Robust Adaptive Control by Petros A. Ioannou, Jing Sun - HTML preview

t

− 1

P ( t) ≤

− 1 α

¯

mI,

∀t > T

T

0 T 0

0

(4.8.41)

0

Because P ( t) ≥ 0 for all t ≥ 0 and the right-hand side of (4.8.41) goes to zero as

t → ∞, we can conclude that P ( t) → 0 as t → ∞. Hence, ˜

θ( t) = P ( t) P − 1 ˜

0

θ(0) → 0

as t → ∞.

✷

4.8.6

Proof of Corollary 4.3.2

Let us denote Γ = P − 1( t), then from (4.3.78) we have

˙

φφ

Γ = −βΓ +

,

Γ(0) = Γ (0) = Γ

m 2

0 = P − 1

0

or

t

Γ( t) = e−βtΓ0 +

e−β( t−τ) φ( τ ) φ ( τ ) dτ

0

m 2

Using the condition that φ( t) is PE and m ∈ L∞, we can show that for all t ≥ T 0

t

Γ( t) ≥

e−β( t−τ) φφ dτ

0

m 2

t

t−T 0

=

e−β( t−τ) φ( τ ) φ ( τ ) dτ +

e−β( t−τ) φ( τ ) φ ( τ ) dτ

t−T

m 2

m 2

0

0

α

≥ e−βT

0 T 0

0

I

(4.8.42)

¯

m

where ¯

m = sup t m 2( t). For t ≤ T 0, we have

Γ( t) ≥ e−βtΓ0 ≥ e−βT 0Γ0 ≥ λmin(Γ0) e−βT 0 I

(4.8.43)

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CHAPTER 4. ON-LINE PARAMETER ESTIMATION

Conditions (4.8.42), (4.8.43) imply that

Γ( t) ≥ γ 1 I

(4.8.44)

for all t ≥ 0 where γ 1 = min { α 0 T 0 , λ

β

min(Γ0) }e−βT 0 .

1

On the other hand, using the boundedness of φ, we can establish that for some

constant β 2 > 0

t

Γ( t) ≤ Γ0 + β 2

e−β( t−τ) dτ I

0

β

≤ λ

2

max(Γ0) I +

I ≤ γ

β

2 I

(4.8.45)

where γ 2 = λmax(Γ0) + β 2 > 0.

β

Combining (4.8.44) and (4.8.45), we conclude

γ 1 I ≤ Γ( t) ≤ γ 2 I

for some γ 1 > 0 , γ 2 > 0. Therefore,

γ− 1

2 I ≤ P ( t) ≤ γ− 1

1 I

and consequently P ( t) , P − 1( t) ∈ L∞. Because P ( t) , P − 1( t) ∈ L∞, the exponential convergence of θ to θ∗ can be proved using exactly the same procedure and arguments as in the proof of Theorem 4.3.2.

✷

4.8.7

Proof of Theorem 4.5.1(iii)

Consider the following differential equations which describe the behavior of the

adaptive law (see (4.5.6) and (4.5.7)):

˙ e = Ace + Bc( −ρ∗ ˜

θ φ − ˜

ρξ − n 2 s)

˙˜ θ = Γ φ sgn( ρ∗)

(4.8.46)

˙ ρ = γ ξ

= Cc e

Because ξ, ρ ∈ L∞ and ξ ∈ L 2, we can treat ξ, ρ as external input functions and

write (4.8.46) as

˙ xa = Aaxa + Ba( −˜

ρξ)

(4.8.47)

where

e

A

B

x

c − n 2

sBcCc

−ρ∗Bcφ

c

a =

˜

,

A

,

B

θ

a =

Γsgn( ρ∗) φC

a =

c

0

0

4.8. PARAMETER CONVERGENCE PROOFS

243

In proving Corollary 4.3.1, we have shown that when φ is PE and φ, ˙ φ ∈ L∞, the

system ˙ x = Aax is e.s. Therefore, the state transition matrix Φ a( t, t 0) of (4.8.47)

satisfies

Φ a( t, 0) ≤ α 0 e−γ 0 t

(4.8.48)

for some constants α 0 , γ 0 > 0, which together with −Ba ˜

ρξ ∈ L 2 imply that xa( t) →

0 as t → ∞.

4.8.8

Proof of Theorem 4.6.1 (iii)

From the proof of Theorem 4.6.1 (i) to (ii), we have the inequality (see (4.6.9))

tk+1

V ( k + 1) − V ( k) ≤ −(2 − T

2

sλm)

( τ ) m 2( τ ) dτ

(4.8.49)

tk

Using inequality (4.8.49) consecutively, we have

tk+ n

V ( k + n) − V ( k) ≤ −(2 − T

2

sλm)

( τ ) m 2( τ ) dτ

tk

n− 1

tk+ i+1

= −(2 − T

2

sλm)

( τ ) m 2( τ ) dτ

(4.8.50)

i=0

tk+ i

for any integer n. We now write

2

t

˜

k+ i+1

tk+ i+1

θ

φ( τ )

2

k+ i

( τ ) m 2( τ ) dτ

=

dτ

t

m 2( τ )

k+ i

tk+ i

2

t

˜

k+ i+1

θ φ( τ ) + (˜

θ

k

k+ i − ˜

θk) φ( τ)

=

dτ

t

m 2( τ )

k+ i

Using the inequality ( x + y)2 ≥ 1 x 2 − y 2, we write

2

2

2

t

˜

k+ i+1

tk+ i+1 θ φ( τ )

tk+ i+1 (˜

θk+ i − ˜

θk) φ( τ)

2

1

k

( τ ) m 2( τ ) dτ ≥

dτ −

dτ

t

2

m 2( τ )

m 2( τ )

k+ i

tk+ i

tk+ i

(4.8.51)

Because φ( τ ) /m( τ ) is bounded, we denote c = sup |φ( τ) | 2 and have m 2( τ )

2

tk+ i+1

(˜

θk+ i − ˜

θk) φ( τ)

dτ ≤ cTs|˜

θk+ i − ˜

θk| 2

t

m 2( τ )

k+ i

From the hybrid adaptive algorithm, we have

tk+ i

˜

θk+ i − ˜

θk =

( τ ) φ( τ ) dτ, i = 1 , 2 , . . . , n

tk

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CHAPTER 4. ON-LINE PARAMETER ESTIMATION

therefore, using the Schwartz inequality and the boundedness of |φ( t) |/m( t),

t

2

2

k+ i

tk+ i

|φ( τ ) |

|˜

θk+ i − ˜

θk| 2 ≤

| ( τ ) ||φ( τ ) |dτ

=

| ( τ ) |m( τ )

dτ

t

m( τ )

k

tk

tk+ i

tk+ i |φ( τ) | 2

≤

2( τ) m 2( τ) dτ

dτ

t

m 2( τ )

k

tk

tk+ i

≤ ciT

2

s

( τ ) m 2( τ ) dτ

tk

tk+ n

≤ ciT

2

s

( τ ) m 2( τ ) dτ

(4.8.52)

tk

Using the expression (4.8.52) in (4.8.51), we have

2

t

˜

k+ i+1

tk+ i+1

θ φ( τ )

tk+ n

2

1

k

( τ ) m 2( τ ) dτ ≥

dτ − c 2 iT 2

2

s

( τ ) m 2( τ ) dτ

t

2

m 2( τ )

k+ i

tk+ i

tk

which leads to

t

n− 1

k+ n

tk+ i+1

2( τ) m 2( τ) dτ =

2( τ) m 2( τ) dτ

tk

i=0

tk+ i





2

n− 1

˜

 1

tk+ i+1

θ φ( τ )

k

tk+ n



≥



dτ − c 2 iT 2

2( τ) m 2( τ) dτ

2

m 2( τ )

s

i=0

tk+ i

tk

1

n− 1

t

n− 1

k+ i+1 φ( τ ) φ ( τ )

tk+ n

≥

˜

θ

dτ ˜

θ

c 2 iT 2

2( τ) m 2( τ) dτ

2 k

m 2( τ )

k −

s

i=0

tk+ i

i=0

tk

tk+ n φ( τ) φ ( τ)

n( n − 1)

tk+ n

=

˜

θ

2

k

dτ ˜

θk −

c 2 T 2

s

( τ ) m 2( τ ) dτ

t

2 m 2( τ )

2

k

tk

or equivalently

tk+ n

tk+ n

2

1

φ( τ ) φ ( τ )

( τ ) m 2( τ ) dτ ≥

˜

θk

dτ ˜

θk (4.8.53)

t

2(1 + n( n − 1) c 2 T 2

m 2( τ )

k

s / 2)

tk

Because φ is PE and 1 ≤ m < ∞, there exist constants α 1 , α 2 and T 0 > 0 such that t+ T 0 φ( τ) φ ( τ)

α 2 I ≥

dτ ≥ α 1 I

t

m 2

for any t. Therefore, for any integer k, n where n satisfies nTs ≥ T 0, we have

tk+ n

˜

φ( τ ) φ ( τ )

V ( k)

θ

˜ ˜

k

dτ ˜

θk ≥ α 1 θk θk ≥

α 1

(4.8.54)

t

m 2

λ

k

m

4.9. PROBLEMS

245

Using (4.8.53), (4.8.54) in (4.8.50), we obtain the following inequality:

(2 − T

V ( k + n) − V ( k) ≤ −

sλm) α 1

V ( k)

(4.8.55)

λm(2 + n( n − 1) c 2 T 2 s)

hold for any integer n with n ≥ T 0 /Ts. Condition (4.8.55) is equivalent to

V ( k + n) ≤ γV ( k)

with

(2 − T

γ = 1 −

sλm) α 1

< 1

λm(2 + n( n − 1) c 2 T 2 s)

Therefore,

V (