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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
state tracking error)
(Infinite steady
state tracking error)
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
state tracking error)
(Infinite steady
state tracking error)
(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
state tracking error)
(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 ))((