Control of Manufacturing Control of Manufacturing ProcessesProcesses

Subject 2.830 Spring 2004Spring 2004Lecture #22Lecture #22

““Discrete System ClosedDiscrete System Closed--Loop Dynamics"Loop Dynamics"May 3, 2004May 3, 2004

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 2

Difference Equations and DynamicsDifference Equations and Dynamics

• Consider the Discrete Transfer Function:

G(z) =Y (z)U (z)

=K

(z − p)Y(z)(z − p) = KU (z)

The corresponding Difference Equation is:

yi+1 − pyi = Kui ⇒ yi+1 = pyi + Kui

next output = p*current output + K* current input

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 3

Step DynamicsStep Dynamicsyi+1 = pyi + pui for p=0.5 and K=0.5 and u=unit step:

⇒ yi+1 = 0.5(yi + ui)I ui y i

1 1 0.0002 1 0.5003 1 0.7504 1 0.8755 1 0.9386 1 0.9697 1 0.9848 1 0.9929 1 0.996

10 1 0.99811 1 0.99912 1 1.00013 1 1.000

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 4

Step ResponseStep Response

0

0.2

0.4

0.6

0.8

1

1.2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

First Order

Transfer Function!G(z) =

Y (z)U (z)

=K

(z − p)=

0.5z −0.5

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 5

Effect of Root Effect of Root pp

Time (samples)

TextEnd

Step Response

0 2 4 6 8 100

0.02

0.04

0.06

0.08

0.1

Time (samples)

Ampl

itude

TextEnd

Step Response

0 2 4 6 8 100

0.02

0.04

0.06

0.08

0.1

p=0.1 p=0.25

Ampl

itude

Time (samples)

TextEnd

Step Response

Time (samples)

TextEnd

Step Response

0 50 100 1500

50

100

150

0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

Ampl

itude

Ampl

itude

p=0.5p=1.0

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 6

Effect of Root Effect of Root pp

Time (samples)

Am

plitu

de

TextEnd

Step Response

0 2 4 6 8 100

0.05

0.1

0.15

0.2

0.25

Time (samples)

Ampl

itude

TextEnd

Step Response

0 2 4 6 8 100

0.02

0.04

0.06

0.08

0.1

p=-0.1 p=-0.25

Time (samples)

TextEnd

Step Response

Time (samples)

TextEnd

Step Response

0 10 20 30 400

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10 120

0.1

0.2

0.3

0.4

0.5

Ampl

itude

p=-1.0

Am

plitu

de

p=-0.5

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 7

What’s Next?What’s Next?

•• ZZ-- domain modeling of control systemdomain modeling of control system•• Root locus in zRoot locus in z--domain (it’s the same!)domain (it’s the same!)•• Cycle to Cycle Stability LimitsCycle to Cycle Stability Limits

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 8

Analysis of Dynamics of Analysis of Dynamics of Discrete Feedback SystemsDiscrete Feedback Systems

•• Given the Discrete Transfer Function Given the Discrete Transfer Function For the Process For the Process GGpp(z(z))

•• What are the Dynamics of the resulting What are the Dynamics of the resulting Closed Loop System?:Closed Loop System?:

Gc(z) Gp(z)

H(z)

R(z) Y(z)

-

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 9

ClosedClosed--Loop DynamicsLoop Dynamics

Gc(z) Gp(z)

H(z)

R(z) Y(z)

-

Y (z )R(z)

= T (z) =GcGp

1+ GcGp H

and the characteristic equation is:

1 + Gc(z)Gp (z)H(z) = 0

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 10

ZZ--Domain Root LocusDomain Root Locus

1+GcGp H(z) = 0

In general the CE:

will be a polynomial in z:

anzn + an−1z

n −1 + an −2 zn− 2 + ...a1z + a0 = 0

and the roots of this polynomial will define the dynamics of our system

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 11

SS--Z Plane MappingZ Plane Mapping

z = esTRecall the Definition:

s = σ + jωAnd that

then z in polar coordinates is given by:

e(σ + jω )T = eσTejωT

z = eTσ ∠z = ωT

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 12

MappingMapping

s

0

s = 0

z = e0 = 1 z

+1

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 13

Mapping Mapping jjωω AxisAxis

s

0

s = + jω

z = e jωT = unit length vector at angle ωTz

+1

ωT

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 14

Mapping Mapping jjωω AxisAxis

jω z

0 +1

Note: when ωT = π, we have reached -1 on z-plane

π /T

− π /T

And when ωT = -π, we have come around the other way

Settling Time MappingSettling Time Mappings = −p + jω

z = e− pT + jωT = e− pTe jωT

S-plane lines of constant settling:

0

ωT

s

ts=4/p

-p

z

+1

circle of radius e-pT

r=e-pT

5/4/04

ts = −4T

1ln(r)

Settling Time MappingSettling Time Mapping

s z

0 +1

ωT

circle of radius e-pT

ts

CLOSER TO Z=0 = FASTER SETTLING

Damping Ratio MappingDamping Ratio Mapping

s z

0

β

∠s = β = constant0 ≤ s ≤ ∞

s

+1

5/4/04

Stability?Stability?

jw axis maps to the “unit circle”zs

0

π /T

−π /T

+1

roots inside the unit circle are stable, outside are unstable

Mapping: Negative Real AxisMapping: Negative Real Axis

s z

0+1

+jω

∠z = ωT = 0

0 ≤ s ≤ −∞⇒ e0 ≤ z ≤ e−∞ ⇒1 ≤ z ≤ 0

-jω

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 20

Mapping: What’s leftMapping: What’s left

s z

0

π /T

−π /T

−∞

oscillatory region

real-root

+1

ZgridZgrid from Matlabfrom MatlabLines of constant damping ratio (ζ=0.7 shown)

Lines of constant natural frequency-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

-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

ZZ--Plane Plane -- ReponseReponseRelationshipRelationship

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 23

Simple First Order Control System Simple First Order Control System

KC Gp(z)-R(z) Y(z)

G p(z) =K

(z − p) open loop root = +p

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 24

Root LocusRoot LocusROOTS OF 1+KCG-P(z) = 0 ?

Same rules as before:

Starts at open loop pole = +p

Real axis part to left of

odd number of poles

Ends at zero at infinityp

z

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 25

Root LocusRoot Locus-- InterpretationInterpretation

p

z

+1

slow, no oscillation

faster, no oscillation

fastest w/o overshoot

begins oscillating

very oscillatory

unstable

Now for Now for

-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

Real Axis

Imag

Axi

s

k r

0.22 0.39

0.73 0.13

1 0

1.5 -0.27

2.8 -0.92

3 -1

5/4/04

G(z) =K

(z − p)=

0.5z − 0.5 K=0.5

p=0.5

Response?Response?k r

0.22 0.39

0.73 0.13

1 0

1.5 -0.27

2.8 -0.92

3 -1

Response?Response?k r

0.22 0.39

0.73 0.13

1 0

1.5 -0.27

2.8 -0.92

3 -1

5/4/04

Response?Response?k r

0.22 0.39

0.73 0.13

1 0

1.5 -0.27

2.8 -0.92

3 -1

Response?Response?k r

0.22 0.39

0.73 0.13

1 0

1.5 -0.27

2.8 -0.92

3 -1

Response?Response?k r

0.22 0.39

0.73 0.13

1 0

1.5 -0.27

2.8 -0.92

3 -1

Response?Response?k r

0.22 0.39

0.73 0.13

1 0

1.5 -0.27

2.8 -0.92

3 -1

5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 33

Steady State Unit Step Error?Steady State Unit Step Error?

unit step

For s - domain y(∞) = lims→0

s G(s) 1s

For z - domain y(∞) = limz→1

(z −1) G(z) zz −1

T (z) =

KcK

z − p

1 +K

cK

z − p

=KcK

z − p + KcKOur closed-loop TF =

Steady State Error?Steady State Error?

Kc p y(∞)

0.22 0.39 0.18

0.73 0.13 0.44

1 0 0.5

1.5 -0.27 0.6

2.8 -0.92 0.74

3 -1 0.75

Steady State Error is Minimized as K increases

limz→1

(z −1) T (z) z

z −1= (z −1)

KcKz − p + KcK

z

(z −1)

=KcK

1− p + KcK

K=0.5p=0.5