Modeling and Analysis of Dynamic Systems · Solution of Linear ODE Stability of Linear Systems Remark:(from script p.89) For those students who followed the courses “Control Systems

Modeling and Analysis of Dynamic Systems

Dr. Guillaume Ducard

Fall 2017

Institute for Dynamic Systems and Control

ETH Zurich, Switzerland

G. Ducard c© 1 / 62

Outline

1 Lecture 11: Analysis of Linear SystemsNormalizationLinearizationSolution of Linear ODEStability of Linear Systems

2 Lecture 11: Reachability and ObservabilityReachability ConditionsObservability ConditionsExample: Ball on Wheel

3 Lecture 11: Balanced Realization and Order ReductionIntroduction / ExampleGramian Matrices


Lecture 11: Analysis of Linear SystemsLecture 11: Reachability and Observability

Lecture 11: Balanced Realization and Order Reduction

NormalizationLinearizationSolution of Linear ODEStability of Linear Systems

Outline








Remark:(from script p.89) For those students who followed thecourses “Control Systems I and II,” most of the material presentedin this chapter will be a repetition. However, since several pointsare discussed in more detail and since additional material isintroduced, it is not recommended to skip these classes.

Motivations

Once the modeling of a system is done, we are interested inanalysing this system for:

stability

controllability, observability

performance

Methodology?





Introduction

Models are usually nonlinear with physical (“non-normalized”):

states x, inputs u,

and outputs y.

according to dynamic and output equations:

d

dtx(t) = f(x(t), u(t), t), y(t) = g(x(t), u(t), t)

However, difficulties arise for systems which:

1 are nonlinear: difficult to analyze

2 have non-normalized variables:

are prone to numerical problemsthe state variables are hard to compare to one another.

Solution for analysis: Normalize and linearize around a chosen setpoint.G. Ducard c© 5 / 62




Analysis of Linear Systems

Questions

1 Assuming known input signals and initial conditions:⇒ what output signals are to be expected?

2 Which points in the state space may be:

reached by appropriate input signals?

observed by analyzing the associated output signals?

3 What parts of a dynamic system are relevant for the input/outputcharacteristics?





NormalizationConsider the scaling factors: xi,0, uj,0, yk,0 such that

xi(t) =xi(t)

xi,0, uj(t) =

uj(t)

uj,0, yk(t) =

yk(t)

yk,0,

The new normalized variables xi(t), uj(t) and yk(t) will have nophysical units, and are usually close to 1.Since xi,0 is constant

d

dtxi(t) = xi,0

d

dtxi(t)

→ no change in the dynamics (only scaled).The normalization may be expressed in vector notation as

x = T · x, T = diag{x1,0, . . . , xn,0}

“Similarity transformation” matrix does not change the systemscharacteristics (stability, input-output behavior, etc.).





Outline








Linearization

The system after normalization has the form

1 Dynamics equation

d

dtx(t) = x(t) = f0(x(t),u(t), t),

2 Output equation

y(t) = g0(x(t),u(t), t)

with x(t) ∈ Rn,u(t) ∈ R

m,y(t) ∈ Rp,

where f0 and g0 nonlinear normalized functions.

Remark: here we can use, the true variable or the normalized. In the

following we will simply use the notations x, y, u.G. Ducard c© 9 / 62




System dynamics in vector form

x(t) = f(x, u, t)

x1x2...xn

=

f1(x, u)f2(x, u)

...fn(x, u)





Notion of Neighborhood

Linearization behavior in a “small” neighborhood

1 operating point {xo, uo}

Br := {x ∈ Rn | ||x− xo||

2 + ||u− uo||2 ≤ r}

2 equilibrium point {xe, ue}

Br := {x ∈ Rn | ||x− xe||

2 + ||u− ue||2 ≤ r}

around a chosen equilibrium point {xe, ue}, f0(xe, ue, t) = 0.





Geometric Interpretation

In the 1-D case, the dynamics of the system are x = f(x, u)

x

u

x&

0 0,x u

0 0( , )f x u

0 0,x u

f

u

¶

¶

0 0,x u

f

x

¶

¶






x

u

x&

0 0,x u

0 0( , )f x u

0 0,x u

f

u

¶

¶

0 0,x u

f

x

¶

¶

x = f(x, u) = f(x0, u0)+∂f

∂x|x0,u0 (x− x0)︸︷︷︸

δx

+∂f

∂u|x0,u0 (u− u0)︸︷︷︸

δu

+1

2

∂2f

∂x2|x0,u0 (x− x0)

2 + . . .G. Ducard c© 13 / 62





x

u

x&

0 0,x u

0 0( , )f x u

0 0,x u

f

u

¶

¶

0 0,x u

f

x

¶

¶

˙δx = f(x, u)− f(x0, u0) ≈∂f

∂x|x0,u0 (x− x0)︸︷︷︸

δx

+∂f

∂u|x0,u0 (u− u0)︸︷︷︸

δu





Deviation Around Equilibrium Point

x(t) = x(t)− xe,

u(t) = u(t)− xe,

y(t) = y(t)− g0(xe, ue, t)

such that in the new coordinates the ODEs are

d

dtx(t) = f0(x(t), u(t), t),

y(t) = g0(x(t), u(t), t)

with f0(0, 0, t) = 0.G. Ducard c© 15 / 62




Only small deviations from the setpoint are considered. Thefollowing new variables are introduced

x(t) = xe + δx(t) with |δx| ≪ |xe|,

u(t) = ue + δu(t) with |δu| ≪ |ue|,

y(t) = ye + δy(t) with |δy| ≪ |ye|

Taylor series neglecting all terms of second and higher order yields

d

dtδx(t) =

∂f0∂x

|xe,ueδx(t) +∂f0∂u

|xe,ueδu(t)

and

δy(t) =∂g0∂x

|xe,ueδx(t) +∂g0∂u

|xe,ueδu(t)





Notationddtx(t) = Ax(t) +Bu(t)

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

System can be described using (infinitely) many other coordinatesystems (similarity transformations)

x = T x, T ∈ Rn×n, det(T ) 6= 0

The columns of T are the unit vectors of the new coordinate frameexpressed in the old coordinate system.In the new coordinates

ddt x(t) = T−1ATx(t) + T−1Bu(t) = Ax(t) + Bu(t)

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





The input-output behavior (i.e., the transfer function) is notaffected by the specific choice of coordinates

P (s) = CT [sI − T−1AT ]−1T−1B +D

= C[sI −A]−1B +D = P (s)

Fundamental system properties (stability, controllability, etc.) areindependent of the coordinates chosen. However, if the systemsODEs are derived by physical arguments there are reasons forsticking to these “natural” coordinates.





Outline








Solution of Linear Ordinary Differential Equations (ODE)

Objective:prediction of the state x and the output y of a linear systemdescribed by its matrices {A,B,C,D},

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

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

with

initial conditions x(0),

control signal u,

are known.





Matrix exponential

eAt = I +1

1!At+

1

2!(At)2 +

1

3!(At)3 + · · ·+

1

n!(At)n + · · ·

Main property

deAt

dt= A eAt = eAtA (commute)

Properties

1 In general, eA · eB 6= eA+B.

2 Only if A and B commute (AB = BA), eA · eB = eA+B.

3 Accordingly, eAte−At = eA(t−t) = e0 = I ⇒ (eAt)−1 = e−At.





Solution to the initial value problem

x(t) = A x(t) +B u(t), x(0) = x0

Multiplying (1) on the left-hand side by the term e−At yields

e−Atx(t) = e−AtAx(t) + e−AtBu(t) ,

⇐⇒ e−Atx(t)− e−AtAx(t) = e−AtBu(t) . (1)

The left-hand side of (1) is equal to d(e−Atx(t))dt ,indeed

d(e−Atx(t))

dt= −Ae−Atx(t) + e−Atx(t) (2)

therefore (1) is rewritten as

d(e−Atx(t))

dt= e−AtBu(t) . (3)





Integrating the above equation on the time interval [0, t] yields

∫ t

0

d(e−Aτx(τ))

dτdτ =

∫ t

0e−AσBu(σ)dσ,

e−Atx(t)− x(0) =

∫ t

0e−AσBu(σ)dσ ,

e−Atx(t) = x(0) +

∫ t

0e−AσBu(σ)dσ ,

x(t) = eAtx(0) +

∫ t

0eA(t−σ)Bu(σ)dσ .





x(t) = eAtx(0) +

∫ t

0eA(t−σ)Bu(σ)dσ

The continuous-time transition matrix is defined as follows:

Φ(t) = eAt = I +At+(At)2

2!+ ...+

(At)n

n!+ ...

and finally the continuous solution for x(t) is written as

x(t) = Φ(t)x(0) +

∫ t

0Φ(t− σ)B u(σ)dσ

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





Outline








Introduction

Stability = most important concept in analysis of dynamic systems

Stability is always connected to a “metric,” i.e., the size of vectorsis important.⇒ The symbol ||x|| will be used to denote this operation, andany norm will be acceptable.

For instance the Euclidean length:

x ∈ Rn, ||x||2 :=

n∑

i=1

x2i





Lyapunov Stability

Linear and time-invariant systems:

d

dtx(t) = A · x(t), x(0) = x0, 0 < ||x0|| <∞ (4)

The input u(t) = 0 for stability analysis, only A is important.

For stability: three possible cases. The system (4) is

1 asymptotically stable if limt→∞

||x(t)|| = 0;

2 stable if ||x(t)|| <∞ ∀ t ∈ [0,∞]; and

3 unstable if limt→∞

||x(t)|| = ∞,

where the norm is the Euclidean length (defined before).Remark: A general definition for nonlinear systems will be seen later.





For linear systems, the stability of the system

d

dtx(t) = Ax(t), x(0) = x0 6= 0

is tightly related to the eigenvalues of the matrix A.

Recall Matrix A characterizes linear mapping Rn → R

n

y = A x, x, y ∈ Rn, A ∈ R

n×n

Eigenvalue - eigenvector

An eigenvector vi of A is a vector which is mapped onto itself,through the scaling parameter λi called the eigenvalue of vi,according to:

A vi = λi vi

Remark: Even for real matrices A ∈ Rn×n, λi and vi can be complex

(always complex conjugate pairs).G. Ducard c© 28 / 62




If n linearly independent eigenvectors exist, then the matrix

T = [v1 . . . vn]

which – by definition – diagonalizes A, i.e.,

A T = T Λ ⇒ T−1 A T = Λ

where

Λ =

λ1 0 . . . 0...

......

...0 . . . 0 λn

Remark:If all eigenvalues are distinct, i.e., λi 6= λj for all i 6= j, then nindependent eigenvectors always exist.





If there are multiple (same) eigenvalues ⇒ situation more complex.

Two scalars are important:

multiplicity of eigenvalue λi: ri

rank loss of the matrix [λiI −A]: ρi, associated with theeigenvalue λi.

In the general case, three distinct situations arise:

1 ρi = 1, ⇒ only 1 eigenvector exists for the (ri > 1) identicaleigenvalues λi,

2 ρi < ri, ⇒ there are less than ri independent eigenvectors,

3 ρi = ri, ⇒ exactly r independent eigenvectors exist.

The third case is similar to the regular case.

Remark: When all eigenvalues are distinct: ri = ρi = 1 for alli = 1, . . . , n and the matrix A is “diagonal.”





Jordan Form

In the first and in the second case, A is not diagonalizable but canbe brought to what is known as a “Jordan Form”

Ji =

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

...... . . . . . . . . . . . .

...0 . . . . . . 0 λi 10 . . . . . . . . . 0 λi









Stability of Linear Systems

If the matrix Λ is diagonal three situations can be encountered:

1 all eigenvalues have negative real parts σi < 0 ⇒ system isasymptotically stable;

2 some eigenvalues have zero real parts σi = 0, but none has apositive real part⇒ system is Lyapunov stable,

3 At least one eigenvalue has a positive real part σi > 0 ⇒system is unstable.

Remark: For cyclic or mixed matrices A, this is no longer true. In fact,

systems with multiple eigenvalues on the imaginary axis can be stable

only if all Jordan blocks associated to multiple eigenvalues with real part

zero are diagonal.





Example: “series double integrator structure”

d

dt

[x1x2

]=

[0 01 0

]·

[x1x2

]

The continuous-time transition matrix is defined as follows:

Φ(t) = eAt = I +At+(At)2

2!+ ...+

(At)n

n!+ ...

and finally the continuous solution for x(t) is written as

x(t) = Φ(t)x(0) +

∫ t

0Φ(t− σ)B u(σ)dσ

Transition matrix (stability analysis u = 0).

x(t) =

{

I +1

1!A t +

1

2!A

2t2+

1

3!A

3t3+ · · ·

}

x(0) =

1 0

t 1

x(0)




Reachability ConditionsObservability ConditionsExample: Ball on Wheel

Outline








Reachability of Linear Systems

x(t) = eAtx(0) +

∫ t

0eA(t−σ)Bu(σ)dσ

Investigate all states that can be reached at time τ starting atx(0) = 0

x(τ) = eAτ∫ τ

0e−AσBu(σ) dσ

Matrix eAτ regular for all τ <∞, therefore use x∗(τ) = e−Aτx(τ)

x∗(τ) =

∫ τ

0e−AσBu(σ) dσ

Using the definition of the matrix exponential

x∗(τ) =

∫ τ

0

{I −

1

1!Aσ +

1

2!(Aσ)2 −

1

3!(Aσ)3 + · · ·

}Bu(σ) dσ





and from that

x∗(τ) = B

∫ τ

0u(σ) dσ−AB

∫ τ

0

1

1!σu(σ) dσ+A2B

∫ τ

0

1

2!σ2u(σ) dσ+· ·

Let’s define the integral basis as:

vi =

∫ τ

0

1

i!σiu(σ) dσ, i = 0, 1, . . .

The reachable states x∗ are then defined by the linear equation

x∗(τ) =[B AB A2B A3B · · ·

]

v1

v2

v3

v4...

= Rv





Define

Rn =[B AB A2B A3B · · · An−1B

]

Therefore, condition rank(Rn) = n guarantees that for arbitraryx(0) = x∗ 6= 0 a suitable control signal u∗(x∗) exists. The system is saidto be reachable.

Read page 105 in the script, last 2 paragraphs.





Outline








Observability of Linear SystemsIs it possible to reconstruct x(0) using the output signal y(t) only?Input u(t) = 0 for all times t, therefore

y(t) = Cx(t) ddty(t) = CAx(t) d2

dt2y(t) = CA2x(t) etc.

Evaluating this equation for t = 0 yields the linear equation

y(t)

ddty(t)

ddt2

y(t)...

(t=0)

= w(0) =

C

CA

CA2

...

x(0) = Ox(0)

By “measuring” the vector w(0), the unknown initial state vectorx(0) can be found uniquely if O is invertible ⇐⇒ det(O) 6= 0,⇐⇒ O is full rank. G. Ducard c© 40 / 62




Cayley-Hamilton theorem

On =

C

CA...

CAn−1

If rank(On) = n, the system is observable.





Outline








Ball on Wheel

y(t)

u(t)

r,m, ϑχ(t)

R,Θ

ϕ(t)

ψ(t)

mg

wheel: mass moment of inertia around c.o.g. is Θ, radius R,

ball: mass m, mass moment of inertia around c.o.g. ϑ, radius rG. Ducard c© 43 / 62




Ball-on-WheelLinearizing system at equilibrium ψ = ψ = χ = χ = 0 yields

δx(t) = A · δx(t) + b · δu(t), δy(t) = c · δx(t)

with

δx(t) =

δψ(t)

δψ(t)

δχ(t)

δχ(t)

A =

0 1 0 0

0 0 a1 0

0 0 0 1

0 0 a2 0

b =

0

b1

0

b2

cT =

0

0

c1

0





Parameters defined by

a1 = mgRϑΓ , a2 =

mg(R2ϑ+ r2Θ)(R + r)Γ

b1 = mr2 + ϑΓ , b2 = Rϑ

(R+ r)Γ, c1 = R+ r

To simplify the notation

Θ = 5ϑ, R = 3 r, ϑ = r2m, g = 10m2/s, r = 0.2m, m = 1kg

Controllability matrix

Rn =

0 50

190 75

76

50

190 75

760

0 5625

7220 13125

1444

5625

7220 13125

14440

,

det(Rn) = 263.444 ( 6= 0), not rank deficient, i.e., the system is completely controllable.G. Ducard c© 45 / 62




Observability matrix

On =

0 0 45 0

0 0 0 45

0 0 14019 0

0 0 0 14019

det(On) = 0 (first 2 columns are equal to 0).

For this special case it is easy to see that the two state variables ψand ψ cannot be estimated using the output signal y as the onlyinformation.





The linearized system is unstable. Its eigenvalues are

λ1 = λ2 = 0, λ3 = −λ4 = 3.03488

According to the Lyapunov principle, the nonlinear system must beunstable as well.However, this does not mean that the system cannot meet thedesired objective, i.e., to balance the ball at the equilibriumposition!

The question is now, can the system be stabilized by anappropriate feedback action?⇒ As it will be shown below, this is not possible.





Transfer function from u→ y

P (s) =15

19 s2 − 175

The two observable states are those whose dynamics aredescribed by the two eigenvalues λ3,4.

The dynamics described by the two eigenvalues λ1,2, cannotbe influenced by feedback action. This subsystem has adouble eigenvalue at zero with rank loss only 1.

Therefore, this subsystem is unstable (intuition confirmed)!For simulations use simple feedback (lead-lag) controller

C(s) =k2 s+ k1τ s+ 1

=2.2417 s + 17.05

0.01 s + 1





time t(s)

closed-loop step response y(t); solid=nolinear, dashed=linear

0 1 2 3 4 5 6 7 8 9 100

0.05

0.1

0.15

0.2

0.25

Figure: Reference step responses of the closed-loop “ball-on-wheel”system; all initial conditions equal to zero.





time t(s)

closed-loop step response, hidden variables; solid=ψ, dashed=dψ/dt

0 1 2 3 4 5 6 7 8 9 10-160

-140

-120

-100

-80

-60

-40

-20

0

20

Figure: Behavior of the “hidden state variables” for the same referencestep as illustrated in previous Figure.





time t(s)

solid=χ, dashed=dχ/dt

time t(s)

solid=ψ, dashed=dψ/dt

0 2 4 60 2 4 6-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

0

1

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

Figure: Behavior of the closed-loop system for an initial condition ofχ(0) = π/12 rad, χ(0) = π/12 rad/s and reference values χr = χr = 0.




Introduction / ExampleGramian Matrices

Outline








Balanced Realization and Order ReductionWith the matrices R and O one obtains “yes/no” answers. Whatabout quantitative criteria?Is a specific subspace “well or hardly” reachable?Closely linked to that question is the notion of system orderreduction.

System order reduction

consists in :

1 identifying those states that do not influence significantly theinput/output behavior of the system,

2 finding a smart way eliminate those “not so important” states.





Outline








Gramian Matrices

Controllability Gramian

WR =

∫∞

0eAσBBT eA

T σ dσ

The closer WR is to a singular matrix (det close to 0), the lesscontrollable the corresponding system will be.

Observability Gramian

WO =

∫∞

0eA

T σCT C eAσ dσ

The closer WO is to a singular matrix, the less observable thecorresponding system will be.





Gramian Matrices

The computation of the two Gramians appears to be difficult. Thefollowing result, however, simplifies those steps considerably.

If the system is Hurwitz, the two Gramian matrices are the

solutions of the two Lyapunov equations

AWR +WRAT = −BBT

andATWO +WOA = −CTC





Bad idea:

simply delete those system parts that do not significantlycontribute to the controllability or observability Gramian matrices(poor controllability compensated by excellent observability).

Better idea:

look for a coordinate transformation T · xb = x, which transformsthe original system into a system whose controllability andobservability Gramians are equal and diagonal, i.e.,

WR,b =WO,b = diag(σi), i = 1, . . . , n

Proper normalization is very important for this procedure to workcorrectly!





Assume: the last ν elements σj with j = n− ν + 1, . . . , n aresubstantially smaller than the other first n− ν elements σi withi = 1, . . . n− ν. Then the contribution of the last ν balancedmodes to the system’s IO behavior may be neglected.

First step: system partitioning

ddt

[x1(t)x2(t)

]=

[A11 A1,2

A2,1 A2,2

] [x1(t)x2(t)

]+

[B1

B2

]u(t)

y(t) = [C1 C2]

[x1(t)x2(t)

]+Du(t)

where x1 ∈ Rn−ν and x2 ∈ R

ν .





Since the contribution of the last ν states is small, the system canbe reduced by simply omitting the corresponding elements, i.e.,

ddtx1(t) = A11x1(t) +B1u(t)

y(t) = C1x1(t) +Du(t)

Typically, this will yield good agreement in the frequency domainbut, in general the DC gains of the original system and of thereduced-order system will be different.





If this is to be avoided, a “singular perturbation” approach isbetter suited where:

1 the dynamics of the last ν states is neglected,2 but not their DC contributions.

The details of that approach are

d

dtx2(t) = 0 → x2(t) = −A−1

2,2[A2,1x1(t) +B2u(t)]

and

ddtx1(t) = [A11 −A1,2A

−12,2A2,1]x1(t) + [B1−A1,2A

−12,2B2]u(t)

y(t) = [C1 − C2A−12,2A2,1]x1(t) + [D − C2A

−12,2B2]u(t)

Always feasible if the original system was asymptotically stable (notrestrictive, otherwise Gramians do not exist).The next figure shows the step responses for increasing ν, red = originaland blue = reduced-order outputs, system order n = 7. G. Ducard c© 60 / 62




0 5 10 15 20 0 5 10 15 20

-0.6

-0.6-0.6

-0.4

-0.4-0.4

-0.2

-0.2-0.2

0

00

0.2

0.20.2

-0.5

0

0.5

00 55 1010 1515 2020

ν = 3 ν = 4

ν = 5 ν = 6

Time (sec)Time (sec)

Time (sec)Time (sec)

Amplitude

Amplitude

Amplitude

Amplitude

Step ResponseStep Response

Step ResponseStep Response





Next lecture + Upcoming Exercise

Next lecture

Case Study

Next exercises:

LinerarizingSystem balancing and order reduction


Documents

Modeling and Analysis of Dynamic Systems · Solution of Linear ODE Stability of Linear Systems Remark:(from script p.89) For those students who followed the courses “Control Systems