MODERN CONTROL SYS-LECTURE III.pdf

7/25/2019 MODERN CONTROL SYS-LECTURE III.pdf

1/24

MODERN CONTROL

SYSTEMS ENGINEERING

COURSE : CS421

INSTRUCTOR:

DR. RICHARD H. MGAYA

Date: October 25th, 2013


2/24

Analysis of Steady-State Error

Final Value Theorem

Examination of the asymptotic behavior of a discrete system

without the need for the evaluation of thez-transforms

associated with transfer functions

Recap: Sequences that have a limiting value for large kmust be

stable functions All poles must lie inside the unit circle

Function should be expandable into the form

Inverse transformation

Dr. Richard H. Mgaya

11)( 1 i

n

z i

i

ppz

zB

z

Az

zF

n

i

k

iipBAkf1

)(

... i


3/24


Final Value Theorem Cont

As kgets large the terms contributed by the summation sign

will approach zero.

Evaluation of Aby partial fraction expansion:

From eqn. i

Multiply by and letzapproaches unity

Formal mathematical theorem:


z

z 1

)(1

lim 1 zFz

zA z

)(1

lim)(lim 1 zFz

zkf zk


4/24


Effect of Sampling on the Steady-State Error

No general conclusion about the steady-state error

Error depend on the placement of the sampler

Continuous Systems:

Steady-state error is based on the open-loop transfer function

Static error constants

Discrete Systems:

Placement of the sampler changes the open-loop transfer

function


,)(lim1

1)(

0 sGe

s

step

and

ssGe

s

ramp ,)(lim

1)(

0

)(lim

1)(

2

0 sGse

s

parabola

vKpK aK


5/24


Effect of Sampling on the Steady-State Error Cont

Consider the following digital system:

Digital computer is represented as asamplerand azero-order hold

Block reduction techniquesDr. Richard H. Mgaya


6/24


Effect of Sampling on the Steady-State Error Cont

From the figure:

Final value for discrete signal

e*() is the final sampled value of e(t) or e(kT)

e*() for unity feedback system


)()()()( zGzEzRzE

)(1

)()(

zG

zRzE

)(1lim)( 11* zEze z

)(1

lim)( 1*

zEz

ze z

)(1

)(1lim)( 11

*

zG

zRze z

... i i


7/24


Unit Step

Unit step input:

Substitution to eqn. ii

Thus,


ssR 1)(

1)(

z

zzR

)(lim 1 zGK zp

)(lim1

1)(

1

*

zGe

z

pKe

1

1)(*


8/24


Unit Ramp

Unit ramp input:

From unit step procedureeqn. ii

Thus,


2

1)(s

sR

21)(

z

TzzR

)()1(lim1

1 zGz

TK

zv

)()1(lim1

1)(

1

*

zGzT

e

z

vKe

1)(*


9/24


Unit Parabolic

Unit parabolic input:

Similarly

Thus,


3

1)(s

sR

32

12

)1()(

z

zzTzR

)()1(lim1 2

12 zGz

TK

za

aKe

1)(*

)()1(lim1

1)(

2

12

*

zGzT

e

z


10/24


Conclusion:

The error constants for digital systems are similar to those ofanalog systems

Multiple pole placement at the origin of thes-plane reduces the error tozero in case of analog systems

Multiple placement of pole atz= 1 of the zplane reduces the steady-

state error to zero for the case of discrete system of the type discussed

Thus,

Stability

Continuous systems stability can be determined by theRouthHurwitzcriterion

Number of unstable poles

The number of sign changes indicates the number of unstable poles andnot their location


Tsezzs under1tomaps0


11/24

Stability of Digital Control System

Jurytest is employed to asses the stability of discrete systems

Consider the characteristic equation s a sampled-data system:


0)( 011

1

azazazazQ n

nn

n

0

2

1

1

2

0

0123

3210

0321

1210

0121

1210

1210

m

m

m

m

m

mllll

llll

bbbb

bbbb

aaaaa

aaaaa

zzzzz

nnn

n

nnn

nn

nn

kn

kn

k aa

aa

b

0

kn

kn

kbb

bbc

1

10

kn

kn

kcc

ccd

2

20


12/24


Necessary and sufficient conditions for stability

No roots outside or on the unit circle

Condition1: Q(1) > 0

Condition2: (-1)nQ(-1) > 0

Condition3:

Conditionn:


naa 0

10 nbb

20 ncc

20 mm


13/24


Example

Given the characteristic equation below find the value ofKto

make the system just unstable

Solution:

Condition1: Q(1) > 0

Condition2: (-1)nQ(-1) > 0


0368.0368.1

)066.0092.0(1

2

zz

zK

0)066.0368.0()368.1092.0(2 KzKz

0)066.0368.0()368.1092.0(1)1( KKQ 0if0)1( KQ

0)066.0368.0()368.1092.0(1{)1()1( 2 KKQ

23.105or0026.0736.2 KK


14/24


Example Cont

Condition3:

The system is marginal stable whenK= 9.58 andK= 105.23


1066.0368.0 K

naa 0

1066.0368.0 K

58.9066.0

368.01

K


15/24


Digital System Stability via s-Plane

Mapping ofjaxis points on thes-plane to unit circle points

on thez-plane, i.e., bilinear transformation

Transformation maps right-half plane points on the s-plane to point

outside the unit circle on the z-plane. It also maps the left-half points on

the s-plane to the point inside the unit circle on the z-plane.

Transformation of the denominator of the pulse transfer

functionD(z) to the denominator of a continuous transfer

function G(s)

Routh-Hurwitz stability criterion can then be applied to the

transformed system



16/24



Bilinear transformation:

Provides tools for applyings-plane design and analysis to

digital system

If

and the inverse

Expansion of the logarithmic function

Bilinear transforms

Dr. Richard H. Mgaya1

12

z

z

Ts

Ts

ez

zT

s ln1

3

1

1

3

1

1

12ln

z

z

z

zz


17/24



Inverse

Substitutes = +j

Thus,


1

1

s

sz

j

jz

)1(

)1(

22

22

)1(

)1(

z

0 when1

0 when1

0 when1

z

z

z


18/24



Example: Given T(z) =N(z)/D(z), where

Use Routh-Hurwitz criterion to find the number of z-pane poles

of T(z) inside, outside, on the unit circle. Is the system stable?

Substitute:Inverse of bilinear transform intoD(z)= 0

Routh-Hurwitz table:


1.02.0)( 23 zzzzD

0174519 23 sss

017-

045.89-

17-19-

45-1

0

1

2

3

s

s

s

s


19/24



Example: Cont

Routh-Hurwitz table shows one root on the right half-plane

and two roots on the left half-plane.

Thus, T(z) has one pole outside the unit circle, two poles in the

unit circle and no pole on the unit circle

Conclusion: System is unstablepoles outside the unit circle



20/24

Root locus Analysis in z-plane

Root Locus Construction Rules

1. Starting points (K= 0): The root loci starts at open-loop poles

2. Terminating points (K=): The root loci terminate at open-loop zeros

when they exists, otherwise at infinity.

3. Number of distinct root loci: This is equal to the order of the characteristicpolynomial

4. Symmetry of root loci: The root loci are symmetrical about the real axis

5. Root locus location on real axis: A point on the real axis is part of the loci

if the sum of the open-loop poles and zeros to the right of the point is zero

6. Break away points: The points where a locus breaks away from real axis

are the root of the equation


0)( zGHdz

d


21/24



7. Unit circle crossover: Can be obtained by determining the

value ofKfor marginal stability using theJurytest

Example: For a given transfer function G(s), determine the

breakaway point, the value ofKfor marginal stability and the unitcircle cross-over for T= 0.5 seconds

z-Transform:


368.0368.1

066.0092.0)(

2 zz

zKzG

)2(

11)(

sss

eKsG

Ts


22/24



Characteristic equation:

Breakaway points:


0)}({ zGHdz

d

0368.0368.1

)066.0092.0(1

2

zz

zK

0)066.0368.0()368.1092.0(2 KzKz

0)368.12)(066.0092.0()368.0368.1( 2 zzKzz

01239.0132.0092.0 2 zz

084.2

647.0

2

1

z

z


23/24



K for marginal stability:Jury test

Unit circle crossover:K into the characteristic equation

Roots


01487.02 zz

23.105

58.9

2

1

K

K

97.0244.01 jz


24/24



Root locus analysis of a discrete system is the plot of the roots

of the characteristic equation 1+G(z)=0 inz-plane as a function

ofK.

Dr Richard H Mgaya

Date post:	26-Feb-2018
Category:	Documents
Upload:	mtanzania-mtanzania
View:	235 times
Download:	0 times

MODERN CONTROL SYS-LECTURE III.pdf

Documents