Control Systems I - ETH Z · Control Systems I Lecture 4: Diagonalization, Modal Analysis, Intro to...

Control Systems ILecture 4: Diagonalization, Modal Analysis, Intro to Feedback

Readings:

Emilio Frazzoli

Institute for Dynamic Systems and ControlD-MAVT

ETH Zurich

October 13, 2017

E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/2017 1 / 26

Tentative schedule

# Date Topic

1 Sept. 22 Introduction, Signals and Systems2 Sept. 29 Modeling, Linearization

3 Oct. 6 Analysis 1: Time response, Stability4 Oct. 13 Analysis 2: Diagonalization, Modal coordi-

nates.5 Oct. 20 Transfer functions 1: Definition and properties6 Oct. 27 Transfer functions 2: Poles and Zeros7 Nov. 3 Analysis of feedback systems: internal stability,

root locus8 Nov. 10 Frequency response9 Nov. 17 Analysis of feedback systems 2: the Nyquist

condition

10 Nov. 24 Specifications for feedback systems11 Dec. 1 Loop Shaping12 Dec. 8 PID control13 Dec. 15 Implementation issues14 Dec. 22 Robustness


Recap of the previous lecture

LTI system:

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

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

Time response:

x(t) = eAtx0 +

∫ t

0

eA(t−τ)Bu(τ) dτ,

y(t) = CeAtx0 + C

∫ t

0

eA(t−τ)Bu(τ) dτ + Du(t).

Easy to compute if the matrix A is diagonal, in which case:

xi (t) = eλi tx0,i +

∫ t

0

eλi (t−τ)biu(τ) dτ,

y(t) =n∑

i=1

cixi (t) + Du(t).

The eigenvalues of A, λi , can be real or complex-conjugate, giving rise to simpleexponentials, or oscillations with exponentially changing magnitude, respectively.

Asymptotically stable if Re(λi ) < 0 for all i .


Today’s learning objectives

After today’s lecture, you should be able to:

Diagonalize a matrix using similarity transformations.

Describe the behavior of an LTI system in modal coordinates.

Understand concepts like controllability and observability.

Understand the basic effects of feedback control on the closed-loop dynamics.


Similarity Transformations

The choice of a state-space model for a given system is not unique.

For example, let T be an invertible matrix, and consider a coordinatetranspormation x = Tx , i.e., x = T−1x . This is called a similaritytransformation.

The standard state-space model can be written as{x = Ax + Bu,y = Cx + Du.

⇒{

T ˙x = ATx + Bu,y = CT x + Du.

i.e.,

˙x = (T−1AT )x + (T−1B)u = Ax + Bu

y = (CT )x + Du = C x + Du.

You can check that the time response is exactly the same for the two models(A,B,C ,D) and (A, B, C , D)!


Diagonalization

Let λi , vi be respectively an eigenvalue and an eigenvector of A, i.e.,

Avi = λivi .

Now assume we have n (=dim. of x and A) independent eigenvectors; thenwe can assemble the eigenvectors into an invertible matrix V whose columnsare the eigenvectors vi . Then

AV = A

v1 v2 . . . vn

=

λ1v1 λ2v2 . . . λnvn

= VΛ.

In other words, if a square matrix A has a full set of independenteigenvectors, then it is diagonalizable (and vice-versa), with the similaritytransformation given by a matrix whose columns are the eigenvectors.


Modal decomposition

The entries in the diagonal matrix A = Λ are the eigenvalues λ1, . . . , λn ofthe matrix A.

Since x(t) = eAt x(0), we get that each component of the homogeneoussolution is given by

xi (t) = eλi t xi (0).

Furthermore, if x(0) = vi for some i = 1, . . . , n, then by the definition ofmatrix exponential and eigenvalues/eigenvectors

x(t) = eAtvi = eλi tvi .

If the eigenvectors vi , i = 1, . . . , n form a basis, then we can always expressany initial condition as a linear combination of eigenvectors, i.e.,x(0) = V x(0), with x0 = V−1x0 , and write

x(t) =n∑

i=1

eλi t xi (0) vi .


Modal coordinates

Eigenvalues and eigenvectors of A define the modes of the system; thetransformed coordinates x = Vx are also called the modal coordinates.

The eigenvector vi defines the shape of the i-th mode;

The modal coordinate xi scales the mode (e.g., at the initial condition);

The eigenvalue λi defines how the amplitude of the mode evolves over time.

1 As an exponential eλi tx0 for real λi

2 As an sinusoid with exponentially changing amplitude eσt sin(ωit + φ0)x0 forcomplex-conjugate λi = σi + jωi .

3 As polynomially-scaled versions of the above tpeλi t for repeated λi .

The amplitude of the i mode goes to zero as t increases if and only ifRe(λi ) < 0.


Example: Simple Pendulum

Recall the model for a simple pendulum, assuming no damping (and noinput/output for now):

x =

[0 1−g/l 0

]x

Eigenvalues are solutions of the characteristic equation:

det(λI − A) = det

[λ −1g/l λ

]= λ2 + g/l = 0,

i.e.,

λ1,2 = ±j√

g

l.

Eigenvectors are obtained as solutions of (λI − A)v = 0, e.g.,

v1,2 =

[1

±j√

gl

]


Pendulum mode shape

Assume x(0) = (1, 0), i.e., x(0) = (1/2, 1/2).

Re

Im

after time t:multiply by eλi t

Re

Im

Combining the two modes, we get

x1(t) = cosωt,

x2(t) = −ω sinωt.


Cartesian vs. polar form of complex numbers

Complex numbers can be represented in basically two ways:

Cartesian form: z = a + jb

Polar form: z = |z |e j∠z

In the above formulas, |z | =√a2 + b2, and ∠z = arctan(b/a)

(arctan understood as the four-quadrant version).

The Cartesian form makes addition easy:

(a1 + jb1) + (a2 + jb2) = a1 + a2 + j(b1 + b2)

The polar form makes multiplication easy:

m1ejφ1 ·m2e

jφ2 = (m1 + m2)e j(φ1+φ2)


General evolution for complex-conjugate modes

Re

Im

after time t:multiply by eλi t

Re

Im

Combining the two modes, we get

x1(t) = c1eσt sin(ωt + φ1,

x2(t) = c2eσt sin(ωt + φ2).


Towards feedback control

So far we have looked at how a given system, represented as a state-spacemodel or as a transfer function, behaves given a certain input (and/or initialcondition).

Typically the system behavior may not be satisfactory (e.g., because it isunstable, or too slow, or too fast, or it oscillates too much, etc.), and onemay want to change it. This can only be done by feedback control!

However, we need to understand to what extent we can change the systembehavior. More precisely, for control design, one must also understand

how the control input can affect the state of the system;

how the state of the system affects the output.


Controllability and Observability

An LTI system of the form x = Ax + Bu is said to be controllable if for anygiven initial state x(0) = xc there exists a control signal that takes the stateto the origin x(t) = 0 for some finite time t.

An LTI system of the form x = Ax + Bu, y = Cx + Du is said to beobservable if any given initial condition x(0) = xo can be reconstructedbased on the knowledge of the input and output signal only, over a finite timeinterval [0, t].


Intuition from the Modal Form

Recall that if the matrix A has a complete set of independent (right)eigenvectors {v1, . . . , vn}, with eigenvalues {λ1, . . . , λn} it can bediagonalized by the matrix T =

[v1 v2 . . . vn

].

The transformed state in the diagonalized system is such that x = Tx .

The transformed model is (A, B, C , D) = (T−1AT ,T−1B,CT ,D).

Component-wise, the dynamics of each modal coordinate xi are given by

d

dtxi (t) = λi xi (t) + biu(t), i = 1, . . . , n.

The output is given by

y = c1x1(t) + c2x2(t) + . . .+ cnxn(t) + Du(t).


Controllability/observability for diagonal systems

From the previous slide, we can deduce

The i-th coordinate can be controlled if and only if bi 6= 0.

The i-th coordinate appears in the output if and only if ci 6= 0.

Hence:

An LTI system in diagonal form is controllable if bi 6= 0, i = 1, . . . , n.

An LTI system in diagonal form is observable if ci 6= 0, i = 1, . . . , n.

An LTI system is stabilizable if all unstable modes are controllable.

An LTI system is detectable if all unstable modes are observable.


An example

A =

1.618j

−1.618j0.618j

−0.618j

, B =

0.4472j−0.4472j

0.44720.4472

C =

[0.276j −0.276j 0.724 0.724

], D = [0].


An example

A =

j−j

j−j

, B =

0.707j−0.707j

00

C =

[0 0 0.707 0.707

], D = [0].


More general conditions

Consider the derivatives of the output:

y(0) = Cx(0), y ′(0) = CAx(0), y ′′(0) = CA2x(0), . . .

One can reconstruct x(0), i.e., the system is observable, as long as theobservability matrix

CCACA2

. . .CAn−1

has full rank.Note that it is sufficient to compute the observability matrix only up to thepower n − 1 of A (Cayley-Hamilton theorem).A similar condition can be obtained for controllability, with a somewhat morecomplicated proof: a system is controllable if the controllability matrix[

B AB A2B . . . An−1B]

has full rank.


Pole placement — first order systems

Consider a first-order control system with dynamics x = ax + bu, and assumethat we are not happy about its behavior (e.g., it is unstable since a > 0, ormaybe stable but “slow” because |a| is small).

Can we change the behavior by choosing u in a clever way?

Feedback control: choose u = −kx ; then the dynamics would become

x = (a− bk)x

As long as b 6= 0 (i.e., if the system is controllable), by choosingk = (a− a∗)/b, we can place the ”closed-loop” eigenvalue at a desired valuea∗, or anywhere we want on the real axis!

This is the simplest example of a general technique called “pole placement”.


Example

Assume x = 0.5x + u, i.e., the system is unstable with time constant 2.

We would like the system to be stable, with time constant 1/2.

Choose u = −(0.5 + 2)x = −2.5x ; the closed loop will be x = −2x asdesired.

t

open-loop

closed-loop

u


Effect on feedback for closed-loop dynamics

If we have an open-loop LTI system

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

y(t) = Cx(t),

by choosing a linear feedback u = −ky = −kCx , we can transform it intoanother, closed-lopp LTI system:

x(t) = (A− BkC )x(t),

y(t) = Cx(t),

In general “negative feedback” (i.e., u = −ky ) has “stabilizing” effects, andthe bigger k , the faster the closed-loop system is, and the smaller the errorsare.

However this is not generally the case.

In the rest of the course, we will look at ways to analyze the behavior of theclosed-loop system, and choosing the feedback control law, withoutnecessarily lots of computation — but rather using primarily “graphical”methods.


Examples

x(t) = 0.5x(t) + u(t),

y(t) = x(t),

-5 -4 -3 -2 -1 0 1-0.3

-0.2

-0.1

0

0.1

0.2

0.3Root Locus

Real Axis (seconds-1)

Imag

inar

y Ax

is (s

econ

ds-1

)


Examples

x(t) =

[1 1−1 1

]x(t) +

[0

1.5

]u(t),

y(t) =[1.34 0.67

]x(t),

-7 -6 -5 -4 -3 -2 -1 0 1 2-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5Root Locus


Imag

inar

y Ax

is (s

econ

ds-1

)


Examples

x(t) =

−1 1 0−1 −1 10 0 −2

x(t) +

005

u(t),

y(t) =[1 0 0

]x(t),

-6 -5 -4 -3 -2 -1 0 1-4

-3

-2

-1

0

1

2

3

4Root Locus


Imag

inar

y Ax

is (s

econ

ds-1

)


Today’s learning objectives

After today’s lecture, you should be able to:

Diagonalize a matrix using similarity transformations.

Describe the behavior of an LTI system in modal coordinates.

Understand concepts like controllability and observability.

Understand the basic effects of feedback control on the closed-loop dynamics.


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