modeling and analysis of dynamic systems - eth z and analysis of dynamic systems dr. guillaume...

Modeling and Analysis of Dynamic Systems

Dr. Guillaume Ducard

Fall 2017

Institute for Dynamic Systems and Control

ETH Zurich, Switzerland

G. Ducard c© 1 / 57

Outline

1 Lecture 13: Linear System - Stability AnalysisZero Dynamics: DefinitionsZero Dynamics: AnalysisExample and Summary

2 Lecture 13: Nonlinear Systems - IntroductionDefinitionsEquilibrium Sets

3 Lecture 13: Nonlinear Systems - First & Second OrderFirst-order SystemsSecond-order SystemsLyapunov PrincipleLyapunov Theory


Lecture 13: Linear System - Stability AnalysisLecture 13: Nonlinear Systems - Introduction

Lecture 13: Nonlinear Systems - First & Second Order

Zero Dynamics: DefinitionsZero Dynamics: AnalysisExample and Summary

Outline








Zero Dynamics

The dynamic behavior of linear system described as

ddtx(t) = x(t) = Ax(t) +Bu(t)

y(t) = Cx(t) +Du(t)

can be studied through its poles (eigenvalues of A) for the stabilityof the state vector x. See previous lecture.Let’s consider the dynamic behavior when the output equation isconsidered y(t) through the output matrix C.

In the Laplace domain, the relationship between input and outputcan be represented by a transfer function matrix:

P (s) = C · [s I −A]−1 ·B +DG. Ducard c© 4 / 57




SISO Case: Transfer Function

In the SISO case, the transfer function matrix:

P (s) = C · [s I −A]−1 ·B +D

is a scalar rational transfer function which can always be written inthe following form

P (s) =Y (s)

U(s)= k

sn−r + bn−r−1sn−r−1 + . . .+ b1s+ b0

sn + an−1sn−1 + an−2sn−2 + . . .+ a2s2 + a1s+ a0





SISO Case: Transfer Function

P (s) =Y (s)

U(s)= k

sn−r + bn−r−1sn−r−1 + . . .+ b1s+ b0


Discussions:

The order of the highest power of s is n.

Input gain: k

The relative degree r:

difference between highest power of s at denominator and thehighest power of s at numerator.r plays an important role in the discussion of system zeros.

A dynamic system can possess - not only poles - but also zeros.Question: What is the influence of the zeros?





There is an equivalence between the transfer function

P (s) =Y (s)

U(s)= k

sn−r + bn−r−1sn−r−1 + . . .+ b1s+ b0


and its state-space representation (controller canonical form).

d

dtx(t) =

0 1 0 . . . 00 0 1 . . . 0· · · · ·0 0 0 . . . 1

−a0 −a1 −a2 . . . −an−1

x(t) +

00·0k

u(t)

y(t) = [ b0 . . . bn−r−1 1 0 . . . 0 ] x(t) = Cx(t)

The state vector is x(t) =[x(t) x(t) x(t) . . . x(n)(t)

]

Controller canonical form with gain k (min. number of parameters)

The terms involved in the numerator are those of the C outputvector (“transmission zeros”).





Zero Dynamics - Definition

The Zero Dynamics of a system:

corresponds to its behavior for those special

non-zero inputs u∗(t)

and initial conditions x∗

for which its output y(t) is identical to zero for a finite interval.

1 Study of the influence of the zeroson the dynamic properties of the system.

2 Study of the “internal dynamics”:analyze the stability of the system states, which are not directlycontrolled by the input u(t).





Zero Dynamics - Problem

In all the reference-tracking control problems, the controller triesto force the error to zero.

ǫ = yref − y

If yref = 0, then y(t) is to be zero for all times, ⇒ all itsderivatives are to be zero as well.

If a plant has internal dynamics, which are unstable, but not visibleat the system’s output, problems are to be expected.

Let’s see the form of the derivatives.





The relative degree r

is the number of differentiations needed to have the input u(t)explicitly appear in the “output” y(r)(t)

y(t) = Cx(t),

y(t) = Cx(t) = CAx(t) +CBu(t) = CAx(t),

...

y(r−1)(t) =d

dty(r−2)(t) = CAr−1x(t) +CAr−2Bu(t) = CAr−1x(t),

y(r)(t) =d

dty(r−1)(t) = CArx(t) +CAr−1Bu(t) = CArx(t) + ku(t)

where k 6= 0 and r ≤ n.G. Ducard c© 10 / 57




The relative degree r

z1 = y = Cx = [b0x1 + b1x2 + . . .+ bn−r−1xn−r + xn−r+1]

z2 = y = CAx = [b0x2 + b1x3 + . . .+ bn−r−1xn−r+1 + xn−r+2]

...

zr = y(r−1) = CAr−1x = [b0xr + b1xr+1 + . . .+ bn−r−1xn−1 + xn]

y(r) = CArx+ ku = [b0xr+1 + b1xr+2 + . . .+ bn−rxn + xn]

xn is found from the state-space representation:

d

dtx(t) =

0 1 0 . . . 00 0 1 . . . 0· · · · ·0 0 0 . . . 1

−a0 −a1 −a2 . . . −an−1

x(t) +

00·0k

u(t)





Outline








Zero Dynamics: Analysis Formulation

The following coordinate transformation z = Φx is introduced

z1 = y = Cx = [b0x1 + b1x2 + . . .+ bn−r−1xn−r + xn−r+1]

z2 = y = CAx = [b0x2 + b1x3 + . . .+ bn−r−1xn−r+1 + xn−r+2]

...

zr = y(r−1) = CAr−1x = [b0xr + b1xr+1 + . . .+ bn−r−1xn−1 + xn]






The remaining n− r coordinates are chosen such that the transformationΦ is regular and such that their derivatives also do not depend on theinput u. Obviously the choice

zr+1 = x1

zr+2 = x2

. . .

zn = xn−r

satisfies both requirements.To simplify notation the vector z is partitioned into two subvectors

z =

[ξ

η

], ξ =

z1. . .zr

, η =

zr+1

. . .zn






In the new coordinates the system has the form

[ξ

η

]=

0 1 0 . . . 00 0 1 . . . 00 . . . . . . 0 1

− − rT − −

0 . . . . . . . . . 00 . . . . . . . . . 00 . . . . . . . . . 0

− − sT − −

0 . . . . . . . . . 00 . . . . . . . . . 00 . . . . . . . . . 0

− − pT − −

0 1 0 . . . 00 0 1 . . . 00 . . . . . . 0 1

− − qT − −

[ξ

η

]+

0. . .

0k

0. . .

. . .

0

u

and obviously y = z1 = ξ(1).In order to have an identically vanishing output it is therefore necessaryand sufficient to choose the following control and initial conditions

ξ∗ = 0, u∗(t) = −1

ksT η∗(t)

where the initial condition η∗0 6= 0 can be chosen arbitrarily.






The internal states (zero dynamics states) evolve according to theequation

d

dtη(t) =

0 1 0 . . . 00 0 1 . . . 00 . . . . . . 0 1− − qT − −

η∗(t) = Qη∗(t), η∗(0) = η∗0

Minimum phase system

If the matrix Q is asymptotically stable (all eigenvalues withnegative real part) ⇒ then the system is minimum phase.

Equivalence: a minimum phase system, is a system whose zeroshave all negative real parts.

These two definitions are consistent: see definition of the vector qT

qT = [−b0, −b1, . . . , −bn−r−2, −bn−r−1]G. Ducard c© 16 / 57




Unstable Zero dynamics

As soon as there is a zero with positive real part:

system is non-minimum phase,

system has its zero dynamics unstable,

its internal states η can diverge without y(t) being affected.

Consequences:

the input u(t) may not be chosen such that the output y(t) is(almost) zero before the states η associated with the zero dynamicsare (almost) zero.

Feedback control more difficult to design.

This imposes a constraint of the bandwidth of the closed-loopsystem:⇒ significantly slower than the “slowest” non-minimum phase zero.






This equation can be derived using the definition of zn, the originalsystem equation, the definition of z1 (1), again the coordinatetransformation,

zn = xn−r

= xn−r+1

= z1 − b0x1 . . .− bn−r−1xn−r

= z1 − b0zr+1 . . .− bn−r−1zn

= z1 + qT η

Therefore, the eigenvalues of Q coincide with the transmissionzeros of the original system and with the roots of the numerator ofits transfer function.






∫∫∫

∫∫∫ ηn−r−1 η1sT

qT

k

u

ξr ξr−1 ξ1 =

rT

pT





Outline








Zero Dynamics Analysis on a Small SISO System

P (s) =Y (s)

U(s)= k

s2 + b1s+ b0s4 + a3s3 + a2s2 + a1s+ a0

Step 1: Convert the plant’s transfer function into a state-spacecontroller canonical form

number of states n = 4, relative degree r = 2.

d

dtx(t) =

0 1 0 00 0 1 00 0 0 1

−a0 −a1 −a2 −a3

· x(t) +

000k

· u(t)

y(t) = [ b0 b1 1 0 ] · x(t) + [ 0 ] · u(t)G. Ducard c© 21 / 57




Step 2: Coordinate transformation

Relative degree r = 2, therefore

y(t) = b0x1(t) + b1x2(t) + x3(t)

y(t) = b0x2(t) + b1x3(t) + x4(t)

y(t) = −a0x1(t)− a1x2(t) + (b0 − a2)x3(t) + (b1 − a3)x4(t) + k u(t)

The coordinate transformation z = Φ−1 · x has the form

z1 = y = b0x1 + b1x2 + x3

z2 = y = b0x2 + b1x3 + x4

z3 = x1

z4 = x2





Step 3: Find the transformation matrices Φ−1, such that z = Φ−1 · x

and then compute Φ

Φ−1 =

b0 b1 1 0

0 b0 b1 1

1 0 0 0

0 1 0 0

and

Φ =

0 0 1 0

0 0 0 1

1 0 −b0 −b1

−b1 1 b0 b1 b21 − b0

Remark: Notice that, by construction, det(Φ) = det(Φ−1) = 1G. Ducard c© 23 / 57




Step 4: Build a new state-space representation in z =

[ξ

η

]

ξ =

[z1z2

], η =

[z3z4

]

in the new coordinates the system

d

dtz(t) = Φ

−1AΦ z(t) +Φ−1B u(t), y(t) = C Φ z (t)

d

dt

ξ1(t)

ξ2(t)

η1(t)

η2(t)

=

0 1 0 0

r1 r2 s1 s2

0 0 0 1

1 0 −b0 −b1

·

ξ1(t)

ξ2(t)

η1(t)

η2(t)

+

0

k

0

0

· u(t)





The coefficients r1, r2, s1, s2 are listed below

r1 = b0 − a2 − b1(b1 − a3)

r2 = b1 − a3

s1 = b0 b1(b1 − a3)− a0 − b0(b0 − a2)

s2 = (b1 − a3)(b21 − b0)− a1 − (b0 − a2)b1





∫∫∫∫ η2 η1 s1

s2

r1

r2

−b0

−b1

k

u

ξ1ξ2

Figure: System structure of the example’s zero dynamics.





Step 5: Study the submatrix Q of A = Φ−1AΦ corresponding to the

zero-dynamics vector η

Choosing the following initial conditions ξ∗1(0) = ξ∗2(0) = 0 and controlsignal u∗(t) = − 1

k[s1η

∗

1(t) + s2η∗

2(t)] yields a zero output y(t) = 0 for allt ≥ 0. The initial conditions η∗1(0) 6= 0 and η∗2(0) 6= 0 may be chosenarbitrarily.

The trajectories of state variables η1(t) and η2(t), in this case, aredefined by the equations

d

dtη∗(t) =

[0 1

−b0 −b1

]· η∗(t) = Q · η∗(t)





Step 6: Conclude on the conditions to have Q asymptotically stable

d

dtη∗(t) =

[0 1

−b0 −b1

]· η∗(t) = Q · η∗(t)




DefinitionsEquilibrium Sets

Outline








Lyapunov Stability

Nonlinear differential equation

d

dtx(t) = f(x(t), u(t), t), x(t0) = x0 6= 0 (1)

Equilibrium xe

is such that f(xe, t) = 0, ∀ t.





Stability of Equilibrium Points

The point xe is Uniformly Lyapunov Stable (ULS)

if for each scalar R > 0, there is a scalar r(R) > 0,such that if the initial condition x0 satisfies

||x0|| < r, (2)

the corresponding solution of (1) will satisfy the condition

||x(t)|| < R (3)

for all times t greater than t0.

Remark: The notion “uniformly” indicates that neither r not Rdepend on t0.





Lyapunov Stability

For isolated equilibrium points: Lyapunov stability is

the most useful stability concept.

always connected to an equilibrium point xe of a system.

The same point xe is asymptotically stable

if it is ULS and attractive, i.e.

limt→∞

x(t) = xe





Lyapunov Stability

A system is exponentially asymptotically stable

if there exist constant scalars a > 0 and b > 0 such that

||x(t)|| ≤ a · e−b·(t−t0) · ||x0||

Remarks:

1 In general, only an exponentially asymptotically stableequilibrium is acceptable for technical applications,

2 ⇒ this form of stability is “robust” with respect to modelingerrors.





Outline








Equilibrium Sets

For linear systems an equilibrium set can be:

one isolated point;

entire subspaces;

periodic orbits (same frequency but arbitrary amplitude).





Equilibrium Sets

Nonlinear systems can have (infinitely) many isolated equilibriumpoints.An equilibrium point can:

have a finite region of attraction;

be non-exponentially asymptotically stable; and

be unstable ⇒ the state of the system can “escape toinfinity” in finite time.




First-order SystemsSecond-order SystemsLyapunov PrincipleLyapunov Theory

Outline








Stability of Nonlinear First-Order Systems

First-order nonlinear systems are easy

d

dtx(t) = f(x(t), u(t)), x, u ∈ R

because they can be separated for u(t) = 0

∫dx

f(x)=

∫dt = t+ c

The constant c is needed to satisfy the initial condition x(t0) = x0.





Example: Non-Exponential Asymptotic Stability

System

d

dtx(t) = −x3(t) + u(t), x(0) = x0

Explicit solution for u(t) = 0∫

dx

−x3=

∫dt ⇒

1

2x2= t+ c

and, thereforex(t) = x0 · (2 · t · x

20 + 1)−1/2

⇒ Therefore, the system is asymptotically stable.





Example: Non-Exponential Asymptotic Stability

But the solution approaches the equilibrium slower thanexponentially.

To see this use the inequality

||x(t)|| = ||x0|| · (2 · t · x20 + 1)−1/2 ≤ a · e−b t · ||x0||

which yields1 ≤ a · e−b t · (2 · t · x20 + 1)1/2

For lim t → ∞, the right-hand side of this inequality tends to 0 forall possible 0 < a, b < ∞.Contradiction! ⇒ No exponential asymptotic stability.





Outline








Stability of Second-Order Systems

Two variables involved: x1 and x2

ddtx1(t) = f1(x1, x2), x1(0) = x1,0

ddtx2(t) = f2(x1, x2), x2(0) = x2,0

(4)





Taylor expansion

ddtx1(t) = a11x1(t) + a12x2(t) + f1(x1, x2), x1(0) = x1,0

ddtx2(t) = a21x1(t) + a22x2(t) + f2(x1, x2), x2(0) = x2,0

ai,j =∂fi∂xj

, i, j = 1, 2 lim||x||→0

fi(x1, x2)

||x||= 0

A2×2 =

[a11 a12

a21 a22

]

Linear system is given by:

d

dtδx(t) = A · δx(t), δx(t) = [δx1(t), δx2(t)]

T (5)





Outline








Lyapunov Principle – 2nd-Order SystemsExcluding the case where the matrix A has two eigenvalues withzero real part, the local behavior of

the nonlinear system (4)

and the behavior of the linear system (5)

are topologically equivalent (i.e., they have the same “geometriccharacteristics”).

eigenvalues linearized system nonlinear system

λ1 ∈ C−, λ2 ∈ C− Stable Focus Stable Focusλ1 ∈ R−, λ2 ∈ R− Stable Node Stable Nodeλ1 ∈ R+, λ2 ∈ R− Saddle Saddleλ1 ∈ R+, λ2 ∈ R+ Unstable Node Unstable Nodeλ1 ∈ C+, λ2 ∈ C+ Unstable Focus Unstable FocusRe(λ1,2) = 0 Center ? ? ?





-1-1-1

-1-1-1

000

000

111

111

-1-1-1

-1-1-1

-0.5-0.5-0.5

-0.5-0.5-0.5

000

000

0.50.50.5

0.50.50.5

111

111

x1x1x1

x1x1x1

x2

x2

x2

x2

x2

x2

Stable Focus Stable Node Saddle

Unstable Node Unstable Focus Center





Example Critical Nonlinear System

ddtx1 = −x1 + x2

ddtx2 = x32

System has only one isolated equilibrium at xe,1 = xe,2 = 0.

Linearization has one eigenvalue at −1 and one at 0.

Observations:

The solution of the linear system is stable (although notasymptotically stable),while the nonlinear system is unstable (it has even finiteescape times).

x2(t) =x2,0√

1− 2 t x22,0

With lim t → 1/(2x22,0) ⇒ x2(t) → “escapes to infinity.”G. Ducard c© 47 / 57




Lyapunov Principle – General Systems

1 The Lyapunov Principle is valid for all finite-order systems: aslong as the linearized system has no eigenvalues on theimaginary axis,

2 The local stability properties of an arbitrary-order nonlinearsystem are fully understood once the eigenvalues of thelinearization are known.

3 Particularly, if the linearization of a nonlinear system aroundan isolated equilibrium point xe is asymptotically stable(unstable resp.), then this equilibrium is an asymptoticallystable (unstable resp.) equilibrium of the nonlinear system aswell.





Lyapunov Principle – General Systems

Remarks:

1 Nothing is said about the “large signal behavior.”

2 In the critical case (Re{λi} = 0), the nonlinear system’sbehavior is not defined by its first-order approximation, but bythe higher-order terms f(x).





Outline








Lyapunov Theory

Local stability properties of equilibrium x = 0 of system

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

fully described by

A =∂f

∂x

∣∣∣∣x=0

provided A has no eigenvalues with zero real part.If this is not true or if one is interested in non-local results for moregeneral systems, then use Lyapunov’s direct method.





Definitions

A scalar function α(p) with α : R+ → R+ is a nondecreasingfunction if α(0) = 0 and α(p) ≥ α(q) ∀ p > q.

A function V : Rn+1 → R is a candidate global Lyapunov functionif

the function is strictly positive, i.e., V (x, t) > 0 ∀ x 6= 0, ∀ tand V (0) = 0; and

there are two nondecreasing functions α and β that satisfy theinequalities β(||x||) ≤ V (x, t) ≤ α(||x||).

Obviously, if these conditions are met only in a neighborhood ofthe equilibrium point x = 0 only local assertions can be made.





Theorem 1

The system

x(t) = f(x(t), t), x(t0) = x0 6= 0

is globally/locally stable in the sense of Lyapunovif there is a global/local Lyapunov function candidate V (x, t) forwhich the following inequality holds true ∀x(t) 6= 0 and ∀ t

V (x(t), t) =∂V (x, t)

∂t+

∂V (x, t)

∂xf(x(t), t) ≤ 0

Theorem 2

The system (6) is globally/locally asymptotically stableif there is a global/local Lyapunov function candidate V (x, t) suchthat −V (x(t), t) satisfies all conditions of a global/local Lyapunovfunction candidate.





In general, it is very difficult to find a suitable function V (x, t)Hint: use physical insight, Lyapunov functions can be seen asgeneralized energy functions.

Linear systems are simple

V (x) = xTPx

where P = P T > 0 is the solution of the Lyapunov equation

PA+ATP = −Q

For arbitrary Q = QT > 0, a solution to this equation exists iff Ais Hurwitz matrix.

Remark: Lyapunov theorems provide sufficient but not necessaryconditions. Many extension proposed (LaSalle, Hahn, etc.).





Example

Systemddtx1 = x1(x

21 + x22 − 1)− x2

ddtx2 = x1 + x2(x

21 + x22 − 1)

(7)

One isolated equilibrium at x1 = x2 = 0. Linearizing the systemaround this equilibrium yields

A =

[−1 −1

1 −1

]

Eigenvalues are λ1,2 = −1± j, system is locally asymptoticallystable (Lyapunov principle).





Example

System has a finite region of attraction. Use a Lyapunov analysisto obtain a conservative estimation of the region of attraction.

Candidate Lyapunov function

V (x) = x21 + x22

Time derivative of V along a trajectory of (7)

V (t) = 2(x21 + x22)(x21 + x22 − 1)

Therefore, at least the region ||x||2 = x21 + x22 < 1 must be part ofthe region of attraction.





x1

x2

Vector field 0.1 · f(x), x=end point

-1.5 -1 -0.5 0 0.5 1 1.5

-1.5

-1

-0.5

0

0.5

1

1.5


modeling and analysis of dynamic systems - eth z and analysis of dynamic systems dr. guillaume...

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