1ME 233 Advanced Control II
Lecture 20
Stability Analysis of a discrete-timeSeries-Parallel Least Squares
Parameter Identification Algorithm

2

ARMA Model
Consider the following systemWhere•

known bounded input•

measured output

3

ARMA Model(anti-Schur)•Orders n and m are known•

a’s and b’s are unknown but constantcoefficients

4

ARMA Model
ARMA model can be written as:

Where:

5

Series-parallel estimation modelA-posteriori series-parallel estimation modelWhere•

a-posteriori estimate of•estimate ofat sampling time k•

estimate ofat sampling time k6Series-parallel estimation modelA-posteriori series-parallel estimation model

Where

•

a-posteriori estimate of7Series-parallel estimation modelA-priori series-parallel estimation model

Where

•

a-priori estimate of8Additional Notation• Unknown parameter vector:• Parameter vector estimate:• Parameter error estimate:• Regressor vector:

9Additional Notation• A-posteriori output estimation error:• A-priori output estimation error:

10

Parameter Adaptation Algorithm (PAA)
A-posteriori version• Parameter estimate update• Gain update• We make the restriction

11

PAA Special Cases• Least squares• Least squares with forgetting factor• Constant gain

12Example• Plant:

13

Example: Constant gainu(k) : zero mean uniform white noise between [-1,1]
parameter convergenceapriori error, series parallel predictor
0.3

2

1

1

00.2

000.1

500

1000

-10500

1000

500

1000

0-0.1

0.5

0.5

0

0

-0.2
-0.3
-0.4
0a-priori error
200

400
600
input:random

800

1000

-0.5
0500

1000

-0.50

14

Example: Least Squares
u(k) : zero mean uniform white noise between [-1,1]
parameter convergence
210.5

1

apriori error, series parallel predictor
0.15

00.1

000.05500

1000-0.50500

1000

500

1000

00.2

0.2

0.1

0-0.05a-priori error

-0.1

-0.15
-0.2
0200

400
600
input:random

800

1000

00500

1000

-0.2
0

15

Example: Least Squares & forgetting factor
u(k) : zero mean uniform white noise between [-1,1]
parameter convergenceapriori error, series parallel predictor
0.1

0.05

2110.5000-10-0.050.2

0.2

0.10-0.1-0.150500

1000-0.50500

1000

500

1000a-priori error
200

400
600
input:random

800

1000

00500

1000

-0.2
0

16Theorem
Under the following conditions:
1. The input u(k) is bounded, i.e.
2.
3.

is anti-Schur
Maximum singular value of F(k) is uniformly boundedand17

Parameter Adaptation Algorithm (PAA)
Since the unknown parameters are constant andthe PAA

implies that

18

A-posteriori dynamics• Error dynamics

• PAA

19

Equivalent Feedback Loop0+m(k+1)w(k+1)G(q)1NLe(k+1)

20Equivalent Feedback Loop0+m(k+1)w(k+1)NL:

1G(q)NLe(k+1)

21

Stability analysis using Hyperstability
0+m(k+1)w(k+1)G(q)1e(k+1)NL

1. Verify that the LTI dynamics are SPR
2 Verify that the PAA dynamics are P-class22

Good News: LTI “very” SPR
0+m(k+1)w(k+1)G(q)1e(k+1)NL

1. Verify that the LTI dynamics are SPR

Always SPR

23

Bad News: NL is not P-class0+m(k+1)w(k+1)•NL:G(q)1e(k+1)NL

Unfortunately the NL block is not P-class24

Solution: Modify the NL block•

Add a feedback term to NL to make it P-classNL1:

2 (k)
2w(k+1)e(k+1) NL

NL1+s(k+1)

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Modifying the NL block•

Add a feedback term to NL to make it P-class2 (k)NL1:

2w(k+1)e(k+1) -+s(k+1)

NL

NL1

Proof: See Additional Material at end of this lecture(the class notes on bSpace are incorrect)

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What happens to the feedback structure?
•Add and subtract the same blocks:
0+m(k+1)1G(q)-0+m(k+1)-G(q)1e(k+1)w(k+1)

+2v(k+1)

=e(k+1)+-+++=02

2 (k)NL2

2 (k)
2w(k+1)NLs(k+1)

-+=0

27

What happens to the feedback structure?
•0+

Rearranging blocks,m(k+1)1G(q)-e(k+1)0+m(k+1)

1G(q)

-


2v(k+1)-2

2 (k)++ +=w(k+1)

s(k+1)-NL1

2NL-+ s(k+1)

1(  -  (k) )
22+L2

NL2NL

2 (k)+2w(k+1)

2+22 (k)v(k+1)-

L1+e(k+1) +28What happens to the feedback structure?
•0+

Rearranging blocks,m(k+1)1G(q)-e(k+1)0+m(k+1)

1G(q)

-


2v(k+1)-2

2 (k)+

2 (k)w(k+1)

s(k+1)-NL1

NL-+ s(k+1)

L21(  -  (k) )
22+-+
NL2NL

NL1

2+ +2
w(k+1)2+2

2 (k)v(k+1)-

L1+e(k+1) +29What happens to the feedback structure?
•0+

Rearranging blocks,m(k+1)1G(q)-e(k+1)0+m(k+1)

1G(q)

-


2v(k+1)-2

2 (k)L2+

2 (k)w(k+1)

s(k+1)-NL-+ s(k+1)

L21(  -  (k) )
22+-+
NL2NL

NL1

2+ +2
w(k+1)2+2

2 (k)v(k+1)-

L1+e(k+1) +30What happens to the feedback structure?
•0+

Rearranging blocks,m(k+1)1G(q)-e(k+1)0+m(k+1)

1G(q)

-L1


2v(k+1)-2

2 (k)+2 (k)w(k+1)

s(k+1)-NL-+ s(k+1)

L21(  -  (k) )22+-+NL2NL

NL1

2+ +2w(k+1)

2+2

2 (k)v(k+1)-L1+e(k+1) +31

Can we now use Hyperstability Theory?
0+m(k+1)

1G(q)-e(k+1) +-L1v(k+1)

For Asymptotic
Hyperstability:2

1. L1 must be SPR
2 (k)

NL12w(k+1)

NL-+ s(k+1)

L2NL21(  -  (k) )22+2. NL2 must beP-class32

Linear Block L10+m(k+1)1G(q)-e(k+1) +-L1v(k+1)

Since:

2L1:
2 (k)NL1

2w(k+1)

NL-+ s(k+1)

L2

NL21(  -  (k) )22+L1 is SPR iff

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Nonlinear Block NL2
0+m(k+1)

1G(q)-e(k+1) +-L1v(k+1)

NL2 :2Feedback combinationof two blocks:2 (k)NL1

1. NL1 : P-class

2w(k+1)

NL-+ s(k+1)+-

2. L2 : P-class

L2NL21(  -  (k) )22

Therefore NL2is P-class34Hyperstability Theorem
If
0+m(k+1)

1G(q)-e(k+1) +-L1v(k+1)

2for some2 (k)Then

NL12w(k+1)

NL-+ s(k+1)

L2

NL21(  -  (k) )22+

1. L1 is SPR
2. NL2 is P-class
Therefore:
The interconnection
is asymptotically
hyperstable

35

Hyperstability Theorem
If
0+m(k+1)

1G(q)-e(k+1) +-L1v(k+1)

2for some2 (k)Then

NL12w(k+1)

NL-+ s(k+1)

L2

NL21(  -  (k) )22+36A-posteriori error convergence
We have concluded that the a-posteriori output
error converges to zero:

whereWhat about the a-priori output error?

37

A-posteriori error convergence• Notice that• Therefore,
imply that

does not necessarily• To prove, we need to first show

38

Boundedness of the regressor vector
Remember that:

Therefore,

By assumption,

39

Boundedness of the regressor vector
Since

and,

we only need to show that

40

Boundedness of the regressor vector
Remember that:

andis anti-Schur.

Therefore LTI system
is BIBO
Thus,

Additional Material(you are not responsible for this)• Proof that NL1 is P-class

41

42

Equivalent feedback loop (review)
0+m(k+1)w(k+1)•NL:G(q)1e(k+1)NL

Recall that the NL block is not P-class43

Solution: Modify the NL block (review)•Add a feedback term to NL to make it P-classNL1:

2 (k)
2w(k+1)e(k+1) NL

NL1

We want to show:+s(k+1)

44

Simplified Notation

45

Proof that NL1 is P-class2 (k)
2w(k+1)e(k+1) -+s(k+1)

NLNL1• From the summing junction, we have

46

Proof that NL1 is P-class
NL:• Multiply the input of NL1 by its output

47

Proof that NL1 is P-class• From the previous slide

Define48

Proof that NL1 is P-class• From the previous slide

49

Proof that NL1 is P-class• From the previous slide

because• Therefore

because50

Proof that NL1 is P-class• Now check the Popov inequality