Linear State-Space Control Systems
Prof. Kamran Iqbal
College of Engineering and Information Technology
University of Arkansas at Little Rock
[email protected]
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 StateSpace 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 =𝐿
𝑑𝑖
𝑑𝑡
+ 𝑅𝑖
𝜏 = 𝑘𝑖 𝑖
• Mechanical subsystem
𝑑𝜔
𝐽
𝑑𝑡
=𝜏
𝑣 = 𝑘𝜔 𝜔
Assume
𝐿=0
𝑘𝑖 = 𝑘𝜔 = 𝑘
DC Motor
• Motor equation
𝐽
𝑑𝜔
𝑑𝑡
= 𝜏 = 𝑘𝑖 (𝑒 − 𝑘𝜔 𝜔)/𝑅
Or
𝑑𝜔
𝑑𝑡
Let
𝐾2
𝐽𝑅
Then
=
𝑘2
− 𝜔
𝐽𝑅
= 𝛼,
𝑑𝜔
𝑑𝑡
𝐾
𝐽𝑅
+
𝑘
𝑒
𝐽𝑅
=𝛽
= −𝛼𝜔 + 𝛽𝑒
• State variables: 𝜃, 𝜔
𝑑
𝑑𝑡
0
𝜃
=
0
𝜔
0
1 𝜃
+
𝑒
𝛽
−𝛼 𝜔
DC Motor
• State-space model
Let 𝑥 =
Then
𝑑𝑥
𝑑𝑡
𝜃
𝜔
= 𝑥 = 𝐴𝑥 + 𝐵 𝑢
Let 𝑦 = 𝜃
𝑦= 1 0
𝜃
= 𝐶𝑥
𝜔
• 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 + 𝑚𝑙 2 𝜃 − 𝑚𝑔𝑙 sin 𝜃 = 0
• Linearization (𝜃 ≈ 0, sin 𝜃 ≅ 𝜃, cos 𝜃 ≅ 1)
𝑀 + 𝑚 𝑥 + 𝑚𝑙𝜃 = 𝑓
𝑚𝑙𝑥 + 𝑚𝑙 2 𝜃 − 𝑚𝑔𝑙𝜃 = 0
Inverted Pendulum on Cart
• State variables: 𝑥, 𝜃, 𝑥 , 𝜃
• State equations:
𝑑
𝑑𝑡
0
𝑥
0
𝜃
= 0
𝑥
𝜃
0
0
0
𝑚
𝑔
𝑀
𝑀+𝑚
𝑔
𝑀𝑙
−
1
0
0
0
• Output variables: [𝑥, 𝜃]
• Output equations:
𝑦
0
=
𝜃
0
0
0
1
0
𝑦
0 𝜃
1 𝑦
𝜃
0
𝑥
0
𝜃
+ 1 𝑓
𝑥
𝑀
1
− 𝑀𝑙
0 𝜃
0
1
0
Inverted Pendulum on Motor-Driven Cart
• Let
– 𝑥 – cart displacement
– 𝜃 – pendulum displacement
– 𝑟 – wheel radius
• Then, 𝑓 =
𝑓=
𝜏
,
𝑟
𝑘2
− 2𝑥
𝑅𝑟
𝜔=
+
𝑥
𝑟
𝑘
𝑒
𝑅𝑟
• Dynamic equations
𝑀 + 𝑚 𝑥 + 𝑚𝑙𝜃 +
𝑥 + 𝑙𝜃 − 𝑔𝜃 = 0
𝑘2
𝑥
𝑅𝑟 2
=
𝑘
𝑒
𝑅𝑟
Inverted Pendulum on Motor-Driven Cart
• Solve for accelerations
1
𝑥
𝑙
= 𝑀𝑙
−1
𝜃
−𝑚𝑙
𝑀+𝑚
𝑘2
− 𝑅𝑟 2 𝑥
𝑘
+ 𝑅𝑟 𝑒
𝑔𝜃
• State variables: 𝑥, 𝜃, 𝑥 , 𝜃
State equations:
𝑑
𝑑𝑡
0
0
0
0
1
0
𝑥
0
1
2
𝜃
𝑚
= 0 − 𝑔 − 𝑘 2 0
𝑥
𝑀
𝑀𝑅𝑟
𝑀+𝑚
𝑘2
𝜃
0 𝑀𝑙 𝑔
0
𝑀𝑅𝑟 2 𝑙
Or 𝑥 = 𝐴𝑥 + 𝐵𝑢
0
𝑥
0
𝜃
𝑘
+
𝑒
𝑥
𝑀𝑅𝑟
𝑘
𝜃
− 𝑀𝑅𝑟𝑙
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:
𝐽𝑥 𝜔𝑥𝐵 + 𝐽𝑧 − 𝐽𝑦 𝜔𝑦𝐵 𝜔𝑧𝐵 = 𝜏𝑥𝐵
𝐽𝑦 𝜔𝑦𝐵 + 𝐽𝑥 − 𝐽𝑧 𝜔𝑥𝐵 𝜔𝑧𝐵 = 𝜏𝑦𝐵
𝐽𝑧 𝜔𝑧𝐵 + 𝐽𝑦 − 𝐽𝑥 𝜔𝑥𝐵 𝜔𝑦𝐵 = 𝜏𝑧𝐵
Two-Axis Gyro
• Assume that 𝑧-axis is the spin axis and 𝜔𝑧 is constant; let
𝐻𝑧 = 𝐽𝑧 𝜔𝑧 , (angular momentum); 𝐻 = 𝐻𝑧 1 −
𝐽𝑑
𝐽𝑧
gyro constant
𝐽𝑥 = 𝐽𝑦 = 𝐽𝑑 (diametrical moment of inertia)
• Dynamic Equations:
𝜔𝑥𝐵
1
0
+
𝜔𝑦𝐵
𝐽𝑑 −𝐻
1 𝜏𝑥
𝐻 𝜔𝑥𝐵
=
𝜔
𝐽𝑑 𝜏 𝑦
𝑦𝐵
0
• Gyro equations including the spring and damping terms:
1 𝐵
𝜔𝑥𝐵
𝜔𝑦𝐵 + 𝐽𝑑 −𝐻
1 𝐵
𝐻 𝜔𝑥𝐵
−
𝐵 𝜔𝑦𝐵
𝐽𝑑 0
1 𝐾𝐷
0 𝜔𝑥𝐸
+
𝐵 𝜔𝑦𝐸
𝐽𝑑 −𝐾𝑄
𝐾𝑄 𝛿𝑥
1 𝜏𝑥
=
𝐾𝐷 𝛿𝑦
𝐽𝑑 𝜏𝑦
Two-Axis Gyro
• Angular displacements (gyro pick off) are:
𝛿𝑥 = 𝜔𝑥𝐵 − 𝜔𝑥𝐸
𝛿𝑦 = 𝜔𝑦𝐵 − 𝜔𝑦𝐸
𝜏
′
𝜔
𝑥𝐸
• Define 𝑥 = 𝛿𝑥 , 𝛿𝑦 , 𝜔𝑥𝐵 , 𝜔𝑦𝐵 , 𝑢 = 𝜏𝑦𝑥 , 𝑥0 = 𝜔𝑦𝐸
Let 𝑏1 =
𝐴2 =
𝐵
,
𝐽𝑑
−𝑏1
𝑏2
𝑏2 =
𝐻
,
𝐽𝑑
𝑐1 =
𝐾𝐷
,
𝐽𝑑
𝑐2 =
𝐾𝑄
𝐽𝑑
−𝑏2
−𝑏1
0
𝐼
−𝐼
0
Then 𝑥 = 𝐴 𝐴 𝑥 + 𝑏 𝐼 𝑥0 + 𝛽𝐼 𝑢
1
2
1
Or 𝑥 = 𝐴𝑥 + 𝐵𝑢 + 𝐸𝑥0
, 𝛽=
1
,
𝐽𝑑
−𝑐1
𝐴1 = 𝑐
2
−𝑐2
−𝑐1 ,
Two-Axis Gyro
• The characteristic equation of the gyroscope is:
𝑠𝐼 − 𝐴 = 𝑠 2 + 𝑏1 𝑠 + 𝑐1
2
𝑏2 𝑠 + 𝑐2
2
• The precession and nutation frequencies are given as:
𝑠 = 𝛼𝑝 + 𝜔𝑝 , 𝛼𝑝 = −
𝑏1 𝑐1 −𝑏2 𝑐2
,
𝑏12 +𝑏22
𝜔𝑝 =
𝑏2 𝑐1 −𝑏1 𝑐2
𝑏12 +𝑏22
𝑠 = 𝛼𝑛 + 𝜔𝑛 , 𝛼𝑛 = 𝛼𝑝 − 𝑏1 , 𝜔𝑛 = 𝜔𝑝 + 𝑏2
• The transfer function of a free gyro is given as:
𝜔𝑥𝐸
𝛿𝑥
= 𝐻(𝑠) 𝜔
; 𝐻 𝑠 =
𝛿𝑦
𝑦𝐸
𝑠 2 +𝑏1 𝑠+𝑐1 − 𝑏2 𝑠+𝑐2
𝑏2 𝑠+𝑐2
𝑠 2 +𝑏1 𝑠+𝑐1
𝑠 2 +𝑏1 𝑠+𝑐1 2 𝑏2 𝑠+𝑐2 2
Two-Axis Gyro
• An ideal gyro is one with zero damping and stiffness
• Then
𝐻 𝑠 =
𝑠 2 −𝑏2 𝑠
𝑏2 𝑠
𝑠2
𝑠 2 +𝑏1 𝑠+𝑐1 2 𝑏2 𝑠+𝑐2 2
• Assume a step input 𝜔𝑥𝐸 = 1, 𝜔𝑦𝐸 = 0
1
𝛿𝑥 𝑡 = 𝑏2 1 − cos 𝑏2 𝑡
2
1
𝛿𝑦 𝑡 = − 𝑏
2
1
𝑡 − 𝑏 sin 𝑏2 𝑡
2
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 = 𝑋𝑢 Δ𝑢 + 𝑋𝛼 𝛼 − 𝑔𝜃 + 𝑋𝐸 𝛿𝐸
𝛼=
𝑍𝑢
Δ𝑢
𝑉
+
𝑍𝛼
𝛼
𝑉
+𝑞+
𝑍𝐸
𝛿
𝑉 𝐸
𝑞 = 𝑀𝑢 Δ𝑢 + 𝑀𝛼 𝛼 + 𝑀𝑞 𝑞 + 𝑀𝐸 𝛿𝐸
𝜃=𝑞
Constant-Altitude Autopilot
• The simplified dynamics of an aircraft at constant speed are
described as:
𝛼=
𝑍𝛼
𝛼
𝑉
+𝑞+
𝑍𝐸
𝛿
𝑉 𝐸
𝑞 = 𝑀𝛼 𝛼 + 𝑀𝑞 𝑞 + 𝑀𝐸 𝛿𝐸
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
𝑉
𝛽+
𝑌𝑝
𝑉
𝑝+
𝑌𝑟
𝑉
𝑔
𝑉
−1 𝑟+ 𝜙+
𝑌𝐴
𝛿
𝑉 𝐴
𝑝 = 𝐿𝛽 𝛽 + 𝐿𝑝 𝑝 + 𝐿𝑟 𝑟 + 𝐿𝐴 𝛿𝐴 + 𝐿𝑅 𝛿𝑅
𝑟 = 𝑁𝛽 𝛽 + 𝑁𝑝 𝑝 + 𝑁𝑟 𝑟 + 𝑁𝐴 𝛿𝐴 + 𝑁𝑅 𝛿𝑅
𝜙=𝑝
𝜓=𝑟
+
𝑌𝑅
𝛿
𝑉 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
𝑍
𝑍𝛿
1 𝛼
+ 𝑉 𝛿
𝑞
𝑀𝑞
𝑀𝛿
𝛼𝑁 = 𝑍𝛼 𝛼 + 𝑍𝛿 𝛿
𝛼
𝛼
= 𝑉
𝑞
𝑀𝛼
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
1
𝜆 = 0 𝑉𝑇 2 𝜆 + 0 𝑎
𝑇 𝑁
𝑧
0 0 𝑧
i.e., the state equations are time varying