7360w1a Overview Linear Algebra

8/8/2019 7360w1a Overview Linear Algebra

1/67

Lecture Notes: Week 1a

ECE/MAE 7360

Optimal and Robust Control(Fall 2003 Offering)


2/67

Control Systems Area

Fall'03 Course Offering

ECE/MAE 7360 Optimal and Robust Control. Advanced methods of control systemanalysis and design. Operator approaches to optimal control, including LQR/LQG/LTR,

mu-analysis, H-infinity loop shaping and gap metric etc. Prerequisite: ECE 6320 or

instructor approval. (3 cr) (alternate Fall).

Day/Time/Venue : MW 2:30-3:45 PM. EL-112 (Control Lab)

Instructor: Dr YangQuan Chen, CSOIS, ECE Dept., (435)797-0148.

Text: Kemin Zhou, with John Doyle,Essentials of Robust Control, Prentice-Hall, 1998.

Course Description:Robust control is concerned with the problem of designing control

systems when there is uncertainty about the model of the system to be controlled or whenthere are (possibly uncertain) external disturbances influencing the behavior of the

system. Optimal control is concerned with the design of control systems to achieve a

prescribed performance (e.g., to find a controller for a given linear system that minimizesa quadratic cost function). While optimal control theory was originally derived using the

techniques of calculus of variation, most robust control methodologies have been

developed from an operator-theoretic perspective. In this course we will mainly use anoperator approach to study the basic results in robust control that have been developed

over the last fifteen years However mathematical programming based techniques for


3/67

ECE/MAE 7360: Optimal and Robust Control

Course Syllabus - Fall 2003

From http://www.ece.usu.edu/academics/graduate_courses.html

***7360. Optimal and Robust Control.Advanced methods of control system analysis and design. Operator approaches to optimalcontrol, including LQR, LQG, and L1 optimization techniques. Robust control theory, including

QFT, H-infinity, and interval polynomial approaches. Prerequisite: ECE/MAE 6320 orinstructor approval. Also taught as MAE 7360. (3 cr) (Sp)

Instructor: YangQuan Chen, Center for Self-Organizing and Intelligent SystemsDepartment of Electrical and Computer Engineering, Utah State UniversityRoom EL152; Tel.(435)797-0148, [email protected]

Lecture Day/Time/Venue: MW 2:30-3:45 PM. EL-112 (Control Lab)

Ofice Hours: MW 1:15-2:30 PM.

Text: Kemin Zhou, with John Doyle,Essentials of Robust Control, Prentice-Hall, 1998.

References: Will be give by the Instructor via email/web/ftp.

Software: (1) MATLAB Control Systems Toolbox (2) MATLAB mu-Synthesis Toolbox (3)RIOTS_95: MATLAB Toolbox for solving general optimal control problems.

Course Requirements:Homework 40 pointsMid-term take home exam 10 pointsFocused Individual Study Project/presentation 10 points

Design project 40 points


4/67

Course Description:

Robust controlis concerned with the problem of designing control systems when there isuncertainty about the model of the system to be controlled or when there are (possibly uncertain)external disturbances influencing the behavior of the system. Optimal controlis concerned withthe design of control systems to achieve a prescribed performance (e.g., to find a controller for agiven linear system that minimizes a quadratic cost function). While optimal control theory wasoriginally derived using the techniques of calculus of variation, most robust controlmethodologies have been developed from an operator-theoretic perspective. In this course we willmainly use an operator approach to study the basic results in robust control that have beendeveloped over the last fifteen years. However, mathematical programming based techniques for

solving optimal control problems will also be briefly covered. This course provides a unifiedtreatment of multivariable control system design for systems subject to uncertainty andperformance requirements.

Course Topics and Approximate Schedule:

Course Topics:

1. Review of multivariable linear control theory and balanced model realization/reduction.2. Signal/system norms and / spaces and internal stability.3. Performance specification and limitations.4. Modeling uncertainty and robustness.5. LFT and mu synthesis.6. Parameterization of controllers.7. -optimal control (LQR/Kalman Filter /LQG/LTR.)8. -optimal control (for unstructured perturbations).9. Gap metric10 S l i ti l t l bl i ll


5/67

7 Oct. 6 Chapter 11Controller parameterization

(Youla-paramterization)

Oct. 8 Chapter 12,13LQR/H2 control

Project#2: Space-shuttlerobustness analysis

(stability andperformance)

8 Oct. 13 Lecturer's NotesLQG/LTR

Oct. 15 Chapter 14H-infinity Control

HW#5

9 Oct. 20 -- Chapter 14H-infinity Control

Oct. 22 Chapter 14H-infinity Control

mid-term take homeexam

10 Oct. 27 -- Chapter 15H-infinity Controller order-reduction

Oct. 29 Chapter 16H-infinity loop shaping

HW#6

11 Nov. 3 Chapter 16H-infinity loop shaping

Nov. 5 Chapter 16H-infinity loop shaping

Project#3: H-infinitycontrol (performance)design of high-maneuvering airplane

12 Nov. 10 Chapter 17Gap metric

Nov. 12 Chapter 17nu-Gap metric

HW#7

13 Nov. 17 Instructor's notes

Mathematical foundation ofRIOTS_95

Nov. 19 Instructor's notes

Sample applications ofRIOTS_95

HW#8

14 Nov. 24 FISP presentations(3 students)

Nov. 26 No class.Thanksgiving

Project #4: Solvingoptimal control problems(you define your ownOCP!) using RIOTS_95

15 Dec. 1 FISP presentations (2students)

Dec. 3 - FISPpresentations (2 students)

16 D 8 N l D 10 N l N Fi l E


6/67

3

Classical control in the 1930s and 1940s

Bode, Nyquist, Nichols, . . .

Feedback amplifier design

Single input, single output (SISO)

Frequency domain Graphical techniques

Emphasized design tradeoffs

Effects of uncertainty

Nonminimum phase systems

Performance vs. robustness


7/67

4

The origins of modern control theory

Early years

Wiener (1930s - 1950s) Generalized harmonic analysis, cybernetics,filtering, prediction, smoothing

Kolmogorov (1940s) Stochastic processes Linear and nonlinear programming (1940s - )

Optimal control

Bellmans Dynamic Programming (1950s) Pontryagins Maximum Principle (1950s)

Li i l l (l 1950 d 1960 )


8/67

5

The diversification of modern control

in the 1960s and 1970s

Applications of Maximum Principle and Optimization Zoom maneuver for time-to-climb

Spacecraft guidance (e.g. Apollo)

Scheduling, resource management, etc.

Linear optimal control

Linear systems theory Controllability, observability, realization theory

Geometric theory, disturbance decoupling

Pole assignment

Al b i h


9/67

6

Modern control application: Shuttle reentry

The problem is to control the reentry of the shuttle, from orbit to

landing. The modern control approach is to break the problem into two

pieces:

Trajectory optimization Flight control

Trajectory optimization: tremendous use of modern control principles

State estimation (filtering) for navigation

Bang-bang control of thrusters

Digital autopilot

N li i l j l i


10/67

7

The 1970s and the return of the frequency domain

Motivated by the inadequacies of modern control, many researchers

returned to the frequency domain for methods for MIMO feedback control.

British school

Inverse Nyquist Array

Characteristic Loci

Singular values

MIMO generalization of Bode gain plots MIMO generalization of Bode design

Crude MIMO representations of uncertainty

Multivariable loopshaping and LQG/LTRA il d d l i l h d


11/67

8

Postmodern Control

Mostly for fun. Sick of modern control, but wanted a name equallypretentious and self-absorbed.

Other possible names are inadequate:

Robust ( too narrow, sounds too macho)

Neoclassical (boring, sounds vaguely fascist )

Cyberpunk ( too nihilistic )

Analogy with postmodern movement in art, architecture, literature,social criticism, philosophy of science, feminism, etc. ( talk about

pretentious ).

The tenets of postmodern control theory


12/67

9

Issues in postmodern control theory

More connection with data Modeling

Flexible signal representation and performance objectives Flexible uncertainty representations

Nonlinear nominal models

Uncertainty modeling in specific domains

Analysis System Identification

Nonprobabilistic theory

System ID with plant uncertainty


13/67

10

Chapter 2: Linear Algebra

linear subspaces eigenvalues and eigenvectors

matrix inversion formulas invariant subspaces vector norms and matrix norms

singular value decomposition

generalized inverses semidefinite matrices


14/67

11

Linear Subspaces

linear combination:1x1 + . . . + kxk, xi Fn, F

span{x1, x2, . . . , xk} := {x = 1x1 + . . . + kxk : i F}. x1, x2, . . . , xk Fn linearly dependent if there exists 1, . . . , k F

not all zero such that 1x2 + . . . + kxk = 0; otherwise they are

linearly independent.

{x1, x2, . . . , xk} S is a basis for S if x1, x2, . . . , xk are linearlyindependent and S = span{x1, x2, . . . , xk}.

{x1, x2, . . . , xk} in Fn are mutually orthogonal if xi xj = 0 for alli = j and orthonormalifxi xj = ij.

h l l f b S F


15/67

12

The rank of a matrix A is defined by

rank(A) = dim(ImA).

rank(A) = rank(A). A Fmn is full row rank if m n andrank(A) = m. A is full column rank ifn m and rank(A) = n.

unitary matrixUU = I = UU.

Let D Fnk (n > k) be such that DD = I. Then there exists amatrix D Fn(nk) such that

D D

is a unitary matrix.

Sylvester equationAX+ XB = C

with A Fnn, B Fmm, and C Fnm has a unique solutionX Fnm if and only if i(A) + j(B) = 0, i = 1, 2, . . . , n and

j = 1, 2, . . . , m.

Lyapunov Equation: B = A.

L A F d B F k h


16/67

13

Eigenvalues and Eigenvectors

The eigenvalues and eigenvectors of A Cnn: , x Cn

Ax = x

x is a right eigenvector

y is a left eigenvector:

yA = y.

eigenvalues: the roots of det(I

A).

the spectral radius: (A) := max1in |i| Jordan canonical form: A Cnn, T

A = T JT1

h


17/67

14

where tij1 are the eigenvectors of A,

Atij1 = itij1,

and tijk = 0 defined by the following linear equations for k 2(A iI)tijk = tij(k1)

are called the generalized eigenvectors ofA.A Rnn with distinct eigenvalues can be diagonalized:

A

x1 x2 xn

=

x1 x2 xn

12

. . .

n

.

and has the following spectral decomposition:

A =n

i=1ixiy

i


18/67

15

Matrix Inversion Formulas

A11 A12

A21 A22

=

I 0

A21A111 I

A11 0

0

I A

111 A12

0 I

:= A22 A21A111 A12

A11 A12

A21 A22

= I A12A122

0 I

0

0 A22

I 0

A122 A21 I

:= A11 A12A122 A21

A11 A12

A21 A22

1

=

A111 + A

111 A12

1A21A111 A111 A121

1A21A111 1

and

A11 A12

A21 A22

1

=

1 1A12A122A122 A211 A122 + A122 A211A12A122

.


19/67

16

Invariant Subspaces

a subspace S Cn is an A-invariant subspace if Ax S for everyx S.For example,

{0

}, Cn, and KerA are all A-invariant subspaces.

Let and x be an eigenvalue and a corresponding eigenvector of

A Cnn. Then S := span{x} is an A-invariant subspace sinceAx = x S.In general, let 1, . . . , k (not necessarily distinct) and xi be a set of

eigenvalues and a set of corresponding eigenvectors and the generalizedeigenvectors. Then S = span{x1, . . . , xk} is an A-invariant subspaceprovided that all the lower rank generalized eigenvectors are included.

An A-invariant subspace S Cn is called a stable invariant subspaceif all the eigenvalues of A constrained to S have negative real parts.


20/67

17

However, the subspaces

S2 = span{x2}, S23 = span{x2, x3}S24 = span{x2, x4}, S234 = span{x2, x3, x4}

are not A-invariant subspaces since the lower rank generalized eigen-

vector x1 ofx2 is not in these subspaces.

To illustrate, consider the subspace S23. It is an A-invariant subspaceifAx2 S23. Since

Ax2 = x2 + x1,

Ax2 S23 would require that x1 be a linear combination of x2 andx3, but this is impossible since x1 is independent ofx2 and x3.


21/67

18

Vector Norms and Matrix Norms

X a vector space. is a norm if(i) x 0 (positivity);

(ii)

x

= 0 if and only ifx = 0 (positive definiteness);

(iii) x = || x, for any scalar (homogeneity);(iv) x + y x + y (triangle inequality)for any x X and y X.

Let x Cn

. Then we define the vector p-norm ofx as

xp := n

i=1|xi|p

1/p

, for 1 p .

In particular, when p = 1, 2, we have

n

| |


22/67

19

A = max1imn

j=1|aij| (row sum) .

The Euclidean 2-norm has some very nice properties:

Let x Fn and y Fm.1. Suppose n m. Then x = y iff there is a matrix U Fnm

such that x = Uy and UU = I.

2. Suppose n = m. Then |xy| x y. Moreover, the equalityholds iffx = y for some F or y = 0.

3. x y iff there is a matrix Fnm with 1 such thatx = y. Furthermore, x < y iff < 1.

4. U x = x for any appropriately dimensioned unitary matrices U.Frobenius norm

AF :=

Trace(AA) =

mi=1

nj=1

|aij|2 .


23/67

20

Singular Value Decomposition

Let A Fmn. There exist unitary matricesU = [u1, u2, . . . , um] FmmV = [v1, v2, . . . , vn] Fnn

such that

A = UV, = 1 0

0 0

where

1 =

1 0

0

0 2 0... ... . . . ...0 0 p

and

1 2 p 0, p = min

{m, n

}.


24/67

21

Geometrically, the singular values of a matrix A are precisely the lengths

of the semi-axes of the hyper-ellipsoid E defined by

E = {y : y = Ax, x Cn, x = 1}.Thus v1 is the direction in which y is the largest for all x = 1; whilevn is the direction in which y is the smallest for all x = 1.

v1

(vn

) is the highest (lowest) gain input direction

u1 (um) is the highest (lowest) gain observing direction

e.g.,

A =

cos 1 sin 1

sin 1 cos 1

1

2

cos 2 sin 2

sin 2 cos 2

.

A maps a unit disk to an ellipsoid with semi-axes of1 and 2.

alternative definitions:

(A) := maxx=1

Ax


25/67

22

Some useful properties

Let A

F

mn and

1 2 r > r+1 = = 0, r min{m, n}.Then

1. rank(A) = r;

2. KerA = span{vr+1, . . . , vn} and (KerA) = span{v1, . . . , vr};3. ImA = span{u1, . . . , ur} and (ImA) = span{ur+1, . . . , um};4. A Fmn has a dyadic expansion:

A =r

i=1iuivi = UrrVr

where Ur = [u1, . . . , ur], Vr = [v1, . . . , vr], and r = diag (1, . . . , r);

5. A2F = 21 + 22 + + 2r ;


26/67

23

Generalized Inverses

Let A Cmn. X Cnm is a right inverse if AX = I. one of theright inverses is given by X = A(AA)1.

Y A = I then Y is a left inverse ofA.

pseudo-inverseor Moore-Penrose inverse A+:

(i) AA+A = A;

(ii) A+AA+ = A+;

(iii) (AA+) = AA+;

(iv) (A+A) = A+A.

pseudo-inverse is unique.

A = BC

B has full column rank and C has full row rank. Then


27/67

24

Semidefinite Matrices

A = A is positive definite (semi-definite) denoted by A > 0 ( 0),ifxAx > 0 ( 0) for all x = 0.

A Fnn and A = A 0, B Fnr with r rank(A) such thatA = BB.

Let B Fmn and C Fkn. Suppose m k and BB = CC. U Fmk such that UU = I and B = U C.

square rootfor a positive semi-definite matrix A, A1/2 = (A1/2) 0,by A = A1/2A1/2.

Clearly, A1/2 can be computed by using spectral decomposition or

SVD: let A = UU, then

A1/2 = U1/2U


28/67

3

Reference Textbooks

G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic

Systems,3rd Edition, Addison Wesley, New York, 1994.

B. D. O. Anderson and J. B. Moore, Optimal Control, Prentice Hall, London, 1989.

F. L. Lewis, Applied Optimal Control and Estimation, Prentice Hall, Englewood Cliffs,

New Jersey, 1992.

A. Saberi, B. M. Chen and P. Sannuti, Loop Transfer Recovery: Analysis and Design,

Springer, London, 1993.

A. Saberi, P. Sannuti and B. M. Chen, H2Optimal Control,Prentice Hall, London, 1995.

B. M. Chen,Robust and HControl,Springer, London, 2000.

Prepared by Ben M. Chen


29/67

6

Revision: Basic Concepts



30/67

7

What is a control system?

System to be controlledController

Desiredperformance:

REFERENCE

INPUT

to the

system

Information

about thesystem:

OUTPUT

+

Difference:

ERROR

Objective:To make the system OUTPUTand the desired REFERENCE as close

as possible, i.e., to make the ERRORas small as possible.

Key Issues: 1) How to describe the system to be controlled? (Modelling)

2) How to design the controller? (Control)

aircraft, missiles,

economic systems,cars, etc



31/67

8

Some Control Systems Examples:

System to be controlledController+

OUTPUTINPUTREFERENCE

Economic SystemDesired

Performance

Government

Policies



32/67

9

A Live Demonstration on Control of a Coupled-Tank System through Internet Based

Virtual Laboratory Developed by NUS

The objective is to control the flow levels of two coupled tanks. It is a reduced-scalemodel of some commonly used chemical plants.



33/67

10m

uv

m

bv =+&

Modelling of Some Physical Systems

A simple mechanical system:

By the well-known Newtons Law of motion:f= m a, wherefis the total force applied to an

object with a massmandais the acceleration, we have

A cruise-control

system

force u

friction

forcebx&

x displacement

accelerationx&&

mass

m

m

ux

m

bxxmxbu =+= &&&&&&

This a 2nd orderOrdinary Differential Equation with respect to displacementx. It can be

written as a 1st orderODE with respect to speedv = :x&

model of the cruise control system,uis input force, vis output.



34/67

11

Controller+


A cruise-control system:

?+

speed vu90 km/h

m

uv

m

bv =+&



35/67

12

Basic electrical systems:

v

i

R

resistor

Riv =

capacitor

Cv (t)

i (t)

dt

dvCi =

inductor

Lv (t)

i (t)

dt

diLv =

Kirchhoffs Voltage Law (KVL):

The sum of voltage drops around any

close loop in a circuit is 0.

v5

v1

v4

v3

v2

054321 =++++ vvvvv

Kirchhoffs Current Law (KCL):

The sum of currents entering/leaving a

note/closed surface is 0.

i i

ii

i

1

23

4

5i i

ii

i

1

23

4

5

054321 =++++ iiiii



36/67

13

Modelling of a simple electrical system:

i

viR

C vo

To find out relationship between the input (vi) and the output (vo) for the circuit:

dtdvRCRivR

o==

dt

dvCi o=

By KVL, we have 0io =+ vvv R

0io

oio =+=+ vdt

dvRCvvvv R

iooioo vvvRCvv

dt

dvRC =+=+ & A dynamic model

of the circuit



37/67

14

Controller+


Control the output voltage of the electrical system:

?+

vovi230 Volts

viR

C vo

ioo vvvRC =+&



38/67

15

Ordinary Differential Equations

Many real life problems can be modelled as an ODE of the following form:

This is called a 2nd order ODE as the highest order derivative in the equation is 2. The ODE

is said to behomogeneousifu(t) = 0. In fact, many systems can be modelled or

approximated as a 1st order ODE, i.e.,

)()()()( 01 tutyatyaty =++ &&&

An ODE is also called the time-domainmodel of the system, because it can be seen the above

equations thaty(t) andu(t) are functions of time t. The key issue associated with ODE is: howto find its solution? That is: how to find an explicit expression fory(t) from the given equation?

)()()( 0 tutyaty =+&



39/67

16

State Space Representation

Recall that many real life problems can be modelled as an ODE of the following form:

)()()()( 01 tutyatyaty =++ &&&

If we define so-called state variables,

yx

yx

&=

=

2

1

uxaxauyayayx

xyx

+=+==

==

1021012

21

&&&&

&&

[ ]

==

+

=

2

1

1

2

1

102

101,

1

010

x

xxyu

x

x

aax

x

&

&

We can rewrite these equations in a more compact (matrix) form,

This is called thestate space representation of the ODE or the dynamic systems.



40/67

17

Laplace Transform and Inverse Laplace Transform

Let us first examine the following time-domainfunctions:

0 1 2 3 4 5 6 7 8 9 10-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

TIME (Second)

Magnitude

A cosine function with a frequencyf= 0.2Hz.

Note that it has a periodT= 5seconds.

( ) ( ) ( )ttttx 6.1cos8.0sin4.0cos)( +=

What are frequencies of this function?

0 1 2 3 4 5 6 7 8 9 10-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

TIME (Second)

Magnitude

Laplace transform is a tool to convert a time-domain function into a frequency-domain one

in which information about frequencies of the function can be captured. It is often much

easier to solve problems in frequency-domain with the help of Laplace transform.



41/67

18

Laplace Transform:

Given a time-domain functionf(t), its Laplace transform is defined as follows:

{ }

==0

)()()( dtetftfLsF st

Example 1: Find the Laplace transform of a constant functionf(t) = 1.

0)(,1

11

01111

)()(0

000>=

=

====

ssssesesesdtedtetfsFststst

Re

Example 2:Find the Laplace transform of an exponential functionf(t) = e a t.

( ) ( )as

ase

asdtedteedtetfsF

tastasstatst >+

=+

====

+

+

)(,

11)()(

0000

Re



42/67

19

Inverse Laplace Transform

Given a frequency-domain functionF(s), the inverse Laplace transform is to convert it back

to its original time-domain functionf(t).

( )2

2

1

1

1

1

1

)()(

aste

ase

st

s

sFtf

at

at

+

+

( )

( ) 22

22

22

22

cos

sin

cos

sin

)()(

bas

asbte

bas

bbte

as

sat

as

a

at

sFtf

at

at

+++

++

+

+

Here are some very useful Laplace and inverse Laplace transform pairs:



43/67

20

Some useful properties of Laplace transform:

{ } { } { } )()()()()()( 221122112211 sFasFatfLatfLatfatfaL +=+=+

1. Superposition:

2. Differentiation: Assume thatf(0) = 0.

{ } { } )()()()( ssFtfsLtfLdt

tdfL ===

&

{ } { } )()()()( 222

2

sFstfLstfLdt

tfdL ===

&&

3. Integration:

( ) { } )(1)(10

sFs

tfLs

dfL

t

==



44/67

21

Re-express ODE Models using Laplace Transform (Transfer Function)

Recall that the mechanical system in the cruise-control problem with m = 1 can be

represented by an ODE:

ubvv =+&

Taking Laplace transform on both sides of the equation, we obtain

{ } { } { } { } { }uLbvLvLuLbvvL =+=+&&

{ } { } { } )()()( sUsbVssVuLvbLvsL =+=+

( )bssU

sVsUsVbs

+==+

1

)(

)()()(

This is called the transfer function of the system model

)(sG=



45/67

22

Controller+


A cruise-control system in frequency domain:

driver? auto?+

speed V(s)U(s)R (s)

bssG

+=

1)(



46/67

23

In general, a feedback control system can be represented by the following block diagram:

+

U(s)R (s))(sG)(sK

Y(s)

E(s)

Given a system represented by G(s) and a referenceR(s), the objective of control system

design is to find a control law (or controller) K(s) such that the resulting output Y(s) is as

close to referenceR(s) as possible, or the errorE(s) =R(s) Y(s) is as small as possible.However, many other factors of life have to be carefully considered when dealing with real-

life problems. These factors include:

R (s)

+ U(s))(sG)(sK

Y(s)

E(s)

disturbances noisesuncertainties

nonlinearities



47/67

24

Control Techniques A Brief View:

There are tons of research published in the literature on how to design control laws for various

purposes. These can be roughly classified as the following:

Classical control:Proportional-integral-derivative (PID) control, developed in 1940s and used

for control of industrial processes.Examples: chemical plants, commercial aeroplanes.

Optimal control: Linear quadratic regulator control, Kalman filter, H2control, developed in

1960s to achieve certain optimal performance and boomed by NASA Apollo Project.

Robust control: Hcontrol, developed in 1980s & 90s to handle systems with uncertainties

and disturbances and with high performances. Example: military systems.

Nonlinear control: Currently hot research topics, developed to handle nonlinear systems

with high performances. Examples: military systems such as aircraft, missiles.

Intelligent control: Knowledge-based control, adaptive control, neural and fuzzy control, etc.,researched heavily in 1990s, developed to handle systems with unknown models.

Examples: economic systems, social systems, human systems.Prepared by Ben M. Chen


48/67

25

Classical Control

Let us examine the following block diagram of control system:

+

U(s)R (s))(sG)(sK

Y(s)

E(s)

Recall that the objective of control system design is trying to match the output Y(s) to the

referenceR(s). Thus, it is important to find the relationship between them. Recall that

)()()()(

)()( sUsGsY

sU

sYsG ==

Similarly, we have , and .)()()( sEsKsU = )()()( sYsRsE = Thus,

[ ])()()()()()()()()()( sYsRsKsGsEsKsGsUsGsY ===

[ ] )()()()()()(1)()()()()()()( sRsKsGsYsKsGsYsKsGsRsKsGsY =+=

)()(1

)()(

)(

)()(

sKsG

sKsG

sR

sYsH

+== Closed-loop transfer function fromR toY.



49/67

26

as

bsG

+

=)(

s

ksk

s

kksK

ipip

+=+=)(

Well focus on control system design of some first order systems with a

proportional-integral (PI) controller, . This implies

Thus, the block diagram of the control system can be simplified as,

)()(1

)()()(

sKsG

sKsGsH

+=

R (s) Y(s)

The whole control problem becomes how to choose an appropriateK(s) such that the

resultingH(s) would yield desired properties betweenR andY.

ip

ip

bksbkas

bksbk

sKsG

sKsGsH

++++

=+

=)()()(1

)()()(

2

The closed-loop systemH(s) is a second order system as its denominator is a polynomial s

of degree 2.



50/67

27

Stability of Control Systems

Example 1: Consider a closed-loop system with,

1

1)(

2 =

ssH

R (s) = 1 Y(s)

We have

1

5.0

1

5.0

)1)(1(

1

1

1)()()(

2 +

=

+=

==

ssssssRsHsY

Using the Laplace transform table, we obtain

ase at

+

1

1

5.0

5.0 +

set

1

5.05.0

set

)(5.0)(

tt

eety

=

This system is said to be unstablebecause the

output responsey(t) goes to infinity as time tis

getting larger and large. This happens because

the denominator ofH(s) has one positive root at

s = 1.

0 2 4 6 8 100

2000

4000

6000

8000

10000

12000

Time (secon ds)

)(ty

&



51/67

28

Example 2: Consider a closed-loop system with,

23

1)(

2 ++=

sssH

R (s) = 1 Y(s)

We have

2

1

1

1

)2)(1(

1

23

1)()()(

2 +

+=

++=

++==

sssssssRsHsY

Using the Laplace transform table, we obtaintt

eety2)( =

0 2 4 6 8 100

0.05

0.1

0.15

0.2

0.25

Time (sec onds)

)(ty

This system is said to be stablebecause

the output responsey(t) goes to 0 as time

tis getting larger and large. This happens

because the denominator ofH(s) has no

positive roots.



52/67

29

We consider a general 2nd order system,

The system is stable if the denominator of the system, i.e., , has no

positive roots. It is unstable if it has positive roots. In particular,

22

2

2

)(nn

n

ss

sH

++=

R (s) = 0 Y(s)

0222 =++ nn ss

Marginally Stable

Unstable

Stable



53/67

30

Stability in the State Space Representation

Consider a general linear system characterized by a state space form,

Then,

1. It is stable if and only if all the eigenvalues ofA are in the open left-half plane.

2. It is marginally stable if and only ifA has eigenvalues are in the closed left-half

plane with some (simple) on the imaginary axis.

3. It is unstable if and only ifA has at least one eigenvalue in the right-half plane.

u

u

D

B

x

x

C

A

y

x

++

==

&

L.H.P.

Stable Region

R.H.P.

Unstable Region



54/67

31

Lyapunov Stability

Consider a general dynamic system, . If there exists a so-called Lyapunov function

V(x), which satisfies the following conditions:

1. V(x) is continuous inx andV(0) = 0;

2. V(x) > 0 (positive definite);

3. (negative definite),

then we can say that the system is asymptotically stable atx = 0. If in addition,

then we can say that the system is globally asymptotically stable atx = 0. In this case, the

stability is independent of the initial conditionx(0).

)(xfx =&

0)()(


55/67

32

Lyapunov Stabili ty for Linear Systems

Consider a linear system, . The system is asymptotically stable (i.e., the eigenvalues

of matrixA are all in the open RHP) if for any given appropriate dimensional real positive

definite matrixQ = QT > 0, there exists a real positive definite solution P = PT > 0 for the

following Lyapunov equation:

Proof. Define a Lyapunov function . Obviously, the first and second conditions

on the previous page are satisfied. Now consider

Hence, the third condition is also satisfied. The result follows.

Note that the condition, Q = QT > 0, can be replaced by Q = QT 0 and being

detectable.

xAx =&

QPAPA =+T

xPxxV T=)(

( ) 0)()(


56/67

33

Behavior of Second Order Systems with a Step Inputs

Again, consider the following block diagram with a standard 2nd order system,

The behavior of the system is as follows:

22

2

2)(

nn

n

sssH

++=

R (s) = 1/s Y(s)

r= 1

The behavior of the system is

fully characterized by ,

which is called the damping

ratio, andn , which is called

thenatural frequency.



57/67

34

Control System Design with Time-domain Specifications

1% settling time

overshoot

rise time

strt

pM

22

2

2)(

nn

n

sssH

++=

R (s) = 1/s Y(s)

r= 1

tn

rt

8.1

n

st

6.4



58/67

35

ip

ip

bksbkas

bksbk

sKsG

sKsG

sR

sY

sH +++

+

=+== )()()(1)()(

)(

)(

)( 2

+

U(s)R (s))(sG)(sK

Y(s)

E(s)

PID Design Technique:

s

ksk

s

kksK

ipip

+=+=)(with and results a closed-loop system:

as

bsG

+=)(

The key issue now is to choose parameters kpand kisuch that the above resulting system

has desired properties, such as prescribed settling time and overshoot.

Compare this with the standard 2nd order system:

22

2

2)(

nn

n

sssH

++=

in

pn

bk

bka

=

+=2

2

bk

b

ak

ni

np

2

2

=

=



59/67

36

To achieve an overshoot less than 25%, we obtain

from the figure on the right that 4.0>

xTo achieve a settling time of 10 s, we use

767.0106.0

6.46.46.4 =

===s

n

n

st

t

6.0=To be safe, we choose

Cruise-Control System Design

Recall the model for the cruise-control system, i.e., . Assume that the

mass of the car is 3000 kg and the friction coefficient b = 1. Design a PI controller for it

such that the speed of the car will reach the desired speed 90 km/h in10 seconds (i.e., the

settling time is 10 s) and the maximum overshoot is less than 25%.

mbs

m

sU

sV

+=

1

)(

)(



60/67

37

The transfer function of the cruise-control system,

000333.03000

1

3000

13000

11

)(

)()( ===

+=

+== ba

sm

bs

m

sU

sYsG

bk

b

ak

ni

np

2

2

=

=

Again, using the formulae derived,

17653000/1

767.0

27603000/1

3000/1767.06.022

22

===

=

=

=

bk

b

ak

n

i

n

p

The final cruise-control system:

+

SpeedReference

90 km/h

s

17652760 +



61/67

38

Simulation Result:

The resulting

overshoot is

less than 25%

and the settling

time is about 10

seconds.

Thus, our

design goal is

achieved.0 2 4 6 8 1 0 1 2 1 4 1 6 18 20

0

2 0

4 0

6 0

8 0

10 0

12 0

Time in Secon ds

Speed

inkm/h



62/67

39

+

r)(sG)(sK

y

e

Bode Plots

Consider the following feedback control system,

+

r)()( sGsK

y

e

Bode Plots are the the magnitude and phase responses of the open-loop transfer function,

i.e., K(s) G(s), with s being replaced byj. For example, for the ball and beam system we

considered earlier, we have

( )222

3.27.33.27.31023.037.0)()(

+

=+

=+===

=

j

s

s

sssGsK

jsjs

js

o1807.3

3.2tan)()(,

)3.2(7.3)()( 1

2

22

=+=

jGjKjGjK



63/67

40

10-1

100

101

-20

0

20

40

60

Frequency (rad/sec)

Magnitude(dB)

10-1

100

101-180

-160

-140

-120

-100

-80

Frequency (rad/sec)

Phase(degrees)

Bode magnitude and phase plots of the ball and beam system:



64/67

41

10-1

100

101

-60

-40

-20

0

20

Frequency (rad/sec)

Magnitude(dB)

10-1

100

101

-250

-200

-150

-100

-50

0

Frequency (rad/sec)

Phase(degrees)

gain

crossoverfrequency phase

crossover

frequency

gain

margin

phase

margin

Gain and phase margins



65/67

42

-0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

Real Axis

ImagAxis

Nyquist Plot

Instead of separating into magnitude and phase diagrams as in Bode plots, Nyquist plot

maps the open-loop transfer function K(s) G(s) directly onto a complex plane, e.g.,



66/67

43

1

PM

GM

1

Gain and phase margins

The gain margin and phase margin can also be found from the Nyquist plot by zooming in

the region in the neighbourhood of the origin.

o

180)()(,)()(

1

== ppppp

jGjKjGjK thatsuchiswhereGM

Mathematically,

1)()(such thatiswhere,180)()(PM =+= ggggg jGjKjGjK o

Remark:Gain margin is the maximum

additional gain you can apply to the

closed-loop system such that it will still

remain stable. Similarly, phase marginis the maximum phase you can tolerate

to the closed-loop system such that it

will still remain stable.



67/67

44

10-1

100

101

-20

0

20

40

60

Frequency (rad/sec)

Magnitude(dB)

10-1

100

101-180

-160

-140

-120

-100

-80

Frequency (rad/sec)

Phase(degrees)

Example: Gain and phase margins of the ball and beam system: PM = 58, GM =


Documents

7360w1a Overview Linear Algebra