Observer and state-space model

  • 10 avr. 2024
  • 22 nov. 2024
  • 4 min de lecture
  • llms.txt

Problemè 1

Consider the following state-space model:

x˙=[010001−5−60]x+[001]uy=[100]x\begin{aligned} \dot{x} &= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -5 & -6 & 0 \\ \end{bmatrix} x + \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix} u \\ y &= \begin{bmatrix} 1 & 0 & 0 \\ \end{bmatrix} x \end{aligned}

Design an observer to place the observer poles at -10, -10, -15

Solution

The characteristic equation of the observer is given by:

det(sI−(A−LC))=(s+10)(s+10)(s+15)=s3+35s2+350s+1500det(sI - (A - LC)) = (s + 10)(s + 10)(s + 15) = s^3 + 35s^2 + 350s + 1500

From the coefficients of the characteristic equation we get

det(sI−(A−LC))=s3+(l1−6)s2+(l2−5−6l1)s+(l3−5l2)det(sI - (A-LC)) = s^3 + (l_1-6)s^2 + (l_2-5-6l_1)s + (l_3-5l_2)

Solving for the coefficients we get the observer gain matrix:

L=[450560141]L = \begin{bmatrix} 4505 \\ 601 \\ 41 \end{bmatrix}

Thus the observer dynamics are given by:

x^˙=[010001−5−60]x^+[001]u+[450560141](y−y^)\dot{\hat{x}} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -5 & -6 & 0 \\ \end{bmatrix} \hat{x} + \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix} u + \begin{bmatrix} 4505 \\ 601 \\ 41 \\ \end{bmatrix} (y - \hat{y})

Problemè 2

Given the plant

x˙=[−1102]x+[01]uy=[11]x\begin{aligned} \dot{x} &= \begin{bmatrix} -1 & 1 \\ 0 & 2 \\ \end{bmatrix} x + \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} u \\ y &= \begin{bmatrix} 1 & 1 \\ \end{bmatrix} x \end{aligned}

Design an integral controller to yield a 10% overshoot, 0.5 second settling time and zero steady-state error for a step input.

Solution

The code for the integral controller is given by p2.py.

Add an integrator to the plant to ensure zero steady-state error for a step input. The augmented state-space model becomes:

x˙a=[−110020−1−10]xa+[010]u\dot{x}_a = \begin{bmatrix} -1 & 1 & 0 \\ 0 & 2 & 0 \\ -1 & -1 & 0 \\ \end{bmatrix} x_a + \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix} u y=[110]xay = \begin{bmatrix} 1 & 1 & 0 \\ \end{bmatrix} x_a

where

x_a = \begin{bmatrix} x \\ \int e \, dt \\ \end{bmatrix} $$ and $e = r - y$ is the tracking error. Then, design the state feedback gains

K = \begin{bmatrix} k_1 & k_2 & k_i \ \end{bmatrix}

suchthattheclosed−loopsystemmeetsthetransientresponsespecifications.Thecharacteristicequationoftheclosed−loopsystemis:such that the closed-loop system meets the transient response specifications. The characteristic equation of the closed-loop system is:

\left| sI - (A_a - B_aK) \right| = 0

Expandingthisyields: Expanding this yields:

(s + k_1)(s^2 + (1 - k_2)s + k_i) = 0

Thecontrollawthengivenby: The control law then given by:

u = -Kx_a = -k_1x_1 - k_2x_2 - k_i\int{e , dt}

Thecodeyields:‘‘‘prologzeta:0.5911550337988974omegan:13.53282902556064Desiredpoles:[−8.+10.91501083j−8.−10.91501083j−135.32829026+0.j]Plantmodel:<LinearIOSystem>:sys[2]Inputs(1):[′u[0]′]Outputs(1):[′y[0]′]States(2):[′x[0]′,′x[1]′]A=[[−1.1.][0.2.]]B=[[0.][1.]]C=[[1.1.]]D=[[0.]]Augmentedplantmodel:<LinearIOSystem>:sys[3]Inputs(1):[′u[0]′]Outputs(1):[′y[0]′]States(3):[′x[0]′,′x[1]′,′x[2]′]A=[[−1.1.0.][0.2.0.][−1.−1.0.]]B=[[0.][1.][0.]]C=[[1.1.0.]]D=[[0.]]Statefeedbackgains:K=[[−10193.77795361152.32829026−12391.83976888]]Integralcontrollertransferfunction:−1.239e+04−−−−−−−−−−sOpen−looptransferfunction:<LinearICSystem>:sys[6]Inputs(1):[′u[0]′]Outputs(1):[′y[0]′]States(3):[′sys[4]x[0]′,′sys[2]x[0]′,′sys[2]x[1]′]A=[[−0.00000000e+000.00000000e+000.00000000e+00][0.00000000e+00−1.00000000e+001.00000000e+00][−1.23918398e+040.00000000e+002.00000000e+00]]B=[[1.][0.][0.]]C=[[0.1.1.]]D=[[0.]]Closed−looptransferfunction:<LinearICSystem>:sys[9]Inputs(1):[′u[0]′]Outputs(1):[′y[0]′]States(3):[′sys[6]sys[4]x[0]′,′sys[6]sys[2]x[0]′,′sys[6]sys[2]x[1]′]A=[[0.00000000e+00−1.00000000e+00−1.00000000e+00][0.00000000e+00−1.00000000e+001.00000000e+00][−1.23918398e+040.00000000e+002.00000000e+00]]B=[[1.][0.][0.]]C=[[0.1.1.]]D=[[0.]]‘‘‘The code yields: ```prolog zeta: 0.5911550337988974 omega_n: 13.53282902556064 Desired poles: [ -8. +10.91501083j -8. -10.91501083j -135.32829026 +0.j ] Plant model: <LinearIOSystem>: sys[2] Inputs (1): ['u[0]'] Outputs (1): ['y[0]'] States (2): ['x[0]', 'x[1]'] A = [[-1. 1.] [ 0. 2.]] B = [[0.] [1.]] C = [[1. 1.]] D = [[0.]] Augmented plant model: <LinearIOSystem>: sys[3] Inputs (1): ['u[0]'] Outputs (1): ['y[0]'] States (3): ['x[0]', 'x[1]', 'x[2]'] A = [[-1. 1. 0.] [ 0. 2. 0.] [-1. -1. 0.]] B = [[0.] [1.] [0.]] C = [[1. 1. 0.]] D = [[0.]] State feedback gains: K = [[-10193.77795361 152.32829026 -12391.83976888]] Integral controller transfer function: -1.239e+04 ---------- s Open-loop transfer function: <LinearICSystem>: sys[6] Inputs (1): ['u[0]'] Outputs (1): ['y[0]'] States (3): ['sys[4]_x[0]', 'sys[2]_x[0]', 'sys[2]_x[1]'] A = [[-0.00000000e+00 0.00000000e+00 0.00000000e+00] [ 0.00000000e+00 -1.00000000e+00 1.00000000e+00] [-1.23918398e+04 0.00000000e+00 2.00000000e+00]] B = [[1.] [0.] [0.]] C = [[0. 1. 1.]] D = [[0.]] Closed-loop transfer function: <LinearICSystem>: sys[9] Inputs (1): ['u[0]'] Outputs (1): ['y[0]'] States (3): ['sys[6]_sys[4]_x[0]', 'sys[6]_sys[2]_x[0]', 'sys[6]_sys[2]_x[1]'] A = [[ 0.00000000e+00 -1.00000000e+00 -1.00000000e+00] [ 0.00000000e+00 -1.00000000e+00 1.00000000e+00] [-1.23918398e+04 0.00000000e+00 2.00000000e+00]] B = [[1.] [0.] [0.]] C = [[0. 1. 1.]] D = [[0.]] ```

Observer and state-space model

Problemè 1

Consider the following state-space model:

x˙=[010001−5−60]x+[001]uy=[100]x\begin{aligned} \dot{x} &= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -5 & -6 & 0 \\ \end{bmatrix} x + \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix} u \\ y &= \begin{bmatrix} 1 & 0 & 0 \\ \end{bmatrix} x \end{aligned}

Design an observer to place the observer poles at -10, -10, -15

Solution

The characteristic equation of the observer is given by:

det(sI−(A−LC))=(s+10)(s+10)(s+15)=s3+35s2+350s+1500det(sI - (A - LC)) = (s + 10)(s + 10)(s + 15) = s^3 + 35s^2 + 350s + 1500

From the coefficients of the characteristic equation we get

det(sI−(A−LC))=s3+(l1−6)s2+(l2−5−6l1)s+(l3−5l2)det(sI - (A-LC)) = s^3 + (l_1-6)s^2 + (l_2-5-6l_1)s + (l_3-5l_2)

Solving for the coefficients we get the observer gain matrix:

L=[450560141]L = \begin{bmatrix} 4505 \\ 601 \\ 41 \end{bmatrix}

Thus the observer dynamics are given by:

x^˙=[010001−5−60]x^+[001]u+[450560141](y−y^)\dot{\hat{x}} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -5 & -6 & 0 \\ \end{bmatrix} \hat{x} + \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix} u + \begin{bmatrix} 4505 \\ 601 \\ 41 \\ \end{bmatrix} (y - \hat{y})

Problemè 2

Given the plant

x˙=[−1102]x+[01]uy=[11]x\begin{aligned} \dot{x} &= \begin{bmatrix} -1 & 1 \\ 0 & 2 \\ \end{bmatrix} x + \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} u \\ y &= \begin{bmatrix} 1 & 1 \\ \end{bmatrix} x \end{aligned}

Design an integral controller to yield a 10% overshoot, 0.5 second settling time and zero steady-state error for a step input.

Solution

The code for the integral controller is given by p2.py.

Add an integrator to the plant to ensure zero steady-state error for a step input. The augmented state-space model becomes:

x˙a=[−110020−1−10]xa+[010]u\dot{x}_a = \begin{bmatrix} -1 & 1 & 0 \\ 0 & 2 & 0 \\ -1 & -1 & 0 \\ \end{bmatrix} x_a + \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix} u y=[110]xay = \begin{bmatrix} 1 & 1 & 0 \\ \end{bmatrix} x_a

where

x_a = \begin{bmatrix} x \\ \int e \, dt \\ \end{bmatrix} $$ and $e = r - y$ is the tracking error. Then, design the state feedback gains

K = \begin{bmatrix} k_1 & k_2 & k_i \ \end{bmatrix}

suchthattheclosed−loopsystemmeetsthetransientresponsespecifications.Thecharacteristicequationoftheclosed−loopsystemis:such that the closed-loop system meets the transient response specifications. The characteristic equation of the closed-loop system is:

\left| sI - (A_a - B_aK) \right| = 0

Expandingthisyields: Expanding this yields:

(s + k_1)(s^2 + (1 - k_2)s + k_i) = 0

Thecontrollawthengivenby: The control law then given by:

u = -Kx_a = -k_1x_1 - k_2x_2 - k_i\int{e , dt}

Thecodeyields:‘‘‘prologzeta:0.5911550337988974omegan:13.53282902556064Desiredpoles:[−8.+10.91501083j−8.−10.91501083j−135.32829026+0.j]Plantmodel:<LinearIOSystem>:sys[2]Inputs(1):[′u[0]′]Outputs(1):[′y[0]′]States(2):[′x[0]′,′x[1]′]A=[[−1.1.][0.2.]]B=[[0.][1.]]C=[[1.1.]]D=[[0.]]Augmentedplantmodel:<LinearIOSystem>:sys[3]Inputs(1):[′u[0]′]Outputs(1):[′y[0]′]States(3):[′x[0]′,′x[1]′,′x[2]′]A=[[−1.1.0.][0.2.0.][−1.−1.0.]]B=[[0.][1.][0.]]C=[[1.1.0.]]D=[[0.]]Statefeedbackgains:K=[[−10193.77795361152.32829026−12391.83976888]]Integralcontrollertransferfunction:−1.239e+04−−−−−−−−−−sOpen−looptransferfunction:<LinearICSystem>:sys[6]Inputs(1):[′u[0]′]Outputs(1):[′y[0]′]States(3):[′sys[4]x[0]′,′sys[2]x[0]′,′sys[2]x[1]′]A=[[−0.00000000e+000.00000000e+000.00000000e+00][0.00000000e+00−1.00000000e+001.00000000e+00][−1.23918398e+040.00000000e+002.00000000e+00]]B=[[1.][0.][0.]]C=[[0.1.1.]]D=[[0.]]Closed−looptransferfunction:<LinearICSystem>:sys[9]Inputs(1):[′u[0]′]Outputs(1):[′y[0]′]States(3):[′sys[6]sys[4]x[0]′,′sys[6]sys[2]x[0]′,′sys[6]sys[2]x[1]′]A=[[0.00000000e+00−1.00000000e+00−1.00000000e+00][0.00000000e+00−1.00000000e+001.00000000e+00][−1.23918398e+040.00000000e+002.00000000e+00]]B=[[1.][0.][0.]]C=[[0.1.1.]]D=[[0.]]‘‘‘The code yields: ```prolog zeta: 0.5911550337988974 omega_n: 13.53282902556064 Desired poles: [ -8. +10.91501083j -8. -10.91501083j -135.32829026 +0.j ] Plant model: <LinearIOSystem>: sys[2] Inputs (1): ['u[0]'] Outputs (1): ['y[0]'] States (2): ['x[0]', 'x[1]'] A = [[-1. 1.] [ 0. 2.]] B = [[0.] [1.]] C = [[1. 1.]] D = [[0.]] Augmented plant model: <LinearIOSystem>: sys[3] Inputs (1): ['u[0]'] Outputs (1): ['y[0]'] States (3): ['x[0]', 'x[1]', 'x[2]'] A = [[-1. 1. 0.] [ 0. 2. 0.] [-1. -1. 0.]] B = [[0.] [1.] [0.]] C = [[1. 1. 0.]] D = [[0.]] State feedback gains: K = [[-10193.77795361 152.32829026 -12391.83976888]] Integral controller transfer function: -1.239e+04 ---------- s Open-loop transfer function: <LinearICSystem>: sys[6] Inputs (1): ['u[0]'] Outputs (1): ['y[0]'] States (3): ['sys[4]_x[0]', 'sys[2]_x[0]', 'sys[2]_x[1]'] A = [[-0.00000000e+00 0.00000000e+00 0.00000000e+00] [ 0.00000000e+00 -1.00000000e+00 1.00000000e+00] [-1.23918398e+04 0.00000000e+00 2.00000000e+00]] B = [[1.] [0.] [0.]] C = [[0. 1. 1.]] D = [[0.]] Closed-loop transfer function: <LinearICSystem>: sys[9] Inputs (1): ['u[0]'] Outputs (1): ['y[0]'] States (3): ['sys[6]_sys[4]_x[0]', 'sys[6]_sys[2]_x[0]', 'sys[6]_sys[2]_x[1]'] A = [[ 0.00000000e+00 -1.00000000e+00 -1.00000000e+00] [ 0.00000000e+00 -1.00000000e+00 1.00000000e+00] [-1.23918398e+04 0.00000000e+00 2.00000000e+00]] B = [[1.] [0.] [0.]] C = [[0. 1. 1.]] D = [[0.]] ```