Part 8 - Recursive Identification Methods

7/25/2019 Part 8 - Recursive Identification Methods

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System IdentificationControl Engineering B.Sc., 3rd yearTechnical University of Cluj-Napoca

Romania

Lecturer: Lucian Busoniu


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Motivating example Recursive least-squares and ARX Recursive instrumental variables

Part VIII

Recursive identification methods


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Recursive identification: Idea

Recursive methods can be seen as working online, while the systemis evolving.

At stepk, computenew parameter estimate

(k), based on the

previous estimate(k 1)and newly available datau(k),y(k).Remark:To contrast them with recursive identification, the previousmethods, which used the whole data set, will be called batchidentification.


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Features

Recursive methods:

Require less memory, and less computation for each update,than the whole batch algorithm.

Easier to apply in real-time.

When properly modified, can deal withtime-varyingsystems.

If model is used to tune a controller, we obtainadaptive control:


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Disadvantage

Recursive methods are usually approximations of batch techniques guarantees more difficult to obtain.

M ti ti l R i l t d ARX R i i t t l i bl


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Table of contents

1 Motivating example: Estimating a scalar

2 Recursive least-squares and ARX

3 Recursive instrumental variables

Motivating example Recursive least squares and ARX Recursive instrumental variables


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Recall: Estimating a scalar

Recall scalar estimation. Model:

y(k) =b+ e(k) =1 b+ e(k) =(k)+ e(k)

where(k) =1k,=b.

For the data points up to and including k:

y(1) =(1)=1 b

y(k) =(k)=1 b

After calculation, the solution of this system is:

(k) = 1k

[y(1) +. . .+ y(k)]

Intuition:Estimate is the average of all measurements, filtering out

the noise.



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Estimating a scalar: Recursive formulation

Rewrite the formula:

(k) =

1

k[y(1) +. . .+ y(k)]

= 1

k[(k 1) 1

k 1 (y(1) +. . .+ y(k 1))+ y(k)]

= 1

k[(k 1)(k 1)+ y(k)]

= 1

k[k(k 1) + y(k) (k 1)]

=(k 1) + 1k

[y(k) (k 1)]



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Estimating a scalar: Discussion

(k) =(k 1) + 1k

[y(k) (k 1)]Recursive formula: new estimate(k)computed based onprevious estimate(k 1)and new datay(k).[y(k) (k 1)]is a prediction error(k), since

(k 1) =

b=

y(k), a one-step-ahead prediction of the output.

Update applies a correction proportional to(k), weighted by 1k

:

When the error (k)is large (or small), a large (or small)adjustment is made.

The weight 1k

decreases with timek, so adjustments get smaller asbbecomes better.



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Table of contents



General recursive least-squares

Recursive ARX

Matlab example




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g p q

Recall: Least-squares regression

Model:y(k) =(k)+ e(k)

Dataset up tokgives a linear system of equations:

y(1) =(1)

y(2) =(2)

y(k) =(k)

After some linear algebra and calculus, the least-squares solution is:

(k) =

kj=1

(j)(j)

1

kj=1

(j)y(j)



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g p q

Least-squares: Recursive formula

With notationP(k) =kj=1(j)(j):(k) =P1(k)

kj=1

(j)y(j)

=P1(k)k1

j=1

(j)y(j) +(k)y(k)=P1(k)

P(k 1)

(k 1) +(k)y(k)

=P

1

(k) [P(k) (k)(k)](k 1) +(k)y(k)=(k 1) + P1(k) (k)(k)(k 1) +(k)y(k)=

(k 1) + P1(k)(k)

y(k) (k)

(k 1)



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Least-squares: Discussion

(k) =(k 1) + P1(k)(k) y(k) (k)(k 1)

(k) =

(k 1) + W(k)(k)

Recursive formula: new estimate(k)computed based onprevious estimate(k 1)and new datay(k).

y(k) (k)

(k 1)

is a prediction error(k), since

(k)(k 1) =yk1(k) is the one-step-ahead prediction usingthe previous parameter vector.W(k) =P1(k)(k)is a weighting vector: elements of Pgrowlarge for largek, thereforeW(k)decreases.



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Recursive matrix inversion

The previous formula requires matrix inverseP1(k), which iscomputationally costly.

ForP, an easy recursion exists: P(k) =P(k 1) +(k)(k), but

this does not help; the matrix must still be inverted.TheSherman-Morrisonformula gives a recursion for the inverse P1:

P1(k) =P1(k 1) P1(k 1)(k)(k)P1(k 1)

1 +(k)P1(k 1)(k)

Exercise: Prove the Sherman-Morrison formula! (linear algebra)



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Overall algorithm

Recursive least-squares (RLS)

initialize

(0),P1(0)

loopat every stepk=1, 2, . . .

measurey(k), form regressor vector(k)find prediction error(k) =y(k) (k)(k 1)update inverseP1(k) =P1(k 1) P

1(k1)(k)(k)P1(k1)1+(k)P1(k1)(k)

compute weightsW(k) =P1(k)(k)

update parameters(k) =(k 1) + W(k)(k)end loop



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Initialization

The RLS algorithm requires initial parameter(0), and initial inverseP1(0).

Typical choices without prior knowledge:

(0) = [0, . . . , 0]

- a vector ofnzeros.

P1(0) = 1

I, witha small number such as 103.An equivalent initial value ofPwould beP(0) =I.

Intuition: Psmall corresponds to a low confidence in the initialparameters. This leads to large P1(0), so initially the weightsW(k)

are large and large updates are applied to.If a prior valueis available,(0)can be initialized to it, andcorrespondingly increased. This means better confidence in the initialestimate, and smaller initial updates.



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Back to the scalar case

y(k) =b+ e(k) =(k)+ e(k), where(k) =1, =b

Take(0) =0, P1(0) . We have with Sherman-Morrison:P1(k) =P1(k 1)

(P1(k 1))2

1 + P1(k 1)=

P1(k 1)

1 + P1(k 1)

P

1(1) =1

P1(2) = 1

2

P1

(k) = 1

k

Also,(k) =y(k) (k 1)and W(k) =P1(k) = 1k

, leading to:

(k) =(k 1) + W(k)(k) =(k 1) +1

k[y(k) (k 1)] formula for the scalar case was recovered.



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Table of contents




Recursive ARX

Matlab example




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Recall ARX model

A(q1)y(k) =B(q1)u(k) + e(k)

(1+a1q1 + + anaq

na)y(k) =

(b1q1 + + bnbq

nb)u(k) + e(k)

In explicit form:

y(k) + a1y(k 1) + a2y(k 2) +. . .+ anay(k na)

=b1u(k 1) + b2u(k 2) +. . .+ bnbu(k nb) + e(k)

where the model parameters are: a1, a2, . . . , anaandb1, b2, . . . , bnb.



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Linear regression representation

y(k) = a1y(k 1) a2y(k 2) . . . anay(k na)

b1u(k 1) + b2u(k 2) +. . .+ bnbu(k nb) + e(k)

= y(k 1) y(k na) u(k 1) u(k nb)

a1 ana b1 bnb

+ e(k)

=:(k)+ e(k)

Regressor vector: Rna+nb

, previous output and input values.Parameter vector: Rna+nb, polynomial coefficients.



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Recursive ARX

With this representation, just an instantiation of RLS template.

Recursive ARX

initialize(0),P1(0)loopat every stepk=1, 2, . . .

measureu(k),y(k)

form regressor vector(k) = [y(k 1), , y(k na), u(k 1), , u(k nb)]

find prediction error(k) =y(k) (k)(k 1)update inverseP1(k) =P1(k 1) P

1(k1)(k)(k)P1(k1)1+(k)P1(k1)(k)

compute weightsW(k) =P1(k)(k)


Remark:Outputs more thannasteps ago, and inputs more than nbsteps ago, can be forgotten.



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Guarantees idea

With(0) =0, P1(0) = 1

Iand 0, recursive ARX at step k isequivalent to running batch ARX on the datasetu(1), y(1), . . . , u(k), y(k).

the same guarantees as batch ARX:

Theorem

Under appropriate assumption (including the existence of trueparameters0), ARX identification isconsistent: the estimatedparameterstend to the true parameters 0 ask .The condition onPis just to ensure the inverses are the same as inthe offline case.



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Table of contents




Recursive ARX

Matlab example




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System

To illustrate recursive ARX, we take a known system:

y(k) + ay(k 1) =bu(k 1) + e(k), a= 0.9, b=1(Soderstrom & Stoica)

system = idpoly([1 -0.9], [0 1]);

Identification data obtained in simulation:sim(system, u, noise);



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Recursive ARX

model = rarx(id, [na, nb, nk], ff, 1, th0, Pinv0);

Arguments:

1

Identification data.2 Array containing the orders ofA and Band the delaynk.

3 ff, 1selects the algorithm variant presented in this lecture.

4 th0is the initial parameter value.

5 Pinv0is the initial inverseP1(0).



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Results

na= nb=nk=1P1(0) = 1

I, left: =103, right: =100.

Conclusions:

Convergence to true parameter values...

...slower whenis larger (good confidence in the initial solution)



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System outside model class

Take now anoutput-errorsystem (so it is not ARX with white noise):

y(k) = bq1

1 + fq1u(k) + e(k), f = 0.9, b=1

(Soderstrom & Stoica)

system = idpoly([], [0 1], [], [], [1 -.9]);

Identification data:



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Results

na= nb=nk=1, so we attempt the ARX model:

y(k) + fy(k 1) =bu(k 1) + e(k)

Also,P1(0) = 1

I,=103.

Conclusion:

Convergence to the true values is not attained! as expected,

due to colored noise



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Other recursive algorithms

Other recursive identification methods available in Matlab, e.g.:

ARMAX, rarmax

output-error, roe

generic prediction error method, rpem



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Table of contents






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Recall: Instrumental variable method

IV methods find the parameter vector using:

= 1N

Nk=1

Z(k)(k)

1 1

N

Nk=1

Z(k)y(k)

(8.1)

which is the solution to the system of equations:1

N

Nk=1

Z(k)[(k) y(k)]

= 0 (8.2)

The regressor vector:(k) = [y(t 1), , y(t na), u(t 1), , u(t nb)]

Motivation: instrument vectorZ(k)uncorrelated with the noise can handle colored noise.


R i f l i


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Recursive formulation

Rewrite the equation for the recursive case:

=P1

k

j=1Z(j)y(j)

(8.3)

whereP=k

j=1 Z(j)(j).

Note averaging by the number of data points has been removed; itwas needed in batch IV since summations over many data pointscould grow very large, leading to numerical problems.

Recursive IV will add terms one by one, and will therefore benumerically stable.


R i f l ti (2)


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Recursive formulation (2)

With this definition ofP, the recursive update is found similarly to

RLS:

(k) =P1(k) k

j=1

Z(j)y(j)

=P1(k)k1

j=1

Z(j)y(j) + Z(k)y(k)=P1(k)

P(k 1)

(k 1) + Z(k)y(k)

=P

1

(k) [P(k) Z(k)(k)](k 1) + Z(k)y(k)=(k 1) + P1(k) Z(k)(k)(k 1) + Z(k)y(k)=

(k 1) + P1(k)Z(k)

y(k) (k)

(k 1)


R i f l ti (3)


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Recursive formulation (3)

Final formula:

(k) =

(k 1) + P1(k)Z(k)

y(k) (k)

(k 1)

(k) =(k 1) + W(k)(k)with the Sherman-Morrison update of the inverse:

P1(k) =P1(k 1) P1(k 1)Z(k)(k)P1(k 1)

1 +

(k)P1

(k 1)Z(k)


R i IV O ll l ith


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Recursive IV: Overall algorithm

Recursive IV

initialize

(0),P1(0)

loopat every stepk=1, 2, . . .

measurey(k), form regressor vector(k)find prediction error(k) =y(k) (k)(k 1)update inverseP1(k) =P1(k 1) P

1(k1)Z(k)(k)P1(k1)1+(k)P1(k1)Z(k)

compute weightsW(k) =P1(k)Z(k)


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Part 8 - Recursive Identification Methods