Linear State-Space Control Systems

Prof. Kamran Iqbal

College of Engineering and Information Technology

University of Arkansas at Little Rock

kxiqbal@ualr.edu

Course Overview

• State space models of linear systems

• Solution to State equations

• Controllability and observability

• Stability, dynamic response

• Controller design via pole placement

• Controllers for disturbance and tracking systems

• Observer based compensator design

• Linear quadratic optimal control

• Kalman filters, stochastic control

• Linear matrix inequalities in control design

• Course assessment

Learning Objectives

• Formulate and solve state-variable models of linear systems

• Apply analytical methods of controllability, observability, and

stability to system models

• Controller synthesis via pole placement

• Observer based compensator design

• Formulate and solve the optimal control problem

• Design optimal observers and Kalman filters

• LMI based controller design

Resources

• Core Text:

• Bernard Friedland, Control System Design: An Introduction to State-Space Methods, Dover Publications, ISBN: 978-0486442785

• References:

• Professor Raymond Kwong’s notes http://www.control.toronto.edu/people/profs/kwong/

• Professor Jongeun Choi’s notes http://www.egr.msu.edu/classes/me851/jchoi/lecture/

• Professor Perry Li’s notes http://www.me.umn.edu/courses/me8281/notes.htm

• Astrom and Murray, Feedback Systems, An Introduction for Scientists and Engineers, Princeton University Press, 2012, http://www.cds.caltech.edu/~murray/amwiki/

Course Schedule

Session Topic

1. State space models of linear systems

2. Solution to State equations, canonical forms

3. Controllability and observability

4. Stability and dynamic response

5. Controller design via pole placement

6. Controllers for disturbance and tracking systems

7. Observer based compensator design

8. Linear quadratic optimal control

9. Kalman filters and stochastic control

10. LM in control design

State-Space Models of Linear Systems

State-Variable Models

• State variables

– Energy variables, e.g., velocity (KE), position (PE)

– Alternate variables, momentum (KE)

– Flow and across variables, e.g., current, voltage

• Dynamic Equations

– Based on physical principles

– Ordinary differential equations

– Partial differential equations

• State-variable equations

– First order differential equation

Transfer Function Models

• Describe input-output relation

• Restricted to LTI systems

• Can be of lower order than actual system

• Example:

Let 𝑥 + 3𝑥 + 2𝑥 = 𝑢, 𝑦 = 𝑥 + 𝑥

𝐻 𝑠 =1

𝑠+2

Example: dc Motor

• Electrical subsystem

𝑒 − 𝑣 = 𝐿𝑑𝑖

𝑑𝑡+ 𝑅𝑖

𝜏 = 𝑘𝑖𝑖

• Mechanical subsystem

𝐽𝑑𝜔

𝑑𝑡= 𝜏

𝑣 = 𝑘𝜔𝜔

Assume

𝐿 = 0

𝑘𝑖 = 𝑘𝜔 = 𝑘

DC Motor

• Motor equation

𝐽𝑑𝜔

𝑑𝑡= 𝜏 = 𝑘𝑖(𝑒 − 𝑘𝜔𝜔)/𝑅

Or 𝑑𝜔

𝑑𝑡= −

𝑘2

𝐽𝑅𝜔 +

𝑘

𝐽𝑅𝑒

Let 𝐾2

𝐽𝑅= 𝛼,

𝐾

𝐽𝑅= 𝛽

Then 𝑑𝜔

𝑑𝑡= −𝛼𝜔 + 𝛽𝑒

• State variables: 𝜃, 𝜔

𝑑

𝑑𝑡

𝜃𝜔

=0 10 −𝛼

𝜃𝜔

+0𝛽

𝑒

DC Motor

• State-space model

Let 𝑥 =𝜃𝜔

Then 𝑑𝑥

𝑑𝑡= 𝑥 = 𝐴𝑥 + 𝐵 𝑢

Let 𝑦 = 𝜃

𝑦 = 1 0𝜃𝜔

= 𝐶𝑥

• Transfer function model

𝜃 𝑠

𝑒 𝑠=

𝛽

𝑠 𝑠+𝛼,

𝜔 𝑠

𝑒 𝑠=

𝛽

𝑠+𝛼

Example: Inverted Pendulum on Cart

• Let

– 𝑥 – cart displacement

– 𝜃 – pendulum displacement

– 𝑓 – applied force

• Dynamic equations

𝑀 +𝑚 𝑥 + 𝑚𝑙 cos 𝜃 𝜃 − 𝑚𝑙𝜃 2 sin 𝜃 = 𝑓

𝑚𝑙 cos 𝜃 𝑥 + 𝑚𝑙2𝜃 − 𝑚𝑔𝑙 sin 𝜃 = 0

• Linearization (𝜃 ≈ 0, sin 𝜃 ≅ 𝜃, cos 𝜃 ≅ 1)

𝑀 +𝑚 𝑥 + 𝑚𝑙𝜃 = 𝑓

𝑚𝑙𝑥 + 𝑚𝑙2𝜃 − 𝑚𝑔𝑙𝜃 = 0

Inverted Pendulum on Cart

• State variables: 𝑥, 𝜃, 𝑥 , 𝜃

• State equations:

𝑑

𝑑𝑡

𝑥𝜃𝑥 𝜃

=

0 0 1 00 0 0 1

0 −𝑚

𝑀𝑔 0 0

0𝑀+𝑚

𝑀𝑙𝑔 0 0

𝑥𝜃𝑥 𝜃

+

001

𝑀

−1

𝑀𝑙

𝑓

• Output variables: [𝑥, 𝜃]

• Output equations:

𝑦𝜃

=0 00 0

1 00 1

𝑦𝜃𝑦

𝜃

Inverted Pendulum on Motor-Driven Cart

• Let

– 𝑥 – cart displacement

– 𝜃 – pendulum displacement

– 𝑟 – wheel radius

• Then, 𝑓 =𝜏

𝑟, 𝜔 =

𝑥

𝑟

𝑓 = −𝑘2

𝑅𝑟2𝑥 +

𝑘

𝑅𝑟𝑒

• Dynamic equations

𝑀 +𝑚 𝑥 + 𝑚𝑙𝜃 +𝑘2

𝑅𝑟2𝑥 =

𝑘

𝑅𝑟𝑒

𝑥 + 𝑙𝜃 − 𝑔𝜃 = 0

Inverted Pendulum on Motor-Driven Cart

• Solve for accelerations

𝑥 𝜃

=1

𝑀𝑙

𝑙 −𝑚𝑙−1 𝑀 +𝑚

−𝑘2

𝑅𝑟2𝑥 +

𝑘

𝑅𝑟𝑒

𝑔𝜃

• State variables: 𝑥, 𝜃, 𝑥 , 𝜃

State equations:

𝑑

𝑑𝑡

𝑥𝜃𝑥 𝜃

=

0 0 1 00 0 0 1

0 −𝑚

𝑀𝑔 −

𝑘2

𝑀𝑅𝑟20

0𝑀+𝑚

𝑀𝑙𝑔

𝑘2

𝑀𝑅𝑟2𝑙0

𝑥𝜃𝑥 𝜃

+

00𝑘

𝑀𝑅𝑟

−𝑘

𝑀𝑅𝑟𝑙

𝑒

Or 𝑥 = 𝐴𝑥 + 𝐵𝑢

Example: Two-Axis Gyro

• Rigid body dynamics (true in an inertial frame):

𝑑𝑝

𝑑𝑡= 𝑓 ,

𝑑ℎ

𝑑𝑡= 𝜏

• Euler’s equations for a spinning body:

𝐽𝑥𝜔 𝑥𝐵 + 𝐽𝑧 − 𝐽𝑦 𝜔𝑦𝐵𝜔𝑧𝐵 = 𝜏𝑥𝐵

𝐽𝑦𝜔 𝑦𝐵 + 𝐽𝑥 − 𝐽𝑧 𝜔𝑥𝐵𝜔𝑧𝐵 = 𝜏𝑦𝐵

𝐽𝑧𝜔 𝑧𝐵 + 𝐽𝑦 − 𝐽𝑥 𝜔𝑥𝐵𝜔𝑦𝐵 = 𝜏𝑧𝐵

Two-Axis Gyro

• Assume that 𝑧-axis is the spin axis and 𝜔𝑧 is constant; let

𝐻𝑧 = 𝐽𝑧𝜔𝑧, (angular momentum); 𝐻 = 𝐻𝑧 1 −𝐽𝑑

𝐽𝑧 gyro constant

𝐽𝑥 = 𝐽𝑦 = 𝐽𝑑 (diametrical moment of inertia)

• Dynamic Equations:

𝜔 𝑥𝐵𝜔 𝑦𝐵

+1

𝐽𝑑

0 𝐻−𝐻 0

𝜔𝑥𝐵

𝜔𝑦𝐵=

1

𝐽𝑑

𝜏𝑥𝜏𝑦

• Gyro equations including the spring and damping terms:

𝜔 𝑥𝐵𝜔 𝑦𝐵

+1

𝐽𝑑

𝐵 𝐻−𝐻 𝐵

𝜔𝑥𝐵

𝜔𝑦𝐵−1

𝐽𝑑

𝐵 00 𝐵

𝜔𝑥𝐸

𝜔𝑦𝐸+1

𝐽𝑑

𝐾𝐷 𝐾𝑄−𝐾𝑄 𝐾𝐷

𝛿𝑥𝛿𝑦

=1

𝐽𝑑

𝜏𝑥𝜏𝑦

Two-Axis Gyro

• Angular displacements (gyro pick off) are:

𝛿 𝑥 = 𝜔𝑥𝐵 − 𝜔𝑥𝐸

𝛿 𝑦 = 𝜔𝑦𝐵 − 𝜔𝑦𝐸

• Define 𝑥 = 𝛿𝑥, 𝛿𝑦, 𝜔𝑥𝐵, 𝜔𝑦𝐵′, 𝑢 =

𝜏𝑥𝜏𝑦

, 𝑥0 =𝜔𝑥𝐸

𝜔𝑦𝐸

Let 𝑏1 =𝐵

𝐽𝑑, 𝑏2 =

𝐻

𝐽𝑑, 𝑐1 =

𝐾𝐷

𝐽𝑑, 𝑐2 =

𝐾𝑄

𝐽𝑑, 𝛽 =

1

𝐽𝑑, 𝐴1 =

−𝑐1 −𝑐2𝑐2 −𝑐1

,

𝐴2 =−𝑏1 −𝑏2𝑏2 −𝑏1

Then 𝑥 =0 𝐼𝐴1 𝐴2

𝑥 +−𝐼𝑏1𝐼

𝑥0 +0𝛽𝐼

𝑢

Or 𝑥 = 𝐴𝑥 + 𝐵𝑢 + 𝐸𝑥0

Two-Axis Gyro

• The characteristic equation of the gyroscope is:

𝑠𝐼 − 𝐴 = 𝑠2 + 𝑏1𝑠 + 𝑐12 𝑏2𝑠 + 𝑐2

2

• The precession and nutation frequencies are given as:

𝑠 = 𝛼𝑝 + 𝜔𝑝, 𝛼𝑝 = −𝑏1𝑐1−𝑏2𝑐2

𝑏12+𝑏2

2 , 𝜔𝑝 =𝑏2𝑐1−𝑏1𝑐2

𝑏12+𝑏2

2

𝑠 = 𝛼𝑛 + 𝜔𝑛, 𝛼𝑛 = 𝛼𝑝 − 𝑏1, 𝜔𝑛 = 𝜔𝑝 + 𝑏2

• The transfer function of a free gyro is given as:

𝛿𝑥𝛿𝑦

= 𝐻(𝑠)𝜔𝑥𝐸

𝜔𝑦𝐸; 𝐻 𝑠 =

𝑠2+𝑏1𝑠+𝑐1 − 𝑏2𝑠+𝑐2𝑏2𝑠+𝑐2 𝑠2+𝑏1𝑠+𝑐1𝑠2+𝑏1𝑠+𝑐1

2 𝑏2𝑠+𝑐22

Two-Axis Gyro

• An ideal gyro is one with zero damping and stiffness

• Then

𝐻 𝑠 =

𝑠2 −𝑏2𝑠

𝑏2𝑠 𝑠2

𝑠2+𝑏1𝑠+𝑐12 𝑏2𝑠+𝑐2

2

• Assume a step input 𝜔𝑥𝐸 = 1,𝜔𝑦𝐸 = 0

𝛿𝑥 𝑡 =1

𝑏22 1 − cos 𝑏2𝑡

𝛿𝑦 𝑡 = −1

𝑏2𝑡 −

1

𝑏2sin 𝑏2𝑡

Example: Aerodynamics

• Define

– 𝛼 – angle of attack, 𝛽 – side slip angle

– 𝜙 – roll angle, 𝜃 – pitch angle, 𝜓 – yaw angle

– 𝑝 – roll rate, 𝑞 – pitch rate, 𝑟 – yaw rate

– 𝐿 – rolling moment, 𝑀 – pitching moment, 𝑁 – yawing moment

– 𝑋 – longitudinal force, 𝑌 – lateral force, 𝑍 – vertical force

– 𝑉 – aircraft speed,

– Δ𝑢 – change in speed

– 𝛿𝐸 – elevator deflection

– 𝛿𝐴 – aileron deflection

– 𝛿𝑅 – rudder deflection

Aerodynamics

• Longitudinal motion:

Let 𝑥 = Δ𝑢, 𝛼, 𝑞 , 𝑞 ′; 𝑢 = 𝛿𝐸

Then,

Δ𝑢 = 𝑋𝑢Δ𝑢 + 𝑋𝛼𝛼 − 𝑔𝜃 + 𝑋𝐸𝛿𝐸

𝛼 =𝑍𝑢

𝑉Δ𝑢 +

𝑍𝛼

𝑉𝛼 + 𝑞 +

𝑍𝐸

𝑉𝛿𝐸

𝑞 = 𝑀𝑢Δ𝑢 +𝑀𝛼𝛼 +𝑀𝑞𝑞 +𝑀𝐸𝛿𝐸

𝜃 = 𝑞

Constant-Altitude Autopilot

• The simplified dynamics of an aircraft at constant speed are

described as:

𝛼 =𝑍𝛼

𝑉𝛼 + 𝑞 +

𝑍𝐸

𝑉𝛿𝐸

𝑞 = 𝑀𝛼𝛼 +𝑀𝑞𝑞 + 𝑀𝐸𝛿𝐸

𝜃 = 𝑞

• Define

Δℎ = (ℎ − ℎ0)/𝑉

Then Δℎ = 𝛾 = 𝜃 − 𝛼

Aerodynamics

• Lateral motion:

Let 𝑥 = 𝛽, 𝑝, 𝑟, 𝜙, 𝜓 ′; 𝑢 = 𝛿A, 𝛿𝑅′

Then,

𝛽 =𝑌𝛽

𝑉𝛽 +

𝑌𝑝

𝑉𝑝 +

𝑌𝑟

𝑉− 1 𝑟 +

𝑔

𝑉𝜙 +

𝑌𝐴

𝑉𝛿𝐴 +

𝑌𝑅

𝑉𝛿𝑅

𝑝 = 𝐿𝛽𝛽 + 𝐿𝑝𝑝 + 𝐿𝑟𝑟 + 𝐿𝐴𝛿𝐴 + 𝐿𝑅𝛿𝑅

𝑟 = 𝑁𝛽𝛽 + 𝑁𝑝𝑝 + 𝑁𝑟𝑟 + 𝑁𝐴𝛿𝐴 + 𝑁𝑅𝛿𝑅

𝜙 = 𝑝

𝜓 = 𝑟

Missile Dynamics

• Define

𝑉 – missile velocity

𝛼𝑁 – normal acceleration

𝜃 – pitch angle

𝛾 – flight path angle

• Assume that

𝑋𝑢 ≈ 0, 𝑍𝑢 ≈ 0, 𝑀𝛼 ≈ 0

Then

𝛼 𝑞

=

𝑍𝛼

𝑉1

𝑀𝛼 𝑀𝑞

𝛼𝑞 +

𝑍𝛿

𝑉

𝑀𝛿

𝛿

𝛼𝑁 = 𝑍𝛼𝛼 + 𝑍𝛿𝛿

Missile Guidance

• Define

𝜆 – line-of-sight angle

𝑧 – projected miss distance

𝑉 – missile speed

𝑉𝑇 – target speed

𝑇 − 𝑡 = 𝑇 – time to go

• Then

𝜆

𝑧 =

01

𝑉𝑇 2

0 0

𝜆𝑧+

0𝑇 𝑎𝑁

i.e., the state equations are time varying