# ECE 486 Control Systems Lecture 19

## State-Space Design

We finished our discussion of frequency domain design method in Lecture 17 and Lecture 18. In this lecture, we will pick up the introduction to state-space representation of a system in Lecture 2 and further introduce other basic notions of state-space control. We will talk about different state-space realizations of the same transfer function; several canonical forms of state-space systems and controllability matrix.

The state-space approach reveals internal system architecture for a given transfer function. And most importantly, the mathematical machinery is very different — there is a lot of use of linear algebra. As it stands in ECE 486, it is just a short introduction; ECE 515 would be a proper follow-up on theory of linear systems.

Important: Frequency-Domain versus State-Space

• About $90\%$ of industrial controllers are designed using frequency-domain methods since PID is a popular architecture.
• About $90\%$ of current research in systems and control is done in the state-space framework.

To be able to talk to control engineers and follow progress in the field, we need to know both methods and understand the connections between them.

#### A General State-Space Model

Recall from Lecture 2, we introduced state vector $x$, input vector $u$ and output vector $y$,

\begin{align*} \text{state } x &= \left( \begin{matrix} x_1 \\ \vdots \\ x_n \end{matrix}\right) \in \RR^n, \\ \text{input } u &= \left( \begin{matrix} u_1 \\ \vdots \\ u_m \end{matrix}\right) \in \RR^m, \\ \text{output } y &= \left( \begin{matrix} y_1 \\ \vdots \\ y_p \end{matrix}\right) \in \RR^p. \end{align*}

So a system representation can be written as

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

where

• $A$ is the system matrix of size $n \times n$;
• $B$ is the input matrix of size $n \times m$;
• $C$ is the output matrix of size $p \times n$; and
• $D$ is the feed through matrix of size $p \times m$.

#### From State-Space to Transfer Function

Let us derive the transfer function from $u$ to $y$ corresponding to the state-space model

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

In the scalar case where $x,y,u \in \RR$, we took the Laplace transform. We can mimic the same strategy here when working with vectors. We apply Laplace transform component-wise.

Recall matrix-vector multiplication

\begin{align*} \dot{x}_i &= (Ax)_i + (Bu)_i \\ \tag{1} \label{d19_eq1} &= \sum^n_{j=1}a_{ij}x_j + \sum^m_{k=1}b_{ik}u_k, \\ y_\ell &= (Cx)_\ell + (Du)_\ell\\ \tag{2} \label{d19_eq2} &= \sum^n_{j=1}c_{\ell j}x_j + \sum^m_{k=1} d_{\ell k}u_k. \end{align*}

Now we take the Laplace transform on both sides of Equation \eqref{d19_eq1}.

\begin{align*} \dot{x}_i &= \sum^n_{j=1}a_{ij}x_j + \sum^m_{k=1}b_{ik}u_k \\ &\qquad \qquad \downarrow {\mathscr L} \\ sX_i(s) - x_i(0) &= \sum^n_{j=1}a_{ij}X_j(s) + \sum^m_{k=1}b_{ik}U_k(s), \forall~ i = 1,\ldots,n. \end{align*}

Reassemble these $n$ equations in matrix-vector form, we have

\begin{align*} sX(s) - x(0) &= AX(s) + BU(s) \\ (sI-A)X(s) &= x(0) + BU(s) \\ \implies X(s) &= (sI-A)^{-1}x(0) + (sI-A)^{-1}BU(s), \end{align*}

where $I$ is the $n\times n$ identity matrix.

By the same token, we derive the Laplace transform of $y$ using Equation \eqref{d19_eq2}.

\begin{align*} y_\ell &= \sum^n_{j=1}c_{\ell j}x_j + \sum^m_{k=1} d_{\ell k}u_k \\ &\qquad \qquad \downarrow {\mathscr L} \\ Y_\ell(s) &= \sum^n_{j=1}c_{\ell j}X_j(s) + \sum^m_{k=1} d_{\ell k}U_k(s), \forall~ \ell = 1,\ldots,p. \end{align*}

Reassemble these $p$ equations in matrix-vector form, we have

\begin{align*} Y(s) &= CX(s) + DU(s) \\ &= C\left[ (sI-A)^{-1}x(0) + (sI-A)^{-1}BU(s)\right] + DU(s) \\ &= C(sI-A)^{-1}x(0) + \left[C(sI-A)^{-1}B + D\right]U(s). \end{align*}

To find the input-output relationship, i.e., transfer function, we set the initial conditions to $0$,

\begin{align*} Y(s) &= G(s)U(s), \, \text{where } G(s) = C(sI-A)^{-1}B + D, \end{align*}

i.e., the transfer function from $u$ to $y$ associated with state-space model $(A,B,C,D)$ is

$G(s) = C(Is-A)^{-1}B + D. \tag{3} \label{d19_eq3}$

Note that $G(s)$ is composed of all four matrices that contain information about the state-space model.

Remarks:

• $G(s)$ would be undefined if the $n\times n$ matrix $sI-A$ is singular or noninvertible, i.e., precisely when $\det(sI-A) = 0$.
• $A$ is $n\times n$, $\det(sI-A)$ is a polynomial of degree $n$, called the characteristic polynomial of $A$.

$\det(sI-A) = \det \left( \begin{matrix} s-a_{11} & -a_{12} & \ldots & -a_{1n} \\ -a_{21} & s-a_{22} & \ldots & -a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ -a_{n1} & -a_{n2} & \ldots & s-a_{nn} \end{matrix} \right),$

The roots of characteristic polynomial are the eigenvalues of $A$.

• $G(s)$ is open loop stable if all eigenvalues of $A$ lie in the Left Half Plane.

Example 1: Consider the state-space model in Controllable Canonical Form (CCF)$^*$

\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} 1 & 1 \end{matrix}\right)}_{C} \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right). \end{align*}

Compute the transfer function from $u$ to $y$.

$^*$ We will explain this terminology later.

Solution: This is a single-input, single-output (SISO) system with $u,y \in \RR$ even though the state $x$ is two-dimensional.

Here $D = 0$ is a zero matrix. By our derivation for $G(s)$ in Equation \eqref{d19_eq3} above,

\begin{align*} \tag{4} \label{d19_eq4} G(s) &= C(Is-A)^{-1}B, \hspace{2.5cm}\text{($D=0$)} \\ \text{ where }\, sI-A &= \left( \begin{matrix} s & -1 \\ 6 & s+5 \end{matrix}\right). \end{align*}

To compute $(sI-A)^{-1}$, recall the inverse formula for $2\times2$ matrix $M$,

\begin{align*} M = \left( \begin{matrix} a & b \\ c & d \end{matrix}\right), \, \det M \neq 0 \quad \Longrightarrow \quad M^{-1} &= \frac{1}{\det M}\left( \begin{matrix} d & - b \\ -c & a \end{matrix}\right). \end{align*}

Applying the formula, we get

\begin{align*} (sI-A)^{-1} &= \frac{1}{\det(sI-A)} \left( \begin{matrix} s+5 & 1 \\ -6 & s \end{matrix}\right) \\ &= \frac{1}{s^2 + 5s + 6} \left( \begin{matrix} s+5 & 1 \\ -6 & s \end{matrix}\right). \end{align*}

Substitute $B, C$ matrices, $G(s)$ in Equation \eqref{d19_eq4} becomes

\begin{align*} G(s) &= C(sI-A)^{-1}B \\ &= \left(\begin{matrix} 1 & 1 \end{matrix}\right) \frac{1}{s^2 + 5s + 6} \left( \begin{matrix} s+5 & 1 \\ -6 & s \end{matrix}\right) \left(\begin{matrix} 0 \\ 1\end{matrix}\right) \\ &= \frac{1}{s^2 + 5s + 6} \left(\begin{matrix} 1 & 1 \end{matrix}\right) \left(\begin{matrix} 1 \\ s\end{matrix}\right) \\ &= \frac{s+1}{s^2 + 5s + 6}. \end{align*}

The given state-space model in this example is called a realization of this resulting transfer function.

Further we notice the coefficients $5$ and $6$ in $A$ matrix appear in reverse order in $G(s)$.

#### State-Space Realizations of Transfer Functions

In the previous section, we computed the transfer function based off of given state-space model. Conversely, we can also realize a state-space model according to a transfer function.

Using the above example, say

$G(s) = \frac{s+1}{s^2 + \color{red}{5}s + \color{blue}{6}}$

is given, then the state-space model

\begin{align*} \left(\begin{matrix} \dot{x}_1 \\ \dot{x}_2 \end{matrix}\right) &= \underbrace{\left(\begin{matrix} 0 & 1 \\ -\color{blue}{6} &-\color{red}{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} 1 & 1 \end{matrix}\right)}_{C} \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right) \\ \end{align*}

is one realization. At least in this example, information about the state-space model $(A,B,C)$ is contained in $G(s)$.

Are there other realizations for such a given transfer function $G(s)$?

There are infinitely many. (They are the same up to a linear transform.)

For example, with the one realization that is already obtained,

\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} 1 & 1 \end{matrix}\right)}_{C} \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right) \end{align*}

consider a new state-space model with transposed $A, B, C$ matrices

\begin{align*} \dot{x} &= \bar{A} x + \bar{B}u, \\ y &= \bar{C}x, \end{align*}

where

\begin{align*} \bar{A} = A^T = \left( \begin{matrix} 0 & -6 \\ 1 & -5 \end{matrix}\right), \, \bar{B} = C^T = \left(\begin{matrix} 1 \\ 1 \end{matrix}\right), \, \bar{C} = B^T = \left( \begin{matrix} 0 & 1 \end{matrix}\right). \end{align*}

This is a different state-space model but it is easy to check that $G(s) = \bar{C}(sI-\bar{A})^{-1}\bar{B}$.

Indeed, we can prove the following claim.

Claim: The state-space model

\begin{align*} \dot{x} &= \bar{A} x + \bar{B}u, \\ y &= \bar{C}x, \end{align*}

where

\begin{align*} \bar{A} = A^T, \, \bar{B} = C^T, \, \bar{C} = B^T \end{align*}

has the same transfer function as the original model with $(A,B,C)$.

Proof: This is just a simple exercise of algebra. Indeed, for a single-input, single-out system,

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

Therefore, by the above claim, for the state-space model in controller in Example 1, a state-space model is derived, called Observable Canonical Form (OCF).

\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} 1 \\ 1\end{matrix}\right) u, \\ y &= \left(\begin{matrix} 0 & 1 \end{matrix}\right) \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right). \end{align*}

Yet another realization of $G(s) = \dfrac{s+1}{s^2+5s+6}$ can be extracted from the partial fractions decomposition.

\begin{align*} G(s) &= \frac{s+1}{(s+2)(s+3)} \\ &= \frac{2}{s+3} - \frac{1}{s+2}. \tag{5} \label{d19_eq5} \end{align*}

The interpretation of Equation \eqref{d19_eq5} is that we treat $G(s)$ as the sum of two first order transfer functions, each of them can be realized by scalars in $(A, B, C)$ model.

This is called Modal Canonical Form (MCF) since different modes (by eigenvalues of $A$) are decoupled.

\begin{align*} \left(\begin{matrix} \dot{x}_1 \\ \dot{x}_2 \end{matrix}\right) &=\left(\begin{matrix} -3 & 0 \\0 &-{2}\end{matrix}\right)\left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right) + \left(\begin{matrix} 1 \\ 1\end{matrix}\right) u, \\ y &= \left(\begin{matrix} 2 & -1 \end{matrix}\right) \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right). \end{align*}

It can be verified that it indeed gives the same transfer function as Controllable and Observable Canonical Forms do.

\begin{align*} C(Is-A)^{-1}B &= \left(\begin{matrix} 2 & -1 \end{matrix}\right) \left(\begin{matrix} s+3 & 0 \\0 &s+{2}\end{matrix}\right)^{-1} \left(\begin{matrix} 1 \\ 1\end{matrix}\right) \\ &= \left(\begin{matrix} 2 & -1 \end{matrix}\right) \left(\begin{matrix} \frac{1}{s+3} & 0 \\0 &\frac{1}{s+2}\end{matrix}\right) \left(\begin{matrix} 1 \\ 1\end{matrix}\right) \\ &= \left(\begin{matrix} 2 & -1 \end{matrix}\right) \left( \begin{matrix} \frac{1}{s+3} \\ \frac{1}{s+2}\end{matrix}\right) \\ &= \frac{2}{s+3}-\frac{1}{s+2}. \end{align*}

Upshot: From the above discussion, we notice

• A given transfer function $G(s)$ can be realized using infinitely many state-space models.
• Certain properties make some realizations preferable to others, e.g., controllable, observable, modal etc.

## Controllability Matrix

Consider a single-input system with $u \in \RR$, $x \in \RR^n$.

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

The Controllability Matrix is defined as

\begin{align*} {\cal C}(A,B) = \left[ B \, | \, AB \, | \, A^2 B \, | \, \ldots \, | \, A^{n-1}B \right]. \end{align*}

Recall that $A$ is $n \times n$ and $B$ is $n \times 1$, so ${\cal C}(A,B)$ is $n \times n$. The controllability matrix only involves $A$ and $B$, not $C$.

We say that the above system is completely controllable if its controllability matrix ${\cal C}(A,B)$ is invertible.

This definition is only true for the single-input case. It can be generalized to the multiple-input case using the rank of controllability matrix ${\cal C}(A,B)$.

• As we will see later, if the system is completely controllable, then we may assign arbitrary closed loop poles by state feedback controller of the form $u = -Kx$.
• Whether or not the system is controllable depends on its state-space realization.

Example 2: Computing the controllability matrix ${\cal C}(A,B)$ of the state-space model given in Example 1.

\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} 1 & 1 \end{matrix}\right)}_{C} \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right). \end{align*}

Solution: Here $x \in \RR^2$ and $A \in \RR^{2 \times 2}$ $\implies$ ${\cal C}(A,B) \in \RR^{2 \times 2}$.

\begin{align*} {\cal C}(A,B) &= [B\, | \, AB], \\ \text{ where }\, AB &= \left(\begin{matrix} 0 & 1 \\ -{6} &-{5}\end{matrix}\right)\left(\begin{matrix} 0 \\ 1\end{matrix}\right) = \left(\begin{matrix}1 \\ -5\end{matrix}\right) \\ \implies {\cal C}(A,B) &=\left( \begin{matrix} 0 & 1 \\ 1 &-5\end{matrix}\right). \end{align*}

Check system controllability by testing the invertibility of $\mathcal{C}(A, B)$,

\begin{align*} \det {\cal C} = -1 \neq 0 \implies \text{system is completely controllable.} \end{align*}

### Controllable Canonical Form

A single-input state-space model

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

is said to be in Controller Canonical Form (CCF) if the matrices $A,B$ are of the form

\begin{align*} A &= \left( \begin{matrix} 0 & 1 & 0 & \ldots & 0 & 0 \\ 0 & 0 & 1 & \ldots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & 0 & 1 \\ * & * & * & \ldots & * & * \end{matrix} \right), \, B = \left( \begin{matrix} 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{matrix}\right) \end{align*}

Note that a state-space model in CCF is always completely controllable.

The proof of this for $n > 2$ uses the Jordan canonical form, we shall not worry about this. Interested readers can follow the underlined link.

### CCF with Arbitrary Zeros

In our example, we had $G(s) = \dfrac{s+1}{s^2 + 5s + 6}$ with a minimum-phase zero at $z=-1$.

Let’s consider a general zero location $s=z$,

\begin{align*} G(s) = \frac{s-z}{s^2 + 5s+6}. \end{align*}

This gives us a CCF realization

\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*}

Since $A,B$ are the same as before as in Example 2, ${\cal C}(A,B)$ is the same $\implies$ the system is still completely controllable. (Zeros only enter $C$ matrix which is not involved in controllability matrix.)

Upshot: A system state-space model in CCF is controllable for any locations of the zeros.

Last updated: 2018-03-22 Thu 17:20