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)u(t) can transform the initial state x(t0)x(t_0) of the system to any required final state x(tf)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 (0ttf)(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)=Adx(k)x(k+1) = A_d x(k) x(0)=x0x(0) = x_0 (unknown)

With output measurement:

y(k)=Cdx(k)y(k) = C_d x(k)

Where:

  • x(k)Rnx(k) \in \mathbb{R}^n

  • y(k)Rpy(k) \in \mathbb{R}^p

  • AdA_d and CdC_d are constant matrices of appropriate dimensions.

Since the nn-dimensional vector x(0)x(0) has nn unknown components, it is expected that nn measurements are sufficient to determine it. By taking k=0,1,,n1k = 0, 1, \dots, n-1, a sequence of equations is generated:

y(0)=Cdx(0)y(0) = C_d x(0) y(1)=Cdx(1)=CdAdx(0)y(1) = C_d x(1) = C_d A_d x(0) y(2)=Cdx(2)=CdAd2x(0)y(2) = C_d x(2) = C_d A_d^2 x(0) \dots y(n1)=Cdx(n1)=CdAdn1x(0)y(n-1) = C_d x(n-1) = C_d A_d^{n-1} x(0)

In matrix form, this is represented as:

(y(0)y(1)y(2)y(n1))=(CdCdAdCdAd2CdAdn1)x(0)\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)x(0) exists if and only if the system matrix has rank nn. This leads to the definition of the observability matrix QoQ_o:

Qo=[CdT,(CdAd)T,(CdAd2)T,,(CdAdn1)T]TQ_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 QoQ_o has rank equal to nn.

Controllability of Discrete-Time Systems

Consider a linear discrete-time invariant control system:

x(k+1)=Adx(k)+Bdu(k)x(k+1) = A_d x(k) + B_d u(k) x(0)=x0x(0) = x_0

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

x(1)=Adx(0)+bdu(0)x(1) = A_d x(0) + b_d u(0) x(2)=Adx(1)+bdu(1)=Ad2x(0)+Adbdu(0)+bdu(1)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)=Adnx(0)+Adn1bdu(0)++bdu(n1)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)Adnx(0)=[bd,Adbd,,Adn1bd](u(n1)u(n2)u(0))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 [bd,Adbd,,Adn1bd][b_d, A_d b_d, \dots, A_d^{n-1} b_d] is the controllability matrix, denoted as QcQ_c. For the general vector input case, the matrix is defined as:

Qc=[Bd,AdBd,,Adn1Bd]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 Qc=nQ_c = n.

Controllability and Observability of Continuous Systems

For continuous systems defined by the state-space model:

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

Where:

  • xx = State vector

  • uu = Input vector

  • yy = Output vector

  • A,B,C,DA, B, C, D = System matrices

Controllability Matrix (QcQ_c): Qc=[B,AB,A2B,,An1B]Q_c = [B, AB, A^2 B, \dots, A^{n-1} B]

Observability Matrix (QoQ_o): Qo=[CT,ATCT,(AT)2CT,,(AT)n1CT]TQ_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(Qc)=nrank(Q_c) = n.

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

Kalman's Tests

Controllability Test
  • Step 1: Construct the matrix Qc=[B,AB,A2B,,An1B]Q_c = [B, AB, A^2 B, \dots, A^{n-1} B].

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

Observability Test
  • Step 1: Construct the matrix Qo=[CT,ATCT,,(AT)n1CT]TQ_o = [C^T, A^T C^T, \dots, (A^T)^{n-1} C^T]^T.

  • Step 2: Find the determinant of QoQ_o. If det(Qo)0\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)H(s) for a single-input single-output (SISO) system:

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

Theorem:

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

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

Worked Examples

Example 1: Discrete-Time System

Given: x(k+1)=(1amp;23amp;4)x(k)x(k+1) = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} x(k) y(k)=[1,2]x(k)y(k) = [1, 2] x(k) Bd=(10)B_d = \begin{pmatrix} 1 \\ 0 \end{pmatrix}

  • Observability Check:

    • n=2n = 2.

    • Cd=[1,2]C_d = [1, 2].

    • CdAd=[1,2](1amp;23amp;4)=[7,10]C_d A_d = [1, 2] \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = [7, 10].

    • Qo=(1amp;27amp;10)Q_o = \begin{pmatrix} 1 & 2 \\ 7 & 10 \end{pmatrix}.

    • det(Qo)=(1×10)(2×7)=1014=4\det(Q_o) = (1 \times 10) - (2 \times 7) = 10 - 14 = -4.

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

  • Controllability Check:

    • Bd=(10)B_d = \begin{pmatrix} 1 \\ 0 \end{pmatrix}.

    • AdBd=(1amp;23amp;4)(10)=(13)A_d B_d = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}.

    • Qc=(1amp;10amp;3)Q_c = \begin{pmatrix} 1 & 1 \\ 0 & 3 \end{pmatrix}.

    • det(Qc)=30=3\det(Q_c) = 3 - 0 = 3. (Note: A separate calculation in notes suggested Bd=(00)B_d = \begin{pmatrix} 0 \\ 0 \end{pmatrix} leading to uncontrollability; however, using the provided Bd=(10)B_d = \begin{pmatrix} 1 \\ 0 \end{pmatrix} yields rank 2).

Example 2: Continuous-Time System

Given: x˙=(3amp;21amp;1)x+(12)u\dot{x} = \begin{pmatrix} 3 & 2 \\ -1 & 1 \end{pmatrix} x + \begin{pmatrix} 1 \\ 2 \end{pmatrix} u y=[1,1]xy = [1, 1] x

  • Controllability:

    • B=(12)B = \begin{pmatrix} 1 \\ 2 \end{pmatrix}.

    • AB=(3amp;21amp;1)(12)=(71)AB = \begin{pmatrix} 3 & 2 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \end{pmatrix}.

    • Qc=(1amp;72amp;1)Q_c = \begin{pmatrix} 1 & 7 \\ 2 & 1 \end{pmatrix}.

    • det(Qc)=114=130\det(Q_c) = 1 - 14 = -13 \neq 0. System is controllable.

  • Observability:

    • C=[1,1]C = [1, 1] implies CT=(11)C^T = \begin{pmatrix} 1 \\ 1 \end{pmatrix}.

    • ATCT=(3amp;12amp;1)(11)=(23)A^T C^T = \begin{pmatrix} 3 & -1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}.

    • Qo=(1amp;12amp;3)Q_o = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}.

    • det(Qo)=32=10\det(Q_o) = 3 - 2 = 1 \neq 0. System is observable.

Example 3: 3rd Order System

Given: A=(3amp;1amp;02amp;3amp;20amp;1amp;3),B=(010),C=[0,1,0]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=(0amp;1amp;0)TB = \begin{pmatrix} 0 & 1 & 0 \end{pmatrix}^T.

    • AB=(131)AB = \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix}.

    • A2B=A(AB)=(3amp;1amp;02amp;3amp;20amp;1amp;3)(131)=(6136)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}.

    • Qc=(0amp;1amp;61amp;3amp;130amp;1amp;6)Q_c = \begin{pmatrix} 0 & 1 & -6 \\ 1 & -3 & 13 \\ 0 & 1 & -6 \end{pmatrix}.

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

  • Observability:

    • CT=(010)C^T = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}.

    • ATCT=(232)A^T C^T = \begin{pmatrix} 2 \\ -3 \\ 2 \end{pmatrix}.

    • (AT)2CT=(121312)(A^T)^2 C^T = \begin{pmatrix} -12 \\ 13 \\ -12 \end{pmatrix}.

    • Qo=(0amp;1amp;02amp;3amp;212amp;13amp;12)Q_o = \begin{pmatrix} 0 & 1 & 0 \\ 2 & -3 & 2 \\ -12 & 13 & -12 \end{pmatrix}.

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

Example 4: Transfer Function to State-Space

Given: H(s)=2s3+6s2+11s+6H(s) = \frac{2}{s^3 + 6s^2 + 11s + 6}

  • Equation derivation:

    • (s3+6s2+11s+6)y(s)=2u(s)(s^3 + 6s^2 + 11s + 6)y(s) = 2u(s).

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

    • Let x1=y,x2=y˙,x3=y¨x_1 = y, x_2 = \dot{y}, x_3 = \ddot{y}.

    • x˙1=x2\dot{x}_1 = x_2

    • x˙2=x3\dot{x}_2 = x_3

    • x˙3=6x111x26x3+2u\dot{x}_3 = -6x_1 - 11x_2 - 6x_3 + 2u

  • State Matrices:

    • A=(0amp;1amp;00amp;0amp;16amp;11amp;6),B=(002),C=[1,0,0]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:

    • Qc=(0amp;0amp;20amp;2amp;122amp;12amp;50)Q_c = \begin{pmatrix} 0 & 0 & 2 \\ 0 & 2 & -12 \\ 2 & -12 & 50 \end{pmatrix}.

    • det(Qc)=80\det(Q_c) = 8 \neq 0. System is fully controllable.

    • Qo=(1amp;0amp;00amp;1amp;00amp;0amp;1)Q_o = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}.

    • det(Qo)=10\det(Q_o) = 1 \neq 0. System is fully observable.

Verification Problems

  1. Verify the controllability of a system represented by:     A=(2amp;41amp;1),B=(31),C=[1,1]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=(0amp;12amp;3),B=(01),C=[2,1]A = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix}, B = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, C = [2, 1].

  3. Find controllability for: A=(1amp;45amp;10),B=(12),C=[1,2]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)=1s2+3s+2H(s) = \frac{1}{s^2 + 3s + 2}.