EEE8072 Subsea Control and Communications Systems Dr Damian Giaouris Senior Lecturer in Control of Electrical Systems [email protected],

EEE8072Subsea Control and Communications Systems

Dr Damian GiaourisSenior Lecturer in Control of Electrical [email protected],http://www.staff.ncl.ac.uk/damian.giaouris/

IntroductionSystem: 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 a certain 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):

f(t)

x(t)

friction

2

2

dt

xdm

dt

dxBfmaBf

maffmaF friction

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

First order ODEs

First order ODEs: ),( 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

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

General form:

Autonomouscbxax

autonomousNontcxtbxta

,'

),()(')(

ukxx ' t

ktktkt udteexex0

110

Output Input

Analytic solution

0 0.5 1 1.5 2-0.5

0

0.5

1

TotalTransient

Input component

0 0.5 1 1.5 2-0.5

0

0.5

1

Total

Transient

Input component

u=0, x(0)=1

k=2 k=5

Analytic solution

t

ktktkt udteexex0

110

0 0.5 1 1.5 20

10

20

30

40

50

60

Total

Transient

Input component

u=0, x(0)=1

k=-2 k=-5

0 0.5 1 1.5 20

0.5

1

1.5

2

2.5x 10

4

Total

Transient

Input component

Analytic solution

t

ktktkt udteexex0

110

k=5, u=0

x0=2 x0=5

0 0.5 1 1.5 20

0.5

1

1.5

2

Total

Transient

Input component

0 0.5 1 1.5 20

1

2

3

4

5

Total

Transient

Input component

Analytic solution

t

ktktkt udteexex0

110

k=5

u=-2 u=2

0 0.5 1 1.5 2-0.5

0

0.5

1

Total

Transient

Input component

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

Total

TransientInput component

Analytic solution

t

ktktkt udteexex0

110

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

Total

Transient

Input component

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

TotalTransient

Input component

u=1

k=2 k=5

Exercise

t

ktktkt udteexex0

110

33' xxA system is given by:

Using Predict how the system will behave for x0=2

Second 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

Overdamped 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

06'7'' xxx

Example

A 2nd order system is given by

1. Find the general solution2. Find the particular solution for x(0)=1, x’(0)=23. Describe the overall response

Critically 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

Underdamped 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

Underdamped 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

Undamped

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

NonHomogeneous Case

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.

Example

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

Previous analysis => Can be complicated => Polynomial expressions

10tydt

tdy

10 sYssY

Laplace Transform

0fssFdt

dtdfL

ssY

dt

dyL

Used only on LTI systems

Domain is a set of values that describe a function

The LT is transforming a DE from the time domain

to another complex domain (i.e. the variable has a real and imaginary part)

Transfer Functions

dttfetfLsF st )()()( js

Use formula tables

0fssFdt

dtdfL

dt

dfsfsFs

dt

dtfdL

002

2

2

ssFfs

ss0

lim

Final Value Theorem

Transfer Functions

Transfer functions

The ratio of the Laplace transform of the output over the Laplace transform of the input.

xout(t)xin(t)maf 2

2

dt

xdmf out

Spring

LT

IC

outoutin

dt

xdmxxK

02

2

sXmssKXsKX outoutin2

KmssXsKX outin2

Kms

K

sX

sX

in

out

2

Transfer function

of the system

Transfer Functions

2

2

dt

xdmKxKx out

outin outout

in Kxdt

xdmKx

2

2

outout Kx

dt

xdm

2

2

0 02 Kmr But before Kms

K

sX

sX

in

out

2

Exactly as the denominator of TF

Transfer Functions

v(t)

R

Lf

Km

x

B

2

2

x

ikf A ikf A

)()()()(

)()(1

)(

2 sXmssBsXskXsIk

RsIsVL

ssI

A

xmxBkxik

iRvLdt

di

A

1

)()()()(

)()()(2 sXmssBsXskXsIk

sVRsIssLI

A

)()()()(

)()(

2 sXmssBsXskXRsL

sVk

RsL

sVsI

A

)()( 2 sXkBsmsRsL

sVkA

kBsmsRsL

k

sV

sX A

2)(

)(

Transfer Functions

Automatic Control – EEE2002

v(t)

R L

EMF

Tm,θm

r1

r2

N1

N2

θoJ2

B

Tm,θm

To,θo

mT

aaaaa Kdt

diLRiv

aTm

m

iKT

Tn

nT

BTJ

1

20

000

01

20

BiK

n

nJ aT

sKssILRsIsV

sBsIKsJs

aaaaa

aLT

nn

KKnn

KK TT 01

0202

,2

12

1

21

sKsIsLRsV

sK

BJssI

aaaa

a

01

02

2

sKsK

BJssLRsV aaa 010

2

2

12

20 1

KK

BJssLR

sV

s

aaa

2122

KKBJssLR

K

aa

01

2

n

nK

dt

diLRiv T

aaaaa

Block diagrams

System

X(t) Y(t)

Input OutputG(s)

X(s) Y(s)

Block Diagram Algebra

X1(s)

X2(s)

X1(s)+X2(s) X1(s)

X2(s)

X1(s)-X2(s)

X1(s) X1(s)

X1(s)

G1(s)Y(s)X(s)

G2(s)Z(s)

G1(s) G2(s)Z(s)X(s)

G1(s)Y(s)

X(s)

G2(s)Z(s)

X(s)

X(s)

C(s)

G1(s)+G2(s)X(s) C(s)

Block Diagram Algebra

Pole location / s-plane

CE of ODE => Characterises the system

Den of TF = CE of ODE => CE of system

Order of ODE => Order of system

Order of Den => Order of system

2

2

dt

xdmxxK out

outin Kms

K

sX

sX

in

out

2

The order of the ODE is 2 = order of the denominator = order of the system

2

22

22212

1

21

KsmKKKsm

KK

sY

sZ

=> order = 4.

What about the num? Later…

s-plane

sD

sNsG )(

roots of the numerator=> zeros

roots of the denominator => poles (roots of CE!!!)

32

1

ss

s

One zero at s=-1 and two poles at s=-2 and s=+3

101 ssZeros:

3

2032

s

sssPoles:

0

j

-1-2 3

Poles, Zeros…

1

1)(

2

sssG

0

j

-0.5

+0.866j

-0.866j

roots( )pzmap()

Poles, Zeros…

In reality all systems have a forcing input signal (which we want to control)

0

r(t)

t

ttr

1sR

0

r(t)

t

Atr

s

AsR

0

r(t)

t

Attr

2s

AsR

0

r(t)

t

2Attr

3s

AsR

We need to study these more extensively

4 main types of inputs

Time domain => s domain=> Time domain to find soln.

Inputs

iRdt

diLV

LsRsV

sI

1 V(s) I(s)

LsR 1

0

j

LR

s

K

sV

sI

1

K=1/R and τ=L/R

1

s

KsVsI

1

1

ss

VKsI

Step response

t

eR

Vti 1

R

V

R

Ve

R

Vi

t

tss

01lim

R

VVK

s

K

s

VsI

sss

1lim

0

Pole at -R/L

tLR

eR

Vi 1

First order systems

Step Response

Time (sec)

Am

plitu

de

0 T 2T 3T 4T 5T 6T0

0.1(V/R)

0.2(V/R)

0.3(V/R)

0.4(V/R)

0.5(V/R)

0.6(V/R)

0.7(V/R)

0.8(V/R)

0.9(V/R)

1 (V/R)

First order systems

Step Response

Time (sec)

Am

plitu

de

0 T 2T 3T 4T 5T 6T0

0.1(V/R)

0.2(V/R)

0.3(V/R)

0.4(V/R)

0.5(V/R)

0.6(V/R)

0.7(V/R)

0.8(V/R)

0.9(V/R)

1 (V/R)

0.632 (V/R)

t

eR

Vti 1

eR

Vi 1

11 eR

Vi

11 eR

Vi

0.63211 1 e

0.6321R

Vi

So 63.21% of final value

First order systems

Time constant: How fast is the system

• The smaller the time constant • The faster the system • The further the pole is in the left hand side

0

j

100 0

j

10

0, ae at a Faster system

01 a 1

First order systems

xout(t)xin(t)

Friction

LT

IC

outoutoutin dt

xdm

dt

dxBxxK

02

2

sXmssBsXsKXsKX outoutoutin2

sXmsBsKsKX outin2

KBsms

K

sX

sX

in

out

2

mKsm

Bsm

K

sX

sX

in

out

2

22

2

2 nn

n

sssR

sC

m

Km

Bnn 2,2

12 nns

Second order systems

Case 1: 1

0

j

s1 s2

212

21

121

s

e

s

etc

tstsn

trtr eCeCxCxCx 21

212211

TransientSteady state

n

n

s

s

1

1

22

21

Second order systems

0 2 4 6 8 10 120

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Step Response

Time (sec)

Am

plitu

de

Overdamped system

Example: z=1.5, wn=2rad/s

1

tn

etc

12

1

Hence the component with s1 will converge very fast to 0

22

2

012

1s

etc

tsn

Second order systems

1ns

rtrt teCeCxCxCx 212211

tetc ntn 11

Step Response

Time (sec)

Am

plitu

de

0 2 4 6 8 10 120

0.5

1

1.5

Overdamped system

Critically damped system

12 nns

Example: z=1, wn=2rad/s

Second order systems

10 21 nn js dn js

0

j

n

dj

dj

The line between the origin and the pole is:

222ndd nnn 2222 1

1coscos n

n

21

2

1tansin

11 t

etc d

tn

btGex at cos r=a+bj

Second order systems

21

2

1tansin

11 t

etc d

tn

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

Sinusoidal term

Exponential term

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

Second order systems

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

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

-0.5

0

0.5

1

1.5

2

Second order systems

Step Response

Time (sec)

Am

plitu

de

0 2 4 6 8 10 120

0.5

1

1.5

Overdamped system

Critically damped systemUnderdamped system

Second order systems

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8Step Response

Time (sec)

Am

plitu

de

Example: z=0.5, wn=2rad/s

Second order systems

0

0

j

nj

nj

njs 0

The system is called marginally stable because the solutions do not diverge to infinity

ttc ncos1

Example: z=0, wn=2rad/s

Second order systems

0 5 10 15 20 250

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Am

plitu

de

Marginally stable

Underdamped

Critically damped

Overdamped

Second order systems

0

21

2

1tansin

11 t

etc d

tn

0 1 2 3 4 5-10

-5

0

5

10

1coscos n

n

0

j

n

dj

dj

Example: z=-0.5, wn=2rad/s

Second order systems

0

j

1

0

j

1

0

j

n

dj

dj

1,0cos

0

j

nj0

0

j

dj

dj

nj

0,1cos

0

j

1

j

1

0

j

n

dj

dj

nj

nj

1,0cos

1

0

1

0,1cos

1

1

Second order systems

Second order systems in s-plane

0

j

0 1 2 3 4 5 60

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5Step Response

Time (sec)

Am

plit

ud

e

0 1 2 3 4 5 60

0.005

0.01

0.015

0.02

0.025

0.03Step Response

Time (sec)

Am

plit

ud

e

0 1 2 3 4 5 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7Step Response

Time (sec)

Am

plit

ud

e

0 1 2 3 4 5 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7Step Response

Time (sec)

Am

plit

ud

e

0 1 2 3 4 5 60

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018Step Response

Time (sec)

Am

plit

ud

e

0 1 2 3 4 5 60

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018Step Response

Time (sec)

Am

plit

ud

e

0 1 2 3 4 5 60

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018

0.02Step Response

Time (sec)

Am

plit

ud

e

0 1 2 3 4 5 60

0.05

0.1

0.15

0.2

0.25Step Response

Time (sec)

Am

plit

ud

e

0 1 2 3 4 5 60

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018

0.02Step Response

Time (sec)

Am

plit

ud

e

0 1 2 3 4 5 60

0.05

0.1

0.15

0.2

0.25Step Response

Time (sec)A

mp

litu

de

0 1 2 3 4 5 60

2

4

6

8

10

12

14

16

18Step Response

Time (sec)

Am

plit

ud

e

0 1 2 3 4 5 6-3

-2

-1

0

1

2

3

4Step Response

Time (sec)

Am

plit

ud

e

0 1 2 3 4 5 6-3

-2

-1

0

1

2

3

4Step Response

Time (sec)

Am

plit

ud

e

0 1 2 3 4 5 60

0.5

1

1.5

2

2.5

3

3.5

4

4.5x 10

5 Step Response

Time (sec)

Am

plit

ud

e

0 1 2 3 4 5 60

0.5

1

1.5

2

2.5

3

3.5

4

4.5x 10

5 Step Response

Time (sec)

Am

plit

ud

e

Time Domain Characteristics

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

td

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

td

tr

drt

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

td

tr

tp

dpt

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

td

tr

tp

Mp

21eM p

nst

3%5

nst

4%2

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

td

tr

tp

Mp

1.05

0.95

ts

nnnn

mmm

asasasa

bsbsb

sR

sC

11

10

10

...

...

)(

)(n>m for a real system

r

k kkk

kkkkkkq

j j

j

ss

csb

ps

a

ssC

122

1 2

11

nrq 2

i.e. combination of first and second order systems

edssfscbsass

223

11 edssfscbsass 223

fesfdesfdscbsass 2323

fec

fdeb

fda

11

Extra Poles

r

k kkk

kkkkkkq

j j

j

ss

csb

ps

a

ssC

122

1 2

11

The response of a higher order system is the sum of exponential and damped sinusoidal curves.

Assuming that all poles are at the left hand side then the final value of the output is “1” since all exponential terms will converge to 0.

Let’s assume that some poles have real parts that are far away from the imaginary axis=>

r

kkk

tk

r

kkk

tk

q

j

tpj tectebeatc kkkkj

1

2

1

2

1

1sin1cos1

21

2

1tansin

11 t

etc d

tn

0 tne

Extra Poles

Overall performance is characterised by the isolated (far away from zeros) poles that are close to the imaginary axis.

If we have only one pole (or a pair for complex roots) that is closed to the real axis then we say that this pole (or pair of poles) is (are) the DOMINANT pole(s) for the system.

A simple rule is that the dominant poles must be at least five to ten times closer to the imaginary axis than the other ones.

r

kkk

tk

r

kkk

tk

q

j

tpj

tec

tebeatc

kk

kkj

1

2

1

2

1

1sin

1cos1

The values of b (numerator coefficients) determine the amplitude of the oscillations of the system but not its stability properties.

r

k kkk

kkkkkkq

j j

j

ss

csb

ps

a

ssC

122

1 2

11

Extra Poles

A system has two poles at -1 and -2, and two complex poles at -10+/-40j (num=1)1. Find the damping factors and natural frequencies of the system2. Do you expect the system to have oscillations?3. Find the step response of the system.4. Approximate the system by using two dominant poles.5. Find the new step response, natural frequency and damping factor.6. Compare that response with the response of the original system.7. Repeat the previous exercise by assuming that the second set of poles is at

-0.5+/-1.5j.8. At the system with the two poles at -0.5+/-1.5j add two zeros at -0.51+/-1.51j.

Extra Poles

Closed loop systems

The input to the system sees the output: Feedback

This is a CLOSED LOOP system.

General closed loop systems

UC

GGc+

-

H

Feedback

Feedback TF

Controller System

RE GGcG '

Gc= TF of controller

G TF of plant Open Loop (OL) TF

GUC

cEGU cGEGC

CHRE CHRGC ' RGHGC ''1

HG

G

R

C

'1

'

)()('1

)('

)(

)(

sHsG

sG

sR

sC

Negative feedback

Our Task: DESIGN Gc and H (if applicable).)()('1

)('

)(

)(

sHsG

sG

sR

sC

Assume H=1 and Gc=K=const.

KRLS

K

KG

KG

R

C

1 KR

K

KRLS

K

ssC

sss

1lim

0

If K>>R then Css=1

KRLS

K

R

C

1

SKR

LKR

K

R

C

newnew KRL faster system.

We can influence1. Steady state2. Transient

General closed loop systems

0 1 2 30

0.2

0.4

0.6

0.8

1Step Response

Time (sec)

Am

plit

ud

e

Step Response

Time (sec)

Am

plitu

de

0 1 2 30

0.2

0.4

0.6

0.8

1

Step Response

Time (sec)

Am

plitu

de

0 1 2 30

0.2

0.4

0.6

0.8

1

General closed loop systems

Summary

1st order systems with feedback. Þ the steady state and the time constant. Þ We moved the pole further into “minus infinity” area. Þ Hence we changed the s-plane of the system. Þ What is it going to happen if we use a 2nd order system???

2nd order systems

21

1

ssG U

CGK

+

-

Feedback

Controller System

RE

211

ssK

K

G

GGCL

Step Response

Time (sec)

Am

plitu

de

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4

10K

• Faster system • Smaller steady state error • Oscillations!

kss

k

kss

kGCL

2321 2 0232 kssCE:

02 22 nnss CE:

32 n

kn 22

12122 nn

433.03122

2nd order systems

The feedback and the controller

can completely change the location of the poles in the s-plane.

321

sss

KG

Step response for K=1, 10 & 100

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-4

-3

-2

-1

0

1

2

3

4

5

Step Response

Time (sec)

Am

plitu

de

3rd order systems

Summary

Properties of feedback systems:• Minimise steady state error.• Faster system.• Less sensitive to system uncertainties. • Introduce instability (even for negative feedback).• Expensive (we need to feedback the signal, i.e. use a sensor).

PI control

Steady state error problemINCREASE K => INCREASE the oscillations=>instability

321

1

sssGp K=10

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

error

output 321

sss

KGOL

KGc

PI control

KGc :s

KK ip1 321

1)(

ssss

KsKsG ip

OL

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

Ki=0

Ki=2

Ki=6

KGc :s

KK ip1 321

1)(

ssss

KsKsG ip

OL

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

Kp=10, K

i=6

Kp=20, K

i=10

PI control

0 2 4 6 8 10-1.5

-1

-0.5

0

0.5

1

De

e

PI control

PI control

PI control

Tuning of PID controllers

1.Trial and error.2.Ziegler Nichols I3.Ziegler Nichols II4.Root locus5.Frequency response6.Other advanced control methods

Tuning of PID controllers

P: Faster system, in some cases reduces the error (can cause instability).I: Reduces the steady state error, increases the number of oscillations.D: Reduces the oscillations.

Trial and error

Ziegler Nichols I

1)(

)(

Ts

Ke

sU

sC Ls

Ziegler Nichols I

1)(

)(

Ts

Ke

sU

sC Ls

Tuning of PID controllers

Ziegler Nichols II

Initially assume Ki=Kd=0.0

t,s

c(t)

Pcr

Kp=Kcr

Tuning of PID controllers

Root Locus

Target a pole location

3411

1)(

2

sssG 34113411

1)(

22

sss

KsK

sss

KKsG ipi

pOL

ip

ipCL

KsKsss

KsKsG

3411)(

2 This is a 3rd order system = 2nd order x 1st order:

222 23411 nnip ssasKsKsss

222323 223411 nnnnip asasasKKsss

2

2234

211

ni

nnp

n

aK

aK

a

5.0

6

n

180

100

5

36

36634

611

i

p

i

p

K

K

a

aK

aK

a

Tuning of PID controllers

Other controllers - 1

R(s)

C(s)

C(s) bs

sK

21

22

s

K

?

?21

b

KKo

n

60

6

221 2

s

K

bs

sKsG

2

22

21

sKsbs

sK

sR

sC KKK

02: 23 KKsbssCE

2223 22: nnssasKKsbssCE

222323 222: nnnn asasasKKsbssCE

2

2

2

2

2

n

nn

n

aK

aK

ab

54

3

9

K

a

b

15.01

sss

KsGOL Unity feedback and input: r(t)=5t

a) If K=1.5, find the steady state errorb) The system must have steady state error, Ess<0.1 find the value of K

C(s)

C(s)

15.01 sss

K

2

5

s sR

Ksss

K

sR

sC

15.01

Ksss

K

ssC

15.01

52

sCsRsE

Ksss

K

ssE

15.011

52

Ksss

ss

ssE

15.01

15.015

Ksss

ss

ssE

sss 15.01

15.015lim

0 KEss

5 a) ...33.3

5.1

5 ssE

1.0ssE 1.05 K

50 Kb)

Other controllers - 2

101.0102.0

sss

KsGOL

a) Find the value of K such as the system is marginally stable

b) Find the frequency of oscillations at that point

Ksss

K

sR

sC

101.0102.0 0101.0102.0: KsssCE For marginally stable system:

js 0

0101.0102.0: KjjjCE

jKjCE 0002.001.001.002.0: 223

003.0

001.002.02

3

K

srad /71.70

0

150K

Other controllers - 3

E(s)R(s)

C(s)

C(s)2

1 sK

1sK2

The control system of a space telescope

The targeting system must have

zero steady state error when

the targeting angle is 0.01o

Overshoot of 5%

Calculate the characteristic equation.

Find the closed loop poles and zeros.

Find the gains K1 and K2.

Determine the maximum allowed velocity of objects that the telescope can follow if the steady state error should not be greater than 0.5o/s. The objects are assumed to move with a constant velocity of Ao/s.

Other controllers - 4

Calculate the characteristic equation.

21

s

ksG 12 sksH 122

1 sks

ksGOL

21212

1

kkskks

ksGCL

021212 kkskks

Find the closed loop poles and zeros. No zeros

2

4 212

2121 kkkkkks

Find the gains K1 and K2.

The targeting system must have zero steady state error when the targeting angle is 0.01o

r=0.01 Css = 0.01 R=0.01/s 12 k2121

21

0

01.0lim

kkskks

k

ssCss

s

Overshoot of 5%

05.021

eH 6903.0 nkk 221 2

21 nkk

sradn /38.1nn 22 2n

9.11 k

9.19.1

9.12

ss

ss

sR

sEAEss

9.19.1

9.1

9.19.1

9.1

9.19.1

9.12222

sss

sA

ss

ss

s

A

ss

sssRsE

A=0.5

Other controllers - 4

Block diagram of a servo unit

Find the closed loop transfer function.

Create the block diagram of the closed loop system

Find the characteristic equation of the closed loop.

Prove that when K is constant the steady state error will increase if λ is increased.

Find the values of K and λ so that the system has a maximum overshoot of 40% and a peak time of 1s.

E(s)R(s)

C(s)

C(s)

1ss

K

λs1

Other controllers - 5

kkss

kGCL

12 R(s) C(s)

kkss

k

12

012 kkss kkss

kGCL

12 CLRGRCRE

kkss

k

s 11

122 CLGR 1

kkss

kss

s

1

112

2

2

kkss

kss

s

1

1122

kkss

kss

ssE

sss

1

11lim

220 k

k1

k

k

kkss

kG

n

nCL 22

12

1

279.04.021

eH

446.08.127.31 2

ktp n

n

Other controllers - 6