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Transcript of 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), . . .
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 3/15
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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), . . .
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
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 3/15
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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), . . .
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
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 3/15
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Persistence of Excitation (Def 2):
Definition 2: Given a signal u(t), i.e. a sequence of numbers
u(0), u(1), u(2), . . .
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 4/15
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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.
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 5/15
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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.
Proof: with A(q) = a0 qn−1
+ · · · + an−1L = [a0, . . . , an−1] C n [a0, . . . , an−1]
T
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 5/15
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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.
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
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 7/15
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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
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
.
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 7/15
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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
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.
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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Examples 3: order of PE for sinusoid
Example 3: consider the signal
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.
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
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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Examples 3: order of PE for sinusoid
Example 3: consider the signal
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.
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
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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Examples 3: order of PE for sinusoid
Example 3: consider the signal
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.
Let us compute c(k):
c(k) = limN →∞
1
N
N i=1
sin(ω0 i) sin(ω0 i − ω0 k)
= 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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Examples 3: order of PE for sinusoid
Example 3: consider the signal
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.
Let us compute c(k):
c(k) = limN →∞
1
N
N i=1
sin(ω0 i) sin(ω0 i − ω0 k)c(k) =
1
2cos(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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Examples 3: order of PE for sinusoid
Example 3: consider the signal
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.
Let us compute c(k):
c(k) = limN →∞
1
N
N i=1
sin(ω0 i) sin(ω0 i − ω0 k)
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.
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 8/15
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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.
Note: R u(k) = R u(−k) and R u(k) = c(k) is calledcovariance of u(t).
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 8/15
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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.
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
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 8/15
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Frequency domain characterization of PE signals
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)
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 9/15
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Frequency domain characterization of PE signals
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.
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 9/15
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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 .
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 10/15
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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 .
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
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 10/15
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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)
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 11/15
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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)
=
−q−1y(t). . .
−q−ny(t)
q−n+m−1u(t)
. . .
q−nu(t)
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 11/15
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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)
=
−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)
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 11/15
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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)
=
−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)
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 11/15
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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)
=
−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)
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 11/15
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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)
= 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.
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 11/15
A( ) ( ) ( ) ( )
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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)
= 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
−v(t − 1). . .
−v(t
− n)
v(t−n−1)
. . .
v(t−n−m)
ψ(t−1)
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.
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 11/15
A( ) ( ) B( ) ( )
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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)
= 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
−v(t − 1). . .
−v(t
− n)
v(t−n−1)
. . .
v(t−n−m)
ψ(t−1)
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.
limt→∞
1
t Φ(t−1)T Φ(t−1)
= S
limt→∞
1
t Ψ(t − 1)T Ψ(t − 1)
S T > 0
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 11/15
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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) .
Theorem (Persistent Excitation for ARMA model):
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.
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 12/15
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.
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 13/15
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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.
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 14/15
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.
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 14/15
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8/17/2019 Lecture 05 EAILC
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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.
c Leonid Freidovich. A ril 15, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 5 – . 15/15