Chapter1 EEE3001 & EEE8013

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Chapter 1 EEE3001- 8013

Chapter #1

EEE3001 & EEE8013

Design of Linear Control Systems

Introduction

1st order dynamics

2nd order dynamics

Nonhomogeneous differential equations

Module Leader: Dr Damian Giaouris - [email protected] 1/25

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Introduction

System: is a set of objects/elements that are connected or related to each

other in such a way that they create and hence define a unity that performs acertain objective.

Control: means regulate, guide or give a command.

Task: To study, analyse and ultimately to control the system to produce a

satisfactory performance.

Model: Ordinary Differential Equations (ODE):

2

2

dt

xdm

dt

dxBfmaBfmaffmaF friction ====

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


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First order ODEs: ),( txfdt

dx=

Analytical solution: Explicit formula forx(t) (a solution which can be found

using separate variables, integrating factor) which satisfies ),( txfdt

dx=

Example:

The ODE adt

dx= has as a solution ( ) attx = .

First order Initial Value Problem : ( ) 00),,( xtxtxfdt

dx==

Analytical solution: Explicit formula for x(t) which satisfies ),( txfdt

dx=

and passes through when0x 0tt=

Example:

( ) ( ) ( ) 00 xattxCxCattxadtdxadt

dx+==+===

INFINITE curves (for all Initial Conditions (ICs)).

For that reason a different symbol is used for the above solution: ( )00 ,, xtt .

You must be clear about the difference to an ODE and the solution to an

IVP! From now on we will just study IVP unless otherwise explicitly

mentioned.


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Numerical solution: Using Simulink it is possible to see how the solution

behaves. In order to do that we place one integrator, set the IC (in the

integrator block) and then we create the signal

( )tx

( )txf , at the input of the

integrator. Remember the set the properties of the Simulation parameters

(I would say set the maximum step size to 1e-3) and of the Scope (change

in the Data History the option Limit data points to last:


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Example: Numerical solution of ( )txx cos2=& , ( ) 1.00 =x

In order to improve the presentation of our results I set in the Save data to

workspace option the following:

That way, once we run the simulation, we have in the command window the

matrix s which has in tits first column the values of the time and in the

second the values of our solution:


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>> plot(s(:,1),s(:,2))

0 1 2 3 4 5 6 7 8 9 100.09

0.095

0.1

0.105

0.11

0.115


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

General form:

=+

=+

Autonomouscbxax

autonomousNontcxtbxta

,'

),()(')(

Example: ukxx =+'

Numerical Solution:

k=5, u=0.5

Scope

1s

Integrator

5

Gain

0.5

Constant

0 1 2 3 40

0.02

0.04

0.06

0.08

0.1


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k=0.5, u=sin(10t)

Sine Wave Scope

1s

Integrator

0.5

Gain

0 5 10-0.1

-0.05

0

0.05

0.1

0.15

0.2


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Analytical Solutions:

Integrating factor (NOT ASSESSED MATERIAL):

ktconstk

kdtee

==

( ) ( ) uexeuekxxe ktktktkt ==+ ''

( ) = udtedtxe ktkt '

ceudteexcudtexe

ktktktktkt

+=+=

or: ( ) +=

tktktktudteexex

0

110

The equation that we derived is:

( ) +=t

ktktkt udteexex

0

110

Assuming that k>0 the first part is called transient and the second is called

steady state solution.


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Response to a sinusoidal input (Not assessed)

( )tkkyy cos' 11 =+

( ) ( )tkjkjyjytkkyy sin'sin' 2222 =+=+

( ) ( )tktkjkyykjyjy cossin'' 1122 +=+++

tjkeyky

=+ ~'~

Integrating factor: =>kte ( ) ( )tjkkt keey +='~

( )

+

=

+

= +

tj

tjkkt

etjk

ky

etjk

key

~

~

( )

+=

ktj

e

k

y

1tan

2

21

1~

( )

( )( )

+

=

+

=

k

tj

t

k

e

k

y

k

1

2

2

tan

2

2tancos

1

1

1

1Re~Re

1

( )

+

kt

k

1

2

2tancos

1

1

= magnified/attenuated amplitude and phase shifted.


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Example:

( tu )2cos= k=1, x(0)=0.1:

0.1

X0

Sine Wave1

Scope1

Product1

Product

eu

MathFunction1

eu

MathFunction

1s

Integrator1

1

Gain3

-1

Gain2

Clock1

Clock

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

Exercise: Find the response for k=0.5, u=1 and u=-1, initial conditions: 0, 1,

-1. Describe the systems behaviour.

Autonomous1st order ODEs => Liner Time Invariant (LTI) systems

)(xfdt

dx= (not t on RHS).

Analytic solution: Can be solved as before: Transient and steady state part.

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Second order ODEs: ),,'(2

2

txxfdt

xd=

Second order linear ODEs with constant coefficients: uBxAxx =++ '''

u=0 => Homogeneous ODE; I need two representative solutions

0''' =++ BxAxx , assume rtex = => rtrex =' & rterx 2'' = =>

=++=++ 00''' 2 rtrtrt BeAreerBxAxx

02

=++ BArr ; Characteristic or Eigenvalue equation => Check its roots.

2

42

BAAr

= , these are the Characteristic values or Eigenvalues.

Roots are real and unequal: r1 and r2 ( BA 42 > =>Overdamped system)

trex 1

1= and are solutions of the ODE =>trex 2

2=

trtreCeCxCxCx 21 212211 +=+= . If r1 and r2

C1=-0.5, C2=3/2 =>tt eex

+=2

35.0 3 :


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Numerical Solution:

Check ICs

Scope

1s

Integrator3

1s

Integrator2

4

Gain5

3

Gain1

0 5 100

0.2

0.4

0.6

0.8

1


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Analytical Solution:

Scope3

Scope2

Scope1

eu

MathFunction1

eu

MathFunction

3/2

Gain6

-1

Gain4

-0.5

Gain3

-3

Gain2

Clockt

0 5 10-0.5

0

0.5

1

1.5

OverallPart 2

Part 1

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Roots are real and equal: r1=r2 ( BA 42 = Critically damped system)

rtex =1 and =>

rttex =2

rtrtteCeCxCxCx 212211 +=+=

Example:

A=2, B=1, x(0)=1, x(0)=0 => c1=c2=1

Numerical Solution

Check ICs

Scope

1s

Integrator3

1s

Integrator2

2

Gain5

1

Gain1

Analytical Solution

Scope1

Product

eu

MathFunction

1

Gain6

1

Gain3

-1

Gain2

Clock

t*exp()

t


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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


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Roots are complex: r=a+bj BA 42 <

o Underdamped system 0A

So( )

( ))sin()cos( btjbteeeeeexatjbtatbjtattbjart

+=====

++

=Re+jIm.

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 btexbtexatat == =>

( ) =+=

+=+=

btGebtcbtcebtecbtecxcxcx

atat

atat

cos)sin()cos()sin()cos(

21

212211

( )

where

=

= 1

21

1

21

1 tan&,

tancosc

c

cc

cG

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

Numerical Solution

Check ICs

Scope

1s

Integrator3

1s

Integrator2

1

Gain5

1

Gain1

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Analytical Solution

sin

TrigonometricFunction1

cos

TrigonometricFunction

Scope1Product

eu

MathFunction

1/sqrt(3)

Gain6

1

Gain4

sqrt(3)/2

Gain3

-0.5

Gain2

Clock

bt

t

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


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o Undamped system 0=A

BrBeerBxxrtrt ==++=++ 22 0000'' => Imaginary roots (If Bjbr= ( )=+= btGbtcbtcx cos)sin()cos( 21

A=0, B=1, x(0)=1, x(0)=0 =>c1=1, c2=0:

Check ICs

Scope

1s

Integrator3

1s

Integrator2

0

Gain5

1

Gain1

sin

TrigonometricFunction1

cos

TrigonometricFunction

Scope1Product

eu

MathFunction

0

Gain6

1

Gain4

1

Gain3

0

Gain2

Clockt

bt


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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

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.


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Root Space

jb

a

jb

a

Name Oscillations? Components of solution

Overdamped No Two exponentials:

tktkee 21 , , 0, 21

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Natural frequency, damping frequency, damping factor

(NOT ASSESSED MATERIAL)

2nd order systems very important with rich dynamic behaviour

So =>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

4222nnnBAA

r

=

=

222nnnr =

1. Real and unequal 110

2222 >>>>

nn => Overdamped system

implies that 1> ; 2222,1 nnnr = => replace at

tr.

treCeCx 21 21 +=

2. Real and equal 1 => Critically damped12222 === nn

system implies that 1= ; nr = =>tt nn teCeCx

+= 1 2 .

3. Complex 110

2222 Underdamped systems

implies 1 ( ) = tGex dtn cos ; d is called damped

frequency or pseudo-frequency.

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4. Imaginary roots 0= and therefore the solution is njr = =>

( ) ; so when there is no damping the frequency of the= tGx ncos

oscillations = natural frequency. ( ) = tGx ncos

5. In all the previous cases if 0> then the transient part tends to zero. If

0

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NonHomogeneous (NH) differential equations

uBxAxx =++ '''

u=0 => Homogeneous => x1 & x2.

Assume a particular solution of the nonhomogeneous ODE: xp

o 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.

Example:

22''' ==++ Pxxxx

( ))sin()cos(22 212211 btcbtcexcxcxat ++=++=

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


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cos

Trigonometric

Function1

sin

Trigonometric

Function

Scope

1

s

Integrator1

1

s

Integrator

1

Gain12

Scope5

Scope4

Scope3

Scope2

Scope1

Product

eu

Math

Function

-1/sqrt(3)

Gain5

sqrt(3)/2

Gain4

-1

Gain3

-0.5

Gain2

Constant

Clockt

bt sin(bt)

cos(bt)

exp(-at)

Numerical Part

0 2 4 6 8 10-1

-0.5

0

0.5

1

1.5

2

2.5

Overall

Homo geneous solution

Particular solution

Documents

Chapter1 EEE3001 & EEE8013