Post on 11-Mar-2018
EECE 574 - Adaptive ControlModel-Reference Adaptive Control - An Overview
Guy Dumont
Department of Electrical and Computer EngineeringUniversity of British Columbia
January 2013
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 1 / 73
Model-Reference Adaptive Systems
The MRAC or MRAS is an important adaptive control methodology 1
1see Chapter 5 of the Åström and Wittenmark textbook, or H. Butler,”Model-Reference Adaptive Control-From Theory to Practice”, Prentice-Hall, 1992
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 2 / 73
Model-Reference Adaptive Systems
The MIT rule
Lyapunov stability theory
Design of MRAS based on Lyapunov stability theory
Hyperstability and passivity theory
The error model
Augmented error
A model-following MRAS
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 3 / 73
MIT Rule The Basics
The MIT Rule
Original approach to MRAC developed around 1960 at MIT foraerospace applications
With e = y− ym, adjust the parameters θ to minimize
J(θ) =12
e2
It is reasonable to adjust the parameters in the direction of the negativegradient of J:
dθ
dt=−γ
∂J∂θ
=−γe∂e∂θ
∂e/∂θ is called the sensitivity derivative of the system and is evaluatedunder the assumption that θ varies slowly
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 4 / 73
MIT Rule The Basics
The MIT Rule
The derivative of J is then described by
dJdt
= e∂e∂ t
=−γe2(
∂e∂θ
)2
Alternatively, one may consider J(e) = |e| in which case
dθ
dt=−γ
∂J∂θ
=−γ∂e∂θ
sign(e)
The sign-sign algorithm used in telecommunications where simpleimplementation and fast computations are required, is
dθ
dt=−γ
∂J∂θ
=−γsign(
∂e∂θ
)sign(e)
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 5 / 73
MIT Rule Examples
MIT Rule: Example 1
Process: y = kG(s) where G(s) is known but k is unknown
The desired response is ym = k0G(s)uc
Controller is u = θuc
Then e = y− ym = kG(p)θuc− k0G(p)uc
Sensitivity derivative
∂e∂θ
= kG(p)uc =kk0
ym
MIT ruledθ
dt= γ
′ kk0
yme =−γyme
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 6 / 73
MIT Rule Examples
MIT Rule: Example 1
Figure: MIT rule for adjustment of feedforward gain (from textbook).
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 7 / 73
MIT Rule Examples
MIT Rule: Example 1
Figure: MIT rule for adjustment of feedforward gain: Simulation results (fromtextbook).
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 8 / 73
MIT Rule Examples
MIT Rule: Example 2
Consider the first-order system
dydt
=−ay+bu
The desired closed-loop system is
dym
dt=−amym +bmuc
Applying model-following design (see lecture notes on pole placement)
degA0 ≥ 0 degS = degR = degT = 0
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 9 / 73
MIT Rule Examples
MIT Rule: Example 2
The controller is thenu(t) = t0uc(t)− s0y(t)
For perfect model-following
dydt
= −ay(t)+b[t0uc(t)− s0y(t)]
=−(a+bs0)y(t)+bt0uc
=−amym(t)+bmuc
This implies
s0 =am−a
b
t0 =bm
b
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 10 / 73
MIT Rule Examples
MIT Rule: Example 2
With the controller we can write2
y =bt0
p+a+bs0uc
With e = y− ym, the sensitivity derivatives are
∂e∂ t0
=b
p+a+bs0uc
∂e∂ s0
=b2t0
(p+a+bs0)2 uc =b
p+a+bs0y
2p is the differential operator d(.)/dtGuy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 11 / 73
MIT Rule Examples
MIT Rule: Example 2
However, a and b are unknown3. For the nominal parameters, we know that
a+bs0 = am
p+a+bs0 ≈ p+am
Then
dt0dt
= −γ′b(
1p+am
uc
)e =−γ
(am
p+amuc
)e
ds0
dt= γ
′b(
1p+am
y)
e = γ
(am
p+amy)
e
with γ = γ ′b/am, i.e. b is absorbed in γ and the filter is normalized with staticgain of one.
3after all, we want to design an adaptive controller!Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 12 / 73
MIT Rule Examples
MIT Rule: Example 2
Figure: MIT rule for first-order (from textbook)
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 13 / 73
MIT Rule Examples
MIT Rule: Example 2
Figure: Simulation of MIT rule for first-order with a = 1, b = 0.5, am = bm = 2 andγ = 1(from textbook)
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 14 / 73
MIT Rule Examples
MIT Rule: Example 2
Figure: Simulation of MIT rule for first-order with a = 1, b = 0.5, am = bm = 2.Controller parameters for γ = 0.2, 1 and 5(from textbook)
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 15 / 73
MIT Rule Examples
MIT Rule: Example 2
Figure: Simulation of MIT rule for first-order with a = 1, b = 0.5, am = bm = 2.Relation between controller parameters θ1 and θ2. The dashed line is θ2 = θ1−a/b.(from textbook)
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 16 / 73
MIT Rule Stability of the Adaptive Loop
Stability of the Adaptive Loop
Consider again the adaptation of a feedforward gain
G =1
s2 +a1s+a2
e = (θ −θ0)Guc
∂e∂θ
= Guc =ym
θ0
dθ
dt= −γ
′e∂e∂θ
=−γ′e
ym
θ0=−γeym
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 17 / 73
MIT Rule Stability of the Adaptive Loop
Stability of the Adaptive Loop
Thus the system can be described as
d2ym
dt2 +a1dym
dt+a2ym = θ0uc
d2ydt2 +a1
dydt
+a2y = θuc
dθ
dt= −γ(y− ym)ym
This is difficult to solve analytically
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 18 / 73
MIT Rule Stability of the Adaptive Loop
Stability of the Adaptive Loop
d3ydt3 +a1
d2ydt2 +a2
dydt
+ γucymy = θduc
dt+ γucy2
m
Assume uoc and yo
m constant, equilibrium is
y(t) = yom =
θ0uoc
a2
which is stable ifa1a2 > γuo
cyom =
γ
a2(uo
c)2
Thus, if γ or uc is sufficiently large, the system will be unstable
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 19 / 73
MIT Rule Stability of the Adaptive Loop
Stability of the Adaptive Loop
Figure: Influence on convergence and stability of signal amplitude for MIT rule(from textbook)
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 20 / 73
MIT Rule Modified MIT Rule
Modified MIT Rule
Normalization can be used to protect against dependence on the signalamplitudes
With ϕ = ∂e/∂θ , the MIT rule can be written as
dθ
dt=−γϕe
The normalized MIT rule is then
dθ
dt=−γϕe
α2 +ϕTϕ
For the previous example,
dθ
dt=−γeym/θ0
α2 + y2m/θ 2
0
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 21 / 73
MIT Rule Modified MIT Rule
Modified MIT Rule
Figure: Influence on convergence and stability of signal amplitude for modified MITrule (from textbook)
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 22 / 73
Lyapunov Design of MRAC Lyapunov Theory
Lyapunov Stability Theory
Aleksandr M Lyapunov(1857-1918, Russia)
Classmate of Markov, taught byChebyshev
Doctoral thesis The generalproblem of the stability of motionbecame a fundamentalcontribution to the study ofdynamic systems stability
He shot himself three days afterhis wife died of tuberculosis
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 23 / 73
Lyapunov Design of MRAC Lyapunov Theory
Lyapunov Stability Theory
Consider the nonlinear time-varying system
x = f (x, t) with f (0, t) = 0
If at time t = t0 ‖x(0)‖= δ , i.e. if the initial state is not the equlibriumstate x∗ = 0, what will happen?
There are four possibilities
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 24 / 73
Lyapunov Design of MRAC Lyapunov Theory
Lyapunov Stability Theory
The system is stable. For a sufficiently small δ , x stays within ε of x∗. Ifδ can be chosen independently of t0, then the system is said to beuniformly stable.
The system is unstable. It is possible to find ε which does not allow anyδ .
The system is asymptotically stable. For δ < R and an arbitrary ε , thereexists t∗ such that for all t > t∗, ‖x−x∗‖< ε . It implies that ‖x−x∗‖→ 0as t→ ∞.
If asymptotic stability is guaranteed for any δ , the system is globallyasymptotically stable.
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 25 / 73
Lyapunov Design of MRAC Lyapunov Theory
Lyapunov Stability Theory
A scalar time-varying function is locally positive definite if V(o, t) = 0and there exists a time-invariant positive definite function V0(x) such that∀t ≥ t0, V(x, t)≥ V0(x). In other words, it dominates a time-invariantpositive definite function.
V(x, t) is radially unbounded if 0 < α‖x‖ ≤ V(x, t), α > 0.
A scalar time-varying function is said to be decrescent if V(o, t) = 0 andthere exists a time-invariant positive definite function V1(x) such that∀t ≥ t0, V(x, t)≤ V1(x).
In other words, it is dominated by a time-invariant positive definitefunction.
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 26 / 73
Lyapunov Design of MRAC Lyapunov Theory
Lyapunov Stability Theory
Theorem (Lyapunov Theorem)
Stability: if in a ball BR around the equilibrium point 0, there exists a scalar functionV(x, t) with continuous partial derivatives such that
1 V is positive definite2 V is negative semi-definite
then the equilibrium point is stable.
Uniform stability and uniform asymptotic stability: If furthermore,
3 V is decrescent
then the origin is uniformly stable. If condition 2 is strengthened by requiring that V benegative definite, then the equilibrium is uniformly asymptotically stable.
Global uniform asymptotic stability: If BR is replaced by the whole state space, andcondition 1, the strengthened condition 2, condition 3 and the condition
4 V(x,t) is radially unbounded
are all satisfied, then the equilibrium point at 0 is globally uniformly asymptoticallystable.
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 27 / 73
Lyapunov Design of MRAC Lyapunov Theory
Lyapunov Stability Theory
The Lyapunov function has similarities with the energy content of the system and mustbe decreasing with time.
This result can be used in the following way to design a stable a stable adaptivecontroller
1 First, the error equation, i.e. a differential equation describing either the outputerror y− ym or the state error x− xm is derived.
2 Second, a Lyapunov function, function of both the signal error e = x− xm and theparameter error φ = θ −θm is chosen. A typical choice is
V = eT Pe+φT
Γ−1
φ
where both P and Γ−1 are positive definite matrices.3 The time derivative of V is calculated. Typically, it will have the form
V =−eT Qe+ some terms including φ
Putting the extra terms to zero will ensure negative definiteness for V if Q ispositive definite, and will provide the adaptive law.
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 28 / 73
Lyapunov Design of MRAC Lyapunov Design of MRAC
Stability of Adaptive Loop
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 29 / 73
Lyapunov Design of MRAC Lyapunov Design of MRAC
Lyapunov Design of MRAC
Determine controller structure
Derive the error equation
Find a Lyapunov equation
Determine adaptation law that satisfies Lyapunov theorem
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 30 / 73
Lyapunov Design of MRAC Lyapunov Design of MRAC
Adaptation of Feedforward Gain
Process modeldydt
=−ay+ ku
Desired responsedym
dt=−aym + k0uc
Controlleru = θuc
Introduce error e = y− ym
The error equation is
dedt
=−ae+(kθ − k0)uc
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 31 / 73
Lyapunov Design of MRAC Lyapunov Design of MRAC
Adaptation of Feedforward Gain
System model
dydt
= −ay+ ku
dθ
dt= ??
Desired equilibrium
e = 0
θ = θ0 =k0
k
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 32 / 73
Lyapunov Design of MRAC Lyapunov Design of MRAC
Adaptation of Feedforward Gain
Consider the Lyapunov function
V(e,θ) =γ
2e2 +
k2(θ −θ0)
2
dVdt
= −γae2 +(kθ − k0)(dθ
dt+ γuce)
Choosing the adjustment rule
dθ
dt=−γuce
givesdVdt
=−γae2
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 33 / 73
Lyapunov Design of MRAC Lyapunov Design of MRAC
Adaptation of Feedforward Gain
Lyapunov rule: dθ
dt =−γuce MIT rule: dθ
dt =−γyme
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 34 / 73
Lyapunov Design of MRAC Lyapunov Design of MRAC
Adaptation of Feedforward Gain
Lyapunov rule: MIT rule:
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 35 / 73
Lyapunov Design of MRAC Lyapunov Design of MRAC
First-order System
Process modeldydt
=−ay+bu
Desired responsedym
dt=−amym +bmuc
Controlleru = θ1uc−θ2y
Introduce error e = y− ym
The error equation is
dedt
=−ame− (bθ2 +a−am)y+(bθ1−bm)uc
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 36 / 73
Lyapunov Design of MRAC Lyapunov Design of MRAC
First-order System
Candidate Lyapunov function
V(t,θ1,θ2) =12
(e2 +
1bγ
(bθ2 +a−am)2 +
1bγ
(bθ1−bm)2)
Derivative
dVdt
=dedt
+1γ(bθ2 +a−am)
dθ2
dt+
1γ(bθ1−bm)
dθ1
dt
= −ame2 +1γ(bθ2 +a−am)(
dθ2
dt− γye)+
1γ(bθ1−bm)(
dθ1
dt+ γuce)
This suggests the adaptation law
dθ1
dt= −γuce
dθ2
dt= γye
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 37 / 73
Lyapunov Design of MRAC Lyapunov Design of MRAC
First-order System
Lyapunov Rule MIT Rule
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 38 / 73
Lyapunov Design of MRAC Lyapunov Design of MRAC
First-order System
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 39 / 73
Lyapunov Design of MRAC Lyapunov Design of MRAC
Lyapunov-based MRAC Design
The main advantage of Lyapunov design is that it guarantees aclosed-loop system.
For a linear, asymptotically stable governed by a matrix A, a positivesymmetric matrix Q yields a positive symmetric matrix P by the equation
ATP+PA =−Q
This equation is known as Lyapunov’s equation.
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 40 / 73
Lyapunov Design of MRAC Lyapunov Design of MRAC
Lyapunov-based MRAC Design
The main drawback of Lyapunov design is that there is no systematicway of finding a suitable Lyapunov function V leading to a specificadaptive law.
For example, if one wants to add a proportional term to the adaptive law,it is not trivial to find the corresponding Lyapunov function.
The hyperstability approach is more flexible than the Lyapunovapproach.
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 41 / 73
Lyapunov Design of MRAC State Feedback
The Lyapunov Equation
Let the linear systemx = AX
be stable
Let Q be an arbitrary positive definite matrix
Then the Lyapunov equation
ATP+PA =−Q
always has a unique solution where P is positive definite.
The functionV(x) = xTPx
is then a Lyapunov function
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 42 / 73
Lyapunov Design of MRAC State Feedback
State Feedback
Process: x = Ax+Bu
Desired response: xm = Amxm +Bmuc
Control law: u = Muc−Lx
Closed-loop system
x = Ax+Bu = (A−BL)x+BMuc = Ac(θ)x+Bc(θ)uc
The parametrization is Ac(θ) = Am and Bc(θ) = Bm
For compatibility we need A−Am = BL and Bm = BM
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 43 / 73
Lyapunov Design of MRAC State Feedback
Error Equation
The error e = x− xm satisfies
e = x− xm = Ax+Bu−Axm−Bmuc
Hence
e = Ame+(A−Am−BL)x+(BM−Bm)uc
= Ame+(Ac(θ)−Am)x+(Bc(θ)−Bm)uc
= Ame+Ψ(θ −θ0)
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 44 / 73
Lyapunov Design of MRAC State Feedback
The Lyapunov Function
Try
V(e,θ) =12(γeTPe+(θ −θ0)
T(θ −θ0))
Differentiating V
V =γ
2(eTPe+ eTPe)+(θ −θ0)
Tθ
Then with e = Ame+Ψ(θ −θ0) we can write
V =γ
2[(
eTATm +Ψ(θ −θ0)
)Pe+ eT (Ame+Ψ(θ −θ0)
]+(θ −θ0)
Tθ
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 45 / 73
Lyapunov Design of MRAC State Feedback
The Lyapunov Function
Now, consider the Lyapunov equation ATmP+PAm =−Q (possible
because reference model always stable)
V = −γ
2eTQe+ γ(θ −θ0)
TΨ
TPe+(θ −θ0)T
θ
= −γ
2eTQe+(θ −θ0)
T(θ + γΨTPe)
If the parameter adjustment law is chosen as
dθ
dt=−γΨ
TPe
We obtaindVdt
=−γ
2eTQe
which is negative semi-definiteNote that this assumes knowledge of the state.
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 46 / 73
Lyapunov Design of MRAC Adaptation of Feedforward Gain
Adaptation of Feedforward Gain
Consider the error
e = (kG(p)θ − k0G(p))uc = kG(p)(θ −θ0)uc
With
x = Ax+B(θ −θ0)uc
e = Cx
A candidate Lyapunov function is
V =12(γxTPx+(θ −θ0)
2)
LetATP+PA =−Q
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 47 / 73
Lyapunov Design of MRAC Adaptation of Feedforward Gain
Adaptation of Feedforward Gain
Differentiating V
V =γ
2(xTPx+ xTPx
)+(θ −θ0)θ
Which after using the model for x and the Lyapunov equation gives
V =−γ
2xTQx+(θ −θ0)
(θ + γucBTPx
)Using the adaptation law
dθ
dt=−γucBTPx
givesdVdt
=−γ
2xTQx
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 48 / 73
Lyapunov Design of MRAC Adaptation of Feedforward Gain
Output Feedback
The adaptation lawdθ
dt=−γucBTPx
assumes knowledge of the state x
If we can find P such thatBTP = C
then the adaptation law becomes
dθ
dt=−γuce
i.e. this is now output feedback and the state x is not required
When is it possible to do so?
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 49 / 73
Lyapunov Design of MRAC Kalman-Yakubovich Lemma
Kalman-Yakubovich Lemma
Definition: Positive real transfer functionA rational transfer function G with real coefficients is positive real (PR) if
Re G(s)≥ 0 for Re s≥ 0
A rational transfer function G with real coefficients is strictly positive real(SPR) if G(s− ε) is PR for some ε > 0
G(s) = 1/(s+1) is SPR
G(s) = 1/s is PR bur not SPR
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 50 / 73
Lyapunov Design of MRAC Kalman-Yakubovich Lemma
Kalman-Yakubovich Lemma
Theorem (Kalman-Yakubovich Lemma)Let the linear time-invariant system
x = Ax+Bu
y = Cx
be completely controllable and completely observable. The transfer function
G(s) = C(sI−A)−1B
is strictly positive real if and only if there exist positive definite matrices P andQ such that
ATP+PA = −Q
BTP = C
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 51 / 73
The Small Gain Theorem Definitions
Operator View of Dynamical Systems
L2 norm
‖u‖=(∫
∞
−∞
u2(t)dt) 1
2
L∞ norm
‖u‖= sup0≤t<∞
|u(t)|
Truncation operator:
xT(t) ={
x(t) 0≤ t ≤ T0 t > T
L2e ={
x | ‖xT‖2 = 〈xT |xT〉< ∞}
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 52 / 73
The Small Gain Theorem Definitions
The Notion of Gain
Gain of a nonlinear system
Let the signal space be Xe. The gain γ(S) of a system S is defined as
γ(S) = supu∈Xe
‖Su‖‖u‖
where u is the input signal to the system.The gain is thus the smallest value γ such that
‖Su‖ ≤ γ(S)‖u‖ for all u ∈ Xe
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 53 / 73
The Small Gain Theorem Definitions
The Notion of Gain
For a linear system G with signals in L2e,
γ(G) = maxω|G(iω)|
Static nonlinear system y(t) = f (u(t)), the gain of the system is
γ = maxu
|f (u)||u|
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 54 / 73
The Small Gain Theorem Definitions
BIBO Stability
DefinitionA system is bounded-input, bounded-output (BIBO) stable if it has boundedgain
A BIBO stable system is a system for which the outputs will remainbounded for all time, for any finite initial condition and input.
A continuous-time linear time-invariant system is BIBO stable if andonly if all the poles of the system have real parts less than 0.
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 55 / 73
The Small Gain Theorem Definitions
The Small Gain Theorem
Theorem (The Small Gain Theorem)Consider the system
Let γ1 and γ2 be the gains of the systems H1 and H2. The closed-loop system isthen BIBO stable if
γ1γ2 < 1 and its gain is less than γ =γ1
1− γ1γ2
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 56 / 73
Positivity and Passivity Background
Positivity and Passivity
Positivity and passivity are nearly equivalent concepts.
Positivity is a property of linear systems, while passivity is a moregeneral concept, applicable to both linear and nonlinear systems.
The scalar, controllable system
x = Ax+bu
y = cTx
with a transfer function
H(s) = cT(sI−A)−1b
is said to be positive real (PR) if Re[H(s)]≥ 0 for all Re(s)≥ 0.
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 57 / 73
Positivity and Passivity Background
Positivity and Passivity
If s = jω , it means that the Nyquist diagram of H(jω) must lies in theright half plane, including the imaginary axis.
Thus, the phase shift must be between −90deg and +90deg.
The Kalman-Yakubovich lemma states that the system H(s) is strictlypositive real if there exist positive definite matrices P and Q such that:
ATP+PA = −Q
Pb = c
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 58 / 73
Positivity and Passivity Background
Positivity and Passivity
Passivity is a general form of positivity
A passive linear system must be positive real and vice-versa.
Consider a system with input u and output y. The energy of the system issuch that
ddt
[Stored energy]=[external power input]+[internal power generation]
If the internal power generation is negative, then the system isdissipative or strictly passive. If the internal power generation is lessthan or equal to zero the system is passive.
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 59 / 73
Positivity and Passivity Background
Positivity and Passivity
If the input u and output y are such that the external power input isexpressed as uTy, then
The system is passive if 〈y|u〉 ≥ 0The system is input strictly passive (ISP) if there exists ε > 0 such that〈y|u〉 ≥ ‖u‖2
The system is output strictly passive (OSP) if there exists ε > 0 such that〈y|u〉 ≥ ‖y‖2
Examples include electrical networks (power proportional to product ofcurrent and voltage), mass-spring systems (mechanical power is force ×velocity).
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 60 / 73
Positivity and Passivity Background
The Passivity Theorem
Theorem (Passivity Theorem)Consider the system
Let H1 be strictly output passive and H2 be passive. The closed-loop system isthen BIBO stable
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 61 / 73
Positivity and Passivity Background
Positive Real Transfer Functions
A rational transfer function G(s) with real coefficients is PR if and only if thefollowing conditions hold
1 The function has no poles in the right half-plane2 If the function has poles on the imaginary axis or at infinity, they are
simple poles with positive residues3 The real part of G is nonnegative along the iω axis
Re(G(iω))≥ 0
G is SPR if conditions 1 and 3 hold and if G(s) has no poles or zeros on theimaginary axis
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 62 / 73
Positivity and Passivity Background
Positivity and Passivity
Positive real (PR): Re G(jω)≥ 0
Input strictly passive (ISP): Re G(jω)≥ ε > 0
Input strictly passive (ISP): Re G(jω)≥ ε|G(jω)|2
ExamplesG(s) = s+1 is SPR and ISP but not OSPG(s) = 1/(s+1) is SPR and OSP but not ISPG(s) = (s2 +1)/(s+1)2 is OSP and ISP but not SPRG(s) = 1/s is PR, but not SPR, OSP or ISP
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 63 / 73
Positivity and Passivity Background
Positive Real Transfer Functions
Theorem (Lemma)
Let r be a bounded square integrable function, and let G(s) be a positive realtransfer function. The system whose input-output relation is given by
y = r(G(p)ru)
is then passive.
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 64 / 73
Positivity and Passivity Application to Adaptive Control
Adaptation of Feedforward Gain
MIT rule:
Lyapunov rule:
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 65 / 73
Positivity and Passivity Application to Adaptive Control
Analysis
Redraw (b) as:
Because γ/s is PR, H is passiveThus, if G is SPR, the passivity theorem implies that the system is L2 (BIBO) stableIf non-zero initial conditions, e(t) is still in L2 and thus goes to zero as t→ ∞
This is stable for all γ > 0, so the adaptation can be made arbitrarily fast as long as G isSPRIf G is not SPR, then may be possible to introduce a compensator Gc such that GGc isSPR
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 66 / 73
Positivity and Passivity Application to Adaptive Control
Analysis
The Kalman-Yakubovich lemma can be used to find Gc such that GGc isPR
However, when relative degree of G is greater than 1, Gc will containderivatives
Augmented error
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 67 / 73
Positivity and Passivity Application to Adaptive Control
Augmented Error
Factor G asG = G1G2
where G1 is SPR
e = G(θ −θ0)uc = (G1G2)(θ −θ0)uc
= G1 (G2(θ −θ0)uc(θ −θ0)G2uc− (θ −θ0)G2uc)
Introduce the augmented error
ε = e+η
where η is the error augmentation defined by
η = G1(θ −θ0)G2uc−G(θ −θ0)uc = G1(θG2uc)−Gθuc
The correction term η vanishes when θ is constant
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 68 / 73
Positivity and Passivity Application to Adaptive Control
Augmented Error
The adaptation lawθ =−γεG2uc
gives a closed-loop system where e(t) goes to zero as t goes to infinity
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 69 / 73
Positivity and Passivity Application to Adaptive Control
Augmented Error
Error augmentation is a fundamental idea in MRAC
Details can sometimes be messy...See example in Section 5.8 for an output feedback design
Error signal
e =b0
A0Am(Ru+Sy−Tuc)
Key is introduction of a filter Q such that b0Q/A0Am is SPR andef = Qe/PThen the closed-loop system is BIBO stable
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 70 / 73
Positivity and Passivity Example
Active Suspension
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 71 / 73
Positivity and Passivity Example
Active Suspension
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 72 / 73
Positivity and Passivity Example
Active Suspension
Guy Dumont (UBC EECE) EECE 574: MRAC - 1 January 2013 73 / 73