Chapter 2 State Space Models · 2014. 1. 21. · Chapter 2 State Space Models General format :...

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Chapter 2

State Space Models

General format :

(valid for any nonlinear causal system)

CT : x = f(x, u), DT : xk+1 = f(xk, uk),

y = h(x, u). yk = h(xk, uk).

where

x = the state of the system, n× 1-vector

u = the input of the system, m× 1-vector

y = the output of the system, p× 1-vector

f = state equation vector function

h = output equation vector function

n = number of states⇒ n-th order system

m = number of inputs

p = number of outputs

Single-Input Single-Output system (SISO) : m = p = 1,

Multi-Input Multi-Output system (MIMO) : m, p > 1,

Multi-Input Single-Output system (MISO) :m > 1, p = 1,

Single-Input Multi-Output system (SIMO) :m = 1, p > 1.

ESAT–SCD–SISTA CACSD pag. 34

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Linear Time Invariant (LTI) systems(only valid for linear causal systems)

Continuous-time system:

x = Ax +Bu

y = Cx +Du

u Bx x

C

D

A

∫y

Discrete-time system:

xk+1 = Axk +Buk

yk = Cxk +Duk

uk B

xk+1 xk

C

D

A

z−1 yk

A: n× n system matrix

B: n×m input matrix

C: p× n output matrix

D: p×m direct transmission matrix


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An example of linear state space modeling:

Example: Tape drive control - state space modeling

Process description:

p1 p3 p2

r r

θ1 θ2

K

D

K

D

Kt Kt

ββJ J

i1 i2

+ +

−−e1 e2

The drive motor on each end of the tape is independently

controllable by voltage resources e1 and e2. The tape is

modeled as a linear spring with a small amount of viscous

damping near to the static equilibrium with tape tension

6N. The variables are defined as deviations from this equi-

librium point.


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The equations of motion of the system are:

capstan 1: Jdω1

dt= Tr − βω1 +Kti1,

speed of x1: p1 = rω1,

motor 1: Ldi1dt

= −Ri1 −Keω1 + e1,

capstan 2: Jdω2

dt= −Tr − βω +Kti2,

speed of x1: p2 = rω2,

motor 2: Ldi2dt

= −Ri2 −Keω2 + e2,

Tension of tape: T =K

2(p2 − p1) +

D

2(p2 − p1),

Position of the head: p3 =p1 + p2

2.


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Description of variables and constants:

D = damping in the tape-stretch motion

= 20 N/m·sec,e1,2 = applied voltage, V,

i1,2 = current into the capstan motor,

J = inertia of the wheel and the motor

= 4×10−5kg·m2,

β = capstan rotational friction, 400kg·m2/sec,

K = spring constant of the tape, 4×104N/m,

Ke = electrical constant of the motors = 0.03V·sec,Kt = torque constant of the motors = 0.03N·m/A,

L = armature inductance = 10−3 H,

R = armature resistance = 1Ω,

r = radius of the take-up wheels, 0.02m,

T = tape tension at the read/write head, N ,

p1,2,3 = tape position at capstan 1,2 and the head,

p1,2,3 = tape velocity at capstan 1,2 and the head,

θ1,2 = angular displacement of capstan 1,2,

ω1,2 = speed of drive wheels = θ1,2.


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With a time scaling factor of 103 and a position scaling fac-

tor 10−5 for numerical reasons, the state equations become:

x = Ax +Bu,

y = Cx +Du,

where

x =

p1

ω1

p2

ω2

i1

i2

, A =

0 2 0 0 0 0

−0.1 −0.35 0.1 0.1 0.75 0

0 0 0 2 0 0

0.1 0.1 −0.1 −0.35 0 0.75

0 −0.03 0 0 −1 0

0 0 0 −0.03 0 −1

B =

0 0

0 0

0 0

0 0

1 0

0 1

, C =

[

0.5 0 0.5 0 0 0

−0.2 −0.2 0.2 0.2 0 0

]

,

D =

[

0 0

0 0

]

, y =

[

p3

T

]

, u =

[

e1

e2

]

.


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State-space model, transfer matrix and impulseresponse

Continuous-time system:

state-space

equations

x = Ax+ Bu

y = Cx+Du

Laplacex(0)=0−→ G(s) =

Y (s)

U(s)= D + C(sI − A)−1B

︸︷︷︸

transfer matrixin practice : D = 0

⇓

G(t) = CeAtB︸︷︷︸

impulse response matrix

Laplace←→ G(s) =∞∑

i=1

CAi−1Bs−i

︸︷︷︸

transfer matrix

Discrete-time system:

state-space

equations

xk+1 = Axk +Buk

yk = Cxk +Duk

z − transfx0=0←→ G(z) =

Y (z)

U(z)= D + C(zI −A)−1B

︸︷︷︸

transfer matrix⇓

G(k) =

D : k = 0

CAk−1B : k ≥ 1︸︷︷︸

impulse response matrix

z − transf←→ G(z) =

∞∑

i=1

CAi−1Bz−i +D

︸︷︷︸

transfer matrix

In case of a SISO system, G(t) or G(k) is the impulse

response.

For MIMO G(t), G(k) are matrices containing the m× pimpulse responses (one for every input-output pair).


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Linearization of a nonlinear systemabout an equilibrium point

Consider a general nonlinear system in continuous time :

dx

dt= f(x, u)

y = h(x, u)

For small deviations about an equilibrium point (xe, ue, ye)

for which

f(xe, ue) = 0

ye = h(xe, ue)

we define

x = xe + ∆x, u = ue + ∆u, y = ye + ∆y,

and obtain

dx

dt=d∆x

dt= f(x, u) = f(xe + ∆x, ue + ∆u)

and

ye + ∆y = h(x, u) = h(xe + ∆x, ue + ∆u).


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By first order approximation we obtain a linear state space

model from ∆u to ∆y :

d∆x

dt= f(xe + ∆x, ue + ∆u)

⇓ f(xe, ue) = 0

d∆x

dt

(1)=∂f

∂x

∣∣∣∣xe,ue︸︷︷︸

n× n⇓A

∆x +∂f

∂u

∣∣∣∣xe,ue︸︷︷︸

n×m⇓B

∆u

and

ye + ∆y = h(xe + ∆x, ue + ∆u)

⇓ ye = h(xe, ue)

∆y(1)=∂h

∂x

∣∣∣∣xe,ue︸︷︷︸

p× n⇓C

∆x +∂h

∂u

∣∣∣∣xe,ue︸︷︷︸

p×m⇓D

∆u

Conclusion : use a linear approximation about equilibrium.


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Example :

Consider a decalcification plant which is used to reduce the

concentration of calcium hydroxide in water by forming a

calcium carbonate precipitate.

The following equations hold (simplified model) :

• chemical reaction : Ca(OH)2 + CO2→ CaCO3 + H2O

• reaction speed : r = c[Ca(OH)2][CO2]

• rate of change of concentration :

d[Ca(OH)2]

dt=k

V− r

V

d[CO2]

dt=u

V− r

V

k and u are the inflow rates in moles/second of calcium

hydroxide and carbon dioxide respectively. V is the tank

volume in liters.

Let the inflow rate of carbon dioxide be the input and the

concentration of calcium hydroxide be the output.


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A nonlinear state space model for this reactor is :

d[Ca(OH)2]

dt=

k

V− c

V[Ca(OH)2][CO2]

d[CO2]

dt=

u

V− c

V[Ca(OH)2][CO2]

y = [Ca(OH)2]

The state variables are x1 = [Ca(OH)2] and x2 = [CO2].

In equilibrium we have :

k

V− c

V[Ca(OH)2]eq[CO2]eq =

k

V− c

VX1X2 = 0

ueq

V− c

V[Ca(OH)2]eq[CO2]eq =

U

V− c

VX1X2 = 0

Y = [Ca(OH)2]eq = X1

For small deviations about the equilibrium :

d∆x1

dt= − c

VX2∆x1 −

c

VX1∆x2

d∆x2

dt= − c

VX2∆x1 −

c

VX1∆x2 +

1

V∆u

∆y = ∆x1

so,

A =

[

−cX2V−cX1

V

−cX2V−cX1

V

]

, B =

[

01V

]

, C =[

1 0]

, D = 0


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If

k = 0.1 molesec , c = 0.5 l2

sec· mole, U = 0.1 molesec ,

X1 = 0.25 molel

, X2 = 0.8 molel

, V = 5 l

then[

A B

C D

]

=

−0.08 −0.025 0

−0.08 −0.025 0.2

1 0 0

The corresponding transfer function is

∆y(s)

∆u(s)=−0.005

s2 + 0.105s

and its bode plot

Frequency (rad/sec)

Pha

se (

deg)

; Mag

nitu

de (

dB)

Bode Diagrams

−50

0

50

10−3

10−2

10−1

100

−400

−350

−300

−250


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One can also obtain a linear state space model for this

chemical plant from linear system identification.

For small deviations about the equilibrium point the dy-

namics can be described fairly well by a linear (A,B,C,D)-

model. Hence, a small white noise disturbance ∆u was

generated and was added to the equilibrium value U . We

applied this signal to the input of a nonlinear model of

the chemical reactor (Simulink model for instance), i.e.

u(t) = U + ∆u.

The following input-output set was obtained :

0 100 200 300 400 500 600 700 800 900 10000.096

0.098

0.1

0.102

0.104u(t)

time (s)

0 100 200 300 400 500 600 700 800 900 10000.249

0.2495

0.25

0.2505

time (s)

y(t)


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By applying a linear system identification algorithm

(N4SID), the following 2nd-order model was obtained :

[

A B

C D

]

=

0.0015 −0.0589 −0.0867

0.0027 −0.1066 0.1713

0.2546 0.129 0

The corresponding transfer function is

y(s)

u(s)=−1.299.10−7s− 0.005

s2 + 0.1051s + 1.346.10−8

It has 2 poles at −1.281.10−7 and −1.051.

As −1.281.10−7 lies close to 0, and taking in account the

properties of the manually derived linear model of page 45,

we conclude that the plant has one integrator pole. Hence,

it might be better to fix one pole at s = 0. In this way

it is guaranteed that the linear model obtained by system

identification is stable.

The transfer function which was obtained using this mod-

ified identification procedure is

y(s)

u(s)=−1.68.10−8s− 0.005

s2 + 0.1051s


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The different models are validated by comparing their re-

sponse to the following input :

∆u(t) = 0.003 if 100 < t < 200

= 0 elsewhere

Four responses are shown :

• the nonlinear system (—)

• the linearised model obtained by hand (page 45) (- -)

• the 2nd-order linear model obtained from N4SID (· · ·)• the linear model obtained from N4SID having a fixed

integrator pole by construction (-.)

0 100 200 300 400 500 600 700 800 900 10000.235

0.24

0.245

0.25

y(t)

mol

e/l

900 905 910 915 920 925 930 935 940 945 9500.2356

0.2358

0.236

0.2362

0.2364y(t) : detail 1

mol

e/l

900 905 910 915 920 925 930 935 940 945 950

0.23572

0.23573

0.23574

mol

e/l

y(t) : detail 2

time(s)


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Digitization

in this course we want to control physical plants using a

digital computer :

Plant

A/D

controllerdigital

sensors

clock

actuators+

D/A

computer

• analog-to-digital converter :

clock

A/Dx(t) x[k] x(t)

filterlowpass sampler quantizer

Qx[k]

– the analog anti-aliasing filter filters out high frequent

components (> Nyquist frequency)

– the filtered analog signal is sampled and quantized

in order to obtain a digital signal. For quantization

10 to 12 bits are common.


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• digital-to-analog converter :

– zero-order hold

0 10 20 30 40 50 60 70 80 90 100−4

−3

−2

−1

0

1

2

3

time

ZOH approximation

∗ introduces a delay

∗ frequency spectrum deformation

– first-order hold

– n-th order polynomial (more general case) : fit a n-

th order polynomial through the n+1 most recent

samples and extrapolate to the next time instance


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Digitization of state space models1. discretization by applying numerical integration rules :

a continuous-time integrator

x(t) = e(t)

⇔sX(s) = E(s)

can be approximated using

• the forward rectangular rule or Euler’s method

xk+1 − xkTs

= ek

⇔z − 1

TsX(z) = E(z)

with xk = x(kTs)

t

e(t)

k-1 k k+1

• the backward rectangular rule

xk+1 − xkTs

= ek+1

⇔z − 1

zTsX(z) = E(z)

t

e(t)

k-1 k k+1


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• the trapezoid rule or bilinear transformation

xk+1 − xkTs

=ek+1 + ek

2

⇔2

Ts

z − 1

z + 1X(z) = E(z)

t

e(t)

k-1 k k+1

Change a continuous model G(s) into a discrete model

Gd(z) by replacing all integrators with their discrete equiv-

alents := projection of the left half-plane

Euler backward rect. bilinear transf.

s→ z−1Ts

s→ z−1zTs

s→ 2Tsz−1z+1

Re

Im

Re

Im

Re

Im

Except for the forward rectangular rule stable continuous

poles (grey zone) are guaranteed to be placed in stable

discrete areas, i.e. within the unit circle.

The bilinear transformation maps stable poles → stable

poles and unstable poles → unstable poles.


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The following continuous-time model is given :

x = Ax +Bu

y = Cx +Du⇔

sX = AX +BU

Y = CX +DU

following Euler’s method s is replaced by z−1Ts

, so

z − 1

TsX = AX +BU

or

zX = (I + ATs)X + BTsU

the output equation Y = CX + DU remains. A similar

calculation can be done for the backward rectangular rule

and the bilinear transformation resulting in the following

table :

Euler backward rect. bilinear transf.

Ad I + ATs (I −ATs)−1 (I − ATs2

)−1(I + ATs2

)

Bd BTs (I −ATs)−1BTs (I − ATs2 )−1BTs

Cd C C(I − ATs)−1 C(I − ATs2

)−1

Dd D D + C(I −ATs)−1BTs D + C(I − ATs2 )−1BTs

2


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2. discretization by assuming zero-order hold ...

x = Ax +Bu

⇒x(t) = eAtx(0) +

∫ t

0

eA(t−τ)Bu(τ )dτ︸︷︷︸

convolution integral

Let the sampling time be Ts, then

x(t + Ts) = eA(t+Ts)x(0) +

∫ t+Ts

0

eA(t+Ts−τ)Bu(τ )dτ

= eATsx(t) + eATs∫ t+Ts

t

eA(t−τ)Bu(τ )dτ︸︷︷︸

approximated

ZOH⇒ xk+1t=kTs= eATsxk + eATsA−1(I − e−ATs)Buk= eATs︸︷︷︸

Ad

xk + A−1(eATs − I)B︸︷︷︸

Bd

uk

One can prove that Gd(z) can be expressed as

Gd(z)ZOH= (1− z−1)Z

L−1

G(s)

s

For this reason this discretization method is sometimes

called step invariance mapping.


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3. discretization by zero-pole mapping :

following the previous method the poles of Gd(z) are

related to the poles of G(s) according to z = esTs. If

we assume by simplicity that also the zeros undergo this

transformation, the following heuristic may be applied:

(a) map poles and zeros according to z = esTs

(b) if the numerator is of lower order than the denomi-

nator, add discrete zeros at -1 until the order of the

numerator is one less than the order of the denomi-

nator. A lower numerator order corresponds to zeros

at∞ in continuous time. By discretization they are

put at -1.

(c) adjust the DC gain such that

lims→0

G(s) = limz→1

Gd(z)


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Example :

given the following SISO system

[

A B

C D

]

=

−0.2 −0.5 1

1 0 0

1 0.4 0

⇒ G(s) = C(sI − A)−1B +D =s + 0.4

s2 + 0.2s + 0.5.

Now compare the following discretization methods (Ts = 1

sec.) :

1. bilinear transformation :

(a) Ad, Bd, Cd and Dd are calculated using the conver-

sion table

(b) Gbilinear(z) = Cd(zI − Ad)−1Bd +Dd.

=0.4898 + 0.1633z−1 − 0.3265z−2

1− 1.4286z−1 + 0.8367z−2

2. ZOH :

(a) Ad = eATs, Bd = A−1(eATs − I)B, Cd = C and

Dd = D

(b) GZOH(z) = Cd(zI − Ad)−1Bd +Dd.

=1.0125z−1 − 0.6648z−2

1− 1.3841z−1 + 0.8187z−2


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3. pole-zero mapping :

(a) the poles of G(s) are −0.1 + j0.7 and −0.1− j0.7(b) there is one zero at −0.4(c)

Gpz = Kz − e−0.4

(z − e−0.1+j0.7)(z − e−0.1−j0.7)

= Kz−1 − 0.6703z−2

1− 1.3841z−1 + 0.8187z−2

hence K =lims→0G(s)

limz→1Gpz(z)= 1.0546

Compare the bode plots :

10−2

10−1

10−1

100

101

frequency (Hz)

freq

uenc

y am

plitu

de s

pect

rum

__ : continuous system

−− : bilinear approximation−. : zero−order hold assumption

.. : pole−zero mapping

10−2

10−1

−150

−100

−50

0

50

frequency (Hz)

phas

e sp

ectr

um (

degr

.)

__ : continuous system

−− : bilinear approximation−. : zero−order hold assumption.. : pole−zero mapping


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Sampling rate selection :

• the lower the sampling rate, the lower the implementa-

tion cost (cheap microcontroller & A/D converters), the

rougher the response and the larger the delay between

command changes and system response.

• an absolute lower bound is set by the specification to

track a command input with a certain frequency :

fs ≥ 2BWcl

fs is the sampling rate and BWcl is the closed-loop sys-

tem bandwidth.

• if the controller is designed for disturbance rejection,

fs > 20BWcl

• when the sampling rate is too high, finite-word effects

show up in small word-size microcontrollers (< 10 bits).

• for systems where the controller adds damping to a

lightly damped mode with resonant frequency fr,

fs > 2fr

In practice, the sampling rate is a factor 20 to 40 higher

than BWcl.


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Advantages of state space models

• More general models: LTI and Nonlinear Time Varying

(NTV).

• Geometric concepts: more mathematical tools (linear

algebra).

• Internal and external descriptions: “divide and con-

quer” strategy.

• Unified framework: the same for SISO and MIMO.


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Geometric properties of linear statespace models

Canonical formsControl canonical form (SISO):

x = Acx + Bcu, y = Ccx.

Ac =

−a1 −a2 · · · · · · −an1 0 · · · · · · 0

0 . . . . . . ...... . . . . . . . . . ...

0 · · · 0 1 0

, Bc =

1

0...

0

,

Cc =[

b1 b2 · · · bn]

Corresponding transfer function:

G(s) =b(s)

a(s)

b(s) = b1sn−1 + b2s

n−2 + · · · + bn

a(s) = sn + a1sn−1 + a2s

n−2 + · · · + an


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Block diagram for the control canonical form

x1 x2 xn−1

b1

b2

bn−1

bn

−a1

−a2

−an−1

−an

u

y

∫ ∫ ∫ ∫

xn


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Modal canonical form:

x = Amx + Bmu, y = Cmx.

Am =

−s1

−s2

. . .

−sn

, Bm =

1

1...

1

Cm =[

r1 r2 · · · rn]

Corresponding transfer function:

G(s) =

n∑

i=1

ris + si

Block diagram for the modal canonical form:

yu

∫

∫

r1

rn

−s1

∫

−s2

−sn

r2

x1

x2

xn


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Similarity transformationA state space model (A,B,C,D) is not unique for a phys-

ical system. Let

x = T x, det(T ) 6= 0,

A = T−1AT, B = T−1B, C = CT, D = D

⇒˙x = T−1ATx + T−1Bu = Ax + Bu,

y = CTx +Du = Cx +Du

T is chosen to give the most convenient state-space descrip-

tion for a given problem (e.g.control or modal canonical

forms).

Example: Eigenvalue decomposition of A:

A = XΛX−1, Λ =

α1 β1

−β1 α1

. . .

λ1

. . .

x = Ax +Bu

y = Cx +Du

X−1x = z=⇒ z = Λz +X−1Bu

y = CXz +Du


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Controllability

An n-th order system is called controllable if one can reach

any state x from any given initial state x0 in a finite time.

Controllability matrix:

C =[

B AB · · · An−1B]

rank(C) = n⇔ (A,B) controllable

Explanation:

Consider a linear (discrete) system of the form

xk+1 = Axk +Buk

Then

xk+1 = Ak+1x0 + AkBu0 + · · · +Buk

such that

xk+1 − Ak+1x0 =[

B AB · · · AkB]

uk

uk−1

...

u0


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Note that

rank[

B AB · · · AkB]

= rank[

B AB · · · An−1B]

for k ≥ n− 1 (Cayley-Hamilton Theorem).

There exists always a vector

uk

uk−1

...

u0

if rank(C) = n.

Remarks: Controllability matrices for a continuous time

system and a discrete time system are of the same form.


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ObservabilityAn n-th order system is called observable if knowledge of

the input u and the output y over a finite time interval is

sufficient to determine the state x.

Observability matrix:

O =

C

CA...

CAn−1

rank(O) = n⇔ (A,C) observable

Explanation:

Consider an autonomous system

xk+1 = Axk

yk = Cxk

Then we find easily

yk = CAkx0


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and

y0

y1

...

yk

=

C

CA...

CAk

x0

Note that

rank

C

CA

· · ·CAk

= rank

C

CA

· · ·CAn−1

for k ≥ n− 1.

One can always determine x0 if rank(O) = n.

Remarks: Observability matrices for a continuous time sys-

tem and a discrete time system are of the same form.


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The Popov-Belevitch-Hautus tests (PBH)

PBH controllability test:

(A,B) is not controllable if and only if there exists a left

eigenvector q 6= 0 of A such that

ATq = qλ

BTq = 0

In other words, (A,B) is controllable if and only if there is

no left eigenvector q of A that is orthogonal to B.

If there is a λ and q satisfying the PBH test, then we

say that the mode corresponding to eigenvalue λ is uncon-

trollable (uncontrollable mode), otherwise it is controllable

(controllable mode).

How to understand (for SISO) ? Put A in the modal canon-

ical form:

A =

∗λi

∗

, B =

∗0

∗

,⇒ q =

0

1

0

Uncontrollable mode: xi = λixi.


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PBH observability test:

(A,C) is not observable if and only if there exists a right

eigenvector p 6= 0 of A such that

Ap = pλ

Cp = 0

If there is a λ and p satisfy the PBH test, then we say that

the mode corresponding to eigenvalue λ is unobservable

(unobservable mode), otherwise it is observable (observable

mode).

How to understand for SISO ? Put A in a modal canonical

form.


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Stability/Stabilizability/Detectability

Stability:

A system with system matrix A is unstable if A has an

eigenvalue λ with real(λ) ≥ 0 for a continuous time system

and |λ| ≥ 1 for a discrete time system.

Stabilizability:

(A,B) is stabilizable if all unstable modes are controllable.

Detectability:

(A,C) is detectable if all unstable modes are observable.


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Example

xk+1 = Axk +Buk

yk = Cxk +Duk

[

A B

C D

]

=

1.1 0 0 0 1

0 −0.5 0.6 0 2

0 −0.6 −0.5 0 −1

0 0 0 2 2

0 1 2 3 1

Modes Stable? Stabilizable? Detectable?

1.1 no yes no

−0.5± 0.6i yes yes yes

2 no yes yes


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Kalman decomposition and minimal realization

Kalman decomposition

Given a state space system [A,B,C,D]. Then we can

always find an invertible similarity transformation T such

that the transformed matrices have the structure

TAT−1 =

r1 r2 r3 r4

r1 Aco 0 A13 0

r2 A21 Aco A23 A24

r3 0 0 Aco 0

r4 0 0 A43 Aco

TB =

m

r1 Bco

r2 Bco

r3 0

r4 0

, CT−1 =(r1 r2 r3 r4

p Cco 0 Cco 0)

where

r1 = rank (OC), r2 = rank (C)− r1,r3 = rank (O)− r1, r4 = n− r1 − r2 − r3.


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The subsystem

[ Aco, Bco, Cco, D ]

is controllable and observable. The subsystem

[

(

Aco 0

A21 Aco

)

,

(

Bco

Bco

)

,(

Cco 0)

, D ]

is controllable. The subsystem

[

(

Aco A13

0 Aco

)

,

(

Bco

0

)

, ( Cco Cco ), D ]

is observable. The subsystem

[ Aco , 0, 0, D ]

is neither controllable nor observable.


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Kalman decomposition

u yCO

CO

CO

CO


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Minimal realization:

A minimal realization is one that has the smallest-size A

matrix for all triplets [A,B,C] satisfying

G(s) = D + C(sI − A)−1B

where G(s) is a given transfer matrix.

[

A B C D]

is minimal ⇔ controllable and observable.

Let[

Ai Bi Ci D]

, i = 1, 2, be two minimal realizations

of a transfer matrix, then[

A1 B1 C1 D]

T ⇓ ⇑ T−1

[

A2 B2 C2 D]

with

T = C1CT2 (C2CT2 )−1


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Input/output properties of state-spacemodels

Transfer matrix

General transfer matrix for a state space model :

G(ξ) = D + C(ξI − A)−1B

ξ can be s (CT) or z (DT).

Poles

Characteristic polynomial of matrix A:

α(ξ) = det(ξI − A)

= αnξn + αn−1ξ

n−1 + · · · + α1ξ + α0

Characteristic equation:

α(ξ) = 0

The eigenvalues λi, i = 1, · · · , n of the system matrix A

are called the poles of the system.

The pole polynomial is defined as

π(ξ) = Πni=1(ξ − λi)

(= characteristic equation up to within a scalar)


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Physical interpretation of a pole :Consider the following second order system :

[

A B

C D

]

=

α β b1

−β α b2

c1 c2 0

The transfer matrix is given by (try to verify it) :

G(ξ) = C(ξI − A)−1B +D

=ξ(b1c1 + b2c2) + β(b2c1 − b1c2)− α(b1c1 + b2c2)

ξ2 − 2αξ + α2 + β2

There are 2 poles at α± jβ.

There is a zero at

α(b1c1 + b2c2)− β(b2c1 − b1c2)b1c1 + b2c2


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In continuous time the impulse response takes on the form:

g(t) = L−1 G(s)

= L−1

s(b1c1 + b2c2) + β(b2c1 − b1c2)− α(b1c1 + b2c2)

s2 − 2αs + α2 + β2

= (Am cos(βt) +Bm sin(βt))eαt = Cmeαt cos(βt + γ)

with Am = b1c1 + b2c2 and Bm = b2c1 − b1c2

⇒ Cm =√

(b21 + b22)(c21 + c22)

⇒ γ = atan

(−Bm

Am

)

Define the damping ratio ζ and natural frequency ωn:

α = −ζωn, β = ωn√

1− ζ2

jω

θ

α + jβ

α− jβ

β

α

ζ = sin θωn

stability degree

r


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For a second order system

G(s) =ω2n

s2 + 2ζωns + ω2n

the time response looks like :

rise time tr ' 1.8ωn

settling time ts = 4.6ζωn

peak time tp = πωd

, ωd = ωn√

1− ζ2

overshoot Mp = e−πζ/√

1−ζ2, 0 ≤ ζ < 1

In general : if a continuous-time system (A,B,C, 0) has

poles at α ± jβ ⇒ the impulse response will have time

modes of the form

Ameαt cos(βt + γ)

Am: amplitude

γ: phase

determined by B and C.


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In discrete time the impulse response matrix Gd(k) takes

on the form :

Gd(k) = CdAk−1d Bd, k ≥ 1

which can be proven to be a sum of terms of the form :

Cmbk−1 cos(ω(k − 1) + γ)

each of which satisfies a second order linear system

[

A B

C D

]

=

α β b1

−β α b2

c1 c2 0

Parameterize A as

A =

[

b cosω b sinω

−b sinω b cosω

]

then

Ak = bk

[

cosωk sinωk

− sinωk cosωk

]


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Now,

CAkB = (Am cosωk +Bm sinωk)bk

= Cmbk cos(ωk + γ)

with Am = Cm cos γ = b1c1 + b2c2

and Bm = −Cm sin γ = b2c1 − b1c2

⇒ Cm =√

(b21 + b22)(c21 + c22)

⇒ b =√

α2 + β2

⇒ ω = atan

(β

α

)

⇒ γ = atan

(−Bm

Am

)


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Transmission zeros

Definition: The zeros of a LTI system are defined as those

values ζ ∈ C for which the rank of the transfer matrixG(ξ)

is lower than its normal rank (=rank of G(ξ) for almost all

values of ξ):

rank(G(ζ)) < normal rank

Property: Let ζ be a zero of G(ξ) (p×m), then

rank (G(ζ)) < normal rank,

⇓ if ζ is not a pole of G(ξ)

∃v 6= 0 s.t.[D + C(ζI − A)−1B

]v = 0,

⇓ define w∆= (ζI − A)−1Bv

Dv + Cw = 0, (ζI − A)w −Bv = 0,

⇓[

ζI − A −BC D

][

w

v

]

= 0


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How to find zeros for square MIMO systems (p = m) with

invertible D ?

(ζI − A)w −Bv = 0, Cw +Dv = 0

⇓Aw +Bv = wζ, v = −D−1Cw

⇓(A−BD−1C)w = wζ

ζ = eigenvalue of A− BD−1C,

w = corresponding eigenvector !

For other cases :

• Generalized eigenvalue problem.

• Use Matlab function “tzero”.

Minimum and non-minimum phase systems:

If a system has an unstable zero (in the right half–plane

(RHP) or outside the unit circle), then it is a non-minimum

phase (NMP) system, otherwise it is a minimum phase

(MP) system.


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Physical explanation (continuous time) :

Let ζ be a real zero, then there will exist vectors x0 and u0

such that: [

ζI − A −BC D

][

x0

u0

]

= 0

This means if we have an input u0eζt, there exists an initial

state x0 such that the response is

y(t) = 0.

For complex zeros :

if ζ is a complex zero, then also its complex conjugate ζ∗

is a zero. Try to prove that if[

ζI − A −BC D

][

x0

u0

]

= 0

the output y(t) will be exactly zero if the input is

u(t) = |u0|eReζt · cos(Imζt + φu0)

and the initial state is chosen to be Rex0. “·” is the

elementwise multiplication, both “cos()” and “e()” are as-

sumed to be elementwise operators,Rex0 is the real part

of x0, Imζ is the imaginary part of ζ and u0 = |u0|·ejφu0.Try to derive equivalent formulas for the discrete-time case.


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Why zeros are important :

• Limited control system performance

k

C G

If G is NMP, k can not go to ∞, since unstable zeros

(which become unstable poles) put a limit on high gains.

• Stability of a stabilizing controller, Parity Interlacing

Property (PIP) :

Let G be unstable. Then, G can be stabilized by a

controller C which itself is stable⇔ between every pair

of real RHP zeros of G (including at ∞), there is an

even number of poles.

Example:

G(s) =s

s2 − 1

× ×−1 10

pole zero pole

∞

zero

-∞

G(s) can not be stabilized by a stable controller.


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RHP zeros can cause undershoot !

Let

G(s) =Πni=1(1− s

ζi)

Πn+ri=1 (1− s

λi),

where r = relative degree of G(s). Let y(t) be the step

response.

Fact :

y(0) and its first r − 1 derivative are 0;

y(r) is the first non-zero derivative;

y(∞) = G(0).

The system has undershoot ⇔ the steady state value

y(∞) has a sign opposite to y(r)(0), i.e.

y(r)(0)y(∞) < 0.

It can be proven that the system has undershoot ⇔ there

is an odd number of RHP zeros.


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An example of state-space analysis

Example: Tape drive control - state space analysis

Controllability/observability

Direct rank test using SVD:

The singular values of C are

4.5762, 4.3861, 1.2940, 1.2758, 0.5737, 0.4787,

so the system is controllable (rank=6).

The singular values of O are

2.5695, 1.4504, 0.7145, 0.7071, 0.5171, 0.2356,

so the system is observable (rank=6).


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PBH test:

There are 6 (independent) left eigenvectors associated with

the 6 eigenvalues of A :

−0.1480± 0.0826i

0.0703± 0.5380i

0.1480∓ 0.0826i

−0.0703∓ 0.5380i

0.2994± 0.2954i

−0.2994∓ 0.2954i

,

0.0766

0.5624

0.0766

0.5624

0.4218

0.4218

0

0.4892

0.0000

0.4892

0.5105

0.5105

,

−0.0046

−0.0227

0.0046

0.0227

−0.7067

0.7067

,

0.0000

−0.0295

0.0000

−0.0295

−0.7065

−0.7065

It can be easily checked that none of them is orthogonal to

B. The result is the same as the rank test above.

The same can be done to investigate the observability.


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Poles and zeros:

Poles: the eigenvalues of A are

−0.2370± 0.5947i, 0, −0.2813, −0.9760, −0.9687

The system is not stable since there is a pole at zero.

The first two oscillatory poles are from the spring-mass

system consisting of the tape and the motor/capstan in-

ertias. The zero pole reflects the pure integrative effect of

the tape on the capstans.


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Zeros: it can be checked in Matlab using function “tzero”

that there is only one zero at -2. The eigenvector of the

matrix [

−2I − A −BC 0

]

associated with the eigenvalue 0 is

[

x0

u0

]

with x0 =

−0.1959

0.1959

0.1959

−0.1959

−0.4571

0.4571

, u0 =

[

0.4630

−0.4630

]

.

Thus, let

x(0) = x0, u(t) = u0e−2t,

then the output y(t) will be always zero, ∀t ≥ 0.


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Physical explanation of the zero:

Since the initial values i1(0) = −i2(0) and the inputs

e1(t) = −e2(t), and the motor drive systems in both sides

are symmetric, the two capstans will rotate at the same

speed, but in opposite directions. So the tape position over

the read/write head will remain at 0 if the initial value is

at 0.

The tension in the tape is the sum of the tensions in the

tape spring and the tape damping. The input voltages vary

such a way that the tensions in the tape spring and the tape

damping cancel each other out.


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Matlab Functions

c2d

c2dm

canon

expm

tf2ss

ss2tf

(d)impulse

(d)step

minreal

tzero

ctrb

obsv

ctrbf

obsvf

(d)lsim

ss

ssdata


Chapter 2 State Space Models · 2014. 1. 21. · Chapter 2 State Space Models General format :...

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Transcript of Chapter 2 State Space Models · 2014. 1. 21. · Chapter 2 State Space Models General format :...