Download - Chapter 3 System ID
-
8/2/2019 Chapter 3 System ID
1/24
Chapter 3 System ID
1. Scalar case, one unknown parameter, intuitive method 12. Scalar case, one unknown parameter, intuitive method 23. Scalar case, one unknown parameter, un-normalized4. Scalar case, one unknown parameter, normalized5. First order linear system-scalar case, un-normalized6. First order linear system-vector case, un-normalized7. First order linear system-vector case, normalized8. SPR Lyapunov approach, normalized9. Gradient method, normalized
10. Least squares, normalized
Chapter 3 System ID for adaptive control
1. Scalar case, one unknown parameter, intuitive method 12. Scalar case, one unknown parameter, intuitive method 23. Scalar case, one unknown parameter, un-normalized method4. Scalar case, one unknown parameter, normalized method5. First order linear system-scalar case, un-normalized6. First order linear system-vector case, un-normalized7. First order linear system-vector case, normalized8. SPR Lyapunov approach, normalized9. Gradient method, normalized
10. Least squares, normalized
-
8/2/2019 Chapter 3 System ID
2/24
1. Scalar case, one unknown parameter, intuitive method 1System : y (t) = .u (t)Where : y (t) = output
u (t) = input
= unknown constantAssume that y (t) and u (t) are measurable t to and it is desired to obtain an estimate of.Solution : Get the 1/0 pair {u (t) , y (t)} over an interval [, + T] so that u (t) 0 forsome t [, + T] and choose as , [, + T]Remark :
u (t) maybe very close to zero = very big! UnboundedRemark :It would be worse if the system is subjected to noise
The system would result in a large spread in the value of at different t [, + T]Example : y (t) = . u (t) + d
Review
y (t) =
. u(t)
if u0 the problem is = of u0 ; may have measuring error/noise by u
2. IDScalar case, one unknown parameter, 2nd intuitive methodLets define an estimator with system : y = . u(t) t [, + T]The solution maybe considered as one that minimizes the performance index (p1) :
Where [ ] Its minimization is evaluated as :
[
]
-
8/2/2019 Chapter 3 System ID
3/24
for u (t) = 0 , t [, + T]Remark : as in the previous method, can only be calculated once if no noise presentRemark : If noise shows up, it might not gine true ofRemark : Increasing of I will improve To anticipate it, we meed RLS : recursive lee square!
Because its impossible to increment the function to another time interval
3. IDScalar case, one unknown parameter, unvormalized (the system)System y(t) = Where u . is an unknown constant and u,y are measurableWe would like to determine from the measured datalets consider the estimated output : Define
output error
= = - ( ) u= -
Define J( performance J( is pointuise of J=
=
J( is convex over and hence the minimizationof J is global convex optimization problemECX is said to be convex if [0,1] such that
= * +=
-
8/2/2019 Chapter 3 System ID
4/24
Stability of the estimator above can be proved as :
Let V (
V(x) is p.d if 0 V (x) = 0 x = 0 0 V (x) > 0 x
parameter error is bounded ( Check (proef)
=
<
Check =
If then 0 as t = y = u
J =
Because of : V =
If
=
as t =
When will as t = ?
as t = (Barbalats lemma)
-
8/2/2019 Chapter 3 System ID
5/24
A necessary and sufficient condition is that u satisfies the PE condition
and for some PE = Persistent ExatationU should be rich enough to have parameter convergence
4. IDScalar case, one unknown parameter, normalizedSystem y(t) = Where is an unknown constant and u,y are measurable u
Since now u could be unbounded, we may not set
Because it would give an unbounded Let us divide the system equation by the normalizing signal m > 0To have Where and , and m2 = 1 + The signal ns is chose so that e.g. pick ns = u
m2
= 1 +
u
2
Consider the estimate Let normalized output error
= =
J( =
=
=
-
8/2/2019 Chapter 3 System ID
6/24
= = m > 0
Using the gradient method
= * += * +
= Let V =
= = -
<
= if
then as t = as t =
For parameter convergence, me need PEPersistent excitation
ID of 1st
order linear systems (unnormalized)
Plant : Where Xp R is the output of the plant and u R is the bounded inputAp and Kp are unknown constants, but me know that ap < 0
The problem is to determine ap and kp from u and xp
-
8/2/2019 Chapter 3 System ID
7/24
Here, we are going to solve this problem with 2 estimators :
1. Parallel estimatoroutput error methodonly for scalar system (not for vector)2. Seriesparallel estimatorequation error methodfor scalar and vector systemsMethod 1 : - output error method
Select the parallel estimator u - error dynamicsWhere e =
, (The output error dynamics is driven by the output error)And parameter error Y (t) = u not discussed yet
5.
ID of 1
st
order linear system (unnormalized)Plant = Where Xp IR , u R are measurableAp and Kp are unknown constants, ap < 0
1. Parallel estimatoroutput error method (only for scalar)2. Seriesparallel estimatorequation error method (also for method)
Method : - output error method
Select the parallel estimator
u - error dynamics
-
8/2/2019 Chapter 3 System ID
8/24
Define a Lyapunov function candidate
Take the time derivate of V along the trajectory of the errors dynamics, we have : Pick
Bounded
U Xp
Stable
LTI System
BI
Bounded
Input
BIBO
Stable
BO
Bounded
Output
-
8/2/2019 Chapter 3 System ID
9/24
We need PE for parameter Convergence
Summary :
For parameter Convergence, we need PE
For parameter Convergence, we need PE
Method 2 : - Equation error method
Select the series-paralel estimator : [ ] [ ]
-
8/2/2019 Chapter 3 System ID
10/24
=[ ] Pick
t < 0 . We need PE for convergence of the parameters.
ID VECTOR CASE UN-FORMALIZED
Plant = Where Assume that
is Hurwitz and u is bonded.
We would like to find and , based on the information of and u.Method : Output error methodDefine the parallel estimator :
The error dynamics can be as : Pick the Lyapinov function candidate : = =
?
Although is known to be stable, its elements are not known and hence a matrix p > 0 cant be found
s.t. with Q > 0.This approach doesnt work for veetos case but its fine for scalar case.
Method 2 : Equation error method
Define the series-parallel estimator Where and and is known Let V =
=
-
8/2/2019 Chapter 3 System ID
11/24
= - Q
is stable s, t,
Method 2 :
Pick
because q is a constant
We need PE for parameter converage!
What about u 7. ID-1st order systems, normalized u
Plant : unknown constansLet + Define e = x - Where s is the laplace variable
Ns is to be designed s.t.
and
with
and ns =
Update laus
-
8/2/2019 Chapter 3 System ID
12/24
(s-am)e = (S-am) . (x - )e. - ) am (x - )e.
-
= (axbu)(am
+ (
-am) x +
= am (x - ) + (a- x + (b-)u (x - ) + (a-x + (b-)uam (x - ) - e. = (am - e + (a-x + (b-)u = (am - e + +
Let V=
-
[ ] (am - ) xe ue (am - E,
||
For some n > 0
=
E = (am - e + + We can not conclude asympt stability of e
e.ns :v
Q.E.D
-
8/2/2019 Chapter 3 System ID
13/24
e. m
= <
-
8/2/2019 Chapter 3 System ID
14/24
= =
In general, we represent a linear parametric model as :
Y = G(s) . Review e = x- updated laws : Positive Linear System SPR : Strictly positive RealPositive linear systems are linear systems with positive real T.F.s
System :
(n-th order SISO linear system)Def : h(s) is positive real (PR) if Re [h(s)] 0 Re [s] 0h(s) is strictly positive real (SPR) if h (s-) is PR for some > 0
e.q. h(s) = > 0, S=
h( ) = =
If Re [h(s)] = Then h(s) is PR
For SPR :
Pick h(s - ) =
is P.R
-
8/2/2019 Chapter 3 System ID
15/24
is SPR
Thm : h(s) is SPR iff
h(s) is strictly stable, and
Re [h(j)] > 0 Some useful necessary conditions to check SPR :1. h(s) strictly stable
e.g h(s) = is not SPR
because the evalue is not in the left hand plane
2. h(s) has relative degree 0 or 1e.g h(s) =
is not SPR
3. h(s) is strictly minimum phasee.g h(s) =
is not SPR4. the ngguist plot of h(j) lies entirely in RHP (RightHand - Plane)5. phaseshift in response to sinusoidal input is always less than 900the kalman-yakuborich lemma (positive real lemma)
lemma :
system :
y = (A,B) controllable
The T.F. h(s) = is SPRIff p > 0 and Q > 0 such that (s.t)
PB = C
Controllable stabilizable uncontrollable
Uncontrollable not stabilizable
PCstabilizable
Costabilizable is continuousstabilizable
stabilizable
stabilizable f is smooth
-
8/2/2019 Chapter 3 System ID
16/24
p.c. stabilizable piece wise continuousmodified k-y-lemma
kaliman-Yakubovich-Meyer Lemma
given x
0 A asymptotically stable, symmetric p.d. matrix L
if the T.F. is SPR, then vector q and p = PT > 0 s.t
PB = C + 8. IDGeneral SISO plantSPR Lyapunov Approach, normalized
Plant : y = G(s) . , y
Choose L(s) s.t. L-1
(s) is a proper stable T.F. and G(s) . L(s)
is a proper SPR T.F
y = G (s) . L(s) . = G (s) . L(s) .
e.q : y =
= * += G(s).
Let L = s+2 L-1
=
Y = G L L-1
= * +
=
An estimator can be constructed as Case 1 : G(s) is minimum phase
Pick L (s) = G-1
(s) s.t . GL = 1
-
8/2/2019 Chapter 3 System ID
17/24
Define
etc
Case 2 :
G(s) is not required to be minimum phase
Define the normalized error
GL e
-
8/2/2019 Chapter 3 System ID
18/24
[ ] [ ]
Realization to state-space function
-
8/2/2019 Chapter 3 System ID
19/24
1DGradient Method - normalized
System:
In parametic model
Two approaches are introduced below based on different cost-function
Case 1 :Instantaneous cost function
Convexity of guarantees the existense of a unique globalMinimum devine by :
-
8/2/2019 Chapter 3 System ID
20/24
(1)Non-recursive algorithm 1 : * +
(2)Non recursive alogarithm 2:
Adapptive control of linear-time-varying system (phD Thesis)
Joannou and Tsaklis
-
8/2/2019 Chapter 3 System ID
21/24
(3)Recursive Algorithm
March 30, 2006
-
8/2/2019 Chapter 3 System ID
22/24
* +
( )
}**it means : to make
-
8/2/2019 Chapter 3 System ID
23/24
Case 2 : Integral Cost Function
[ ] [
]
[ ]
{ } {
}
Homogeneous
-
8/2/2019 Chapter 3 System ID
24/24
Note that R(t) and Q(t) are exactly solutions to the following two
Non-homogenous D.E. (differential equation) respectively
{ }