581 Lecture 36C

May 9, 2007 Feedback Control Systems (II) © Douglas Looze 1

Lecture 36

ECE 581

Feedback Control Systems (II)

Doug Looze


AnnounceProblem Set 6 available

– Due Tuesday, May 15Final exam

– Friday, May 18– 1:30 3:30 PM– Marston 211– 2005 exam on course site

Reading: FPE 8.3


Last Time: Design by EmulationIdea

– Use continuous-time design model and objectives

– Design continuous-time controller

– Approximate continuous-time controller in discrete-time

– Analyze• Bode• Nyquist• Root locations• Simulation


Today

Discrete-time design by emulation (cont.)– Matched pole-zero– Bilinear

• Tustin

• Trapezoidal


Matched Pole-Zero Emulation

Exploit sTz e

– If pole is at in continuous-time, thenp

pole is at in discrete-timepTe

– Match zeros also


Suppose

c

n sK s

d s

Polynomials

1

1

11

11

m ll

ciicp n k

k

cii

s sz

K

s sp

– Poles– Zeros

1n

ci ip 1

mci iz

– In general n m• If n > m

can augment zeros

zeros at n m


Matched pole-zero method– Poles

1ci

np T

ie

– Zeros 1

cinz T

ie

zeros at 1n m

– Pick Kdp

Without0

Integrators/Differentiators

limd k l

c

K zT

K s

“DC Gain” unchanged


where

1 1

1

1 1

1

1 1

1 1

ci

ci

m l lz T

id dp n k kp T

i

z e z

K z K

z e z

Without integrator/differentiators


– What is Kdp?• Each term (non-zero pole or non-zero, finite zero) has factor in both• Zero (non-zero, finite)

• Pole (non-zero)

0

11

11

ci

ci

ci p Tp TsT

s

sp

ee e


• Overall discrete-time proportional gain

1

1

1

1

ci

ci

n kp T

ik ldp cp m l

z T

i

e

K K T

e


Modified matched pole-zero emulation– Used by Matlab c2d

Matched pole-zero emulation– Infinite zeros

• Relative degree n m• Add zeros at 1 in discrete-time for each infinite zero

– No effect if #poles = # zeros


Comparison

5

5cs

K ss

0.1 sT

– 1 pole at origin

– 1 zero

– Gain

1 1n k

5 1 0ciz m l

1 00.5

11 0.1

1dpK

e

0.25

– Controller

0.610.25

1dz

K zz

0.5 0.61ciz Te e


10-1

100

101

Mag

nitu

de (

abs)

10-1

100

101

-90

-60

-30

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

Continuous-Time

Forw ard Diff

Backw ard Diff

Matched PZ

0.2 1c

sK s

s

0.1 sT


Bilinear TransformationIntegral method

– Trapezoidal integration1

1

2 1

1

zs

T z

– Direct substitution

2 1

1d cz

K z KT z

2 1

1

z

T z


– Assume continuous-time controller is rational

1

0 11

1

m mm

c n nn

b s b s bK s

s a s a

– Then

2 1 2 1

1 1

2 1 2 1

1 1

10 1

11

z z

T z T z

z z

T z T z

m mm

d n nn

b b bK z

a a

1

1

n

n

z

z

rational


Mapping 2 1

1

zs

T z

2 2

sz s zT T

2 2s z s

T T

1 12 2

sT sTz

12

12

sT

zsT

2 1

1

zs

T z

Bilinear transform pair


Poles and zeros–

Suppose has a pole at : Re 0c c cK s s p p (OLHP)

12

12

c

dc

p T

pp T

12

12

c

dc

p T

pp T

2

2

c

c

pT

pT

2

2

jT

jT

cp j


2 22 2

T T

2 22 22 2

T T

• Note

22

22

2

2d

Tp

T

1

2 2

T T

Inside unit circle

– OLHP gets mapped to unit disk


s-plane z-plane

StableStable

-j

-1 1

j

12

12

sT

zsT

2 1

1

zs

T z


10-1

100

101

Mag

nitu

de (

abs)

10-1

100

101

-90

-60

-30

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

Continuous-Time

Matched PZ

Bilinear


Summary

Bilinear emulation


Next Time

Example

581 Lecture 36C

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