Class Handout: Chapter 5 Input-Output Stability 2006 Fall I. L

Class Handout: Chapter 5 Input-Output Stability

2006 Fall

I. L Stability

y=Hu

wherey: [0,∞)→R^q andu: [0,∞)→R^mare functions, andH is a mapping (or operator) between two signals.

To measure the size of signals, let us introduce the norm functionk · k, with which it holds that

• kuk= 0 if and only ifu(t)≡0,

• kauk=|a|kuk,

• ku1+u2k ≤ ku1k+ku2k.

For example,

kukL∞:= sup

t≥0

ku(t)k

kukLp:=

µZ _∞

0

ku(t)k^pdt

¶_1/p

(Here, the norm inside the integral is not necessarily equal to thep-norm, but usually it is.)

The set of all piecewise continuous functions whose Lp (p ∈ [1,∞]) norm is finite, is In fact, ‘piecewise continuous’ can be replaced by ‘measurable’.

denoted by L^m_p. Sub-, or super-script may be omitted if clear from the context.

“Extended space”:

L^m_e ={u:uτ ∈ L^m,∀τ ∈[0,∞)}

where uτ is a truncation ofudefined by

u_τ(t) =







u(t), 0≤t≤τ, 0, t > τ

The extended space is a linear space that contains the unextended space as a subset. It allows us to deal with unbounded growing signals (e.g.,u(t) =t belongs toL∞,e).

“Causal” mapping H:

(Hu)τ= (Huτ)τ, ∀τ≥0.

Causality is an intrinsic property of dynamical systems represented by state models.

Definition 5.1

A mappingH :L^m_e → L^q_e isLstable if∃a class-K functionα, andβ >0 s.t.

k(Hu)τkL≤α(kuτkL) +β for allu∈ L^m_e andτ∈[0,∞).

It is finite-gainLstable if∃ nonnegative constantsγand β s.t.

k(Hu)τkL≤γkuτkL+β (1) for allu∈ L^m_e andτ∈[0,∞).

• β is a bias term, which possibly considers the initial condition.

• If (1) holds, then the system has anLgain less than or equal toγ. The smallestγs.t. (1) holds is called the gain of the system.

• For causal,Lstable systems, we have

u∈ L^m⇒Hu∈ L^q and

kHuk_L≤α(kuk_L) +β, ∀u∈ L^m.

• L∞ stability is the familiar notion of BIBO stability.

Example 5.1Let

h(u) =a+btanhcu=a+be^cu−e^−cu e^cu+e^−cu Then

|h(u)| ≤a+bc|u|.

Hence,his finite-gainL∞stable withγ=bcandβ=a.

Ifa= 0, hisLp stable with zero bias, gainγ=bc, for eachp∈[1,∞] since Z _∞

0

|h(u(t))|^pdt≤(bc)^p Z _∞

0

|u(t)|^pdt.

Ifh(u) =u², it isL∞stable with zero bias andα(r) =r², but is not finite-gainL∞stable.

Example 5.2Consider a causal convolution y(t) =

Z _t

0

h(t−σ)u(σ)dσ

where h(t) = 0 fort <0. Suppose thath∈ L1,e; that is, for everyτ ∈[0,∞), kh_τk_L1 =

Z _∞

0

|h_τ(σ)|dσ= Z _τ

0

|h(σ)|dσ <∞.

Ifu∈ L∞,e andτ ≥t, then

|y(t)| ≤ Z _t

0

|h(t−σ)| |u(σ)|dσ

≤ Z _t

0

|h(t−σ)|dσ sup

0≤σ≤τ

|u(σ)|= Z _t

0

|h(s)|ds sup

0≤σ≤τ

|u(σ)|

Consequently,

ky_τk_L∞ ≤ kh_τk_L1ku_τk_L∞, ∀τ∈[0,∞).

This does not meanLstability of the system, becausekhτkL1 is notuniformlybounded with respect to τ. (For example,h(t) may diverge.)

Assume thath∈ L₁; that is,

khkL1 = Z _∞

0

|h(σ)|dσ <∞.

Then, it holds that

kyτkL∞≤ khkL1kuτkL∞, ∀τ ∈[0,∞), which means that the system is finite-gainL∞stable.

In fact, this assumption implies finite-gainLpstability for anyp∈[1,∞]. See the textbook pp. 199 and 200.

Definition 5.2 (Small-signalL stability)

A mappingH :L^m_e → L^q_eis small-signal (finite-gain)Lstable if∃r >0 s.t. it is (finite-gain) L stable for allu∈ L^m_e with sup_0≤t≤τku(t)k ≤r.

Example 5.3See the textbook.

II. L Stability of State Models Consider

˙

x=f(t, x, u), x(0) =x0, y=h(t, x, u)

where x ∈ Rⁿ, u ∈ R^m, y ∈ R^p, f : [0,∞)×D×Du → Rⁿ is piecewise continuous in t and locally Lipschitz in (x, u), h: [0,∞)×D×Du → R^q is piecewise continuous in t and continuous in (x, u),D is a domain containingx= 0,Duis a domain containing u= 0.

* For each fixed x0 ∈ D, the above system defines an operator H that assigns to each input signal u(t) the corresponding output signaly(t).

Assume thatx= 0 of

˙

x=f(t, x,0) is an equilibrium.

Theorem 5.1

Letr >0 andru>0 s.t. {kxk ≤r} ⊂D and{kuk ≤ru} ⊂Du.

• IFx= 0 of ˙x=f(t, x,0) is LES, and∃V(t, x) s.t.

c1kxk²≤V(t, x)≤c2kxk² (2)

∂V

∂t +∂V

∂xf(t, x,0)≤ −c3kxk² (3)

°°

°°∂V

∂x(t, x)

°°

°°≤c4kxk (4)

for all (t, x)∈[0,∞)×D, and, IF

kf(t, x, u)−f(t, x,0)k ≤Lkuk kh(t, x, u)k ≤η1kxk+η2kuk for all (t, x, u)∈[0,∞)×D×Du,

THEN, for eachkx0k ≤rp

c1/c2, the system is small-signal finite-gainLpstable (p∈[1,∞]).

In particular, for each u∈ Lp,e with sup_0≤t≤τku(t)k ≤min{ru, c1c3r/(c2c4L)}, the output y(t) satisfies

kyτkLp≤γkuτkLp+β (5) for allτ∈[0,∞), with

γ=η2+η1c2c4L

c1c3 , β=η1kx0k rc2

c1ρ where

ρ=







1, ifp=∞,

³2c2

c3p

´_1/p

, ifp∈[1,∞) .

• IF the origin is GES andD=Rⁿ,Du=R^m, THEN the above holds for anyx0andu(t).

Proof. Derivative ofV along ˙x=f(t, x, u) is V˙ =∂V

∂t +∂V

∂xf(t, x,0) +∂V

∂x[f(t, x, u)−f(t, x,0)]

≤ −c3kxk²+c4Lkxkkuk

(At time point, the proof for global case is actually done already. Can you see?) (For numerical analysis, we continue...) Take W(t) =p

V(t, x(t)).

IfV(t, x(t))6= 0,

W˙ = V˙ 2√

V ≤ 1 2√ V

¡−c3kxk²+c4Lkxkkuk¢

≤ 1 2√ V

Ã

−c3

c2

V(t, x) +c4L

√V

√c1

kuk

!

=−c3

2c2

W+ c4L 2√

c1

kuk

IfV(t, x(t)) = 0 (i.e.,x(t) = 0), it can be shown that Exercise 5.6 and Exercise 3.24 are employed here.

D⁺W(t)≤ c4L 2√

c1ku(t)k.

Hence, for all V(t, x(t)), we have

D⁺W(t)≤ −1 2

c3

c2

W+ c4L 2√

c1

ku(t)k.

By the comparison lemma,

W(t)≤e^−t2cc32W(0) + c4L 2√

c1

Z _t

0

e^{−(t−τ)2cc32}ku(τ)kdτ.

We then obtain (since√

c1kx(t)k ≤W(t)≤√

c2kx(t)k) kx(t)k ≤

rc2

c1kx0ke^−t2cc32 +c4L 2c1

Z _t

0

e^{−(t−τ)2cc32}ku(τ)kdτ

= rc2

c1

kx0ke^−t2cc32 +c4L 2c1

2c2

c3

³

1−e^−t2cc32´ µ sup

0≤σ≤t

ku(σ)k

¶

(We now check ifx(t)∈Br for allt≥0 so that the whole analysis is valid.) Then, since kx0k ≤r

rc1

c2, sup

0≤σ≤tku(σ)k ≤ c1c3r c2c4L, we have

kx(t)k ≤re^−t2cc32 +

³

1−e^−t2cc32

´ r=r.

(We now obtain (5).) From the assumption, we have ky(t)k ≤k1e^−at+k2

Z _t

0

e^−a(t−τ)ku(τ)kdτ+k3ku(t)k where

k1= rc2

c1kx0kη1, k2=c4Lη1

2c1 , k3=η2, a= c3

2c2. Set

y1(t) =k1e^−at, y2(t) =k2

Z _t

0

e^−a(t−τ)ku(τ)kdτ, y3(t) =k3ku(t)k.

Then, for any p∈[1,∞], we have

ky2,τkLp ≤k2

akuτkLp

sincekhkL1 = 1/aand from Example 5.2, and

ky3,τkLp≤k3kuτkLp. For the term y1,

ky1,τkLp ≤k1ρ where

ρ=







1, ifp=∞,

³1 ap

´_1/p

, ifp∈[1,∞) .

Thus, by the triangular inequality,

γ=k3+k2

a, β =k1ρ.

On the other hand, the global case follows easily.

Corollary 5.1

IF the origin of ˙x=f(t, x,0) is GES(LES) and

kf(t, x, u)−f(t, x,0)k ≤Lkuk kh(t, x, u)k ≤η1kxk+η2kuk

THEN the original system is (small-signal) finite-gain Lp stable for eachp∈[1,∞].

Corollary 5.2

A LTI system is finite-gainLp stable ifAis Hurwitz.

(In particular,

γ=kDk2+2λ²_max(P)kBk2kCk2

λ_min(P) β =kCk2kx0k

s

λmax(P) λmin(P)ρ ρ=







1, p=∞,

³2λmax(P) p

´¹

p, p∈[1,∞).

where A^TP+P A=−I.) Example 5.4

˙

x=−x−x³+u, x(0) =x0, y= tanhx+u

WithV(x) =x²/2, we can show that the system is finite-gainLpstable.

Let us consider LUAS case restricting toL∞ stability.

Theorem 5.2(local version)

Letr >0 andru>0 s.t. {kxk ≤r} ⊂D and{kuk ≤ru} ⊂Du. IF x= 0 of ˙x=f(t, x,0) is LUAS, and∃V(t, x) s.t.

α1(kxk)≤V(t, x)≤α2(kxk)

∂V

∂t +∂V

∂xf(t, x,0)≤ −α3(kxk)

°°

°°∂V

∂x(t, x)

°°

°°≤α4(kxk) for all (t, x)∈[0,∞)×D, and,

IF

kf(t, x, u)−f(t, x,0)k ≤α5(kuk) kh(t, x, u)k ≤α6(kxk) +α7(kuk) +η for all (t, x, u)∈[0,∞)×D×Du,

THEN, for eachkx0k ≤α⁻¹₂ (α1(r)), the system is small-signalL∞ stable.

Proof.

Derivative ofV along ˙x=f(t, x, u) is V˙ =∂V

∂t +∂V

∂xf(t, x,0) + ∂V

∂x[f(t, x, u)−f(t, x,0)]

≤ −α3(kxk) +α4(kxk)α5(kuk)

≤ −(1−θ)α3(kxk)−θα3(kxk) +α4(r)α5

µ sup

0≤t≤τ

ku(t)k

¶

where 0< θ <1. Set

µ=α⁻¹₃

Ãα4(r)α5

¡sup_0≤t≤τku(t)k¢ θ

! .

We consider only suchu(t) that sup_0≤t≤τku(t)kis small enough forµ < α⁻¹₂ (α1(r)). Then, V˙ ≤ −(1−θ)α3(kxk), ∀kxk ≥µ.

From Theorem 4.18,

kx(t)k ≤β(kx₀k, t) +γ µ

sup

0≤t≤τ

ku(t)k

¶

for all 0≤t≤τ. Hence, ky(t)k ≤α₆

µ

β(kx₀k, t) +γ µ

sup

0≤t≤τ

ku(t)k

¶¶

+α₇(ku(t)k) +η

≤α6(2β(kx0k, t)) +α6

µ 2γ

µ sup

0≤t≤τku(t)k

¶¶

+α7(ku(t)k) +η

Therefore, we have

kyτkL∞ ≤γ0(kuτkL∞) +β0

γ0=α6◦2γ+α7, β0=α6(2β(kx0k,0)) +η

Theorem 5.3(global version)

IF D=Rⁿ,Du=R^m, the system ˙x=f(t, x, u) is ISS, and kh(t, x, u)k ≤α1(kxk) +α2(kuk) +η, THEN, the system isL∞stable.

Proof. Trivial. Isn’t it?

* Think about why the global asymptotic stable case needs so strong property (ISS)?

Consider

˙

x=− x 1 +x²+u,

which is GAS (but not GES), and is notL∞ stable (also not ISS).

III. L2 Gain

“L2 stability plays a special role in systems analysis. It is natural to work with square- integrable signals, which can be viewed as finite-energy signals. In many control problems, the system is represented as an input-output map, from a disturbance inpututo a controlled outputy, which is required to be small.”

Here we study how to calculate theL2gain for TI systems.

Theorem 5.4Consider

˙

x=Ax+Bu y=Cx+Du

where Ais Hurwitz. LetG(s) =C(sI−A)⁻¹B+D. Then, theL2 gain is sup

ω∈RkG(jω)k2= sup

ω∈R

q

λmax[G^T(−jω)G(jω)].

Proof. Due to linearity, we setx(0) = 0. From Fourier transform theory, for a causal signal y∈ L2,

Y(jω) = Z _∞

0

y(t)e^−jωtdt, Y(jω) =G(jω)U(jω).

By Parseval’s theorem, kyk²_L2 =

Z _∞

0

y^T(t)y(t)dt= 1 2π

Z _∞

−∞

Y^∗(jω)Y(jω)dω

= 1 2π

Z _∞

−∞

U^∗(jω)G^T(−jω)G(jω)U(jω)dω

≤ µ

sup

ω∈RkG(jω)k2

¶₂ 1 2π

Z _∞

−∞

U^∗(jω)U(jω)dω

= µ

sup

ω∈RkG(jω)k2

¶₂ kuk²_L2

which shows that the L2 gain is less than or equal to sup_ω∈RkG(jω)k2. See Appendix C.10 to show that the L2 gain is, in fact, equal to sup_ω∈RkG(jω)k2.

Consider

˙

x=f(x) +G(x)u, x(0) =x0, y=h(x)

where f is locally Lipschitz,Gandhare continuous, andf(0) = 0 andh(0) = 0.

Theorem 5.5

IF ∃C¹positive semidefinite functionV(x) andγ >0 s.t.

H(V, f, G, h, γ) := ∂V

∂xf(x) + 1 2γ²

∂V

∂xG(x)G^T(x) µ∂V

∂x

¶_T +1

2h^T(x)h(x)≤0, THEN, for eachx0∈Rⁿ, the system is finite-gainL2 stable and the gain≤γ.

* Hamilton-Jacobi inequality

* Compare with Theorem 5.1—no exponential stability, but needs a solution to HJ eq.

Proof.

V˙ =LfV(x) +LGV(x)u=−1

2γ²ku− 1

γ²(LGV)^T(x)k²₂+LfV(x) + 1

2γ²LGV(x)(LGV)^T(x) +1 2γ²kuk²₂

≤1

2γ²kuk²₂−1

2kyk²₂−1

2γ²ku− 1

γ²(LGV)^T(x)k²₂

≤1

2γ²kuk²₂−1 2kyk²₂ Thus,

V(x(τ))−V(x(0))≤1 2γ²

Z _τ

0

ku(t)k²₂dt−1 2

Z _τ

0

ky(t)k²₂dt.

Since V(x)≥0, we have Z _τ

0

ky(t)k²₂≤γ² Z _τ

0

ku(t)k²₂dt+ 2V(x₀).

Then,

kyτkL2 ≤γkuτkL2+p 2V(x0).

Example 5.8

˙ x1=x2

˙

x2=−ax³₁−kx2+u, a >0, k >0, y=x2

LetV(x) =α(ax⁴₁/4 +x²₂/2). Then,LfV =−αkx²₂,LGV =αx2,h(x) =x2, so that H=

µ

−αk+ α² 2γ²+1

2

¶ x²₂. That is, if

γ²≥ α² 2αk−1

then the system isL2 stable. The right-hand side has a minimum 1/k²forα= 1/k, so with γ= 1/k, we conclude that the system is finite-gainL2stable with the gain less than or equal to 1/k.

Example 5.9Consider the system with the property

LfW(x)≤ −kh^T(x)h(x), k >0 LGW(x) =h^T(x)

where W(x) is a C¹ positive semidefinite function.

LetV(x) =αW(x). Then, H=

µ

−αk+ α² 2γ²+1

2

¶

h^T(x)h(x).

By Example 5.8, we conclude that the system is finite-gain L₂ stable with the gain≤1/k.

Example 5.10Consider the system with the property LfW(x)≤0 LGW(x) =h^T(x) where W(x) is a C¹ positive semidefinite function.

Let an output feedback control

u=−ky+v, k >0.

Then, the closed-loop system becomes

˙

x=f(x)−kG(x)G^T(x) µ∂W

∂x

¶_T

+G(x)v y=h(x)

This system is, in fact, the case of Example 5.9. (Verify!)

So, this system is finite-gain L2 stable with the gain ≤1/k from v to y. Note that the gain can be arbitrarily assigned with the feedback.

Example 5.11Consider

˙

x=Ax+Bu y=Cx Suppose∃P ≥0 s.t.

P A+A^TP+ 1

γ²P BB^TP+C^TC= 0 (the Riccati equation), with someγ >0.

Then, V(x) = ¹₂x^TP x satisfies the HJ equation for this system, i.e., H = 0. Thus, the system is finite-gainL2stable with the gain≤γ. (In fact, the Riccati equation has a solution P ≥0 ifand only if the system’sL2 gain is less than or equal toγ.)

• From the proof of Theorem 5.5, we note that, if the assumptions hold only on a finite domain D, then we obtain the same conclusion as long as the solution x(t) stays in D.

(Corollary 5.4.) This possibly restricts the class of input signal, which also depends on the initial conditionx0.

• If ˙x=f(x) is AS and ifkx0k and sup_0≤t≤τku(t)k are sufficiently small (relative toD), then the solutionx(t) remains in the neighborhood of the origin. This leads to the following Lemma.

Consider

˙

x=f(x) +G(x)u, x(0) =x0, y=h(x)

where f isC¹,Gandhare continuous on a setD⊂Rⁿ, andf(0) = 0 andh(0) = 0.

Lemma 5.1

IF the origin of ˙x=f(x) is AS, and∃C¹ positive semidefinite function V(x) and γ >0 s.t.

H(V, f, G, h, γ) := ∂V

∂xf(x) + 1 2γ²

∂V

∂xG(x)G^T(x) µ∂V

∂x

¶_T +1

2h^T(x)h(x)≤0 on a domain D,

THEN ∃k1 > 0 s.t. for each kx0k ≤ k1, the system is small-signal finite-gain L2 stable with the gain ≤γ.

Proof. We can apply Corollary 5.4, if we show thatx(t) stays in a neighborhood of the origin.

By the converse theorem,∃ aC¹ functionW(x) andr0>0 s.t.

α1(kxk)≤W(x)≤α2(kxk)

∂W

∂x f(x)≤ −α3(kxk)

for allx∈Br0, and without loss of generality, we assume thatBr0 ⊂D.

Letkand Lbe an upper bound ofk∂W/∂xk andkG(x)k, respectively. Then, W˙ (x, u)≤∂W

∂x f(x) +∂W

∂x G(x)u≤ −α3(kxk) +kLkuk

≤ −(1−θ)α3(kxk)−θα3(kxk) +kL sup

0≤t≤τku(t)k

≤ −(1−θ)α3(kxk), ∀kxk ≥α⁻¹₃ µ

kL sup

0≤t≤τku(t)k/θ

¶

where 0< θ <1.

This implies that (you may recall Theorem 4.18), ∃k1, k2>0 s.t., for everykx0k< k1and for inputs such that sup_0≤t≤τku(t)k ≤ k2, the solution x(t) ∈Br0. Hence, the conclusion follows from Corollary 5.4.

Lemma 5.2

IF ∃C¹positive semidefinitefunctionV(x) andγ >0 s.t.

H(V, f, G, h, γ) := ∂V

∂xf(x) + 1 2γ²

∂V

∂xG(x)G^T(x) µ∂V

∂x

¶_T +1

2h^T(x)h(x)≤0

on D, and ‘no solution of ˙x= f(x) can stay identically in S ={x ∈D :h(x) = 0} other than the trivial solutionx(t)≡0’,

THEN the origin of ˙x=f(x) is AS, and∃k1 >0 s.t. for eachkx0k ≤ k1, the system is small-signal finite-gain L2 stable with the gain≤γ.

• In this lemma the Lyapunov functionV(x) came from the HJ inequality.

• How to interpret the phrase: ‘no solution of ˙x=f(x) can stay identically inS={x∈D: h(x) = 0}other than the trivial solutionx(t)≡0’ ? (Not clear in the textbook. Discuss it in

the proof.) I would say ‘there is nolocally invariant setinSexcept the origin’. By a locally Locally invariant set⊂ Invariant set

invariant setLinS, I mean that, ifx(0) =x0∈L, there exist nonnegative constantsδ1and δ2, not both zero, s.t.x(t)∈L fort∈[−δ1, δ2].

Proof. The proof is done by Lemma 5.1 if we show that the origin of ˙x=f(x) is AS.

From the HJ inequality,

∂V

∂xf(x)≤ − 1 2γ²

∂V

∂xG(x)G^T(x) µ∂V

∂x

¶_T

−1

2h^T(x)h(x)≤ −1

2h^T(x)h(x),

which seemingly means that the origin is stable, and by employing the LaSalle’s theorem, we have the conclusion. (In particular, the set {x∈ D : ^∂V_∂xf(x) = 0} ⊂ S. In addition, since there is no locally invariant set inS except the origin, the only invariant set inS is the origin. So, the LaSalle’s theorem can be applied.) This is true ifV(x) is positive definite.

However, since V(x) is assumed just positive semidefinite, we now prove that, from the assumption, the function V(x) is in fact positive definite.

Letr >0 be s.t.Br⊂D (strict inclusion). Then, for eachx∈Br,∃δ >0 s.t. the solution fromx,φ(t;x), stays inDfort∈[0, δ]. Integrating the above inequality over [0, δ], we obtain

V(φ(δ;x))−V(x)≤ −1 2

Z _δ

0

kh(φ(t;x))k²₂dt.

Using V(φ(δ;x))≥0, we have

V(x)≥1 2

Z _δ

0

kh(φ(t;x))k²₂dt.

Now suppose that ∃x6= 0 s.t. V(x) = 0 (not positive definite). Then, the above implies thath(φ(t;x)) = 0 fort∈[0, δ]. This means thatx6= 0 is an element of locally invariant set.

However, since the only element of the locally invariant set is the origin,x= 0 (contradiction).

Example 5.12DIY.

IV. Feedback Systems: The Small-Gain Theorem

u ₁

H ₁

H ₂

e ₁ y ₁

y ₂ e ₂ + u ₂ -

+ +

From the textbook,

• The formalism of input-output stability is particularly useful in studying stability of inter- connected systems, since the gain of a system allows us to track how the norm of a signal increases or decreases as it passes through the system.

• See Figure 5.1. H1:L^m_e → L^q_e andH2:L^q_e→ L^m_e . Suppose both system are finite-gainL stable; that is,

ky1,τkL≤γ1ke1,τkL+β1, ∀e1∈ L^m_e,∀τ ∈[0,∞) ky2,τkL≤γ2ke2,τkL+β2, ∀e2∈ L^q_e,∀τ ∈[0,∞) (finite-gain for simplicity).

• Suppose that the feedback system iswell definedin the sense that for every pair of inputs u1∈ L^m_e andu2∈ L^q_e,∃ unique outputse1, y2∈ L^m_e ande2, y1∈ L^q_e.

• Define u= [u1;u2], y = [y1;y2], e= [e1;e2]. We ask if the feedback system, viewed as a mapping from uto y (or to e since they are equivalent in the sense of Exercise 5.21, your homework), is finite-gainL stable.

• If the external input and output of interest is just u1 and y1, why should we consider L stability from both (u1, u2) to both (y1, y2)? The reason is that, by considering all the pairs of inputs and outputs, any possibly hidden internal mode comes out. (See Exercise 5.20.)

Theorem 5.6 (Small-gain Theorem) IF γ1γ2<1,

THEN the feedback system is finite-gain Lstable.

Proof.

e1τ =u1τ−(H2e2)τ, e2τ =u2τ+ (H1e1)τ

Then,

ke1τkL≤ ku1τkL+k(H2e2)τkL≤ ku1τkL+γ2ke2τkL+β2

≤ ku1τkL+γ2(ku2τkL+γ1ke1τkL+β1) +β2

=γ1γ2ke1τkL+ (ku1τkL+γ2ku2τkL+β2+γ2β1) Since γ1γ2<1,

ke1τkL≤ 1

1−γ₁γ₂(ku1τkL+γ2ku2τkL+β2+γ2β1) for allτ∈[0,∞). Similarly,

ke_2τk_L≤ 1 1−γ1γ2

(ku_2τk_L+γ₁ku_1τk_L+β₁+γ₁β₂) for allτ∈[0,∞). The proof completes sincekekL≤ ke1kL+ke2kL.