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II.C. Steady State Tracking and System Type

Ability of a control system to track (or follow) given input?

r(t)System

y(t)

i.e., will for tlarge?0 )()()( tytrte

Some general ideas -- by applying step, ramp, and parabolic

inputs to Unity FeedbackSystem:

R(s) G(s) Y(s)+

-

E

(s)

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Tracking performance given by signal e(t) = r(t) - y(t)

Steady state tracking error using Final Value Theorem:

(perfect tracking if e(t)= 0 ! )

)()(1

1)()()( sR

sGsYsRsE

An Unity Feedback system is of typeN, withNgiven by G(s):

))...((

))...(()(

1

1

n

m

pspss

zszsKsG

N

(N+ve,z1,z2, zm nonzero;p1,p2, pn nonzero)

)(1

)(lim)(lim

00 sG

ssRssEe

ssss

(Assuming closed

loop system is stable)

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Steady state tracking for type 0 system:

))...((

))...(()(

1

1

n

m

psps

zszsksG

- Step Input:s

sRttr 11 )(),()(

)(lim sGK

s

p

0

with (Position error constant)

- Ramp Input:2

1

ssRttr )(,)(

0

1

)(lim

1

)(1

)(lim

0

1

0

2

ssGsG

se

s

s

sss

- Parabolic Input: 32 1

2 ssRttr )(,)(

0

1

)(lim

1

)(1

)(lim

2

0

1

0

3

sGssG

se

s

s

sss

(Finite steady

state tracking error)

(Infinite steady


(Infinite steady


ps

s

sss

KsGsG

se

1

1

1

1

10

1

0 )(lim)(

)(lim

s01

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- Ramp Input:2

1s

sRttr )(,)(

with (Velocity error constant)

- Parabolic Input: 32

1

2 ssRttr )(,)(

0

1

)(lim

1

)(1

)(lim

2

0

1

0

3

sGssG

se

s

s

sss

Steady state tracking for type I system:

- Step Input: ssRttr 11 )(),()(

01

1

)(lim1

1

)(1

)(lim

0

1

0

sGsG

se

s

s

sss

(Finite steady


(Infinite steady


(Zero steady state

tracking error)

)(lim0

ssGKs

v

1

)(lim

1

)(1

)(lim

0

1

0

2

KssGsG

se

Vs

s

sss

))...((

))...(()(

1

1

n

m

psps

zszsksG

s1

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- Ramp Input: 21

ssRttr )(,)(

01

)(lim

1

)(1

)(lim

0

1

0

2

ssGsG

se

s

s

sss

- Parabolic Input: 32 1

2 ssRttr )(,)(

as

s

sss

KsGssG

se

1

)(lim

1

)(1

)(lim

2

0

1

0

3

)(lim

2

0 sGsK sa

with (Acceleration error constant)

Steady state tracking for type II system:

- Step Input:s

sRttr 11 )(),()(

01

1

)(lim1

1

)(1

)(lim

0

1

0

sGsG

se

s

s

sss

(Finite steady


(Zero steady state

tracking error)

(Zero steady state

tracking error)

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

1

1

n

m

pspszszsksG

s2

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Summary of steady state error for unity feedback system

Generally, a feedback system is typeNif ess is finite for r(t)=CtN

sse

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Revisit Design Example: Disk Drive Read System- With Kb=0 and neglecting electrical response

(Textbook Fig. 5.46) DC Motor

Table 5.9

Reduced Model, see

Design Example: Disk

Drive Read System in

Section I.E.

- Transient performance consideration

2025520

5 2

2

nan

a

a KKss

K

sR

sY ,

)(

)(

-- Transfer function

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-- Different Kadifferent different P.O. and Tsand n

-- Performance against disturbance?

Maximum value due to unit disturbance calculated from

transfer functionD(s) to Y(s) (Textbook Fig. 5.48)

- Possible Ka to satisfy specification?

System no longer underdamped for Ka=20,

expressions forTp, P.O., Tr1 and Ts not valid

- Steady state performance consideration

-- System is Unity feedback system System is Type I

-- ess= 0 for step response-- for ramp input,

vss K

e 14)20(

5lim

0

aa

sv

K

ss

KsK

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II.D. Stability

Definition: System is stable (in BIBO sense) if for all

Bounded Input yield Bounded Output.

G(s)R(s) Y(s)

)()()( sYsRsG

System BIBO stable Poles of G(s) inside LHP!

Poles ofR(s) + Poles of G(s)= Poles of Y(s)

For Bounded OutputPoles of Y(s) inside LHP with

possible non-overlapping

ones on imaginary axis

For Bounded InputPoles ofR(s) inside LHP with

possible non-overlapping

ones on imaginary axis

G(s) poles must be inside

LHP for BIBO stability

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Given a transfer function

01

1

1

01

1

1

asasasa

bsbsbsb

sa

sbsG

n

n

n

n

m

m

m

m

...

...

)(

)()(

A First Test

- Poles of G(s) = roots of a(s)

- System BIBO stable iff roots of a(s) all inside LHP

- Question: can we tell if ROOTS of a(s) are all inside LHP?

- Given any polynomial a(s) and after factorization

- For all roots of a(s) in LHP

- a(s) is product of factors of and

ALL with +ve coefficients

])[()())(()( 212

121 edscscscssa r

11 jeds ,,,, 21 rcscscs

(without computing the roots of a(s)!)

real roots complex roots

ve!,,and,,,, 121 dccc r

)( rcs

)](2[(2

1

2

11

2

edsds

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- Hence, if a(s) has roots in LHP, the coefficients of a(s)

must all be positive

If not all , then not all roots of a(s) inside LHP0,...,, 21 naaa

- Example:

LHPinsiderootsallnot

0allnot201023

iassssa ..)(

(Actually, roots of )3497.0,341.0675.0:)( jsa

- a(s) is product of factors of and, where coefficients are all positive

- Multiplying out, the coefficients of

must all be +ve!

)( rcs )](2[( 21

2

11

2edsds

01

1

1 ...)( asasasasa

n

n

n

n

First Test

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- Note: First Test is sufficient but not necessary, i.e.,If not all coefficients of a(s)>0, then not all roots inside LHP

But

If all coefficients of a(s)>0, cannot conclude all roots inside LHP!

201023 ..)( ssssaExample:

All coefficients of a(s)>0, but actual roots are

not all inside LHP!07911429003950 .,.. j

G(s) has unstable rootsIf not all 021 naaa ,...,,- Hence:

Need more detailed test:

Rouths Stability Criterion

If all 0,...,, 21 naaa

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01

1

1 asasasasa

n

n

n

n

...)(Step 1. Given

Routh Stability Test- A sufficient and necessary test

- Procedures:

Step 2. Form Array:

10

531

3

531

2

5311

42

n

nnn

n

nnn

n

nnnn

nnn

n

hs

cccs

bbbs

aaas

aaas

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1

0sAnd so on until row

from rowscomputed

from rowscomputedfrom coefficients ofa(s)sandsRow

nn

21sands

nn 3sRow n

1sands

nn 2sRow n

Step 3. Check the 1st column of array:

* If elements are all +ve, then all roots of a(s) inside LHP

* If some elements are -ve, then

# of sign changes in column = # of roots of a(s) inside RHP

,;51

4

1

3

31

2

1

1

1

1

nn

nn

n

n

nn

nn

n

n

aa

aa

a

b

aa

aa

a

b

,;

51

51

1

3

31

31

1

1

1

1

nn

nn

n

n

nn

nn

n

n

bb

aa

b

b

bb

aa

b

c

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* EXAMPLE

Application of Routh Stability Test

- Case 1: No element in the first column is zero

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- Case 2: A row has 1st element zero but some other

elements nonzero

* Two possibilities:

-- a(s) has some roots on Imaginary Axis, OR

-- a(s) has some roots in RHP

* To proceed:

(a) Replace zero by small positive number(b) Complete Array

(c) Let

and check 1st column:

-- No sign change some roots on Imaginary Axis

-- Sign change some roots in RHP

(# of sign changes = # of roots in RHP)

0

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* EXAMPLE

Two sign change a(s) has two poles in RHP

Possible to divide whole row

by some constant. In this case,

row divided by 3

0

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- Case 3: A row has 1st element zero AND all other elementsin the same row zero

(a) Form Auxiliary polynomialAP(s) with coefficients

from previous row(b) Use coefficients of

(c) Complete Array and check 1st columnds

sdAP )(to replace the all zero row

-- No sign change some roots on imaginary axis

-- Sign change some roots in RHP(# of sign changes = # of roots in RHP)

* Again, only two possibilities:

-- a(s) has some roots on Imaginary Axis, OR-- a(s) has some roots in RHP

* To proceed:

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* EXAMPLE

No sign change a(s) contains roots on imaginary axis

ssAP

ds

d

ssAP

6

123 2

)(

)(

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* In Case 3, additional structure known about a(s) :o AP(s) can always be factorized as:

o a(s) can be factorized byAP(s): )()(~)( sAPsasa

bothor),)((or))(( jsjsss

o Moreover, 1st column

of array can be divided

into before and after

occurrence ofAP(s)!

4

23

31

Occurrence ofAP(s)

1st column of array

)(~ sa

Sign changes before

occurrence ofAP(s)

indicates # of unstable

roots of

Sign changes after

occurrence of

AP(s) indicates #

of unstable roots

of AP(s)

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* EXAMPLE

No sign change a(s) contains roots on imaginary axis

Additional structure on a(s):o yields roots on Imag. Axis

o a(s) factorizablebyAP(s):

o No sign change in first column before occurrence ofAP(s)

roots inside LHP (which are -3, -1 and -1 ))(~ sa

ssAPds

d

ssAP

6

123 2

)(

)(

)(sAP

))(()( 12337531 223 sssssa

)(~ sa

))(()( jsjssAP 22 j2

Occurrence ofAP(s)

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Application to Feedback system

- Example 6.5: Welding control

* Transfer function:

)())()(()(

)()(

asKssssasK

sRsY

321

* Stability of system dictated by:

06116 234 KasKssssq )()(

* Rouths stability test

K

KKaK

36

66060

))((;

(with K > 0, a > 0)

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* Plotting the allowable region on (K-a) plane

* Example: for K=40, a=0.639

Allowable

region

K

KKa

36

660 ))((

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