2.14 Fall 20042.14 Fall 2004

Digital Control - Z-plane analysisDigital Control - Z-plane analysis

11/10/04 2.14 Fall 2004 2

Difference Equations and DynamicsDifference Equations and Dynamics

• Consider the Discrete TransferFunction:

The corresponding Difference Equation is:

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

�

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

=K

(z − p)

�

Y(z)(z − p) = KU (z)

�

yi+1 − pyi = Kui ⇒ yi+1 = pyi + Kui

11/10/04 2.14 Fall 2004 3

Step DynamicsStep Dynamicsyi+1 = pyi + pui

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 0.99610 1 0.99811 1 0.99912 1 1.00013 1 1.000

⇒ yi+1 = 0.5(yi + ui)for p=0.5 and K=0.5 and u=unit step:

11/10/04 2.14 Fall 2004 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

11/10/04 2.14 Fall 2004 5

Effect of Root Effect of Root pp

Time (samples)

Ampl

itude

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

Time (samples)

Ampl

itude

TextEnd

Step Response

0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

Time (samples)

Ampl

itude

TextEnd

Step Response

0 50 100 1500

50

100

150

p=0.1 p=0.25

p=0.5p=1.0

11/10/04 2.14 Fall 2004 6

Effect of Root Effect of Root pp

Time (samples)

Ampl

itude

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 10 120

0.1

0.2

0.3

0.4

0.5

Time (samples)

Ampl

itude

TextEnd

Step Response

0 10 20 30 400

0.2

0.4

0.6

0.8

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

p=-0.5p=-1.0

11/10/04 2.14 Fall 2004 7

WhatWhat’’s Next?s Next?

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

11/10/04 2.14 Fall 2004 8

Analysis of Dynamics ofAnalysis of Dynamics ofDiscrete Feedback SystemsDiscrete Feedback Systems

•• Given the Discrete Transfer FunctionGiven the Discrete Transfer FunctionFor the Process GFor the Process Gpp(z)(z)

•• What are the Dynamics of the resultingWhat are the Dynamics of the resultingClosed Loop System?:Closed Loop System?:

Gc(z) Gp(z)

H(z)

-R(z) Y(z)

11/10/04 2.14 Fall 2004 9

Closed-Loop DynamicsClosed-Loop Dynamics

Gc(z) Gp(z)

H(z)

-R(z) Y(z)

Y (z)R(z)

= T (z) =GcGp

1+GcGpH

and the characteristic equation is:

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

11/10/04 2.14 Fall 2004 10

Z-Domain Root LocusZ-Domain Root Locus

1 +GcGpH(z) = 0

In general the CE:

will be a polynomial in z:

anzn + an−1z

n −1 + an −2zn− 2 + ...a1z + a0 = 0

and the roots of this polynomial willdefine the dynamics of our system

11/10/04 2.14 Fall 2004 11

S-Z Plane MappingS-Z Plane Mapping

Recall the Definition: z = esT

And that s = σ + jω

then z in polar coordinates is given by:

e(σ + jω )T = eσTejωT

z = eTσ ∠z =ωT

11/10/04 2.14 Fall 2004 12

MappingMapping

s

0

s = 0z = e0 = 1 z

+1

z = eTσ ∠z =ωT

11/10/04 2.14 Fall 2004 13

Mapping Mapping jjωω Axis Axis

s

0

s = + jωz = e jωT = unit length vector at angle ωT

z

+1

ωT

11/10/04 2.14 Fall 2004 14

Mapping Mapping jjωω Axis Axis

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 Mapping

z

+1

S-plane lines of constant settling:

0

ωT

s

ts=4/p

-p

s = −p + jωz = e− pT+ jωT = e− pTe jωT

circle of radius e-pT

r=e-pT

11/10/04

ts ≈ −4Tln(r)

Settling Time MappingSettling Time Mapping

s z

0 +1

ωT

circle of radius e-pT

CLOSER TO Z=0 = FASTER SETTLING

ts

Damping Ratio MappingDamping Ratio Mapping

s z

0 +1

β

11/10/04

�

∠s = β = constant0 ≤ s ≤ ∞

�

s

Stability?Stability?

s z

0 +1

jw axis maps to the “unit circle”

roots inside the unit circle are stable, outside are unstable

π /T

−π /T

Mapping: Negative Real AxisMapping: Negative Real Axis

s z

0+1

+jω

-jω

∠z =ωT = 00 ≤ s ≤ −∞⇒ e0 ≤ z ≤ e−∞ ⇒1 ≤ z ≤ 0

11/10/04 2.14 Fall 2004 20

Mapping: WhatMapping: What’’s lefts left

s z

0 +1

π /T

−π /T

−∞

oscillatory region

real-root

-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

Real Axis

Imag

Axis

TextEnd

Pole-zero mapzgrid zgrid from Matlabfrom Matlab Lines of

constantdamping ratio(ζ=0.7 shown)

Lines of constantnatural frequency

-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

Real Axis

Imag

Axis

TextEnd

Pole-zero map

Z-Plane - ResponseZ-Plane - ResponseRelationshipRelationship

11/10/04 2.14 Fall 2004 23

Simple First Order Control SystemSimple First Order Control System

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

�

Gp(z) =K

(z − p) open loop root = +p

11/10/04 2.14 Fall 2004 24

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

Same rules as before:Starts at open loop pole = +pReal axis part to left of

odd number of polesEnds at zero at infinity

p

z

11/10/04 2.14 Fall 2004 25

Root Locus- InterpretationRoot Locus- Interpretation

p

z

+1

slow, no oscillation

faster, nooscillation

fastest w/oovershoot

beginsoscillating

very oscillatory

unstable

-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

Axis

Now forNow for

k r

0.22 0.39

0.73 0.13

1 0

1.5 -0.27

2.8 -0.92

3 -1

11/10/04

�

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

11/10/04

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

11/10/04 2.14 Fall 2004 33

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

Our closed-loop TF =

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 + KcK

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 isMinimizedas K increases

�

limz→1

(z −1) T (z) zz −1

= (z −1) KcKz − p + KcK

z

(z −1)

�

=KcK

1− p + KcK

K=0.5p=0.5