Adv Control & Robotic Lec 5

8/14/2019 Adv Control & Robotic Lec 5

1/31

METR4202/7202AdvancedControl&Robotics,

Semester2,

2006:Page:1

METR4202 Advanced Control & Robotics

Lecture 5

The z-DomainDiscrete Transfer Functions

Discrete Stability (Bilinear Transformation)Discrete Control Design (Tustin Transformation)

Control Course Overview

G. Hovland 2004-2006


Semester2,

2006:Page:2

Continuous and Discrete Systems

Controller System-

yref yu

Continuous Control System:

Discrete Control System:

Controller System-

yref

D/A A/D


2/31


Semester2,

2006:Page:3

Digital to Analog (D/A) Converter

D/A Simple and effectively instantaneous


Semester2,

2006:Page:4

A/D Successive Approximation

D/A Converter

ComparatorCircuit

Analog Input Signal

Higher/Lower

Binary search from MSB to LSB


3/31


Semester2,

2006:Page:5

Analog to Digital (A/D) Converter

Analog Signal Zero-order Sample and Hold

In general,M / 2n voltagelevels.

A/D is not instantaneous

Quantisation

Errors


Semester2,

2006:Page:6

Digital vs Analog Analysis

If A/D could be made instantaneous, there would be noneed to differentiate between digital and analog controlsystems.

In practice, A/D requires time (eg. comparator circuit).We will need to model the sample-and-hold process.

Note that a system that is stable with an analogcontroller, can become unstable with the same controllerbut implemented as a digital controller with a slowsampling rate.


4/31


Semester2,

2006:Page:7

a. switch openingand closing;b. product of timewaveform and

sampling waveform

Two Views of Sampling

Constant sampling times

The waveform view b) will be usedin the following analysis

Whenever you see * in thefollowing, it refers to the sampledfunction!


Semester2,

2006:Page:8

Sample Pulse Function u(t) - u(t-Tw)

t

u(t)

1

0Tw

t

1

0Tw

u(t-Tw)

t

1

0Tw

u(t) - u(t-Tw)

Laplace of a step iss

1

Laplace of a delayed step is

s

esTw

Laplace of a pulse is

s

esTw1


5/31


Semester2,

2006:Page:9

The Sampled Function f*Tw

=

=== kk wTW TkTtukTtutftstftf )()()()()()(

*

k: integer

T: pulse train period


Semester2,

2006:Page:10

The Sampled Function f*Tw

Assumption: Tw is small in comparison to T

f(t) is constant in sampling interval

[ ]=

==

=

k

k wTW TkTtukTtukTftf

kTftf

)()()()(

)()(

*

f(kT) moved inside the summation


6/31


Semester2,

2006:Page:11

The Laplace Transform f*Tw F*Tw(s)

[ ]

kTsk

k

sT

k

k

sTkTskTs

Tw

k

k wTW

es

ekTf

s

e

s

ekTfsF

TkTtukTtukTftf

w

w

=

=

=

=

=

=

=

=

=

1)(

)()(

)()()()(

*

*


Semester2,

2006:Page:12

Taylor Series Expansion of F*Tw(s)

=

=

=

=

=

=

=

=

=

+

=

=

k

kwTw

k

k

kTs

w

kTsk

k

w

w

kTsk

k

sT

Tw

kTtkTfTtf

eTkTf

es

sT

sTkTf

es

ekTfsF

w

)()()(

)(

!2

)(

11)(

1)()(

*

2

*

Inverse Laplace Dirac delta function

Assuming Tw small


7/31


Semester2,

2006:Page:13

Sampler Model: Uniform Rectangular Pulse Train

Ideal Sampler:

this will be used forfurther analysis

Sample waveform dependency

(in this example rectangular

pulse train)


Semester2,

2006:Page:14

Ideal Sample and Hold

Individual step responses with length T

s

esG

Ts

h

=1

)(

Delay T because step response

ends at T


8/31


Semester2,

2006:Page:15

Summary so Far

Ideal Sampler

Zero-order Sample-and-Hold (z.o.h.)

We are now ready to introduce the z-transform

s

esG

kTtkTftf

Ts

h

k

k

=

=

=

=

1)(

)()()(

*


Semester2,

2006:Page:16

The z-Transform (Nise Chapter 13.3)

Ideal Sampler

s-domain (Continuous control systems)

z-domain (discrete control systems)

=

=

=

=

=

=

=

0

0

*

*

)()(

)()(

)()()(

kk

k

kTs

k

k

zkTfzF

ekTfsF

kTtkTftf

Tsez =1

Time delay!


9/31


Semester2,

2006:Page:17

Simple Example z-Transform

f(kT) = u(kT) - 2u([k-1]T) - 3u([k-2]T)

F(z) = 1 - 2z-1 - 3z-2

Each further delay increases exponent of z


Semester2,

2006:Page:18

Example 13.1

Find the z-transform of a sampled unit ramp

Open Form


10/31


Semester2,

2006:Page:19

Example 13.1 - cont'd

What is the Laplace transform F(s) of a ramp?

Note F(z) is closed-form while F*(s) is not. This is one reason whythe new z-transform is introduced. The Laplace F(s) is closed-form,but the sampled F*(s) is not!


Semester2,

2006:Page:20

Partial Table of s- and z-Transforms

These 3 are the main ones you will use in practice


11/31


Semester2,

2006:Page:21

z-Transform Theorems


Semester2,

2006:Page:22

Example 13.4

Given a z.o.h. in cascade with G1(s) = (s+2) / (s+1) or

Find the sampled-data transfer function, G(z), if the sampling

time T=0.5 seconds

1

21)(

+

+=

s

s

s

esG

Ts

G1(s)Zero-order-hold

s

esG

Ts

=1

)(2

By default, assuming sampled inputs and outputs

G(s) G(z)


12/31


Semester2,

2006:Page:23

Example 13.4 - cont'd

Again, partialfractions!

Tez

z

s

z

z

s

+

1

1

1

1


Semester2,

2006:Page:24

z-Domain Stability

1:01:0

1:0

)(

1

>

==

====

=

+

T

T

T

TTjTjTTs

Ts

ee

e

Teeeeez

ez

Region B (marginally stable)

Region C (unstable)

Region A (stable)


13/31


Semester2,

2006:Page:25

z-domain stability directly from s-domain

There is no equivalent Routh-Hurwitz stability criterionfor discrete control systems

A simple transformation allows us to check discretestability by transforming to s-domain and applying theRouth-Hurwitz criterion as normal.

Bilinear transformations

1

1

1

1

+=

+=

s

sz

z

zs


Semester2,

2006:Page:26

Bilinear Transformations

01

01

01

)1(

)1(

)1(

)1(

22

22

==

>>


14/31


Semester2,

2006:Page:27

Example 13.8: Discrete Stability

Closed-Loop Discrete denominator:

Use the Routh-Hurwitz criterion to determine stability

1.02.023

+ zzz

17451911 23

+= sss

ssz

How manypoles outside

the unit circlein z-domaindoes the systemhave?


Semester2,

2006:Page:28

s-domain design, z-domain implementation

We want to re-use all the techniques we have developedfor continuous systems.

Ideally, we want to design everything in s-domain andthen convert to z-domain as the last step beforeimplementation on the real-time controller.

The Tustin approximation will allow us to do this

This transformation yields a digital transfer functionwhose output response at the sampling instants isapproximately the same as the analog transfer function.

sT

sT

zz

z

Ts

21

21

1

12

+

=+

=


15/31


Semester2,

2006:Page:29

s-domain to z-domain: Tustin Approximation

1

12

+

=

z

z

Ts

)1.29(

)6(1977)(

+

+

s

ssGc

746.0

16741778

9.1701.229

)194206(1977

1.291.29200200

)66200200(1977

1.291

)1(200

)61

)1(200(1977

)(

=

=

++

++=

++

++

=

z

z

z

z

zz

zz

z

zz

z

zGc

Sample time, choose T=0.01


Semester2,

2006:Page:30

Example 13.12 - Verification

The effects of three

different samplingtimes


16/31


Semester2,

2006:Page:31

Matlab / Simulink Discrete Control Functions

c2d: Converts continuous to discrete transfer

functions, example Gd = c2d(G,dt,'tustin')


Semester2,

2006:Page:32

Discrete Control Summary

z-transform provides closed-form transfer function forsampled systems, (Laplace transform does usually not)

Stability analysis in z-domain required for sampledsystems

Bilinear transformation allows Routh-Hurwitz stabilitytest for z-domain roots

Tustin approximation allows us to design controllers asusual in s-domain and convert the final result to z-domain just prior to implementation


17/31


Semester2,

2006:Page:33

Main Themes of METR4202/7202

FrequencyDomainAnalysis

(Nise Chapter 10.1-7)

Simple PDControl Design

(Bode Plots, 11.1-11.2)

State-SpaceControl Design(Nise Chapter 12)

State-SpaceObserver Design

(Nise Chapter 12)

DiscreteTime Domain

Analysis(Nise Chapter 13)

BilinearTransformations

Tustin

Overshoot, settling time

phase margin, bandwidth

Signal-Flow charts

Pole placement

Similarity

Transformations

Phase andGain Margins

DesiredPolynomials


Semester2,

2006:Page:34

Advantages / Disadvantages

Frequency Domain: (+) Easy to obtain modelsexperimentally through frequency responses. (-) Modelsdo not reveal the physical structure, only input-outputrelationships.

Continuous Time Domain: (+) Models based on naturallaws (physics, chemistry, etc). Some states are moreimportant than others and controllers can be designedaccordingly. (-) More difficult to estimate modelparameters.

Discrete Time Domain: (+) Allows analysis of sample-and-hold effects. (-) Requires controller design in z-plane. (+) Tustin approximation allows us to re-use toolsfrom Frequency Domain.


18/31


Semester2,

2006:Page:35

Overview: Frequency Domain Analysis

Polar Plots

Asymptotic Bode Plots: 1st and 2nd order zeros and poles

The Nyquist Stability Criterion

Sketching Nyquist Diagrams: Polar Plots positive, negative jw

Gain and Phase Margins from Nyquist Plots

Stability, Gain and Phase Margins from Bode Plots

*

*

*

*

Checklist

* means that examples given in the following slides


Semester2,

2006:Page:36

Common Asymptotes

Figure 10.9Normalised andscaled

Bode plots fora. G(s) = s;b. G(s) = 1/s;c. G(s) = (s+ a);d. G(s) = 1/(s+ a)


19/31


Semester2,

2006:Page:37

The Final Statement of the Nyquist Criterion

The number of closed-looppoles, Z, in the right half-planeequals the number of open-

loop poles, P, that are in theright half-plane minus thenumber of counter-clockwiserevolutions, N, around -1 ofthe mapping GH

Z = P N

Only if Z=0, the system isstable

Remember this !

Zeros of 1 + G(s) H(s) Poles of G(s) / [1 + G(s) H(s) ]


Semester2,

2006:Page:38

Examples

Figure 10.25Mapping examples:

a. contour doesnot encloseclosed-loop poles;b. contour doesencloseclosed-loop poles

Z = P - N

Class Question:Are the closed-loop systems a,b stable or unstable?

Frequency ResponsePolar Plots


20/31


Semester2,

2006:Page:39

Sketching the Nyquist Diagram (10.4)

Figure 10.26a. Turbine andgenerator;b. block diagramof speed controlsystem forExample 10.4


Semester2,

2006:Page:40

Vector Evaluation of the Nyquist Diagram

a. vectors on contourat low frequency;b. vectors on contouraround infinity;c. Nyquist diagram

Class Question:Is the closed-loop system stable?


21/31


Semester2,

2006:Page:41

Example 10.7

22222

22

2

)6()1(16

)6()1(4)(

)2)(22()(

www

wjwwjwG

sss

KsG

+

=

+++=

Double-check

this calculation

yourself!

Class Questions (use only positive jw axis):

a) Find the range of gain for stability and instabilityb) For marginal stability find the radian frequency of oscillation


Semester2,

2006:Page:42

Example 10.7 - Solution

22222

22

2

)6()1(16

)6()1(4)(

)2)(22()(

www

wjwwjwG

sss

KsG

+

=

+++=


22/31


Semester2,

2006:Page:43

Gain and Phase Margin via Nyquist (10.6)

Figure 10.35Nyquist diagramshowing gainand phasemargins


Semester2,

2006:Page:44

Phase and Gain Margin in Bode Plots

PhaseMargin

GainMargin


23/31


Semester2,

2006:Page:45

Overview: State-Space Analysis and Design

State-Space Models: ABCD Form

Controllability by Inspection: Parallel Form

The Controllability Matrix

Similarity Transformations

The Transformation Matrix P via the Controllability Matrices

Observability by Inspection: Parallel Form

The Observability Matrix

The Transformation Matrix P via the Observability Matrices

*

*

*

*

*

*


Semester2,

2006:Page:46

Figure 5.31State-space forms for

)(

)6)(4(

3

)(

)(

tcy

ss

s

sR

sC

=

++

+=

Summary State Space Forms

Controllability /Observability

by inspection


24/31


Semester2,

2006:Page:47

Phase-Variable Form

u

xx

x

x

aaaxx

x

x

n

n

nn

n

+

=

10

0

0

1000

0100

0010

1

2

1

110

1

2

1

A MatrixB Vector

+++

=

)()()(

1000

0100

0010

12110 nn kakaka

BKA

Major

advantage of

the phase-variable andcontrollercanonicalforms


Semester2,

2006:Page:48

Phase-Variable Form

Poles of uncontrolled system:

0011

1 =++++

asasas

n

n

n

n system parameters to adjust

Poles of controlled system:

0)()()( 10211

1 =+++++++

kaskaskas

n

nn

n


25/31


Semester2,

2006:Page:49

P from controllability matrix CM

For the original system

For the transformed system

Hence, P = CMZ * CMX-1

In Matlab: P = ctrb(Az,Bz) * inv(ctrb(Ax,Bx))

][12BABAABBC

n

MZ

=

][

])()([

121

11121111

BABAABBP

BAPPBPAPPBAPPPBPC

n

n

MX

=

=

PP-1

termsdisappear


Semester2,

2006:Page:50

P-matrix that gives phase-variable form

If we have the original system

and the phase-variable form

P = ctrb(Az,Bz) * inv(ctrb(Ax,Bx)) will transform anycontrollable system z to the phase-variable form x!!!!

uBzAz zz +=

uBxAu

x

x

x

x

aaax

x

x

x

xx

n

n

nn

n

+=

+

=

1

0

0

0

1000

0100

0010

1

2

1

110

1

2

1


26/31


Semester2,

2006:Page:51

Controller Design via State-Space (Ch. 12.1-4)

State-space formulation of the uncontrolled system:

State-space formulation of the controlled system:

CxyBuAxx =+=

CxyBrxBKAKxrBAxBuAxx =+=+=+= )()(


Semester2,

2006:Page:52

Example 12.1: State-Space Controller Design

Given the plant

)4)(1(

)5(20)(

++

+=

sss

ssG

design the phase-variablefeedback gains to yield9.5% overshoot and asettling time Ts of 0.74 sec.

= 0.60 0.94==

sn

Tw

Desired poles:

))(2(22

pswsws nn +++

Should not

interfere withdesign requirements!


27/31


Semester2,

2006:Page:53

Example 12.1: Desired Poles

Bode plots

)2(

20)(

)2)(1.5(

)5(20)(

222

221

nn

nn

wwssG

wwss

ssG

++=

+++

+=

and

In general:Use extra poles to cancelout zeros. If no cancellations

required, place poles far away

from 2nd

order pole.


Semester2,

2006:Page:54

Example 12.1: Signal-Flow Diagram

)4)(1(

)5(20)(

++

+=

sss

ssG

]020100[

540

100

010

=

=

C

A

How?


28/31


Semester2,

2006:Page:55

Example 12.1: Controller Gains

++

=

)5()4(

100

010

321kkk

BKA

Characteristic equation:

0)4()5(12

2

3

3=+++++ ksksks

Must match design requirements:

1.41308.1369.15))(2(2322

+++=+++ ssspswsws nn

Controller gains by inspection:

k1 = 413.1, k2 = 132.08, k3 = 10.9


Semester2,

2006:Page:56

Controller design by transformation: Ex 12.4

Convert the signal-flow diagram above to the form

Cxy

BAxx

=

+= u


29/31


Semester2,

2006:Page:57

Ex 12.4: Check Controllability

==

111

310

100

][2BAABBC

M

Is this system state controllable? Check by inspection


Semester2,

2006:Page:58

Ex 12.4: Transform to Phase-Variables

10178

1

)5)(2)(1(

123

+++=

+++ ssssss

==

==

+

=

1710

015001

*

4781

810

100

][

1

0

0

81710

100

010

1

2

3

2

1

3

2

1

MXMZ

M

CCP

BAABBC

u

x

x

x

x

x

x

Phase-variableform


30/31


Semester2,

2006:Page:59

Ex 12.4: Desired Response

20.8% overshoot and settling time Ts=4.0

= 0.447 24.24==

sn

Tw

Desired poles:

20136)4)(52(

))(2(

232

22

+++=+++=

+++

ssssss

pswsws nn

We use the extra poleto cancel the zero


Semester2,

2006:Page:60

State-Feedback Controller

+++

=

)8()17()10(

100

010

321kkk

BKA

20136)4)(52(

))(2(

232

22

+++=+++=

+++

ssssss

pswsws nn

Desired:

By inspection: k1=10, k2=-4, k3=-2


31/31


Semester2,

2006:Page:61

State-Feedback Controller: Original

Korig = K * P-1 = [ -20 10 -2]

DBAIC +==1)(

)(

)()( s

sU

sYsT

Verify design withdesired poles

AdvancedControl&Robotics,

Semester2,

2006:Page:62

Mid-Semester Class Test

Wednesday September 6 at 10:00am - 11:50am

Two Venues: Lecture Theatre (50-2)

GP-South Room 421 (20-25 seats)

Questions and Answers Session, September 5, 12-1pm

GP South: Room 421

Adv Control & Robotic Lec 5

Documents

Transcript of Adv Control & Robotic Lec 5