Time domain response specifications Defined based on unit step response with i.c. = 0 Defined for...
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Transcript of Time domain response specifications Defined based on unit step response with i.c. = 0 Defined for...
Time domain response specifications• Defined based on unit step response with i.c. = 0
• Defined for closed-loop system
Prototype 2nd order system:
22
2
2 nn
n
sssH
10
:dampedUnder
:case Useful
target
Prototype 2nd order system:
22
2
2 nn
n
sssH
2 2
2 2
Given and :
d
nd
d
d
p j
2
2
1
Given an
1
d :n
n n
n
d n
p j
1cos (180 ( ))
Re( )Given p
(
ol :
)
e
n
o
d
p
angle p
p
imag p
p
2d n
= = 1
pt
21σ max 1 1tpy e e
21Overshoot : pM e
21percentag % 100e e
Settling time: 2ln 1 ln( ) 3, 4, 5s
tol tol or ort
Delay time :
0.8 0.9 1.4d
n n
t
4.5( 0.2)
Rise time : 1.8 ~ 2.2 2
rnn n
t
5%, 2%, 1%
5% or 10% or 16% or 25%
0.7 0.6 0.5 0.4
pM
Remember:
Example: Given 0.6, 5, estimate specs.n
sol: 0.6 5 3n 21 4d n
peak time: 0.8sec.pd
t
21 0.096 10%pM e
settling time for 1% :5 5
1.67sec.3st
42% : 1.33sec.
3stol t
1.8rise time: 0.36sec.r
n
t
1
k
s s +
-
Example:
When given unit step input, the output looks like:
i.e. 25%pM 3sec.pt
Q: estimate k and τ.
sol: from 25% 0.4pM
3sec.3p d
p
tt
21 n
2 21.14
1 3 1 0.4d
n
From block diagram:
22 1
k kH s
s s k s s k
2
2 2Match this against
2n
n ns s
2 21.14
12 2 0.4 1.14
n
n
k
1.09
1.42k
Solve for and , get:k
Effects of additional zerosSuppose we originally have:
0H s 1 t 0y t i.e. step response
Now introduce a zero at s = -z
01s
H s H sz
The new step response:
0
1 11
sY s H s H s
s z s
01s
Y sz
0
11
dy t y t
z dt
tytyz
ty 00
1
beginning.at rising is Typically, 0 ty
01
,0 If)1 0 tyz
z
overshoot.larger has and , before
overshoots ,n faster tha rising is
0
0
ty
tyty
01
,0 If)2 0 tyz
z
initially 0 i.e.
direction in wrong offstart may
? overshoot smaller
n slower tha rises 0
ty
tyty
:0 e.g. z
Effects: • Increased speed,• Larger overshoot,
• Might increase ts
pdr ttt ,, i.e.
pM
When z < 0, the zero s = -z is > 0,
is in the right half plane.
Such a zero is called a nonminimum phase zero.
A system with nonminimum phase zeros is called a nonminimum phase system.
Nonminimum phase zero should be
avoided in design.
i.e. Do not introduce such a zero in your controller.
Effects of additional poleSuppose, instead of a zero, we introduce
a pole at s = -p, i.e.
01
10
psHsH
p
s
s
sHs
sHsYp
s
1
1
110
sYps
p0
filter pass loworder first a is ps
p
.at frequency corner with p
L.P.F. has smoothing effect, or
averaging effect
Effects: • Slower,• Reduced overshoot,
• May increase or decrease ts
pdr ttt ,, i.e.pM
Stability• BIBO-stable:
Def: A system is BIBO-stable if any bounded input produces bounded output.
Otherwise it’s not BIBO-stable.
))((
resp. impulse)( where
finite )(stable-BIBO :Thm
1
0
sH
th
dtth
L
cancelled pole/zero allafter plane half
leftopen in the )( of poles"" all stable-BIBO sH
Asymptotically Stable
A system is asymptotically stable if for any arbitrary initial conditions, all variables in the system converge to 0 as t→∞ when input=0.
A system is marginally stable if for all initial conditions, all variables in the system remain finite, but for some initial conditions, some variable does not converge to 0 as t→∞.
A system is unstable if there are initial conditions that can cause some variables in the system to diverge to infinity.
A.S., M.S. and unstable are mutually exclusive.
Asymptotically Stable
Theorem:
A system is A.S. all eigenvalues
have real part 0
A system is M.S. at least one pole is on j -axis
all j -axis pole(s) non-repea
ted
all other pole in OLHP
A system is unstable if there is pole(s) in ORHP
or repeated j -axis poles
Asymptotically Stable vs BIBO-stable
Thm: If a system is A.S.,
then it is BIBO-stable
If a system is not BIBO-stable, then it cannot be A.S., it has to be either M.S. or unstable.
But BIBO-stable does not guarantee A.S. in general.
If there is no pole/zero cancellation, then
BIBO-stable Asymp Stable
1
1 1 0 1 01
n n m
n mn n m
d d d d dy a y a y a y b u b u b u
dt dt dt dt dt
DuCxy
BuAxx
1 01
1 1 0
( )( )
( )
mm
n nn
b s b s bY sH s
U s s a s a s a
Characteristic polynomials
Three types of models:
Assume no p/z cancellation
System characteristic polynomial is:1
1 1 0( ) det( ) n nnd s sI A s a s a s a
A polynomial
is said to be Hurwitz or stable if all of its roots are in O.L.H.P
A system is stable if its char. polynomial is Hurwitz
A nxn matrix is called Hurwitz or stableif its char. poly det(sI-A) is Hurwitz, orif all eigenvalues have real parts<0
11 1 0( ) n n
n nd s a s a s a s a
Routh-Hurwitz MethodFrom now on, when we say stability we mean
A.S. / M.S. or unstable.
We assume no pole/zero cancellation,
A.S. BIBO stable
M.S./unstable not BIBO stable
Since stability is determined by denominator, so just work with d(s)
3
1
541
1
3212
5311
642
011
1
:
:
:
)(
n
n
nnnn
n
nnnnn
nnnn
nnnnn
nn
nn
s
a
aaaa
a
aaaas
aaas
aaaas
asasasasd
:table Routh
polynomialstic characteri the
called is d(s) T.F., c.l. of den. the be
Let
0
Routh Table
Repeat the process until s0 row
Stability criterion:
1) d(s) is A.S. iff 1st col have same sign
2) the # of sign changes in 1st col
= # of roots in right half plane
Note: if highest coeff in d(s) is 1,
A.S. 1st col >0
If all roots of d(s) are <0, d(s) is Hurwitz
Example:
unstable
RHP in roots 2 changes, sign 2
- :sign col first
65.2
065.2:
05.24
64:
64:
11:
64)(
0
1
2
3
23
s
s
s
s
ssssd ←has roots:3,2,-1
unstable
roots unstb 2
changes sign 2
10:
43.6:
107:
51:
1032:
10532)(
0
1
2
3
4
234
s
s
s
s
s
sssssd(1*3-2*5)/
1=-7
(1*10-2*0)/1=10
(-7*5-1*10)/-7
A.S.
change sign no 0,col 1st
1:
2:
11:
42:
131:
1432)(
0
1
2
3
4
234
s
s
s
s
s
sssssd
Remember this
sign same have coeff all iff
A.S.is system order 2nd
:system order 2nd
cs
bs
cas
cbsassd
:
:
:
)(
0
1
2
2
3 2
3
2
1
0
3rd order system:
( )
:
:
:
:
3rd order system is A.S.
, , , all same sign iff
d s as bs cs d
s a c
s b d
bc ads
bs d
a b c d
bc ad
Remember this
A.S.
A.S.
A.S.Not 0coeff all
A.S.Not
A.S.Not
A.S.Not
A.S.Not
A.S.Not
A.S.
943)943(
943532
632632
235
1
13
2
3
15
23
23
23
23
23
23
2
2
2
2
sss
ssssss
sss
sss
ss
ss
s
ss
sse.g.
Routh CriteriaRegular case: (1) A.S. 1st col. all same sign
(2)#sign changes in 1st col. =#roots with Re(.)>0
Special case 1: one whole row=0Solution: 1) use prev. row to form aux. eq. A(s)=0
2) get:
3) use coeff of to replace 0-row 4) continue as usual
)(sAds
d
Example
5 4 3 2
5
4
3
2
1
2
2 2
( ) 4 8 8 7 4
: 1 8 7
: 4 1 8 2 4 1
: 6 1 6 1
: 4 1 4 1
: 0
prev. row: : 4 4
( ) 4 4 4( 1)
d s s s s s s
s
s
s
s
s
s
A s s s
←whole row=0
5 4 3 2
5
4
3
2
1
2
2 2
( ) 4 8 8 7 4
: 1 8 7
: 4 8 4
: 6 6
: 4 4
: 0
prev. row: : 4 4
( ) 4 4 4( 1)
d s s s s s s
s
s
s
s
s
s
A s s s
2
2
1
0
1( )differentiate: 8
2
: 4 4
: 8
: 4
No sign change in 1 col. no roots in R.H.P
But Not A.S. since we did have 0 in 1st col. originally.
marginally stable
sdA ss
ds s
s
s
s
2 2
5 4
Fact: The roots of are all roots of
original ( ) ( ) ( )
e.g.: in prev example:
( ) 4 4 4( 1)
has roots:
&these are roots of ( ).
Indeed ( ) 4 8
A(s)
d s A s
A s s s
s j
d s
d s s s
3 28 7 4
has roots: 1.5 1.3229
0
1
s s s
j
j
-
7
)474)(1()(
0
44
44
7
747
44
874
474478841
232
2
2
3
23
24
234
35
23
23452
sssssd
s-
s
ss-
sss
ss-
sss
ss-
sssssssss
0
11
44
44)(
)(121
00
121
121
122)(
1
2
3
3
244
3
4
5
2345
:s
:s
:s
ssds
sdA
sAsss
:s
:s
:s
ssssssd
:row From
e.g.
0)Re( withroots no
col. 1st in change sign No
:row From
1:
2:
2)(
1)(
0
1
22
s
s
sds
sdA
ssAs
unstable. is
roots. double are &
0))fence.(Re( the on roots areThey
at root double
at root double
of roots are
of roots But
)(
)()()1(12)(
).(
12)(
222224
24
sd
js
js
jsjsssssA
sd
sssA
e.g.
continue & 0by "0" replace :solution
0row wholebut
0col 1st in #an :2 case Special
3:
32:
3:
30:
21:
321:
322)(
0
1
2
2
3
4
234
s
-s
s
s
s
s
sssssd
Replace by
0)Re( have two these
:roots has :Verify
0)Re( i.e. RHP, in roots 2
col. 1st in changes sign 2
0 assume wesince
2928140570
902.009057.0
0)(
03
2
.j.
j
sd
Useful case: parameter in d(s)
How to use: 1) form table as usual
2) set 1st col. >0
3) solve for parameter range for A.S.
2’) set one in 1st col=0
3’) solve for parameter that leads to M.S. or leads to sustained oscillation
Example
s+3
s(s+2)(s+1) Kp
pp
p
p
plc
p
KsKss
sKssssd
sKsss
sKsG
K
3)2(3
)3()1)(2()(
)3()1)(2(
)3()(
23
..
char.poly
:Sol
stability for of range find:Q
+
0
03)2(3
03
0
3:3
3)2(3:
33:
21:
0
1
2
3
p
pp
p
p
pp
p
p
K
KK
K
Ks
KKs
Ks
Ks
col. 1st : A.S.For
:table Routh
03
7)1(
03
42
0463
41)2(3
202
003
4)2(3)(
2
2
2
23
k
kk
kk
kk
kk
kk
skksssd
2)
1)
:need we: A.S.For
prod. outer two mid of prod 2)
0coeff all 1)
criteria? Routh order 3rd remember
e.g.
A.S.for
need weall over
also. and but
or
528.013
7
20
13
7
3
71
13
7
3
71
3
7)1( 2
k
kk
kk
kk
k
)137
(3
4
3
4
3
4
0)(43
)(,
13
7
4)2(3
22
1
k
kjs
sAkss
s
sdk
k
kk
:freq osci
noscillatio sustained to leads
:row From
0row And
M.S. is this At
:get we
set weIf
Q: find region of stability in K- plane.
2
2
3 2
( ) ( 2)( )
( 1) ( ) ( 2)
( ) ( 1) ( ) ( 2)
( ( 2) 1) 2
:
0
1( 2) 1 0
22 0 0
1 2( ( 2) 1) 2
s K sH s
s s s K s
d s s s s K s
s Ks K s K
RouthCriteria
K
K K
K
K K K K
2
K