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
244
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 (