System IdentificationLecture 9: Time-domain correlation methods, persistency of excitation and

parametric models

Roy Smith

2018-11-13 9.1

Example: Ceres orbit determination

1801, 1st January. Guissepe Piazzi finds and makes 22observations of a “planet” over 44 days.

1801, February. Ceres is obscured by the sun. Theobservations correspond to 9 degrees of arc.

1801, June. Zach publishes the observational data.Many astronomers attempt to estimate when andwhere Ceres will reappear.

1801, September. Karl Fredrich Gauss estimates theorbit of Ceres and Zach publishes his empheris (alongwith many others). Those of Gauss differ significantly.

1801, 7th December.Zach locates Ceres within 1/2 adegree of Gauss’s prediction. Gauss refuses to disclosehis method and is accused by some of sorcery.

Dawn mission image of Ceres

(930 km. diameter)

2018-11-13 9.2

Example: Ceres orbit determination

1802. Gauss also estimates the orbit of Pallas.

1805 Legendre publishes a paper on the method of leastsquares.

1807 Gauss becomes director of Gottingen observatory. Olbersdiscovers Vesta and Gauss takes on 10 hours to estimate itsorbit. Cere’s orbit had taken him 100 hours.

1809. Gauss finally publishes details of his method and claimspriority over Legendre.

Gauss’s approach: least squares

Basic assumptions: Kepler’s laws of planetary motion.

§ Planetary orbits are ellipses with the sun at one of the foci.

§ A line drawn between the Sun and any planet sweeps out equal areas inequal time.

§ The ratio of the square of the orbital periods of two planets is equal to theratio of the cubes of their major semi-axes.

2018-11-13 9.3

Example: Ceres orbit determination

Gauss’s approach: least squares

The orbit of a planet is defined by five variables (orbital elements)

§ Two angles that orient the plane of the orbit with respect to the eclipticplane.

§ The size and eccentricity of the orbit. Equivalently, the length of themajor and minor semi-axes.

§ The tilt of the main axis of the orbit with respect to that of Earth’s orbit.

Ecliptic plane

Sun

Ceres

EarthP

P

E

C

§ Specifying the position on the orbita particular time, then determinesthe planet’s position in space.

2018-11-13 9.4

Example: Ceres orbit determination

Gauss’s approach: least squares

§ Select three observations: 1 January, 21 January and 11 February.§ Calculate a nominal orbit matching the observations.

§ Calculate a Taylor series approximation to the nonlinear differentialequations.

§ Iterative refinement of the nominal orbit.

§ Adjust linearized versions of the orbit to minimise the sum of squares ofthe error in all 22 observations.

Problem formulation

y “

»—————–

RA of obs. 1Dec. of obs. 1

...RA of obs. 22

Dec. of obs. 22

fiffiffiffiffiffifl“ Hpθ, tq, where θ “ vector of orbital elements.

2018-11-13 9.5

Example: Ceres orbit determination

Problem formulation

ε “ yobs ´Hpθ, tq, ε is the vector of residuals.

Define a “cost function”: Jpθq “ εTR´1ε

2, with weighting: R ą 0.

Optimisation: θ “ argminθ

Jpθq.

Nominal orbit

1 Jan.

21 Jan.

11 Feb.

Nominal orbit

LS fit candidate orbits

εj

Nominal orbit determination Estimated orbit errors6 equations in 6 unknown parameters. 44 equations in 6 unknown parameters.

2018-11-13 9.6

Example: Ceres orbit determination

Linearisation and iterative refinement

The nominal orbit is specified by θ0. Close to this,

Hpθq « Hpθ0q `Hθpθ ´ θ0q, where Hθ “ BHBθ

ˇˇθ0

.

Define δθ “ θ ´ θ0 and δy “ yobs ´Hpθ0q. Then,

Jlinearpδθq “ pδy ´HθδθqTR´1pδy ´HθδθqT2

« Jpθ0 ` δθq.

The gradient of Jlinearpθq is,

BJlinearpθqBθ “ ´HT

θ R´1pδy ´Hθδθq.

Setting this to zero is equivalent to solving,

HTθ R

´1Hθδθ “ HTθ R

´1δy. (linear least-squares problem)

Solve for δθ and iteratively update: θ0,new “ θ0,old ` δθ.

2018-11-13 9.7

Example: Ceres orbit determination

Gauss’s sketch References

Skizze der Bahnen derKleinplaneten Ceres, Pallas undVesta. “AstronomischeUntersuchen und Rechnungenvornehmlich ber die CeresFerdinandea,” 1802. SUBGottingen: Cod. Ms. GaußHandbuch 4, Bl. 1.

R. Bannister, Am. J. Phys.,71(12), pp. 1268–1275, Dec.2003.

L. Weiss, “Gauss and Ceres,”(www.math.rutgers.edu/~cherlin/History/Papers1999/weiss.html).

2018-11-13 9.8

Time-domain parametrisations

Pulse response

ypkq “8ÿ

i“0

gpiqupk ´ iq

Real-rational transfer functions

G “ BpzqApzq “

b1zn´1 ` b2zn´2 ` . . . bmzn´m

zn ` a1zn´1 ` a2zn´2 ` . . . an

“ b1z´1 ` b2z´2 ` . . . bmz´m

1` a1z´1 ` a2z´2 ` . . . anz´n

State-space representations

xpk ` 1q “ Axpkq `Bupkqypkq “ Cxpkq `Dupkq

2018-11-13 9.9

Parametrised model sets

We are looking for the plant, G, in a parametrised set,

tGpθqu ,that most closely matches the experimental data.

θ P Rd is the parameter vector.

Model structure Parameter vector, θ P Rd

Pulse response: gpkq “gp0q gp1q . . .

‰T

Transfer function:BpzqApzq

“a1 . . . b1 . . .

‰T

State-space:

„A B

C D

“Aij . . . Bij . . . Cij . . .

‰T

2018-11-13 9.10

Identification framework

Measurement data

Define the measurement data set:

ZK “ tup0q, yp0q, . . . , upK ´ 1q, ypK ´ 1qu.

Objective

Define the cost function (includes the model structure)

Jpθ, ZKq

General optimisation formulation

minimiseθ

Jpθ, ZKq or θ “ argmin Jpθ, ZKq.

2018-11-13 9.11

Optimisation objectives

Residual error objectives

error: epk, θq “ ypkq ´Gpθqupkq.Jpθq “ }epθq}22 or }epθq}8, or }epθq}1?

Parametric error objectives

Jpθq “ }θ ´ θ0}2 or Etθ ´ θ0u

Prediction error objective

Jpθq “ Etypk ` 1q ´ ypk ` 1, θ|kqu

2018-11-13 9.12

Correlation-based methods

Autocorrelation (periodic signals)

Rxpτq “ 1

N

N´1ÿ

k“0

xpkqxpk ´ τq, τ “ 0, . . . , N ´ 1.

Note that Rxpτq is also periodic.

Autocorrelation (random stationary signals)

Rxpτq “ Etxpkqxpk ´ τqu

“ limNÝÑ8

1

N

N´1ÿ

k“0

xpkqxpk ´ τq (for x „ N p0, σ2q )

2018-11-13 9.13

Correlation-based methods

Input-output relationship

ypkq “8ÿ

i“0

gpiqupk ´ iq ` vpkq

Etypkqupk ´ τqu “ E

# 8ÿ

i“0

gpiqupk ´ iqupk ´ τq+` Etvpkqupk ´ τqu

“8ÿ

i“0

gpiqEtupk ´ iqupk ´ τqu (if noise is uncorrelated)

“8ÿ

i“0

gpiqRupτ ´ iq

Ryupτq “ gpkq˚Rupτq. (convolution)

2018-11-13 9.14

Correlation-based methods

Matrix representation

The input-output relationship,

Ryupτq “ gpkq˚Rupτqcan be written in matrix form,

»—————–

Ryup0qRyup1qRyup2q

...

fiffiffiffiffiffifl“

»—————–

Rup0q Rup´1q Rup´2q ¨ ¨ ¨Rup1q Rup0q Rup´1qRup2q Rup1q . . .

. . .

.... . .

. . .

fiffiffiffiffiffifl

»—————–

gp0qgp1qgp2q...

fiffiffiffiffiffifl.

2018-11-13 9.15

Correlation-based methods

Finite data estimate

Create finite length estimates from this relationship;

»——–

Ryup0q...

RyupN ´ 1q

fiffiffifl

loooooooomoooooooonRyu

“

»——–

Rup0q ¨ ¨ ¨ Rup´pN ´ 1qq...

...

RupN ´ 1q ¨ ¨ ¨ Rup0q

fiffiffifl

loooooooooooooooooooooooomoooooooooooooooooooooooonRN

»——–

gp0q...

gpN ´ 1q

fiffiffifl .

Note that Rup´τq “ Rupτq.

In the periodic (noise-free) and FIR case this is exact,

if N is the period and gpτq “ 0 for τ ě N .

2018-11-13 9.16

Correlation-based methods

If RN is invertible

g “ R´1N Ryu.

Invertibility of RN is guaranteed if upkq is “persistently exciting”.

Under these conditions g is uniquely determined.

2018-11-13 9.17

Persistency of excitation

A stationary input, upkq, is persistently exciting of order n if,

Rn “

»—–

Rup0q ¨ ¨ ¨ Rup´pn´ 1qq...

...Rupn´ 1q ¨ ¨ ¨ Rup0q

fiffifl

is positive definite (Rn ą 0).

This is sufficient to uniquely determine the first n coefficients of the pulseresponse, gpkq, via the correlation approach.

The definition also applies to deterministic (or quasi-stationary) signals.

A signal is simply called persistently exciting if this holds for all n.

2018-11-13 9.18

Persistency of excitation

Spectra of persistently exciting signals

If upkq is persistently exciting of order n then,

φupejωq ‰ 0, for at least n frequencies.

Moving average (MA) filtering

An nth order MA filter: Mnpzq “ m1z´1 ` ¨ ¨ ¨ `mnz

´n.

If for all nth order MA filters,

ˇˇMnpejωq

ˇˇ2

φupejωq “ 0 ùñ Mpejωq “ 0,

then upkq is persistently exciting of order at least n.

2018-11-13 9.19

Examples

Step function

upkq “ 1, k “ 0, 1, . . . is persistently exciting of order 1.

PRBS signal of period M :

This is persistently exciting of order M .

Sum of sinusoids:

upkq “Sÿ

s“1

αs cospωsk ` φsq

persistently exciting of order

$’&’%

2S if 0 ă ωs ă π, s “ 1, . . . , S

2S ´ 1 if ω “ 0 or ω “ π P tωs, s “ 1, . . . , Su2S ´ 2 if ω “ 0 and ω “ π P tωs, s “ 1, . . . , Su

2018-11-13 9.20

ARX and ARMAX models

Autoregressive moving average models

Model form (without noise):

Gpzq “ b1z´1 ` ¨ ¨ ¨ ` bmz´m

1` a1z´1 ` ¨ ¨ ¨ ` anz´n .

Input output reationship:

ypkq “ Gpzqupkq“ ´a1ypk ´ 1q ´ ¨ ¨ ¨ ´ anypk ´ nq ` b1upk ´ 1q ` ¨ ¨ ¨ ` bmupk ´mq“ ϕT pkqθ

where

ϕpkq “ “´ypk ´ 1q ¨ ¨ ¨ ´ypk ´ nq upk ´ 1q ¨ ¨ ¨ upk ´mq‰T

θ “ “a1 ¨ ¨ ¨ an b1 ¨ ¨ ¨ bm

‰T.

2018-11-13 9.21

ARX and ARMAX models

Autoregressive moving average models

ypkq “ Gpθ, zqupkq “ ϕT pkqθ.where ϕpkq is called the regressor vector

θ is the parameter vector.

Because we can calculate ypkq from the model, with knowledge of past ypkq,upkq and model parameters, θ, the output ypkq is sometimes written as,

ypkq “ ϕT pkqθ “ ypk|θq,to emphasize the parameter dependence.

2018-11-13 9.22

Least squares estimation

Model framework (ARX or ARMAX)

Consider ypkq “ ϕT pkqθ for k “ 0, . . . , N ´ 1.

»—–

yp0q...

ypN ´ 1q

fiffifl

looooooomooooooonY

“

»—–

ϕT p0q...

ϕT pN ´ 1q

fiffifl

loooooooomoooooooonΦ

θ.

or, in matrix form, Y “ Φθ

If N ą pn`mq this system is overdetermined.

Least squares solution

θ “´

ΦTΦ¯´1

ΦTY (Matlab: Theta = Phi\Y;)

2018-11-13 9.23

Least squares estimation

Optimisation framework

The least squares solution, θ “ `ΦTΦ

˘´1ΦTY, solves the problem,

minimiseθ

}ε}2subject to Y “ Φθ ` ε

Define the cost function: Jpθ, ZN q “ }Y ´ Φθ}2 “ }ε}2

General formulation:

θ “ argminθ

Jpθ, ZN q

“ argminθ

}ε}2subject to : ε “ Y ´ Φθ.

2018-11-13 9.24

Example

System: Gpz, θq “ bz´1

p1` az´1q “b

pz ` aq where θ “„ab

.

Difference equation: ypkq “ ´aypk ´ 1q ` bupk ´ 1q.

Least squares solution:

«a

b

ff“

»————–

N´1ÿ

k“0

y2pk ´ 1q ´N´1ÿ

k“0

ypk ´ 1qupk ´ 1q

´N´1ÿ

k“0

ypk ´ 1qupk ´ 1qN´1ÿ

k“0

u2pk ´ 1q

fiffiffiffiffifl

´1

»————–

´N´1ÿ

k“0

ypkqypk ´ 1qN´1ÿ

k“0

ypkqupk ´ 1q

fiffiffiffiffifl

2018-11-13 9.25

Example

Step response experiment

vartvpkqu “ 0.252

ypkq

Gpzqupkq (noise-free)

upkq

´1

´0.5

0

0.5

1

1.5

2

2.5

3

index: k5 10 15 20

2018-11-13 9.26

Example

Least-squares estimates: step response experiment

b “ 0.5

a “ ´0.7

astep

bstep

´1

´0.5

0

0.5

1.0

index: k5 10 15 20

Gpθ, zq “ bz´1

1` az´1.

Estimates (apkq,bpkq) calculated from: (ypiq,upiq), i “ 0, k.

2018-11-13 9.27

Example

PRBS response experiment

PRBS period “ 7, vartvpkqu “ 0.252

ypkq

Gpzqupkq(noise-free)

upkq

´1.5

´1.0

´0.5

0

0.5

1.0

1.5

index: k5 10 15 20

2018-11-13 9.28

Example

Least-squares estimates: step response and PRBS

b “ 0.5

a “ ´0.7

astep

bstep

aprbs

bprbs

´1.0

´0.5

0

0.5

1.0

index: k5 10 15 20

Gpθ, zq “ bz´1

1` az´1.

Estimates (apkq,bpkq) calculated from: (ypiq,upiq), i “ 0, k.

2018-11-13 9.29

Statistical properties of the LS estimates

Model

Y “ Φθ ` ε, ε “

»—–

εp0q...

εpN ´ 1q

fiffifl

Error assumptions,

Etεu “ 0, EtεεT u “ σ2I.

LS estimator properties

Assume that Φ is fixed, but consider ε as a stochastic variable.

Etθu “ θ (unbiased estimator)

covtθu “ E!pθ ´ θqpθ ´ θqT

)“ σ2

´ΦTΦ

¯´1

.

2018-11-13 9.30

Least square estimators

Model with correlated noiseY “ Φθ ` ε, with EtεεT u “ R.

LS estimator properties

Assume that Φ is fixed, but consider ε as a stochastic variable.

Etθu “ θ (unbiased estimator)

covtθu “ E!pθ ´ θqpθ ´ θqT

)“

´ΦTΦ

¯´1

ΦTRΦ´

ΦTΦ¯´1

Note the simplification to the white noise case if R “ σ2I.

2018-11-13 9.31

Best linear unbiased estimator (BLUE or Markov estimate)

Model with correlated noise

Y “ Φθ ` ε, with EtεεT u “ R.

Best linear estimator

Linear estimators: θ “ ZTY

The linear estimator: Z “ R´1Φ´

ΦTR´1Φ¯´1

satisfies:

EtθZu “ θ, (unbiased)

covtθZu “´

ΦTR´1Φ¯´1 ď covtθu for any unbiased estimate.

Note that the BLUE requires knowledge of the error covariance, R.

2018-11-13 9.32

Error statistics and noise statistics

Experimental configuration

Gpθ, zq`ypkq upkqvpkq

Estimation framework»—–

yp0q...

ypN ´ 1q

fiffifl “

»—–

ϕT p0q...

ϕT pN ´ 1q

fiffifl θ `

»—–

vp0q...

vpN ´ 1q

fiffifl ,

But,

ϕT pkq “ “´ypk ´ 1q ¨ ¨ ¨ ´ypk ´ nq upk ´ 1q ¨ ¨ ¨ upk ´mq‰

` “´vpk ´ 1q ¨ ¨ ¨ ´vpk ´ nq 0 ¨ ¨ ¨ 0‰,

and so Φ is a function of ypkq, upkq, and vpkq.So it is not the case that Y “ Φθ ` V . (vpkq ‰ εpkq in the LS problem)

2018-11-13 9.33

Equation error models

Model form

ypkq ` a1ypk ´ 1q ` ¨ ¨ ¨ ` anypk ´ nq “b1upk ´ 1q ` ¨ ¨ ¨ ` bmupk ´mq ` vpkq

ApzqY pzq “ BpzqUpzq ` V pzq

Now we have: Y “ Φθ ` V.

Bpθ, zq`1

Apθ, zqypkq upkq

vpkq

The noise also has the autoregressive dynamics (in Apzq).

2018-11-13 9.34

Example revisted

Equation error system response

PRBS period “ 7, vartvpkqu “ 0.252

ypkq

Gpzqupkq(noise-free)

upkq

´1.5

´1.0

´0.5

0

0.5

1.0

1.5

index: k5 10 15 20

2018-11-13 9.35

Example revisted

Least-squares estimates: equation error structure

b “ 0.5

a “ ´0.7

astep

bstep

aprbs

bprbs

aee

bee

´1.0

´0.5

0

0.5

1.0

index: k5 10 15 20

Gpθ, zq “ bz´1

1` az´1.

Estimates (apkq,bpkq) calculated from: (ypiq,upiq), i “ 0, k.

2018-11-13 9.36

Bibliography

Persistency of excitation

Lennart Ljung, System Identification;Theory for the User, 2nd Ed., Prentice-Hall,1999, [section 13.2].

ARX modelsLennart Ljung, System Identification;Theory for the User, 2nd Ed., Prentice-Hall,1999, [sections 1.3 and 4.2].

LS estimation statisticsLennart Ljung, System Identification;Theory for the User, 2nd Ed., Prentice-Hall,1999, [section 10.1 and appendix II].

2018-11-13 9.37