chapter 5 design using transformation technique – classical method

Robotics Research Labo-ratory

1

Chapter 5

Design using Transformation Technique – Classical Method


2

2

2 2( )

2n

n n

ωH s

s ζω s ω

σ

jω

nω dω

σ 0θ

2

cos

1n n d

θ ζ

s ζω jω ζ σ jω


3

( )y t

1.0 9

pTrT

.0 1

1%

sT

time

( ) 1 cos sinσtd d

d

σy t e ω t ω t

ω

Transient response


4

2

2

/ 1

2

1.8 rise time

4.6 4.6 settling time

1

0 <1 %OS

0.6(1 )

PM (phase margin)100

1 2

rn

sn

pd n

πζ ζp

p

BW n

Tω

Tζω σ

π πT

ω ω ζ

M e ζ

ζ M

PMζ

ω ω ζ

4 24 4 2ζ ζ


5

Steady-state response

G(s)D(s)( )R s ( )Y s( )E s

unity feedback

+ -

0 0

( ) 1 1 or ( ) ( )

( ) 1 ( ) ( ) 1 ( ) ( )

1lim ( ) lim ( )

1 ( ) ( )ss s se sE

E sE s R s

R s D s G s D s G

s s R sD s s

s

G


6

0

0

If the system is stable,

1Type 0 , lim ( ) ( ) for step input

1

1Type 1 , lim ( ) ( ) for ramp input

1Type 2 ,

ss p sp

ss v sv

ssa

e k D s G sk

e k sD s G sk

ek

2

0lim ( ) ( ) for parabolic inputa s

k s D s G s

1 2 3

1 2 3

Remark: Forward transfer function, not the closed-loop transfer function

( )( )( ) ( ) ( )

( )( )( )

If 0, the system type is zero.

If 1, the system ty

n

s z s z s zD s G s

s p s p s p

n

n

s

pe is one.


7

Rule of Thumb

A reasonable choice of is one that results in at least

4 samples in the rise time.

For better and smoother control result, more than 10

samples in the rise ti

T

me.

1.8 1.8 1.8

10 10 2 35

should be 35 times faster than the natural frequency .

s sn

r

s n

ω ωω

T T π

ω ω

ex)


8

1.83 / sec

For 0.125sec, 50 / sec

For 0.25 sec, 25 / sec

For 0.5sec, 12.5 / sec

For 1.0 sec, 6.28 / sec

n

s

s

s

s

ω rad

T ω rad

T ω rad

T ω rad

T ω rad


9

Design by Emulation

) Design of Antenna Angle-Tracker Servo Controller

in p.215 of Frankin's

1 ( ) System type ?

(10 1)

ex

G ss s

Design specifications:

1. Overshoot to a step input less than 16%.2. Settling time to 1% to be less than 10sec. 3. Tracking error to a ramp input of slope 0.01 rad/sec to be less than 0.01 rad.4. Sampling time to give at least 10 samples in a rise time.


10

Transient response

i) %OS 0.16 0.5

4.6 ii) 0.46s

ζ

σ T

-0.46 0

forbidden region

cos 0.5 60oθ θ


11

0 0


iii) Type 1 system

1 lim ( ) ( ) lim ( )

(10 1)

0.01 1

0.01

10 1 ( ) - one of many solutions

1

v s s

vss

k sD s G s sD ss s

Rk

e

sD s

s

( ) ( )

( ) ,( ) ( )

( ))( )

,

. .. .

2

Since 1

1 1

1 3 31 1( 2 2 2 2 Since 1

1 8 1 8 from , 0 18 0 2

10

n

rn n

D s G sH s

D s G s

H ss s s j s j

ω

T Tω ω


12

1

1

( ) 9.15( 0.9802) 9.15(1 0.9802 )( )

( ) 0.8187 1 0.8187

( 0.8187) ( ) 9.15( 0.9802) ( )

Forward

( 1) 0.8187 ( ) 9.15( ( 1) 0.9802 ( ))

U z z zD z

E z z z

z U z z E z

u k u k e k e k

0.11

1

z 1 0

Pole-Zero Mapping

0.9802 iv) ( )

0.8187

lim ( ) lim ( ) 1

0.9802 ( ) 9.15 where 0.2

0.8187

T

T

s

z z z e zD z K K K

z p zz e

D z D s

zD z T

z


13

1 sTe

s

r(t) u(k) c(t)e(t) e(k)

T=0.2

.

. .

0 8187 1

9 15 0 982 1

u k u k

e k e k

1

10 1s s

10 1

1

s

s

1

10 1s s R E C

( ) . ( ) . ( ( ) . ( ))

Backward

0 8187 1 9 15 0 9802 1u k u k e k e k


14

2

ex) Antenna angle-tracker servo with slow sampling in . 220

1 0.1 ( )

( 0.1)

0.99340.00199

( 1)( 0.9802)

1sec

0.9802 ( ) 9.15

0.8187

1 0.( )

0.2sec

p

zG z

z s s

z

z z

zD

z

z

T

T

z

zG z

z

z 2

1 0.96720.0484

( 1)( 0.9048)( 0.1)

0.9048 ( ) 6.64

0.3679

z

z zs s

zD z

z


15

10-1

100

101

10-2

10-1

100

101

Bode plot of the continuous design for the antenna control

Mag

nitu

de

10-1

100

101

-180

-160

-140

-120

-100

-80

Pha

se, d

egre

es

Phase Margin of continuous-time system 51.8PM

cpω


16

0.2 4.5 47.3

1 23 28.8

T φ PM

T φ PM

Remark: ( ) sin2

sample and hold (sec)2

sample and hold ( ) 2

where is the crossover frequency.

e

time delay

phase

x) In this exampl

de

2

lay

e

ho

cp

cp

ωT ωG jω

T

Tω rad

ω

T

, 0.1 sec and 0.8 / sec 45.8 / sec2

ocp

Tω rad


17

T=0.2 T=1

0 2 4 6 8 10 12 14 16 18 20-0.5

0

0.5

1

1.5

2

OU

TPU

T,

Y

and

CO

NTR

OL,

U

/10

TIME (SEC)

Fig. 7.7 Step response of the 1-Hz controller


18

Direct Design by Root Locus

in the z-plane The method for continuous-time systems can be extended w/o modifi-

cation The effects of the system gain and/or sampling period can be investi-

gated

Performance

-1 10 Rez

0ζ

Imz

damping ( %OS )

2

( )

1

d

d

nn

σ jω TsT

jω TσT

jω T ζζω T

z e e

e e

e e

1ζ

0.8ζ

0.6ζ


19

s-plane1ζ ζ

0 -1 1

Overshoot

1σ2σ

1djω1dω

1σ Te

Settling time s-plane z-plane

z-plane

1djω1dω

2σ Te


20

0.5

ex) From the given specification for transient response

0.5, 1 / sec, 0.2sec

For 0.9048

For damping ratio ( ) 0.5

For

n

Ts

p

ζ ω rad T

T r e

M ζ

2 ( ) 1 0.75

p p d nd

πT T ω ω ζ

ω

-1 1

nζω TσTr e e 1dω

1dω


21


22

1

Consider the steady-state error

( ) ( )

1 ( ) ( )

( ) where ( ) 1

Remark: System type of discrete-time control system

the number of open-loop poles at 1

1 ( ) (

( 1)

)N

R zE z

D z G z

G sG z z

s

z

B z

zG z

A

z

( )

where B(z)/A(z) contains neither a pole nor zero at 1

z

z


23

1

For a unit step input

1( )

11 ( ) ( )

1( ) lim( 1)

11 ( )

1 1

1 (1) (1) 1

Type 0 finite error

Type 1 zero

z

p

zE z

z D z G z

ze z

z D z G z

D G k

Position error constant


24

2

21

1 1

For a unit ramp input

1( )

1 ( ) ( )( 1)

1( ) lim( 1)

1 ( ) ( )( 1)

lim lim( 1)(1 ( ) ( )) ( 1) ( ) ( )

1

z

z z

v

TzE z

D z G zz

Tze z

D z G zz

Tz Tz

z D z G z z D z G z

k

Velocity error constant

2

21

2

21

Similarly,

( ) lim( 1) (1 ( ) ( ))

1 lim

( 1) ( ) ( )

z

za

Tze

z D z G z

Tz

kz D z G z

Acceleration error constant


25

1 2

1 2

the overall transfer function

Assume

( )( ) ( ) ( )

( )( ) (

Remark: ( Truxal's Rule )

( )( ) :

( )

and ( ) results

)

from a Type 1 sy

For unit

stem of (

step

).

, st

n

n

z z z z z zH z K

z p z

Y z

p z p

G

H zR z

H zz

eady-state error must be zero

( ) ( ) ( ) ( ) 0

(1) 1 --- (1)

( ) ( ) 1 Remark: ( ), ( ) ( ) ( ), and

( ) ( ) 1 ( )

E z R z H z R z

H

Y z E zH z Y z G z E z

R z R z G z


26

2

21

1

For a ramp input

( ) ( )(1 ( ))

(1 ( ))( 1)

1( ) lim( 1) (1 ( ))

( 1)

1 1 ( ) lim --- (2)

1

We can use L'Hopital's rule.

zv

zv

E z R z H z

TzH z

z

Tze z H z

kz

H z

Tk z


27

1 1

1

1

1

1 ( / )(1 ( )) ( )lim lim

( / )( 1)

1Note : ln ( ) ( ) ( ) since (1) 1

( )

1lim ln ( )

( ) lim ln

( )

lim ln(

z zv

zv

i

zi

iz

d dz H z dH z

Tk d dz z dz

d d dH z H z H z H

dz H z dz dz

dH z

Tk dz

z zdK

dz z p

dz z

dz

1

1 1

) ln( ) ln

1 1 lim ( ) ( )

1 1 1

1 1

i

i

n n

i i i

ii i

v i

z

z p K

d dz z z p

z z dz z p

Tk p z

dz


28

Imz

Rez-1 0 1

pole

zero

pole 0

ze ro 1vk

Large overshoot Poor dynamic response

Small steady-state error against Good transient response

Errors are decreased

1 1

1 1 1

1 1

n n

i iv i iTk p z


29

0.5

Characteristic polynomial

0.9672 1 0.0484 0

( 1)( 0.9048)

Design specifications :

1 , 10sec 0.5 , 1sec

4.6 0.5 0.61

v s

s

zK

z z

k T ζ T

σ r eT

0.9672( ) 0.0484 where 1 sec

( 1)( 0.9048)

( ) ( proportional controller - static gain)

( )( )

1 ( )

zG z T

z z

D z K

KG zH z

KG z

ex) Antenna system in p. 228 of Franklin’s


30

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0.1p/T

0.2p/T

0.3p/T

0.4p/T0.5p/T

0.6p/T

0.7p/T

0.8p/T

0.9p/T

p/T

0.1p/T

0.2p/T

0.3p/T

0.4p/T0.5p/T

0.6p/T

0.7p/T

0.8p/T

0.9p/T

p/T

0.10.2

0.30.40.50.60.70.80.9

Discrete root locus with and without compensation

Real Axis

Imag

inar

y A

xis

1 2 1

1 2 1

without compensator ( )

( 1.0, 0.9048, 0.9672)

0.9048with ( ) 6.64 , 1sec (emulation)

0.3679 ( 1.0, 0.3697, 0.9672)

D Z K

p p z

zD z T

zp p z


31

The system goes unstable at 19 where 0.92.

That means there is no value of gain that meets

the steady-state specification.

"Trial and error" based controller design

vK k

Imz

Rez-1 10 0.61

ζ

r


32

0.9048 ( ) 6.64 1sec (design by emulation)

0.3679

Direct design using root locus (trial and error)

0.8) ( ) 6

0.05

0.88) ( ) 13

0.5

0.8) ( ) 9 (hidden ocillation)

0.8

)

zD z T

z

zi D z

z

zii D z

z

ziii D z

z

iv

0.88 ( ) 13 (delay)

( 0.5)

zD z

z z

( . ~ )Dynamic compensation 228 235 pp


33

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0.1p/T

0.2p/T

0.3p/T

0.4p/T0.5p/T

0.6p/T

0.7p/T

0.8p/T

0.9p/T

p/T

0.1p/T

0.2p/T

0.3p/T

0.4p/T0.5p/T

0.6p/T

0.7p/T

0.8p/T

0.9p/T

p/T

0.10.20.30.40.50.60.70.80.9

Root locus for antenna design

Real Axis

Imag

inar

y A

xis

1 2 3 0 1

0.8 ( ) 6

0.05 ( 1.0, 0.9048, 0.05, 0.8, 0.9672)

zD z

zp p p z z


34

0 2 4 6 8 10 12 14 16 18 20-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

OU

TP

UT

, Y

an

d C

ON

TR

OL,

U

/10

TIME (SEC)


35

-1 -0.5 0 0.5 1

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0.1p/T

0.2p/T

0.3p/T

0.4p/T0.5p/T

0.6p/T

0.7p/T

0.8p/T

0.9p/T

p/T

0.1p/T

0.2p/T

0.3p/T

0.4p/T0.5p/T

0.6p/T

0.7p/T

0.8p/T

0.9p/T

p/T

0.10.2

0.30.40.50.60.70.80.9

Root locus for compensated Antenna Design

Real Axis

Imag

inar

y A

xis

1 2 3 0 1

0.88 ( ) 13

0.5 ( 1.0, 0.9048, 0.5, 0.88, 0.9672)

zD z

zp p p z z


36

0 2 4 6 8 10 12 14 16 18 20-1.5

-1

-0.5

0

0.5

1

1.5

OU

TP

UT,

Y a

nd C

ON

TR

OL,

U/1

0

TIME (SEC)

Step Response of Compensated Antenna


37

-1 -0.5 0 0.5 1

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

10.160.340.50.64

0.76

0.86

0.94

0.985

0.160.340.50.640.76

0.86

0.94

0.985

0.20.40.60.811.21.4

Root locus for Compensated Antenna Design

Real Axis

Imag

inar

y A

xis

1 2 3 0 1

0.8 ( ) 9

0.8 ( 1.0, 0.9048, 0.8, 0.8, 0.9672)

zD z

zp p p z z


38

0 2 4 6 8 10 12 14 16 18 20-1

-0.5

0

0.5

1

1.5

OU

TP

UT,

Y

an

d

CO

NT

RO

L,

U/1

0

TIME (SEC)

Step response of compensated Antenna Design


39

-1 -0.5 0 0.5 1

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

10.160.340.50.64

0.76

0.86

0.94

0.985

0.160.340.50.640.76

0.86

0.94

0.985

0.2

0.4

0.6

0.8

1

1.2

0.2

0.4

0.6

0.8

1

1.2

Root locus for compensated Antenna Design

Real Axis

Imag

inar

y A

xis

1 2 3 4 0 1

0.88 ( ) 13

( 0.5)

( 1.0, 0.9048, 0.5, 0, 0.88, 0.9672)

zD z

z

p p p p z z

z


40

0 2 4 6 8 10 12 14 16 18 20-1

-0.5

0

0.5

1

1.5

2

OU

TP

UT,

Y a

nd C

ON

TR

OL,

U/1

0

TIME (SEC)

Step response for compensated antenna Design


41

Frequency Response Methods1. The gain/ phase curve can be easily plotted by hand.

2. The frequency response can be measured experimentally.

3. The dynamic response specification can be easily interpreted in terms of gain/ phase margin.

4. The system error constants and can be read directly from the low frequency asymptote of the gain plot.

5. The correction to the gain/phase curves can be quickly computed.

6. The effect of pole/ zero gain changes of a compensator can be easily deter-mined.

Note : 1, 5, 6 above are less true for discrete frequency response design

using z - transform.

vkpk


42

Nyquist Stability Criterion Continuous case

zeros of the closed-loop characteristic equation, n(s) + d(s) = poles of the closed-loop system, n(s)+d(s)

( ) ( ) ( )

( ) ( ) ( ) ( ) 1+

( )

( )1

n sKD s G s

H sn sKD s G ss

d s

d

( )

( )( )

(

( )( ) ( ) ( ))

n s n s d

n s

n sd s

d ssd s

known

open-loop system

closed-loop system

characteristic equation


Z (unknown) = # of unstable zeros (same direction) of 1 + K D(s) G(s) ( or # of unstable poles of H(s) )

P (known) = # of unstable poles (opposite direction) of 1 + K D(s) G(s) ( or # of unstable poles of KD(s)G(s))

N(known after mapping) = # of encirclement (same direction) of the origin of 1+KD(s)G(s) ( or -1 of KD(s)G(s) )

Z must be zero for stability

43


44

unstable poles

0 -1

1+KD(s)G(s)-plane KD(s)G(s)-plane

Z – P =N

or Z = P + N

S-plane

-1/K

D(s)G(s)-plane


45

Discrete case ( The ideas are identical )

Unstable region of the z-plane is the outside of the unit circle

Consider the encirclement of the stable region.

N = { # of stable zeros } - { # of stable poles}

= { n – Z } – { n – P }

= P – Z Z = P – N

In summary,

1. Determine the number, P, of unstable poles of KDG.

2. Plot KD(z)G(z) for the unit circle, and .

3. Set N equal to the net number of CCW encirclements of the point

-1 on the plot

4. Compute Z = P – N. This system is stable iff Z = 0.

jωTz e 0 2ωT π


46


47

1( ) , 2

( 1)

1.135 ( 0.523)( )

( 1) ( 0.135)

1

G s ZOH at Ts s

zG z

z z

K

ex) p. 241 (Franklin’s)

The unit feedback discrete system with the plant transfer

function with sampling rate ½ Hz and zero-order hold

0, # of CCW encirclements of the point 1 0

0

P N

Z


48

-2 -1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1 Nyquist plot from Example 1 using contour

Real Axis

Imag

inar

y A

xis


49

Design Spec. in the Frequency Domain

Gain Margin (GM) : The factor by which the gain can be increased before the system to go unstable

Phase Margin (PM): A measure of how much additional phase lag

or time delay can be tolerated in the loop before

instability results.ex) p. 243

2

2

1( ) with ZOH at 0.2 sec

( 1)

( 3.38) ( 0.242)( ) 0.0012

( 1) ( 0.8187)

1

G s Ts s

z zG z

z z

K


50

Bode plot Nyquist plot

GM=1.8, PM=18


51

Sensitivity Function

• Stability robustness – stability

1( ) ( )

1 ( ) ( )

1( ) ( ) ( ) ( )

1 ( ) ( )

sensitivity function

E s R sD s G s

E jω R jω S jω R jωD jω G jω

• Tracking error – performance

Remarks:

( ) ( )i) ( ) complementary sensitivity function

1 ( ) ( )

ii) ( ) 1 ( )

D jω G jωT jω

D jω G jω

T jω S jω


52

-1PM

1

GM

1VGM

Re

Im

max

1min(1 ) the distance of the closest point from -1

| |

1 1 1

1

DGS

S S

VGM S

SVGM

S

VGM : vector gain margin

( )1

1 DGS

( ) ( )ω ωD j G j plane


53

Performance

1

1

1

| | | || | where is an error bound| |

| | 1

| |Define ( ) performance frequency function

| | | | 11

| | 1| |

b b

b

b

E S R e eR

Se

RW ω e

S W

W DGS

1

Remark:

At every frequency, the point on the Nyquist plot lies

outside the disk of the center -1, radius ( ) .

DG

W jω

-1

1| |W

DG


54

1

If the loop gain is large | | 1

1 1 | | where

| | 1

| | | |

DG

S SDG DG

W DG

1

) 0.005, 1 below 100

0 2 200

1 = | | 2000.005

b

b

ex e R Hz

ω πf π

R We

DG

200

2001( )W ω

ωThe higher the magnitude curve at low frequency,

the lower the steady-state errors


55

Robust Stability

0 2

0 2

( ) ( )[1 ( ) ( )]

multiplicative uncertainty

( ) ( ) ( ) ( )

additive uncertainty

G jω G jω w ω jω

G jω G jω w ω jω

0

2

2 2

2

where ( ) : nominal transfer function

( ) : magnitude

( ) : phase

| ( ) | 1

( ) ( ) upper bound

( ) : robust stability frequen

G s

w ω

jω

jω

w ω W ω

W ω

cy function


56

02 2

0 0

2

0

2 2

0 0

1

( )

1

1

1

sB

ωKG s

s s sB

ω ω

s

ωK

s s sB

ω ω

2 ( )

2

0

2

0 0

1

s

ωw ω

s sB

ω ω

ex) p. 250

small for low frequencies

and large for high frequencies


5710-2

10-1

100

101

102

10-2

10-1

100

101

102

103

Fig.7.27 Plot of typical plant uncertainty

10-1

100

101

10-2

10-1

100

101

102

Fig.7.26 Model uncertainty for disk read/write head assembly


58

2 2 2

2 2

Suff & Nec. Cond.

1 not to be zero any

1 1

Tw w W

T w T W

0

0

0

0 2

Define ( ) 1 ( )

complementary sensitivity function1

1Assume that the nominal system is stable, 1 0

1 0 (for stability robustness)

1 [1 ]

T jω S jω

DG

DG

DGS

DG

DG w

00 2

0

0 2

0

(1 ) 1 01

(1 )(1 ) 0

DGDG w

DG

DG Tw


59

0 0

0

2

since is small for high frequencies.

1

T DG DG

DGW

200 ω

Bode plot

0DG

1

robust performance

W

2

1

robust stability

W


60

2

2 0

0

2 0 0

Remark:

1

1 1

1

W T

W DGω

DG

W DG DG

2

Remark:

At every frequency, the critical point, -1, lies

outside the disk of the center , radius ( ) ( ) ( ) .DG W jω D jω G jω

Im

Re

-1

0DG

| 2 0 |W DG


61

Re

Robust Performance

Im

-1

0DG

| 2 0 |W DG

1W

1

2

Remark:

For each frequency , construct two closed disks;

one with center -1, radius radius ( ) ;

the other with center , radius ( ) ( ) ( ) .

Then robust performance holds iff for each

ω

W jω

DG W jω D jω G jω

ω, these two disks are disjoint.


62

Bode plot


63

Remark:

1 2 bilinear rule 1 22 1

1

sT

T sz e z

T s

zω

T z

There exists big difference in the Bode diagram at the high frequency.

Types of Compensator:

phase-lead ( high pass ~ PD) transient response

phase-lag ( low pass ~ PI ) steady-state response

PID : a special case of a phase lead-lag compensator


64

100

Type 0 system

20 log at the low-frequency magnitude in Bode log-magni

Transient response


Remark:

p

PMς

K

tude plot

Type 1 system

is the intersection (extension of the initial -20 dB/decde) with frequency

axis in Bode log-magnitude plot

Ty

vK

pe 2 system

is the intersection (extension of the initial -40 dB/decde) with frequency

axis in Bode log-magnitude plot aK


65

Design Procedure in Frequency

1.

2. Bode plot

3.

4.

1 2 bilinear transformation 1 2

T νz

T ν

( )G shold

( )G z ( )G ν

( )G ν ( )G jων jω

( ) ,

( ).

Assuming that the low frequency gain of is unity

design

D ω

D ω

2 1

1

zω

T z

( ) ( )D ω D z


66

1

ex) . 230- (Ogata's)

1 10( )

10

1 10 10 ( ) (1 )

10 ( 10)

0.6321

0.3679

Ts

Ts

p

eG s

s s

eG z z

s s s s

z

z z

1 sTe

s

10

10s Tδ


67

1 ( / 2) 1 0.05

1 ( / 2) 1 0.05

0.6321 0.6321(1- 0.05 )( )

1 0.05 0.6321 0.068400.36791 0.05

1 0.059.241

9.241

T v vz

T v v

vG v

v vv

v

v


68

0

1

ex) .236- (Ogata's)

1 ( )

( 1)

Design Specifications:

phase margin: 50 gain margin: at least 10db

velocity error constant : 2sec

sampling peri

Ts

v

p

e KG s

s s s

K

2

od 0.2

1 ( )

( 1)

( 0.9356) 0.01873

( 1)( 0.8187)

(0.08173 0.01752)

1.8187 0.8187

Ts

T

e KG z

s s s

K z

z z

K z

z z

z


69

T v vz

T v v

vK

vG v

v v

v vv v

KK v v

v vv v

2

2

2

1 ( / 2) 1 0.1

1 ( / 2) 1 0.1

1 0.10.08173( ) 0.01752

1 0.1( )

1 0.1 1 0.1( ) 1.8187( ) 0.81871 0.1 1 0.1

(1 )(1 )( 0.000333 0.09633 0.9966) 300 10

( 1)0.9969

How to determ Kine the gain ?


70

0

2

2

lim ( ) ( ) 2

2(1 )(1 )2( 0.000333 0.09633 0.9966) 300 10( )

( 1)0.9969

From Bode plot, and GM 14.5dB

PM should be properly

PM 3

adj

0

usted.

What type of compensator is neede

o

v DvK vG v G v K

v vv v

G vv vv v

d?

1 ( ) , 0 1 (pha se-lead compensator)

1D

τvG v α

ατv


71


72

D

D

τvG v

ατv

φ τν ατν

d

τvG v

αv

τ αα α

φαα

τ

φ τ ατ

dν τν ατν

τ

α

v

p

1 1

max

1 1a

2

m

2

x

1( )

1

tan tan

1 ( ) 1 ( )

( and ) (1)

( from t

1

1 1tan

1 break

sin12

h

frequencies from ( )1

e req

Review ( .628 ιn Νιse's):

D

D

DG

jjτω αG jω

jατω j α

α

G jω G j

j

ω ω

ωα

ma

maxmax

m

max max

x

m

x

x

a

a

uired phase) (2)

11

1( )

1 1

(magnitude at the peak of phase curve) (3)

where .

Solving (2), we can obtain

( )

1(

.

( )

U

at

)

sing τ (3) and (1), we can determine .


73

o

max ma

1

x

1max

max

max

How to design ( )?

1 sin 28 0.361

1

120 log 4.425 dB

0.361

At 1.7, ( ) 4.425 dB

11.7 0.9790

1 1 0.( ) ( )

1

1 1tan sin

121

( )

1

D

D

D

D

α αφ

αα

G jωα

v

G v

αα

α

v G jω

ττ α

τvG

τ α

v G vατv

9790.

1 0.3534

v

v


74


75

D

D

D

vG v

vz

zG zz

zz

z

z zG z G z

z z z

make it a difference equatio

1 0.9790( )

1 0.35341

1 0.9790(10 )1( )1

1 0.3534(10 )1

2.3798 1.9387

0.5589

2.3798 1.9387 0.03746( 0.9356)( ) (

n!

)0.5589 ( 1)( 0.8187)

z z

z z z

C z z z

R z z z z

z z

z z j z j

2

3 2

2

3 2

0.0891 0.0108 0.0679

2.377 1.8352 0.4576

( ) 0.0891 0.0108 0.0679

( ) 2.2855 1.8460 0.5255

0.0891( 0.9357)( 0.8145)

( 0.8126)( 0.7379 0.3196)( 0.7379 0.3196)


76


77

Design by Direct Method (Analytic Method)

Ts

n nn

n nn

eG z G s

s

C z D z G zH z

R z D z G z

H zD z

G z H z

b z b z bH z

a z a z a

10 1

10 1

1( ) ( )

( ) ( ) ( )( )

( ) 1 ( ) ( )

1 ( )( )

( ) 1 ( )

where ( )

z


78

n nn

n

nn

a z a z aH z

z

a a z a z

10 1

10 1

( )

Direct Design Method of Ragazzini

i) (transient) determine characteristic equation according to the specification

denominator of ( )

ii) (steady-state) determine coefficient of numerator of ( )

The system sh

H z

H z

1

ould be causal.

(1) 1 for Type 1 system

1 |z

v

H

dHT

dz K

Deadbeat Controller (unique characteristic in digital controller)

– minimum settling time with zero steady-state error


79

Digital PID Controller

0

1 ( )( ) ( ) ( ) ( )

1( ) 1 ( )

1

where 1 is a low-pass filter , 3 10

(0) ( ) ( ) (2 )( ) [ ( ) [

2 2

(( 1) ) ( ) ( ) (( 1)]

2

t

di

d

di

d

i

d

de tu t K e t e t d t T

T dt

T sU s K E s

T sT sN

T s NN

T e e T e T e Tu kT K e kT

T

e k T e kT e kT e k TT

)]

trapezoidal backward T

Positional form


80

1 1

1

1

1 1

1

Define

(( 1) ) ( ) ( ), (0) 0

2(( 1) ) ( )

( )2

(( 1) ) ( )

2

( )

1 ( ) ( ) (0)

1 1

1( ) ( ) ( )

2

h h

k k

h

k

h

k

e k T e kTf kT f

e k T e kTf kT

e k T e kT

f kt

F zF z f

z z

zF z f kt E z

z

z

z


81

10

0

1 Show that ( ) ( )

1

( ) ( ), 0,1,2,

(0) (0)

(1) (0) (1)

( ) (0) (1) (2) ( )

( ) ( 1)

k

h

k

h

Z x h X zz

y k x h k

y x

y x x

y k x x x x k

y k y k

1

1

10

( )

( ) ( ) ( )

1 ( ) ( )

1

1 ( ) ( ) ( ) ( )

1

k

h

x k

Y z z Y z X z

Y z X zz

Z x h Z y k Y z X zz


82

1

10

1 ( 1)

0

1Show that ( ) ( ) ( ) where 1 1

1

( ) ( ) ( ) ( 1) ( )

( ) ( ) ( 1) ( )

Noting that

( ) ( ) ( ) (0) (

k ih

h i h

k

h i

i k

k

k

Z x h X z x h z i kz

y k x h x i x i x k

X z x i z x i z x k z

X z Z x k x k z x x

1 2

1

0

1

1

1 10

1) (2)

( ) ( ) ( )

Since

( ) ( 1) ( ), , 1, 2,

( ) ( ) ( )

1 1 ( ) ( ) ( ) ( ) ( )

1 1

ih

h

k ih

h i h

z x z

X z X z x h z

y k y k x k k i i i

Y z z Y z X z

Z x h Y z X z X z x h zz z


83

( )

,

11

1

11

11

1( ) ( ) ( ) 1 ( )

2 1

(1 ) ( )1

2 2

( )( ) (1 )

( ) 1

d

i

Ip D

Ip

i

dI D

i

ID p D

TT zU z K E z E z z E z

T z T

KK K z E z

z

KKTK K K

T

KTKTK K

T T

KU zG z K K z

E z z

112 1i i

T T

T T z


84

( ) ( ) (( 1) )

[ ( ) (( 1) ] ( )

+ [ ( ) 2 (( 1) (( 2) ]

If ( ) ( ) ( )

( ) [ ( ) (( 1) ) ( ) (( 1) ]

+ [ ( ) (

p I

D

p

I

U kT U kT U k T

K e kT e k T K e kT

K e kT e k T e k T

e kT r kT c kT

U kT K r kT r k T c kT c k T

K r kT c k

If ( ) is a constant reference input

)]

+ [ ( ) ( ) ]

( ) [ ( ) (( 1) )]

[ ( ) ( 1)]

[ ( )

(or step change)

2 (( 1) (( 2

D

p

I

D

T

K r kT c kT

U kT K c kT c k T

K r kT c k

K c

r k

kT k k

T

c T c

) ]

Incremental form or velocity form

T

Velocity form


85

( ) (( ) ) ( )

( ) ( ) ( ) ( )

( ( ) ( ))

( ) ( )

( )( ) ( ) ( ) ( )

)

(

1 1

1 2

11

1

1 1

1 2

11

p

I

D

p I D

U kT U k T U kT

z U z K z C z

K R z C z

K z z C z

C zU z K C z K K z C z

z

R z

velocity form

A drawback of velocity form is that it can not be used for P- or PD-controllers. In these cases the controller will not be unable to keep the reference value.


86

Remarks: Integral action in PID Controller (Astrom’s)

Integrator windup :

A controller with integral action combined with an actuator that becomes saturated can give some undesirable effects.

If the controller error is so large that the integrator saturates the actuator, the feedback path will be broken, because the actuator will remain satu-rated even if the process output changes.

When the error is finally reduced, the integral may be so large that it takes considerable time until the integral assumes a normal value again.

Antiwindup:

i) Stop updating the integral when the actuator is saturated.

ii) Or an extra feedback path is provided by measuring the actuator output and forming error signal as the difference between the actuator output and the controller output and feeding this error back to the integrator through a proper gain(tracking-time constant).


87


Unsaturated

1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101 106 111 116 121 126 131 136 1410

0.5

1

1.5

2

2.5

3

r(t)

u(t)

y(t) [Unsaturated]


Saturated

1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101 106 111 116 121 126 131 136 1410

0.5

1

1.5

2

2.5

3

r(t)

uc(t)

u(t)

y(t) [Saturated]


Anti-windup

1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101 106 111 116 121 126 131 136 1410

0.5

1

1.5

2

2.5

3

r(t)

uc(t)

u(t)

y(t) [Anti-windup]


Comparison (u)

1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101 106 111 116 121 126 131 136 1410

0.5

1

1.5

2

2.5

3

3.5

u(t)

uc(t)

u(t)

uc(t)

u(t)

Unsaturated

Saturated

Anti-Windup


Comparison (y)

1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101 106 111 116 121 126 131 136 1410

0.2

0.4

0.6

0.8

1

1.2

1.4

y(t) [Unsaturated]

y(t) [Saturated]

y(t) [Anti-windup]

r(t)


93

Bumpless transfer:

Practically all PID controllers can run in two modes: manual and automatic.

When there are changes of modes and parameters, it is essential to avoid switching transients.

When the system is in manual mode, the controller produces a control sig-nal that may be different from the manually generated control signal.

It is necessary to make sure that the value of the integrator is correct at the time of switching. This is called bumpless transfer.

Bumpless transfer is easy to obtain for a controller in incremental form (ve-locity form) because the switching only influence the increments , there will not be any large transients. Initialization is not necessary when the opera-tion is switched from automatic to manual.

chapter 5 design using transformation technique – classical method

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