Lecture 05 EAILC

8/17/2019 Lecture 05 EAILC

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Theorem (Persistent Excitation for FIR model):

Assume that data are generated by the following FIR model:y(t) = g1 u(t − 1) + g2 u(t − 2) + · · · + gn u(t − n) + e(t)with E

{e(t)

} = 0, and mutually independent e(t).

Then, one can use Recursive Least Square algorithm with

φ(t, n)T =

u(t − 1), u(t − 2), . . . , u(t − n)

to estimate θ0

(n) = g1, g2, . . . , gnT .If the signal u(t) is persistently exciting of at least order n, theestimate is unbiased. Possible conditions to check is

limN →∞

1

N

φ(n, n)T

φ(n + 1, n)T

...

φ(N, n)T

T

φ(n, n)T

φ(n + 1, n)T

...

φ(N, n)T

> 0

c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 2/15


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Persistence of Excitation (Def 1):

Definition 1: Given a signal u(t), i.e. a sequence of numbers

u(0), u(1), u(2), . . .



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u(0), u(1), u(2), . . .

It is called persistently exciting of order n if the limits exist

c(k) = limN →∞

1

N

N i=1

u(i)u(i − k), k = 0, 1, . . . , n − 1



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u(0), u(1), u(2), . . .

It is called persistently exciting of order n if the limits exist

c(k) = limN →∞

1

N

N i=1

u(i)u(i − k), k = 0, 1, . . . , n − 1and if

C n =

c(0) c(1) . . . c(n

− 1)

c(1) c(0) . . . c(n − 2)...

c(n − 1) c(n − 2) . . . c(0)

> 0



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u(0), u(1), u(2), . . .



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Polynomial Conditions for Persistent Excitation:

The signal with the property that limits

c(k) = limN →∞

1

N

N

i=1 u(i)u(i − k), k = 0, 1, . . . , n − 1exist, is PE of order n if and only if

L = limN →∞

1

N

N k=1

[A(q) u(k)]2 > 0

for all non-zero polynomials of degree n − 1 or less.



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c(k) = limN →∞

1

N

N


L = limN →∞

1

N

N k=1

[A(q) u(k)]2 > 0


Proof: with A(q) = a0 qn−1

+ · · · + an−1L = [a0, . . . , an−1] C n [a0, . . . , an−1]

T



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c(k) = limN →∞

1

N

N


L = limN →∞

1

N

N k=1

[A(q) u(k)]2 > 0


Proof: with A(q) = a0 qn−1

+ · · · + an−1L = [a0, . . . , an−1] C n [a0, . . . , an−1]

T

Corollary: if u(t) satisfies a polynomial equation of degree nfor t ≥ t0 with some t0, then it can be at most PE of order n.c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 5/15


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Examples of computing order of PE:

Example 1: The STEP signal

u(t) = 0 for t


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Examples 3: order of PE for sinusoid

Example 3: consider the signal

u(t) = sin(ω0 t), with 0 < ω0 < π



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u(t) = sin(ω0 t), with 0 < ω0 < π

It is not hard to show [using sin(a + b) = sin(a) cos(b) + sin(b) cos(a)]that

q2 − 2 q cos(ω0) + 1

u(t) = 0

Hence, it can at most be PE of order 2

.



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u(t) = sin(ω0 t), with 0 < ω0 < π


q2 − 2 q cos(ω0) + 1

u(t) = 0

Hence, it can at most be PE of order 2.

Let us compute c(k):

c(k) = limN →∞

1

N

N i=1

sin(ω0 i) sin(ω0 i − ω0 k)

= limN →∞

1

N

N i=1

sin2(ω0 i) cos(ω0 k)−sin(ω0 i) cos(ω0 i) sin(ω0 k)c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 7/15


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u(t) = sin(ω0 t), with 0 < ω0 < π


q2 − 2 q cos(ω0) + 1

u(t) = 0



c(k) = limN →∞

1

N

N i=1


= limN →∞

1

N

N i=1

12 cos(ω0 k)−12 cos(2 ω0 i) cos(ω0 k)−12 sin(2 ω0 i) sin(ω0 k)c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 7/15


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u(t) = sin(ω0 t), with 0 < ω0 < π


q2 − 2 q cos(ω0) + 1

u(t) = 0



c(k) = limN →∞

1

N

N i=1


= limN →∞

1

N

N i=1

12 cos(ω0 k) − 12 cos(2 ω0 i + ω0 k)c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 7/15


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u(t) = sin(ω0 t), with 0 < ω0 < π


q2 − 2 q cos(ω0) + 1

u(t) = 0



c(k) = limN →∞

1

N

N i=1


= limN →∞

1

N constant +N i=1

1

2 cos(ω0 k)c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 7/15


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u(t) = sin(ω0 t), with 0 < ω0 < π


q2 − 2 q cos(ω0) + 1

u(t) = 0



c(k) = limN →∞

1

N

N i=1

sin(ω0 i) sin(ω0 i − ω0 k)c(k) =

1

2cos(ω0 k)



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u(t) = sin(ω0 t), with 0 < ω0 < π


q2 − 2 q cos(ω0) + 1

u(t) = 0



c(k) = limN →∞

1

N

N i=1


c(k) =

1

2 cos(ω0 k) =⇒ C 2 = 1 1

2 cos(ω0)

12 cos(ω0) 1

> 0c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 7/15


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Frequency domain characterization of PE signals

Definition: A signal u(t) is called stationary if the limit

R u(k) = limt→∞

1

t

t+m−1

i=mu(i) u(i − k)

exists uniformly in m.



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R u(k) = limt→∞

1

t

t+m−1

i=mu(i) u(i − k)


Note: R u(k) = R u(−k) and R u(k) = c(k) is calledcovariance of u(t).



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R u(k) = limt→∞

1

t

t+m−1

i=mu(i) u(i − k)


Note: R u(k) = R u(−k) and R u(k) = c(k) is calledcovariance of u(t).

Definition: The Fourier transform of the covariance, Φu(ω), iscalled the frequency spectrum of u(t):

R u(k) = 12 π

π−π

e j ω k Φu(ω) dω, j = √ −1

Φu(ω) =

+∞

k=−∞

R u(k) e− j ω k = 12 π

+∞

−∞

u(t) e− j ω t dt2



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Example: for the signal

u(t) = sin(ω0 t), with 0 < ω0 < πwe have

R u(k) = 1

2cos(ω0 k), Φu(ω) =

π

2

δ(ω−ω0)+δ(ω+ω0)



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Example: for the signal

u(t) = sin(ω0 t), with 0 < ω0 < πwe have

R u(k) = 1

2cos(ω0 k), Φu(ω) =

π

2

δ(ω−ω0)+δ(ω+ω0)

Lemma: A stationary signal is PE of order n if its frequencyspectrum is non zero for at least n points.

Example: white noise is PE of any order, while the signal

u(t) =mi=1

Ai sin(ωi t+ϕi), with 0 < ωi < π, ωi = ω j

is PE of order 2 m.



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Theorem (Persistent Excitation for ARMA model):

Assume that data are generated by the following ARMA model:qn + a1 q

n−1 + · · · + an

A(q)

y(t) =

b1 qm−1 + b2 q

m−2 + · · · + bm

B(q)

u(t)

with n > m. Then, one can use RLS algorithm with

φ(t−1)T = − y(t − 1), . . . ,−y(t − n), u(t − n + m − 1), . . . , u(t − n)to estimate θ0 =

a1, . . . , an, b1, . . . , bm

T .



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Assume that data are generated by the following ARMA model:qn + a1 q

n−1 + · · · + an

A(q)

y(t) =

b1 qm−1 + b2 q

m−2 + · · · + bm

B(q)

u(t)

with n > m. Then, one can use RLS algorithm with

φ(t−1)T = − y(t − 1), . . . ,−y(t − n), u(t − n + m − 1), . . . , u(t − n)to estimate θ0 =

a1, . . . , an, b1, . . . , bm

T .

The estimate is unbiased if the signal u(t) is such that

limt→∞

1

t Φ(t−1)T Φ(t−1)

= lim

t→∞

1

t

φ(n)T

φ(n + 1)T

...

φ(t − 1)T

T

φ(n)T

φ(n + 1)T

...

φ(t − 1)T

> 0



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Persistent Excitation for A(q)y(t) = B(q)u(t)

φ(t−1) =

−y(t − 1). . .

−y(t

− n)

u(t−n+m−1)

. . .

u(t

− n)



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φ(t−1) =

−y(t − 1). . .

−y(t

− n)

u(t−n+m−1)

. . .

u(t

− n)

=

−q−1y(t). . .

−q−ny(t)

q−n+m−1u(t)

. . .

q−nu(t)



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φ(t−1) =

−y(t − 1). . .

−y(t

− n)

u(t−n+m−1)

. . .

u(t

− n)

=

−q−1y(t). . .

−q−ny(t)

q−n+m−1u(t)

. . .

q−nu(t)

= qm

A(q)

−q−1−mA(q)y(t). . .

−q−n−mA(q)y(t)

q−n−1A(q)u(t)

. . .

q−n−mA(q)u(t)



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φ(t−1) =

−y(t − 1). . .

−y(t − n)u(t−n+m−1). . .

u(t

− n)

=

−q−1y(t). . .

−q−n

y(t)q−n+m−1u(t)

. . .

q−nu(t)

= qm

A(q)

−q−1−m A(q)y(t). . .

−q−n−m

A(q)y(t)q−n−1A(q)u(t)

. . .

q−n−mA(q)u(t)



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φ(t−1) =

−y(t − 1). . .

−y(t

− n)

u(t−n+m−1)

. . .

u(t

− n)

=

−q−1y(t). . .

−q−ny(t)

q−n+m−1u(t)

. . .

q−nu(t)

= qm

A(q)

−q−1−m B(q)u(t). . .

−q−n−m

B(q)u(t)q−n−1A(q)u(t)

. . .

q−n−mA(q)u(t)



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φ(t−1) =

−y(t − 1). . .

−y(t

− n)

u(t−n+m−1)

. . .

u(t

− n)

= qm

A(q)

−q−1−mB(q)u(t). . .

−q−n−mB(q)u(t)

q−n−1A(q)u(t)

. . .

q−n−mA(q)u(t)

= S

−q−1v(t). . .

−q−nv(t)

q−n−1v(t)

. . .

q−n−mv(t)

where v(t) = qm

A(q)u(t) and S is not singular in the

case when A(q) and B(q) are relatively prime.


A( ) ( ) ( ) ( )


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φ(t−1) =

−y(t − 1). . .

−y(t

− n)

u(t−n+m−1)

. . .

u(t

− n)

= qm

A(q)

−q−1−mB(q)u(t). . .


q−n−1A(q)u(t)

. . .

q−n−mA(q)u(t)

= S

−v(t − 1). . .

−v(t

− n)

v(t−n−1)

. . .

v(t−n−m)

ψ(t−1)

where v(t) = qm




A( ) ( ) B( ) ( )


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φ(t−1) =

−y(t − 1). . .

−y(t

− n)

u(t−n+m−1)

. . .

u(t

− n)

= qm

A(q)

−q−1−mB(q)u(t). . .


q−n−1A(q)u(t)

. . .

q−n−mA(q)u(t)

= S

−v(t − 1). . .

−v(t

− n)

v(t−n−1)

. . .

v(t−n−m)

ψ(t−1)

where v(t) = qm



limt→∞

1

t Φ(t−1)T Φ(t−1)

= S

limt→∞

1

t Ψ(t − 1)T Ψ(t − 1)

S T > 0



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The conditions for parameter convergence, formulated in the


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The conditions for parameter convergence, formulated in thecase of FIR model for the signal u(t), are similar in the case of

ARMA model, but should be concerning the signal

v(t) = qm

A(q)u(t)

n > m, qm

is stable, qm

/A(q) is stable and minimum phase.

Hence, Φv(ω) =ejmω

A(ejω)

2Φu(ω) .


Suppose that

1. A(q) and B(q) are relatively prime,

2. A(q) is stable (i.e. all roots are inside the unite circle),

3. u(t) is a stationary signal such that Φu(ω) is nonzero for atleast (n + m) points.

Then, RLS algorithm converges to the correct value.


Example


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Example

Consider the system

y(t) + a1 y(t − 1) + a2 y(t − 2) = b1 u(t − 1) + b2 u(t − 2)

Choose a nice (as simple as possible) input signal, which is richenough to ensure parameters convergence.



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Example 2 12 (book)


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Example 2.12 (book)

Consider the system y(t) + a y(t − 1) = b u(t − 1) + e(t)with a = −0.8, b = 0.5, e(t) – zero mean white noise withstandard deviation σ = 0.5.


Example 2 12 (book)


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Example 2.12 (book)

Consider the system y(t) + a y(t − 1) = b u(t − 1) + e(t)with a = −0.8, b = 0.5, e(t) – zero mean white noise withstandard deviation σ = 0.5.


Next Lecture / Assignments:


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Next Lecture / Assignments:

Next meeting (April 19, 13:00-15:00, in A205Tekn): Recitations

Homework problem: repeat the simulation for Example 2.12

shown above (see pages 71–72), taking P (0) = 100 I 2 and

θ̂(0) = 0.


Lecture 05 EAILC

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