Realizability and State Space Representation

Realizations: Controllable, Observable and Modal Forms

So far in the course, we considered the input-output approach to study control systems, which in the convolution (linear time-invariant) system setup, resulted in frequency domain methods through a transfer function analysis (such as in arriving at the root locus / Nyquist stability / Bode plot methods). These are often referred to as classical control design methods.

In this chapter, we will introduce state-space based methods. Control design based on state-space methods is called modern control design.

The notion of a state. Suppose that, given $t \in \mathbb{R}$ (or $\mathbb{Z}$), we wish to compute the output of a system at $t \geq t_0$. In a general causal system, we may need to use all the past applied input terms $u(s)$, $s \leq t$ and/or all the past output values $y(s)$, $s < t$ to compute the output at $t$. The state of a system summarizes all the past relevant data that is sufficient to compute the future paths in the sense that if the state at $t_0$, $x(t_0)$ is given, then to compute $y(t)$, one would only need to use $\{u(s) \; s \in [t_0,t]\}$, $\{y(s) \; s \in [t_0,t)\}$ and $x(t_0)$. In particular, the past $\{y(s), u_s; \; s < t_0\}$ would not be needed.

Some systems admit a finite-dimensional state representation, some do not.

Consider a linear system in state-space form:

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

We will say that such a system is given by the 4-tuple: $(A, B, C, D)$.

In the above $x$ serves as the state variable for the system.

We know that the solution to this system is given with

$x(t) = e^{At}x(0) + \int_0^t e^{A(t-s)}Bu(s)ds$

and

$y(t) = Ce^{At}x(0) + C\int_0^t e^{A(t-s)}Bu(s)ds + Du(t)$

Taking the (one-sided) Laplace transform, we obtain for $s$ in the ROC:

$Y_+(s) = C(sI - A)^{-1}(x(0) + BU_+(s)) + DU_+(s)$

Assuming $x(0) = 0$, we have the following as the transfer function:

$Y_+(s) = (C(sI - A)^{-1}B + D)U(s)$

A transfer function $H(s)$ is state-space realizable if there exists finite dimensional $(A, B, C, D)$ so that we can write for $s \in \mathbb{C}$, whenever well-defined:

$H(s) = C(sI - A)^{-1}B + D$

Controllable canonical realization

Consider a continuous-time system given by:

$\sum_{k=0}^{N} a_k \frac{d^k}{dt^k}y(t) = \sum_{m=0}^{N} b_m \frac{d^m}{dt^m}u(t),$

with $a_N = 1$. Taking the Laplace transform, we know that the transfer function writes as

$H(s) = \frac{\sum_{m=0}^{N} b_m s^m}{\sum_{k=0}^{N} a_k s^k}$

Suppose that the system is strictly proper. Such a system can be realized with the form:

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

with

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

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

$C_c = \begin{bmatrix} b_0 & b_1 & \cdots & b_{N-1} \end{bmatrix}$

If the system is proper, but not strictly proper, then we will also have $D_c = d$, where $d$ is the remainder term in the partial fraction expansion,

$H(s) = d + \sum_{i=1}^{K}\left(\sum_{k=1}^{m_i}\frac{A_{ik}}{(s - p_i)^k}\right)$

where $p_i$ are the roots of $s^n + \sum_{k=0}^{N-1}a_k s^k$ and $m_i$ is the multiplicity of $p_i$.

This is called the controllable canonical form. The key insight is that the characteristic polynomial of $A_c$ is exactly $s^N + a_{N-1}s^{N-1} + \cdots + a_1 s + a_0$, matching the denominator of the transfer function. The companion structure of $A_c$ ensures that the single input $u$ can influence all state components through successive integrations.

Observable canonical realization

Consider the same system. This system can also be realized as

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

with

$A = \begin{bmatrix} -a_{N-1} & 1 & 0 & \cdots & 0 \\ -a_{N-2} & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & & \ddots & \vdots \\ -a_1 & 0 & 0 & \cdots & 1 \\ -a_0 & 0 & 0 & \cdots & 0 \end{bmatrix}$

$B = \begin{bmatrix} b_{N-1} \\ b_{N-2} \\ \vdots \\ b_0 \end{bmatrix}$

$C = \begin{bmatrix} 1 & 0 & \cdots & 0 \end{bmatrix}$

If we reverse the order of the coordinates of $x$, we arrive at the standard observable canonical form:

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

with

$A_o = \begin{bmatrix} 0 & 0 & 0 & \cdots & -a_0 \\ 1 & 0 & 0 & \cdots & -a_1 \\ \vdots & \ddots & & & \vdots \\ 0 & 0 & 0 & \cdots & -a_{N-2} \\ 0 & 0 & 0 & \cdots & -a_{N-1} \end{bmatrix}$

$B_o = \begin{bmatrix} b_0 \\ b_1 \\ \vdots \\ b_{N-1} \end{bmatrix}$

$C_o = \begin{bmatrix} 0 & 0 & \cdots & 1 \end{bmatrix}$

Observe that $A_o = A_c^T$, $B_o = C_c^T$, $C_o = B_c^T$. This is the standard observable canonical form.

The duality $A_o = A_c^T$ between controllable and observable forms is a fundamental structural property. It reflects the fact that observability and controllability are dual concepts -- a connection that will be made precise in Chapter 12.

Exercise. Show that the transfer functions under the controllable and observable realization canonical forms are equivalent directly by comparing $C_c(sI - A_c)^{-1}B_c$ and $C_o(sI - A_o)^{-1}B_o$.

Modal realization

Consider a partial fraction expansion with only simple poles:

$H(s) = \frac{\sum_{m=0}^{N} b_m s^m}{\sum_{k=0}^{N} a_k s^s} = d + \sum_{i}^{N} \frac{k_i}{s - p_i}$

In this case, we can realize the system as the sum of decoupled modes:

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

with

$A = \begin{bmatrix} p_1 & 0 & 0 & \cdots & 0 \\ 0 & p_2 & 0 & \cdots & \vdots \\ \vdots & \vdots & & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & p_N \end{bmatrix}$

$B = \begin{bmatrix} k_1 \\ k_2 \\ \vdots \\ k_N \end{bmatrix}$

$C = \begin{bmatrix} 1 & 1 & \cdots & 1 \end{bmatrix}$

If in the partial fraction expansion is more general (with repeated poles), then the corresponding structure can be realized also: this will lead to a Jordan form for the matrix $A$, since, e.g.,

$\frac{1}{(s - p_i)^2} = \frac{1}{s - p_i}\frac{1}{s - p_i}$

will define a serial connection of two modal blocks; the first one with $x_2' = p_i x_2 + u$ and the second one $x_1' = p_i x_1 + x_2$.

The modal realization makes the system poles (eigenvalues of $A$) directly visible as diagonal entries. Each mode evolves independently, making it easy to see which modes are fast, slow, stable, or unstable.

Discrete-time setup. The above also apply to the discrete-time setup. For example, a discrete-time system of the form

$\sum_{k=0}^{N} a_k y(n-k) = \sum_{m=1}^{N} b_m u(n-m)$

with $a_0 = 1$, can be written in the controllable canonical form

$x(n+1) = Ax(n) + Bu(n), \qquad y_t = Cx_t$

where $x_N(n) = y(n), x_{N-1}(n) = y(n-1), \;\cdots\;, x_1(n) = y(n-(N-1))$

$A = \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & & \ddots & 1 \\ -a_N & -a_{N-1} & -a_{N-2} & \cdots & -a_1 \end{bmatrix}$

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

$C = \begin{bmatrix} b_N & b_{N-1} & \cdots & b_1 \end{bmatrix}$

Observable and modal canonical forms follow similarly.

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Zero-State Equivalence and Algebraic Equivalence

We say that two systems $(A, B, C, D)$ and $(\tilde{A}, \tilde{B}, \tilde{C}, \tilde{D})$ are zero-state equivalent if the induced transfer functions are equal, that is

$C(sI - A)^{-1}B + D = \tilde{C}(sI - \tilde{A})^{-1}\tilde{B} + \tilde{D}$

The quantities $CA^mB$ are called the Markov parameters of the system. Two systems are zero-state equivalent precisely when they share the same Markov parameters and the same feedthrough term $D$.

There is an alternative notion, called algebraic equivalence: Let $P$ be invertible and let us define a transformation through $\tilde{x} = Px$. Then, we can write the state-space model as

$\frac{d\tilde{x}}{dt} = \tilde{A}\tilde{x}(t) + \tilde{B}u(t), \qquad \tilde{y}(t) = \tilde{C}\tilde{x}(t) + \tilde{D}u(t),$

with $\tilde{A} = PAP^{-1}$, $\tilde{B} = PB$, $\tilde{C} = CP^{-1}$, $\tilde{D} = D$. In this case, we say that $(A, B, C, D)$ and $(\tilde{A}, \tilde{B}, \tilde{C}, \tilde{D})$ are algebraically equivalent.

Algebraic equivalence is a stronger condition: it says the two systems are really the same system viewed in different coordinates. Zero-state equivalence only says the input-output maps agree, but the internal structure may differ.

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Discretization

Consider

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

Suppose we apply piece-wise constant control actions $u$ which are varied only at the discrete time instances given with $\{t : t = kT, k \in \mathbb{Z}_+\}$ so that $u(t) = u(kT)$ for $t \in [kT, (k+1)T)$.

We write

$x((k+1)T) = e^{AT}x(kT) + \left(\int_{kT}^{(k+1)T} e^{A((k+1)T - s)}Bu(s)ds\right)$

Writing $\tau = (k+1)T - s$ with $d\tau = -ds$, we arrive at

$x((k+1)T) = e^{AT}x(kT) + \left(\int_0^{T} e^{A\tau}Bd\tau\right)u(kT)$

With $x_k := x(kT)$ and $u_k := u(kT)$, we arrive at

$x_{k+1} = A_d x_k + B_d u_k$

where

$A_d = e^{AT}$

$B_d = \int_0^{T} e^{A\tau}Bd\tau$

If $A$ is invertible, the integration of $\int e^{A\tau}d\tau$ leads to $A^{-1}(e^{AT} - I)B$.

Discretization converts a continuous-time state-space model into an equivalent discrete-time model that is exact at the sampling instants. The key point is that no approximation is involved -- the discrete-time system exactly reproduces the continuous-time state at times $t = kT$, provided the input is held constant between samples (zero-order hold).

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Exercises