Controllability and Observability in Control Systems

Introduction to Controllability and Observability

• The concepts of controllability and observability are fundamental to modern control system theory, having been introduced by R.E. Kalman in 1960.

• While originally theoretical, these concepts are particularly important for practical implementations in engineering.

• Controllability: This property determines if it is possible to manipulate the dynamic system to achieve a desired behavior under a specific control input.

• Observability: This property determines if the internal state of a system can be identified by observing its external outputs.

Definitions and Core Principles

• Controllability Definition: A system is said to be controllable if an input function $u(t)$ can transform the initial state $x(t_0)$ of the system to any required final state $x(t_f)$ within a finite time interval $(t_f - t_0), t > 0$.

• Observability Definition: A system is said to be observable if, from a finite measurement of output carried over an interval of time $(0 \le t \le t_f)$, the initial state of the system can be determined.

• Uncontrollability: If there is no connection between a specific state and the input, the system is called uncontrollable.

• Unobservability: If there is no contact/connection between a certain state and the output, the system is deemed unobservable.

Observability of Discrete-Time Systems

Consider a linear, time-invariant, discrete-time system in state-space form:

$x(k+1) = A_d x(k)$ $x(0) = x_0$ (unknown)

With output measurement:

$y(k) = C_d x(k)$

Where:

• $x(k) \in \mathbb{R}^n$

• $y(k) \in \mathbb{R}^p$

• $A_d$ and $C_d$ are constant matrices of appropriate dimensions.

Since the $n$-dimensional vector $x(0)$ has $n$ unknown components, it is expected that $n$ measurements are sufficient to determine it. By taking $k = 0, 1, \dots, n-1$, a sequence of equations is generated:

$y(0) = C_d x(0)$ $y(1) = C_d x(1) = C_d A_d x(0)$ $y(2) = C_d x(2) = C_d A_d^2 x(0)$ $\dots$ $y(n-1) = C_d x(n-1) = C_d A_d^{n-1} x(0)$

In matrix form, this is represented as:

$\begin{pmatrix} y(0) \\ y(1) \\ y(2) \\ \dots \\ y(n-1) \end{pmatrix} = \begin{pmatrix} C_d \\ C_d A_d \\ C_d A_d^2 \\ \dots \\ C_d A_d^{n-1} \end{pmatrix} x(0)$

From linear algebra, a unique solution for the unknown $x(0)$ exists if and only if the system matrix has rank $n$. This leads to the definition of the observability matrix $Q_o$:

$Q_o = [C_d^T, (C_d A_d)^T, (C_d A_d^2)^T, \dots, (C_d A_d^{n-1})^T]^T$

Theorem: The linear discrete-time system is observable if and only if the observability matrix $Q_o$ has rank equal to $n$.

Controllability of Discrete-Time Systems

Consider a linear discrete-time invariant control system:

$x(k+1) = A_d x(k) + B_d u(k)$ $x(0) = x_0$

Assuming the input $u(k)$ is a scalar, the input matrix $B_d$ is a vector denoted by $b_d$. Taking $k = 0, 1, 2, \dots$:

$x(1) = A_d x(0) + b_d u(0)$ $x(2) = A_d x(1) + b_d u(1) = A_d^2 x(0) + A_d b_d u(0) + b_d u(1)$ $\dots$ $x(n) = A_d^n x(0) + A_d^{n-1} b_d u(0) + \dots + b_d u(n-1)$

Reorganized into matrix form:

$x(n) - A_d^n x(0) = [b_d, A_d b_d, \dots, A_d^{n-1} b_d] \begin{pmatrix} u(n-1) \\ u(n-2) \\ \dots \\ u(0) \end{pmatrix}$

The term $[b_d, A_d b_d, \dots, A_d^{n-1} b_d]$ is the controllability matrix, denoted as $Q_c$. For the general vector input case, the matrix is defined as:

$Q_c = [B_d, A_d B_d, \dots, A_d^{n-1} B_d]$

Theorem: The linear discrete-time system is controllable if and only if the rank of $Q_c = n$.

Controllability and Observability of Continuous Systems

For continuous systems defined by the state-space model:

$\dot{x} = Ax + Bu$ $y = Cx + Du$

Where:

• $x$ = State vector

• $u$ = Input vector

• $y$ = Output vector

• $A, B, C, D$ = System matrices

Controllability Matrix ($Q_c$): $Q_c = [B, AB, A^2 B, \dots, A^{n-1} B]$

Observability Matrix ($Q_o$): $Q_o = [C^T, A^T C^T, (A^T)^2 C^T, \dots, (A^T)^{n-1} C^T]^T$

Theorems:

1. The linear continuous-time system is controllable if and only if $rank(Q_c) = n$.

2. The linear continuous-time system is observable if and only if $rank(Q_o) = n$.

Kalman's Tests

Controllability Test

• Step 1: Construct the matrix $Q_c = [B, AB, A^2 B, \dots, A^{n-1} B]$.

• Step 2: Find the determinant of $Q_c$. If $\det(Q_c) \neq 0$, the control system is controllable (system is non-singular).

Observability Test

• Step 1: Construct the matrix $Q_o = [C^T, A^T C^T, \dots, (A^T)^{n-1} C^T]^T$.

• Step 2: Find the determinant of $Q_o$. If $\det(Q_o) \neq 0$, the system is observable.

Frequency Domain Perspective

In the frequency domain, controllability and observability are related to the transfer function $H(s)$ for a single-input single-output (SISO) system:

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

Theorem:

• If there are no pole-zero cancellations in the transfer function $H(s)$, then the system is both controllable and observable.

• If a pole-zero cancellation occurs in $H(s)$, then the system is either uncontrollable, unobservable, or both.

Worked Examples

Example 1: Discrete-Time System

Given: $x(k+1) = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} x(k)$ $y(k) = [1, 2] x(k)$ $B_d = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$

• Observability Check:

  • $n = 2$.

  • $C_d = [1, 2]$.

  • $C_d A_d = [1, 2] \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = [7, 10]$.

  • $Q_o = \begin{pmatrix} 1 & 2 \\ 7 & 10 \end{pmatrix}$.

  • $\det(Q_o) = (1 \times 10) - (2 \times 7) = 10 - 14 = -4$.

  • Since $\det(Q_o) \neq 0$, the system is observable.

• Controllability Check:

  • $B_d = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$.

  • $A_d B_d = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$.

  • $Q_c = \begin{pmatrix} 1 & 1 \\ 0 & 3 \end{pmatrix}$.

  • $\det(Q_c) = 3 - 0 = 3$. (Note: A separate calculation in notes suggested $B_d = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$ leading to uncontrollability; however, using the provided $B_d = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ yields rank 2).

Example 2: Continuous-Time System

Given: $\dot{x} = \begin{pmatrix} 3 & 2 \\ -1 & 1 \end{pmatrix} x + \begin{pmatrix} 1 \\ 2 \end{pmatrix} u$ $y = [1, 1] x$

• Controllability:

  • $B = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$.

  • $AB = \begin{pmatrix} 3 & 2 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \end{pmatrix}$.

  • $Q_c = \begin{pmatrix} 1 & 7 \\ 2 & 1 \end{pmatrix}$.

  • $\det(Q_c) = 1 - 14 = -13 \neq 0$. System is controllable.

• Observability:

  • $C = [1, 1]$ implies $C^T = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

  • $A^T C^T = \begin{pmatrix} 3 & -1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$.

  • $Q_o = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}$.

  • $\det(Q_o) = 3 - 2 = 1 \neq 0$. System is observable.

Example 3: 3rd Order System

Given: $A = \begin{pmatrix} -3 & 1 & 0 \\ 2 & -3 & 2 \\ 0 & 1 & -3 \end{pmatrix}, B = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, C = [0, 1, 0]$

• Controllability:

  • $B = \begin{pmatrix} 0 & 1 & 0 \end{pmatrix}^T$.

  • $AB = \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix}$.

  • $A^2 B = A(AB) = \begin{pmatrix} -3 & 1 & 0 \\ 2 & -3 & 2 \\ 0 & 1 & -3 \end{pmatrix} \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix} = \begin{pmatrix} -6 \\ 13 \\ -6 \end{pmatrix}$.

  • $Q_c = \begin{pmatrix} 0 & 1 & -6 \\ 1 & -3 & 13 \\ 0 & 1 & -6 \end{pmatrix}$.

  • $\det(Q_c) = 0$. (Two columns/rows are identical). The system is uncontrollable.

• Observability:

  • $C^T = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$.

  • $A^T C^T = \begin{pmatrix} 2 \\ -3 \\ 2 \end{pmatrix}$.

  • $(A^T)^2 C^T = \begin{pmatrix} -12 \\ 13 \\ -12 \end{pmatrix}$.

  • $Q_o = \begin{pmatrix} 0 & 1 & 0 \\ 2 & -3 & 2 \\ -12 & 13 & -12 \end{pmatrix}$.

  • $\det(Q_o) = -1( -24 - (-24)) = 0$. The system is unobservable.

Example 4: Transfer Function to State-Space

Given: $H(s) = \frac{2}{s^3 + 6s^2 + 11s + 6}$

• Equation derivation:

  • $(s^3 + 6s^2 + 11s + 6)y(s) = 2u(s)$.

  • Inverse Laplace: $\dddot{y}(t) + 6\ddot{y}(t) + 11\dot{y}(t) + 6y(t) = 2u(t)$.

  • Let $x_1 = y, x_2 = \dot{y}, x_3 = \ddot{y}$.

  • $\dot{x}_1 = x_2$

  • $\dot{x}_2 = x_3$

  • $\dot{x}_3 = -6x_1 - 11x_2 - 6x_3 + 2u$

• State Matrices:

  • $A = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -6 & -11 & -6 \end{pmatrix}, B = \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix}, C = [1, 0, 0]$.

• Verification:

  • $Q_c = \begin{pmatrix} 0 & 0 & 2 \\ 0 & 2 & -12 \\ 2 & -12 & 50 \end{pmatrix}$.

  • $\det(Q_c) = 8 \neq 0$. System is fully controllable.

  • $Q_o = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$.

  • $\det(Q_o) = 1 \neq 0$. System is fully observable.

Verification Problems

1. Verify the controllability of a system represented by:     $A = \begin{pmatrix} 2 & -4 \\ 1 & -1 \end{pmatrix}, B = \begin{pmatrix} 3 \\ 1 \end{pmatrix}, C = [1, -1]$.

2. Determine if the system is completely controllable and observable given:     $A = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix}, B = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, C = [2, 1]$.

3. Find controllability for: $A = \begin{pmatrix} 1 & 4 \\ -5 & 10 \end{pmatrix}, B = \begin{pmatrix} 1 \\ 2 \end{pmatrix}, C = [1, 2]$.

4. Test controllability and observability for the system: $H(s) = \frac{1}{s^2 + 3s + 2}$.