu,\\ y = Cx, \end{array} \quad \xrightarrow{\quad T \quad} \quad \begin{array}{l} \dot{\bar{x}} = \bar{A}\bar{x} + \bar{B}u, \\ y = \bar{C}\bar{x}, \end{array} \end{align*}

where

\begin{align*} \bar{A} = TAT^{-1}, \, \bar{B} = TB, \, \bar{C} = CT^{-1}. \end{align*}

Then what happens to

• the transfer function? and
• the controllability matrix?

We claim both remain the same.

Claim: The transfer function does not change under linear transformation.

Proof: Indeed,

\begin{align*} \bar{G}(s) &= \bar{C}(sI-\bar{A})^{-1}\bar{B} \\ &= (CT^{-1})\left(sI - TAT^{-1}\right)^{-1} (TB) \\ &= CT^{-1}\left(sTIT^{-1} - TAT^{-1}\right)^{-1} TB \\ &= CT^{-1}\left[T\left(sI - A\right)T^{-1}\right]^{-1} TB \\ &= C\underbrace{T^{-1}T}_{I} \left(sI-A\right)^{-1}\underbrace{T^{-1}T}_{I}B \\ &= C\left(sI-A\right)^{-1}B \\ &= G(s). \end{align*}

Regarding the transfer function under linear transformation, we can say more. In fact,

• the open-loop poles do not change.
• the characteristic polynomial does not change.

\begin{align*} \det(sI-\bar{A}) &= \det(sI - TAT^{-1}) \\ &= \det\left[T(sI-A)T^{-1}\right] \\ &= \det T \cdot \det (sI-A) \cdot \det T^{-1} \\ &= \det (sI-A). \end{align*}

Claim: Controllability does not change under linear transformation.

Proof: For any $k = 0,1,\ldots$, note by induction

\begin{align*} \bar{A}^k \bar{B} &= (TAT^{-1})^k TB \\ &= TA^{k}T^{-1}TB \\ &= TA^kB. \end{align*}

Therefore,

\begin{align*} {\cal C}(\bar{A},\bar{B}) &= [\, TB \, | \, TAB \, | \, \ldots \, | \, TA^{n-1}B\,] \\ & = T [\,B \, | \, AB \, | \, \ldots \, | \, A^{n-1}B\,] \\ & = T{\cal C}(A,B). \end{align*}

Since $\det T \neq 0$, $\det {\cal C}(\bar{A},\bar{B}) \neq 0$ if and only if $\det {\cal C}(A,B) \neq 0$.

Thus, the new system is completely controllable if and only if the original one is.

Note that the controllability matrix may change,

\begin{align*} \underbrace{{\cal C}(\bar{A},\bar{B})}_{\text{new}} &= \underbrace{T}_{\text{coord.} \atop \text{trans.}} \underbrace{{\cal C}(A,B)}_{\text{old}} \\ & \Updownarrow \\ T &= {\cal C}(\bar{A},\bar{B}) \left[\, {\cal C}(A,B)\, \right]^{-1}. \tag{2} \label{d20_eq2} \end{align*}

This is a recipe for going from one controllable realization of a given transfer function to another.

Controllable Canonical Form is the most convenient controllable realization of a given transfer function, so we want to convert a given controllable system state-space model to CCF (if it is useful for control design).

Example: Consider $A = \left( \begin{matrix} - 15 & 8 \\ -15 & 7 \end{matrix}\right), \,\, B = \left( \begin{matrix} 1 \\ 1 \end{matrix}\right)$. Find a linear transform such that it converts this state-space model to CCF if possible. Note $C$ is immaterial because of Equation \eqref{d20_eq2}.

Solution: Step 1: Check for controllability.

\begin{align*} AB &= \left( \begin{matrix} - 15 & 8 \\ -15 & 7 \end{matrix}\right) \left( \begin{matrix} 1 \\ 1 \end{matrix}\right) = \left( \begin{matrix} -7 \\ -8 \end{matrix}\right) \\ \implies {\cal C} &= \left( \begin{matrix} 1 &-7 \\ 1 &-8 \end{matrix}\right). \end{align*}

Controllability matrix is nonsingular $\det {\cal C} = -1$ hence the system is completely controllable.

Step 2: Determine the desired ${\cal C}(\bar{A},\bar{B})$.

We need to figure out $\bar{A}$ and $\bar{B}$.

By $\bar{A}, \bar{B}$ in CCF, we must have

\begin{align*} \bar{A} &= \left( \begin{matrix} 0 & 1 \\ -a_2 & -a_1 \end{matrix}\right), \\ \bar{B} &= \left( \begin{matrix} 0 \\ 1 \end{matrix}\right), \end{align*}

so we need to find two unknown coefficients $a_1,a_2$.

Recall the characteristic polynomial remains invariant under linear transformation.

\begin{align*} \det(Is-A) &= \det(Is - \bar{A}) \\ \implies \det\left(\begin{matrix} s+15 & -8 \\ 15 & s-7 \end{matrix}\right) &= \det\left( \begin{matrix} s & -1 \\ a_2 & s+a_1 \end{matrix}\right) \\ \implies (s+15)(s-7) + 120 &= s(s+a_1) + a_2 \\ s^2 +\color{red}{8}s + \color{blue}{15} &= s^2 + \color{red}{a_1} s + \color{blue}{a_2}. \end{align*}

By matching the coefficients, we got

\begin{align*} \bar{A} = \left( \begin{matrix} 0 & 1 \\ -15 & -8 \end{matrix}\right), \, \bar{B} &= \left( \begin{matrix} 0 \\ 1 \end{matrix}\right) \end{align*}

Therefore, the new controllability matrix should be

\begin{align*} {\cal C}(\bar{A},\bar{B}) = [\,\bar{B} \, | \, \bar{A}\bar{B}\,] = \left( \begin{matrix} 0 & 1\\ 1&-8 \end{matrix}\right). \end{align*}

Step 3: Compute $T$.

By Equation \eqref{d20_eq2}, $T = {\cal C}(\bar{A},\bar{B}) \cdot \left[\, {\cal C}(A,B)\,\right]^{-1}$. And we already have

\begin{align*} {\cal C}(A,B) &= \left( \begin{matrix} 1 &-7 \\ 1 &-8 \end{matrix}\right), \\ \left[\,{\cal C}(A,B)\,\right]^{-1} &= \left( \begin{matrix} 1 &-7 \\ 1 &-8 \end{matrix}\right)^{-1} \\ &= \frac{1}{-1}\left( \begin{matrix} -8 & 7 \\ -1 & 1\end{matrix}\right)\\ &= \left( \begin{matrix} 8 & -7 \\ 1 & -1\end{matrix}\right), \\ {\cal C}(\bar{A},\bar{B}) &= \left( \begin{matrix} 0 & 1\\ 1&-8 \end{matrix}\right). \\ \implies T &= \left( \begin{matrix} 0 & 1\\ 1&-8 \end{matrix}\right) \left( \begin{matrix} 8 & -7 \\ 1 & -1\end{matrix}\right) \\ &= \left( \begin{matrix} 1 & -1 \\ 0 & 1 \end{matrix}\right). \end{align*}

Last updated: 2018-04-19 Thu 15:48