---
date: '2024-04-10'
description: assignment on observer design for pole placement, integral controller design for zero steady-state error, augmented state-space models, and state feedback gains.
id: A5
modified: 2026-06-05 15:08:42 GMT-04:00
tags:
  - sfwr3dx4
title: Observer and state-space model
created: '2024-04-10'
published: '2024-04-10'
pageLayout: default
slug: thoughts/university/twenty-three-twenty-four/sfwr-3dx4/a5/A5
permalink: https://aarnphm.xyz/thoughts/university/twenty-three-twenty-four/sfwr-3dx4/a5/A5.md
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---
## Problemè 1

Consider the following state-space model:

$$
\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:

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

From the coefficients of the characteristic equation we get

$$
det(sI - (A-LC)) = s^3 + (l_1-6)s^2 + (l_2-5-6l_1)s + (l_3-5l_2)
$$

Solving for the coefficients we get the observer gain matrix:

$$
L = \begin{bmatrix}
4505 \\
601 \\
41
\end{bmatrix}
$$

Thus the observer dynamics are given by:

$$
\dot{\hat{x}} = \begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
-5 & -6 & 0 \\
\end{bmatrix} \hat{x} + \begin{bmatrix}
0 \\
0 \\
1 \\
\end{bmatrix} u + \begin{bmatrix}
4505 \\
601 \\
41 \\
\end{bmatrix} (y - \hat{y})
$$

## Problemè 2

Given the plant

$$
\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 [[thoughts/university/twenty-three-twenty-four/sfwr-3dx4/a5/p2.py|p2.py]].

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

$$
\dot{x}_a = \begin{bmatrix}
-1 & 1 & 0 \\
0 & 2 & 0 \\
-1 & -1 & 0 \\
\end{bmatrix} x_a + \begin{bmatrix}
0 \\
1 \\
0 \\
\end{bmatrix} u
$$

$$
y = \begin{bmatrix}
1 & 1 & 0 \\
\end{bmatrix} x_a
$$

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}


$$

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


$$

Expanding this yields:

$$

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


$$

The control law then given by:

$$

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


$$

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.]]


```

$$
$$