autumn 2008 eee8013 revision lecture 1 ordinary differential equations

Autumn 2008

EEE8013

Revision lecture 1

Ordinary Differential Equations

Autumn 2008

Model:

Ordinary Differential Equations (ODE):

ModelingModeling

f(t)

x(t)

frictionmaF

Dynamics: Properties of the system, we have to solve/study the ODE.

2

2

dt

xdm

dt

dxBf

maff friction

maBuf

Autumn 2008

First order ODEs:

First order systems: Study approachesFirst order systems: Study approaches

),( txfdt

dx

Analytic:

Explicit formula for x(t) (a solution – separate variables, integrating factor)

which satisfies ),( txfdt

dx

adt

dx INFINITE curves (for all Initial Conditions (ICs)). Cattx adtdx

Autumn 2008

First order linear equationsFirst order linear equations

First order linear equations - (linear in x and x’)

General form:

Autonomouscbxax

autonomousNontcxtbxta

,'

),()(')(

ukxx '

Numerical Solution: k=5, u=0.5

u

u Scope

1s

Integrator

k

Gain

x

x

Dxu

0 1 2 3 40

0.02

0.04

0.06

0.08

0.1

Autumn 2008

Analytic solution: Step inputAnalytic solution: Step input

ktconstkkdt

ee

uexeuekxxe ktktktkt ''

udtedtxe ktkt '

ceudteexcudtexe ktktktktkt

t

ktktkt udteexex0

110

Autumn 2008

Response to a sinusoidal inputResponse to a sinusoidal input

tkkyy cos' 11 tkkyy sin' 22

tktkjkyykjyjy cossin'' 1122 tjtkjyykjyy sincos'' 2121

tjkeyky ~'~

kte tjkkt keey '~ tjkkt etjk

key

~

ktj

e

k

y

1tan

2

21

1~

ktje

k

y

1tan

2

21

1Re~Re

tkjkjyjy sin' 22

tjetjk

ky

~

kt

k

1

2

2tancos

1

1

Autumn 2008

Response to a sinusoidal inputResponse to a sinusoidal input

Sine Wave

Scope3

Scope2

Scope1

Scope

Product1Product

eu

MathFunction1

eu

MathFunction

1s

Integrator1

1s

Integrator

0.1

Gain3

k

Gain2

-k

Gain1

k

Gain

Clock

total

total

steady state

Transient

0 2 4 6 8 10-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

TransientSteadystate Overall

Autumn 2008

Second order ODEsSecond order ODEs

Second order ODEs: ),,'(2

2

txxfdt

xd

uBxAxx ''' u=0 => Homogeneous ODE

0''' BxAxx rtex rtrex ' rterx 2''

00''' 2 rtrtrt BeAreerBxAxx 02 BArr

21

2

,2

4xx

BAAr

2211 xCxCx

rtex

adt

xd

2

2

1Catdt

dx21

25.0 CtCatx So I am expecting 2 arbitrary constants

Let’s try a

Autumn 2008

Overdamped systemOverdamped system

BA 42 2

42 BAAr

Roots are real and unequal

trex 11 trex 2

2 trtr eCeCxCxCx 21212211

03'4'' xxx

tt eCeCx 23

1

10 x 00' x 10 21 CCx

tt eCeCx 23

13'

030' 21 CCx

tt eex 235.0 3

0130342 rrrr

rtex

0 1 2 3 4 5 6-0.5

0

0.5

1

1.5

Overall solution

x2

x1

Autumn 2008

Critically damped systemCritically damped system

BA 42 2

42 BAAr

Roots are real and equal

rtex 1rttex 2

rtrt teCeC

xCxCx

21

2211

A=2, B=1, x(0)=1, x’(0)=0 => c1=c2=1

0 2 4 6 8 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

e-t

te-t

overall

Autumn 2008

Underdamped systemUnderdamped system

BA 42 2

42 BAAr

Roots are complex Underdamped system

r=a+bj

0A

jbtatee

Theorem: If x is a complex solution to a real ODE then Re(x) and Im(x) are the real solutions of the ODE:

)sin(),cos( 21 btexbtex atat )sin()cos( 212211 btecbtecxcxcx atat

1

21

1

21

1 tan&,tancos

cc

cc

cG

bjtattbjart eeex )sin()cos( btjbteat

btGebtcbtce atat cos)sin()cos( 21

Autumn 2008

Underdamped system, exampleUnderdamped system, example

A=1, B=1, x(0)=1, x’(0)=0 => c1=1, c2=1/sqrt(3)

0 2 4 6 8 10-1.5

-1

-0.5

0

0.5

1

1.5

C1cos(bt)

C2sin(bt)

C1cos(bt)+C

2sin(bt)

0 2 4 6 8 10-1.5

-1

-0.5

0

0.5

1

1.5

C1cos(bt)+C

2sin(bt)

eat

Overall

)sin()cos( 21 btcbtceat btGeat cos

Autumn 2008

UndampedUndamped

0A Undamped system 00'' Bxx

btGbtcbtcx cos)sin()cos( 21 A=0, B=1, x(0)=1, x’(0)=0 =>c1=1, c2=0:

0 2 4 6 8 10-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Overall

BrBeer rtrt 22 00

Autumn 2008

StabilityStability

In all previous cases if the real part is positive then the solution will diverge to infinity and the ODE (and hence the system) is called unstable.

jb

a

Critical or overdamped

underdamped

jb

a

Stable

Unstable

Autumn 2008

Natural frequency, damping frequency, damping factorNatural frequency, damping frequency, damping factor

2,2 nn BA 0'2'' 2 xxx nn

is the damping factor and n is the natural frequency of the system.

2

422

2

4 222nnnBAA

r

222nnnr

110

2222

nn

2222,1 nnnr

=> Overdamped system implies that 1

trtr eCeCx 2121

Case 1:

Autumn 2008

Natural frequency, damping frequency, damping factorNatural frequency, damping frequency, damping factor

112222 nn => Critically damped system implies that 1

nr tt nn teCeCx 21

110

2222

nn => Underdamped systems implies 1

22222,1 1 nnnnn jjr

dn jnd

21

tGex dtn cos d called damped frequency or pseudo-frequency

njr tGx ncos0 No damping the frequency of the oscillations = natural frequency

Case 2:

Case 3:

Case 4:

Autumn 2008

SummarySummary

Autumn 2008

Stability revisedStability revised

0If then cases 1-3 are the same but with 0k

dj

n

Critical or overdamped

underdamped

Stable

Unstable

dj

n

Autumn 2008

NonHomogeneous (NH) differential equationsNonHomogeneous (NH) differential equations

uBxAxx '''

u=0 => Homogeneous => x1 & x2. Assume a particular solution of the nonhomogeneous ODE: xp

If u(t)=R=cosnt => B

RxP

Then all the solutions of the NHODE are 2211 xcxcxx P

So we have all the previous cases for under/over/un/critically damped systems plus a constant R/B.

If complementary solution is stable then the particular solution is called steady state.

Autumn 2008

ExampleExample

22''' Pxxxx )sin()cos(22 212211 btcbtcexcxcx at

x(0)=1, x’(0)=0 => c1=-1, c2=-1/sqrt(3)

0 2 4 6 8 10-1

-0.5

0

0.5

1

1.5

2

2.5

Overall

Homogeneous solution

Particular solution

Autumn 2008

State SpaceState Space

1,...'',', nn xxxxfx

Very difficult to be studied => so we use computers

Computers are better with 1st order ODE

1 nth => n 1st

Powerful tools from the linear algebra

kxxBFxm

xx

xx

2

1

mFx

x

m

B

m

k

x

x 010

2

1

2

1

xx1 2x

xx2 12

1kxBxF

m

Fmx

x

m

B

m

k

x

x

1

010

2

1

2

1 UBAXX

Use sensors: Output = x =>

DUyx

xxy

CX

2

101

Autumn 2008

State SpaceState Space

DUy

U

CX

BAXX Input Output

U yDUy

U

CX

BAXX

U1 y1

DUCX

BUAXX

y

U1

Uq

y2

yp

DUCXy

BUAXX

U Y

pq y

y

y

y

u

u

u

3

2

1

2

1

, YU

)()()()()(

)()()()(

ttttt

ttttt

UDXCY

UBXAX

Autumn 2008

Block DiagramBlock Diagram

)()()()()(

)()()()(

ttttt

ttttt

UDXCY

UBXAX

•X is an n x 1 state vector•U is an q x 1 input vector•Y is an p x 1 output vector•A is an n x n state matrix•B is an n x q input matrix•C is an p x n output matrix•D is an p x m feed forward matrix (usually zero)

)()()(

)()()(

ttt

ttt

DUCXY

BUAXX

BU dt C

DX X Y

A

Autumn 2008

State space rulesState space rules

The state vector describes the system => Gives its state =>

The state of a system is a complete summary of the system at a particular point in time.

If the current state of the system and the future input signals are known then it is possible to define the future states and outputs of the system.

The choice of the state space variables is free as long as some rules are followed:1. They must be linearly independent.2. They must specify completely the dynamic behaviour of the system.3. Finally they must not be input of the system.

Autumn 2008

State spaceState space

The system’s states can be written in a vector form as:

Tx 0,,0,11 x ...0,,,0 22Tx x Tnn x,,0,0 x

A standard orthogonal basis (since they are linear independent) for an n-dimensional vector space called state space.

x1

x2 R2

x2

x3 R3

Matlab definition

Autumn 2008

SolutionSolution

2

1

2

1

52

22

x

x

x

x

067

41025

252

5

2

1

252

125

2

1

52

1

52

1

52

22

222

22222

22222

22222

221

221

212

211

xxx

xxxxx

xxxxx

xxxxx

xxx

xxx

xxx

xxx

067 222 xxx

0672 rr

ta Cex

tb Dex 6

Autumn 2008

Solution IISolution II

2

1

2

1

52

22

x

x

x

x tea

a

2

1X

21

21

2

1

2

1

2

1

52

22

52

22

aa

aa

a

ae

a

ae

a

a tt

X

tea

a

2

1X

How can we solve that ???

Assume is a parameter => A homogeneous linear system

052

022

21

21

aa

aa

0672 052

22

(This last equation is the characteristic equation of the system, why???).

Autumn 2008

Solution IIISolution III

6

1067

2

12

11

042

02

21

21

aa

aa

I assume that a2=1 so a1=2 :1

1

2

2

1

a

a tet

1

2)(1X

61

02

024

21

21

aa

aaI assume that a1=2 so a2=-2 tet 6

2 6

1)(

X

tt eCeCt 621 2

1

1

2

X

Matlab example

Autumn 2008

General SolutionGeneral Solution

AXX

tec eX 0 eAIeAe 0 AI

The roots of this equation are called eigenvalues

ie in ...221

negative eigenvalues => stable

positive eigenvalues => unstable

repeated eigenvalues => eigenvectors are not linearly independent.

Complex eigenvalues => conjugate and the eigenvector will be complex =>solution will consists of sines, cosines and exponential terms

Autumn 2008

Properties of general solutionProperties of general solution

If we start exactly on one eigenvector then the solution will remain on that forever.

Hence if I have some stable and some unstable eigenvalues it is still possible (in theory) for the solution to converge to zero if we start exactly on a stable eigenvector.

tii

ieC e

tie Determines the nature of the time response (stable, fast..)

ie Determines the extend in which each state contributes to tie

iC Determines the extend in which the IC excites the tie

To find the eigenvalues and eigenvectors use the command eig()

Autumn 2008

ExampleExample

2

1

2

1

52

22

x

x

x

x

-1 -0.5 0 0.5 1 1.5-1

-0.5

0

0.5

1

1.5

x1

x 2

state space

0,0

e1

e2

Autumn 2008

ExampleExample

Autumn 2008

State Transition MatrixState Transition Matrix

Until now the use of vector ODEs was not very helpful.

We still have special cases 0 AI

axx

0xetx at AXX

0XX Ate

teA => No special cases are needed then

...!3

1!2

1 32 ttte t AAAA Use the command expm (not exp)!

2

1

2

1

52

22

x

x

x

x

52

22A

X(0)=[1 2]

X(5) =?

4 ways to calculate it!!!

Autumn 2008

0XX Ate

iii λeAe

n

nn

1

2121 eeeeeeA

TΛAT 1TΛA

...!3

1

!2

1 31211 ttt TΛTΛTΛ ...!3

1

!2

1 32 ttte t AAAA

1211 TΛTΛTΛ 21TΛ

...!3

1

!2

1 31321211 ttte t TΛTΛTΛTIA

1132 ...!3

1

!2

1

tettt ΛTΛΛΛT

t

t

t

ne

e

e

1

Λ


Autumn 2008

00 1XTXX ΛA tt ee

012121

1

XeeeeeeX

nt

t

nne

e

02

1

21

1

X

w

w

w

eeeX

n

t

t

nne

e

n

ii

ti bet i

1

0eX ...00 22211121 xexet tt weweX

n

iii

ti xe i

1

0we

)0(

)0(

)0(

2

1

2

1

21

1

nn

t

t

n

x

x

x

e

e

n

w

w

w

eeeX


Autumn 2008

SolutionSolution

buaxxbuaxx

ate txedt

daxxe atat

t

at

a budedxedt

d

00

bue at

t

aat budextxe0

0

t

aatat budeexetx0

0 t

taat budexetx0

0

t

tt deet0

0 BUXX AA

Autumn 2008

SS => TF???SS => TF???

LTttt

)()()( BUAXX )()()0()( ssss BUAXXX

)0()()( 11 XAIBUAIX ssss )0()()( XBUXAI sss

)()()( sss DUCXY )()0()()( 11 sssss DUXAIBUAICY

)0()()( 11 XACUDBACY sIssIs

DBAIC 1s TF

)0(1XAIC s Response to ICs

Autumn 2008

SS => TF???SS => TF???

ILT

ssss )0()()( 11 XAIBUAIX

)0()()( 1111 XAIBUAIX sLssLt

)0()( 11 XAIX sLt

)0()( XX Atet 11 AIA sLe t

DBAICG 1ss AsI CE of the TF AI

BCDAIG

s

ss

The TF is a matrix

)()()(

)()()(

)()()(

)(

21

22221

11211

sGsGsG

sGsGsG

sGsGsG

s

pqpp

q

q

G ...,,, 222

221

1

212

2

111

1

1 GUY

GUY

GUY

GUY

Autumn 2008

Example: Find the TF of 21

0

5.01

10

2

1

2

1

x

x

x

x

2

101x

xy

SS => TF???SS => TF???

Autumn 2008

Basic properties of state space Basic properties of state space

State space transformations

State space representations are not unique

Same input/output properties, => same eigenvalues

)()()( ttt BUAXX

)()()( ttt DUCXY

TZXT is an invertible matrix

Z is the new state vector

)()( 11 tt BUTAXTZ

XTZ 1

XTZ 1

)(~~

)(11 tt UBZAZBUTATZTZ

ATTA 1~ BTB 1~ )(~~

)( tt UDXCY )()( tt DUCTXY

CTC~DD~

Autumn 2008

Do these two systems have the TF? DBAICG 11 ss DBAICG

~~~~ 12

ss

DBTTAITTCG~111

1 ss

DBAICG 11 ss

DBTTAITCTG~111

1 ss

DBTATICG

~~~ 111 ss

sss 21

1~~~~

GDBAICG

Matlab example

Basic properties of state space Basic properties of state space

Autumn 2008

Observability - Controllability

xy

uxx

3

22

X

XX

03

1

2

10

02

y

uNotice the structure of A and C

2

6223

3

22 1

ss

xy

uxx

X

XX

03

1

2

10

02

y

u

1

2

10

0203

1

s

s

1

22

2

0312

2

1203

12

1

2

20

0103

s

sss

s

s

ss

s

s

2

6

s

Observability - ControllabilityObservability - Controllability

Autumn 2008

There is a pole zero cancellation

0

DC

BAIs

pzmap(ss_model) Matlab verification

The cancellation is due to C=[3 0].

We can influence x2 through U but we cannot observe how it behaves

and hence there is no way to feedback that signal to a controller!!!

0

001

110

202

s

s

001

102

01

100

00

112

sss

101 ss

X

XX

03

1

2

10

02

y

u

2

1

2

1

2

1

03

1

2

10

02

x

xy

ux

x

x

x

1

2

1

3

2

2

1

xy

uxx

uxx


Autumn 2008

X

XX

03

1

2

10

02

y

u

21

0

2

20

0123

ss

s

s

X

XX

23

0

2

10

02

y

u

2

6

0

2

1

2

2

3

sss

In this case we can see how both states behave but we can not change U in any way so that we can influence x2 due to the form of B.

Unobservable & uncontrollable Unobservable & uncontrollable Minimal realisation.

Difficult task if the system is nonlinear!!!!


Autumn 2008

1

2

n

O

CA

CA

CA

C

M

BABAABBM 12 nC

Check the rank

>> rank(obsv(A,C)) >> rank(ctrb(A,B))


autumn 2008 eee8013 revision lecture 1 ordinary differential equations

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