slides fall 10

8/8/2019 Slides Fall 10

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Lesson 1

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System model (mathematical description)

Model inversion

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6

Special case linear systems:

Nonlinear systems:

Linearization of nonlinear systems:


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Saugrohr

Drosseklappe

Katalysator (Haupt-)Katalysator

Einspritz-ventile

vK

mL

*

hK2P

hK , NO , CO , ( HC, CO 2)

Motronic

n

vK :BOSCH LSU4

(Breitband-Sonde)

hK2P :BOSCH LSF4

(Zweipunkt-Sonde)

Driver

System: Model:

d dt

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

y( t )= g( x( t ), u(t ), p)

Uncertainty: pi,min p p i,max ,

i = 1, N

8

y

u


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Feedback: powerful, but dangerous

Feedback is everywhere

stabilization model uncertainty not measurable disturbances cost reduction

engineering biology

social systems

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d s

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Lesson 2

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Example Water Tank


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Example Stirred Tank Reactor


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Strecke

Strung

Sollwert

r

Regler-

e

Stellsignal

u

FehlerIstwert y

cylinders

injectors

T e

u

y

Example: cruise control

throttle

disturbance

outputcar ECU

control

error reference

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v(t)

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Example Loudspeaker


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Example Conveyor Belt


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Lesson 3

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nominal point

no physical dimension and around 1 if OK

Normalized system equation

Equilibrium condition

h0

0

0

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Problem: equilibrium point cannot be used for normalization

Standard:

Choice:

z g=

Equilibrium:

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f 0 ( x, u )

x

u

x e

ue

f 0 ( xe + x, ue + u) f 0 ( xe , ue ) + f 0 x xe ,ue

x + f 0 u xe ,ue

u

= 0

x

u

f 0 x xe ,ue

x + f 0 u xe ,ue

u

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x ( t ) = 1 + x (t )u ( t ) = 1 + u (t )

1 + x (t ) = 1 +1

2 x (t ) +

Linearized equation:

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Equilibrium condition:

therefore:

or:

with:

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Equilibrium condition:

therefore:

or:

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Equilibrium z i , e , v e , w e

f ( zi,e

,ve) = 0

we = g( zi,e ,ve )

defined by

Case A z i ,e 0 z i ,0 = z i ,ev e 0 v 0 = v ew e 0 w 0 = w e

Case B z i ,0 =

v 0 =

w 0 =

Reasonable, but arbitrary choice(depends on problem at hand)

Result case A x i,e = 1, u e = 1, ye = 1

Result case B

x i, e = 0, u e = 0, ye = 0

d dt

z( t )= f ( z( t ), v( t )), w( t ) = g( z( t ), v( t ))System model:

Always normalize with z i ,0 , v 0 , w 0

and linearize around x i , e , u e , ye

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m

v 0

(m

, ,v

0 )

k (m , v 0 )

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First-order system (n=1, the most simple case):62

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Lesson 4

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C

y0

0

v

x(t ) = A x(t) + b u(t )

y(t ) = c x(t) + d u(t )

. . .

. .

0u0

. z(t )= f(z (t),v( t ))

w(t )= g(z (t), v(t ))w0

-1 y

y

u

u v w

approximated by:

P

ue y

e

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d dt x( t )

= A x( t ) + b u( t )

y( t ) = c x( t ) + d u( t )

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Pro memoria scalar exponential:

Main feature:

The derivative of an exponential is a linear function of anexponential!

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Generalization matrix exponential:

Key point:

Verify that!

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A few remarks:

if matrices do not commute, but:

Matrix exponentials are not as simple as scalar ones, i.e,

The inverse is well defined because

Proofs see QC

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Application of matrix exponentials to the solution of the

initial-value problem :

For a known initial condition

and a known input

what is the state trajectory ?

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System output

Definition convolution operator ( x0=0 , d =0)

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Quantitative solutions to test signals easy.For the sake of simplicity d =0.

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High-order systems: quantitative solutions to testsignals not easy!

Qualitative solutions?

b l81


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x 1

x n

x (0)

Lyapunov Stability

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Pro memoria I

A vector v that is mapped by a linear mapping A intoa collinear vector

vis called an Eigenvector

The scalar gain is an Eigenvalue of A. In general,both and v are complex-valued objects.

Equation (*) can be written as

and a nontrivial solution exists iff

(*)

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If n linearly independent Eigenvectors vi exist (this isoften the case, for instance if all Eigenvalues aredistinct) then they can be used to define a coordinatetransformation with

T =

with the following property

T 1

A T =

1

0 0

0 2

0 0 n

det( T ) 0

Pro memoria II

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Back to Stability Problem General solution

T x (t )= e A t T x (0)

x (t )= T 1 e A t T x (0)

x (t )= T 1 I + A t + 12

A 2 t 2

+

T x (0)

x (t )= T 1 T + T 1 A T t +1

2T 1 A T T

1 A T t

2+

x (0)

x (t )= I + A t +1

2 A 2 t 2 +

x (0) = e

A t x (0) = e

T 1 A T t x (0)

det( T ) 0

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Assume that transformation T can be chosen such that

In this case is diagonal as well ( 2, 3, diagonal) with

T 1 A T =

1

0 0

0 2

0 0 n


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Solution in the new coordinates

real part imaginary part

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x = 0 x = 0

x = x =

x < x <

det( T ) 0

[T ]ij <

For

Therefore

therefore

What if A is not diagonalizable? 89


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What if A is not diagonalizable?

More details subsequent lectures.

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region of attraction

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Example:

pendulum on a cart

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nominalposition

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nominalposition

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Lesson 5

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A = [ 0 1 00 0 1

-1 -2 -1]c = [ 1 1 1]x0 = [ 1

1

1 ]t=0:0.01:10;y=[];for i=1:max(length(t));

y = [y;c*expm(A*t(i))*x0];end;plot(t,y)

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x2

x1

xn

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x2

x1

xn

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x2

x1

xn

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therefore, analyze

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Definition:

Yields:

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x2

x1

xn

r1

rr

Problem:103


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Solution:

Pro memoria: characteristic polynomial of A:

Therefore:

M i l104


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Main result:

The system { A,b} is completely controllable iff

the matrix defined by

has full rank n. If its rank r


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For linear time-invariant systems the reachableand the controllable subspace are identical.

The system is stabilizable only if the state variablesof the non-reachable subspace are all asymp-totically stable.

+

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T ti l th t l108


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Tangential thruster only:

Rank?

R di l th t l109


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Radial thruster only:

Rank?

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When can two different I C yield the same output?111


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When can two different I.C. yield the same output?

Only when the difference

The system is completely observable iff the kernelis empty, i.e., if the rank of is full.

is in the kernel of

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Can the pendulum be stabilized in its upper position?114


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Can the pendulum be stabilized in its upper position?

bu =

Controllability OK

M i g d l gl i t OK115


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Measuring pendulum angle is not OK

It is not possible to stabilize the system withthis information!

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Measuring tip position is OK

i.e., it is possible to stabilize the system using thisinformation in a feedback loop.

Surprisingly measuring the cart position is OK too!117


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Surprisingly, measuring the cart position is OK too!

Later well see that it is possible, but more difficultto stabilize the system using only this signal.

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yes

no

yes no

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RO

RO

RO

u y

R

O

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~ ~ ~ ~

~~

b

c

~

~

1

2

4

3

Block diagram representations I121


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Block diagram representations Id

dt x

1( t ) = 2 x

1( t ) + 3 u(t )

y( t ) = 4 x1 (t ) + 1 u( t )

Block diagram representations IId 122


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Block diagram representations IIdt

x1( t ) = x 2 ( t )

d

dt x 2 ( t ) = x 3 ( t )

d

dt x

3 ( t ) = x1( t ) 2 x 2 ( t ) 3 x 3 ( t ) + u ( t )

y( t ) = x1 ( t )

Input/Output Formsd 123


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Input/Output Formsdt

x1( t ) = x 2 ( t )

d

dt x 2 ( t ) = x 3 ( t )

d

dt x

3 ( t ) = x1( t ) 2 x 2 ( t ) 3 x 3 ( t ) + u ( t )

y( t ) = x1 ( t )

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Internal description {A,b,c}

Physical coordinates!

I/O description

No coordinates!

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realizationusing

canonical

coordinateslater

without coordinates

with coordinates

Li S S i i P i i l

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Linear Systems Superposition Principle

u y

( u1 + u2) = (u1) + (u2)

u1,

u2 : signals , : real numbers

Controller Canonical Form, Example n=3 and m=1127


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y(3)

(t ) + a 2 y(2)

(t ) + a 1 y(1)

(t ) + a 0 y(t ) = b1 u(1)

( t ) + b0 u (t )

Known IO description of system:

Define auxiliary variable (t):

(3) ( t ) + a 2 (2) ( t ) + a 1

(1) ( t ) + a 0 ( t ) = u ( t )

Find y(t) using superposition principle:

y(t ) = b1 (t ) + b0 ( t )

Define auxiliary variable (t): (3) ( t ) + a 2

(2) ( t ) + a 1 (1) ( t ) + a 0 ( t ) = u

(1) ( t )

x( i ) (t ) = d i x

dt i(t )

y(0) = y(1) (0) = y(2 ) (0) = 0

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Realization of (t) :

(2) (1) (3)u+

a 2

a 1

a 0

-

-

-

(2) (1) (3)u+

(1)ddt

u

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Realization of (t) :a 2

a 1

a 0

-

-

-

(2) (1) (3)u+

a 2

a 1

a 0

-

-

-

ddt

(2) (1) (3)u+

a 2

a 1

a 0

-

-

-

1.

2.

3.

Superposition principle and coordinates

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(2) (1) (3)u+

b1

b0 y

a2

a 1

a0

+

+

-

-

-

x1 x2 x 3

=

0 1 0

0 0 1

a 0 a1 a 2

x1 x2 x 3

+

0

0

1

u

y = b0 b1 0[ ] x1 x2 x 3

+ 0[ ]u

State-space description incanonical coordinates(arbitrary but convenientchoice):

x 1= , x 2 = (1) , x 3= (2)

x 1 x 2 x 3

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Lesson 6


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Is there a shortcut?

Yes, there is Laplace transformation!

General solution powerful, but difcult, in

particular if u(t)=u( x(t)) (feedback):

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Denition of Laplace transformation for a signal x ( t ) 140


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Function x:

Function X :

Transformation is reversible, no information is lost.

s = + j


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Proof:142


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Generalization:

d

dt x (t ) e st dt

0

= x (t ) e st |0 x (t ) (s ) e st dt 0

= x (0) + s x (t ) e st dt 0

= x (0) + s X (s )

The Laplace transforms of themost important signals

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most important signals

Note: all signals = 0 for t


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Ratio of two (real-coefcient) polynomials

Strictly proper, i.e., m < n

The transfer function is the Laplace transform of theimpulse response

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impulse response

Using internal description ( d =0)

Therefore:

(c and b are constant rank-1 vectors)

Therefore:

Laplace transforms and IO system description146


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Assume all initial conditions = 0 (makes sense in an IO setting)

s n Y ( s ) + a n 1 sn 1

Y ( s ) + + a 1 s Y ( s ) + a 0 Y ( s ) =

b m smU ( s ) + + b1 s U ( s ) + b 0 U ( s )

s n + a n 1 sn 1

+ + a 1 s + a 0[ ]Y ( s ) = b m s m + + b 1 s + b 0[ ]U ( s )

Y ( s ) = b m sm

+ + b 1 s + b 0s n + a n 1 s

n 1+ + a 1 s + a 0

U ( s ) = ( s ) U ( s )

Relationship between IO and internal description147


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( s ) =b

m s m + + b

1 s + b

0

s n

+ a n 1 s n 1

+ + a 1 s + a 0=

b (s )

a (s )

Same system, two descriptions of IO dynamics

derived using internaldescription

derived using IO

measurements only

( s ) = ( s ) iff the system is minimal, i.e., completelycontrollable and completely observable

Otherwise pole/zero cancellations must occur. These cancellationsremove the not controllable and/or not observable modes.

Dealing with delays148


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There is no nite dimensional state-space description of that system

d dt

x(t ) = A x(t ) + b u(t ), y(t ) = c x(t )

but with Laplace transform (use shift law of Table 6.1)

Note that (s) is not rational anymore

m

F (t)

F (t)rExample Geostationary Satellite149


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r(t)

M

R (t)

( )

d dt

x1 x2 x3 x4

=

0 1 0 0

3 0

2 0 0 2 0 r0

0 0 0 1

0 2 0 r0 0 0

x1 x2 x3 x4

+

0

1

0

0

u1 +

0

0

0

1/ r0

u 2

y = 0 0 1 0[ ] x

measure azimuth

Transfer function from tangential thruster to azimuth angle:

u 2 y

Two zeros at

1/ 2= 3

0

Four poles at 1/ 2 = j 0 , 3/ 4 = 0

Transfer Function from radial thruster to azimuth angle:150


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u 1 y

Pole-zero cancellation (one pole and one zero at s=0)

What is the system-theoretic explanation for that? The system isnot completely controllable with radial thruster only:

Therefore minimal realization has order < 4 (here 3) and onemode does not inuence the transfer function.

Quick Check : For a system dened by its internal description

1 3 1

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x 1

x 2

=

+ 1 3

0 1

x 1

x 2

+

1

1

u

y = 0 1[ ] x 1

x 2

Compute its transfer function (s).

Is the system minimal?

Eigenvalues and poles?

Is the system stabilizable?


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( )

1

Transfer functions (use superposition principle)154


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y1 = 1 (u 2 y 1 ) y1 =1

1 + 2 1 u

y 2 = 2 1 (u y 2 ) y 2 = 2 11 + 2 1 u

y = y 1 + y 2 y =(1 + 2 ) 11 + 2 1

u =(1 + 2 ) 11 + 2 1

Laplace transformations


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For the design of feedback control systems the

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transfer function

is the most important object.

Using Cramers rule

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QC:

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Q

DC gain of

DC gain of

( s ) =2 s + 3

s2

+ 3 s + 2

( s ) =s + 3

s + 7

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Test for BIBO stability:

Main result (LTI systems):


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The Tent-Pole Theorem

s 2

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Example: (s)= s+2(s2+2s+5)( s+4)

Map: s | (s)|

= -2 = -4, -1 j 2

| (s)|


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Computation residuum169


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Example: i=1, i=-1, i=1

1,1 = lims 11

(1 1)!d (11)

ds (11)s + 2

( s + 1)( s + 3) 2 (s 2 + 1)(s + 1)

=11

1 + 2(1 + 3)2 ((1) 2 + 1)

=1

4 2=

18

Contribution of this part to u(t)1

8

1

(1 1)! t (1

1) e (

1 t )=

1

8 e

t


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Poles:

Parameters:

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(linearity, table 6.2, and damping law)

Y (s)

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Y (s)

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QC

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If the real part of all zeros is negative then the systemis minimum phase . If the real part of at least one zerois positive the system is non-minimum phase.

The zeros are those frequencies for which a non-zeroinput u*(t ) and initial conditions x*(0) exist that producea zero output y*(t)=0

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a zero output y (t) 0.

(s)

M -1 Adj (M)c Adj (sI-A) b

(s)

( s ) =b

1s + b

0

s 2 + a1s + a

0

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Result:184


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< k < +

| k | case no finite zero

| k | 0 case worst finite zero


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m

l

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Lesson 8

Computation of output signals for low-order systems:189


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(partial fraction expansion)

(plus linearity of Laplace transf.)

(for complex poles useEuler theorem)

Example:

(t ) 4 (t ) 6 (t ) 4 (t ) 3 (t ) 6 (t ) 4 (t )

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y(t ) + 4 y(t ) + 6 y(t ) + 4 y(t ) = 3u (t ) + 6u (t ) + 4u (t )

u(t ) = h(t )

Y (s) =3s 2 + 6 s + 4

s 3 + 4 s 2 + 6 s + 4

1

s (all i.c. zero)

Y (s) =1

s

1

s + 2+

1

(s + 1) 2 + 1 (p.f.d.)

y(t ) = h(t ) 1 e 2 t + e t sin( t )[ ] (linearity & damping law)

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Y (s)

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k =193


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20o

70o

10o15 o

40o


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Example: non-minimumphase system196


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"no slip"

x(t )=v(t ) cos( z(t )).d dt

momentarycenter of rotation

model ODE:

v(t )=u (t )d dt 1

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y(t )

x(t )

l

z(t )

v(t )

w(t )

w(t )

l tan( w(t ))

w(t )=u (t )d dt 2

y(t )=v(t ) sin( z(t )).d

dt z(t )= tan( w(t )).d dt l

v(t )


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Experiment

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(s )

astab

Explanation

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Input:

Output:

System astab 201


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therefore

where

m cos( t + ) = m cos( t )cos( ) sin( t )sin( )[ ]

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= m cos( t )cos( ) m sin( t )sin( ) = cos( t ) + sin( t ) = m cos( ), = m sin( )

2 + 2 = m 2 cos 2 ( ) + m 2 sin 2 ( ) = m 2

= m sin( )m cos(

)

= tan( )

Connection and , or m and with (s )?203


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Therefore

Therefore204


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or, with

the final result

Main result

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(s )

astab

Nyquist Diagram, example206


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frequency is animplicit parameter

2 2

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( j ) =

0

( j ) 2 + 2 j + 02 =

0

( 02 2 ) + j (2 )

=

0

2 ( 02 2 ) j (2 )

( 02 2 ) 2 + 4 2 2=

re(

)+

j

im(

)

re(

)=

02

02 2[ ]

( 02 2 ) 2 + 4 2 2 , im(

)=

02 2

( 02 2 ) 2 + 4 2 2

Bode Diagram, example208


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frequency is anexplicit parameter

omega_0=1; !

delta=0.2; !

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omega=logspace(-1,1,1000); !num=omega_0^2; !

den=[1,2*delta*omega_0,omega_0^2]; !

[m,phi]=bode(num,den,omega); !subplot(211); !

semilogx(omega,20*log10(m)); !

subplot(212); !semilogx(omega,phi) !

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1

s2

+ s + 1

1s ( s + 3)

s + 1

s2

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Relative Degree216


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1

s2

+ s + 1

s + 3

s2

+ 3 s + 3

s + 2

s2

+ 3 s + 3

0.05 s 2 + 0.1 s + 0.2

s2

+ 0.2 s + 0.2

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( s ) =1

z

s + 1

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( s ) =

zs + 1

1

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Lesson 9

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Physical interpretation (for a mechanical system, similar for allother areas of system dynamics)

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Remark: small damping added in all subsequent plots/calculationsto avoid numerical problem; this has no inuence on the results

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Nominal model described by TF

True plant described by unknown TF

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Set of possible TF

Uncertainty generator

Uncertainty bound

Quick Check:

100 1 mm

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100 1 mm

What is the nominal model?What is the uncertainty generator?What is the uncertainty bound?Is this family equivalent to the family


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Step 1: fit a nominal model:238


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Step 2: true plant

satisfies

therefore

Step 3: find a low-order bound239


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Where are we now?

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loop gain

Definitions242


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sensitivity

complementarysensitivity

return diference

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If plant and/or controller unstable then CL

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system astab iff all following 9 (4) TF astab:

If both P(s) and C(s) are astab then it is sufcientto check that S(s) or T(s)are astab


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Case n+=n0=0247


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L( j )

T ( j ) astab?

Case n+=n0=0248


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L( j )

T ( j ) astab?


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return difference

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Re

Im-1

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log

||

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0

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0

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| L( j )|

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log 0

Test plant:

dominant NMP zero

dominant unstable poleastab and MP plant,but uncertain


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Interpretation:264


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upper limit for

Example:

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T (s)= k n(s)(s)

ImPoles of the CL system for varying k :

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Re

OL poles ( k =0)

OL zeros ( k =0)

Generalization267


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log

| L( j )|

0

rule of thumb:

Interpretation

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Time delays are similar to NMP zeros.

e-sT 1/T

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k p

e -sT

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Simplest case:

L( j ) = k /(a-j ) = k (a+j )/(a2+ 2)n+ = 1, n0 = 0

k > 0 not OK

k < 0 can be OK271


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| ( )|

Generalization273


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log

| L( j )|

0

rule of thumb:

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Maximum-Modulus Theorem:

Robst Stability Theorem:

Therefore:

Interpretation:

Both NMP zeros and unstable poles:275


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Lesson 11

Good Design278


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Why Not?100 dB

|L(j )|

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log( )

log( )

-100 dB

-45

Arg{L(j )}

If:

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Then:


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r(t ), = h(t )

uP

yrC

-e

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Rise time t 90Max overshoot

time

CL step response

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Step1: CL TD -> CL FD 285


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Step2: CL FD -> OL FD

=>

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L( j ) =

02

j ( j + 2 0 )

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cross-over frequency

phase margin

insert

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uP

yrC

-e

Specs:290


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=>

Controller has only 1 DOF =>

Phase P(j) = Phase L(j) 291


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c=0.65 rad/s

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|P(j)|

| L(j)|

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= 0.53

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Frequency-Domain Closed -Loop Specs 295


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Definition:

Im

L(j )

S max

1/ S max

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Re(j )

1 +L(j )

-1


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Recap: 299


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Easy to communicate with non-experts using closed-loop

and time-domain quantiers, e.g.:

rise time

max. overshoot

t 90

Step 1:

Approximate closed-loop system

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R(s)

because the exact solution y(t)of the latter is known.

time-domain -> frequency domain

{t 90 , }

{ 0 , }

by a second-order system Y(s)

Step 2:

Transform closed-loop specs to open-loop specs !!

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Plot results and derive approximations

{ 0 , } { c , }

Recipe1. Choose desired rise time and maximum

overshoot of the CL system

2 Compute cross-over frequency and phasei f h O i

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2. Compute cross over frequency and phasemargin of the OL system using

3. Check if these specs are realizable

4. Find a controller C(s) that, for a given plant P(s),realizes these specs (Chapter 11, 12, and 13)

5. Check nominal and robust stability of CL system

303An Old Exam Question


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a) Since plant is type = 1

P ( s ) =1s

1s + 10

(1 10 s ) = 10 s + 1s ( s + 10)

+ = 0.1305


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a) Since plant is type 1

b) NMP zero limits crossover frequency!

c) No overshoot requires

Arg{ L( j c )}=

Arg{ P ( j c )}=

110 c 0.03

C (s) = k p

c

< 0.1 t 90 > 1.7 /0.1 = 17

70

| P ( j c ) | 10 dB k p 10 dB 0.3

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c

0.03

t 90 1.7 /0.03 56 s

Arg{ P ( j

c)} = 110

(finally )

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(finally )

First approach: choose a suitable controller structure and optimize the parameter of that controller.

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90-95% of all controllers are PI(D) controllers

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l ( )

dB

Why is PI(D) so common?310


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0

|P ( j )|

log( )

-20 dB/dec

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|L(j )|

c

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|C ( j )|

Performance OK313


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Stability?Robustness?


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Main assumptions:

Real plant is well approximated by

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Ratio of time constants

must be small (below 0.3)


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Interpretation

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1: Set

2:

= small318


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measure

3: Choose


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P ( j ) =1

1 + 2( )5 e

j arctan { }5

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kp*| P( j )|

|P(j)|

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Arg{ P( j )}

|P( j )|

P(s)

d

yC(s)

-

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Limits of Ziegler-Nichols Approach

> 0.3

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Iterate

Objective: improve shape of loop gain

Badly damped resonances, NMP zeros,

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Numerical solution:

y(t)

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t

y(t)

ZN

LS

1 2


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Lesson 13

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Lead

Lag

Lead

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Lag

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Recap: How to Design a PID-Controller?

Ziegler-Nichols

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Parametric loop shaping

Lead/Lag loop shaping

{k p ,T i,T d , }

Plant:

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Specs:

In a second step, a 2nd-order Lead is added

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C 2(s)=C l2(s).C(s)

l2

with

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There is much more in RT II:

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and in the subsequent lessons

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Starting point: {P(s), W 2(s)}

Objective: find a C (s) that takes into account boththe nominal plant and the uncertainty

Possible Model-Error Structures339


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Dangerous!

Rule #1: never cancel unstable poles or non-

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minimumphase zeros!

Rule #2: never apply these methods if the nominaland the true plant have not the samenumber of unstable poles or NMP zeros!

Rule #3: use plant inversion methods with caution.

Case 1:

Known: W 2 (s) with | W 2 ( j ) | < 1, < 2

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Desired loop gain

Plant343


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Resulting controller


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Why not? Example:

Case 2:

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Loop shaping using PID/Lead/Lag/ valid approach

L ( s ) = C (s ) P ( s ) =( s + 1)

( s 1) s

s 1

( s + 1)=

1

s

Why not? Example:

What is wrong?

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2 (not restricting)

First iteration

Choose suchthat cross-over frequency is and 0.2

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that cross over frequency is andloop rolls off with -20 dB/dec around crossover

Problem: phase reserve is not OK, therefore

c

0.2

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P(s)

d

yC(s)

-

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trade-off robustness versus performance

| 1 + L ( j ) |

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| 1 + L1 ( j ) |

| 1 + L2 ( j ) |

| W 2 ( j ) L2 ( j ) |

| W 2( j ) L

1( j ) |

y

Case 3:358


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P(j )

-1

Re

=0+

+

or not feasibleC (0) +feasible, but phase shift must be >180

(plant has -180, controller has -180 at =0, loopmust reach at least -179, )

C (0) C (0)

|P( j )||C ( j )|

| L( j )|

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Therefore to reach more than+180 phase shift

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to make controller realizable

first high-pass than low-pass bevior

1

L( j ) C ( j )

1

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=1P( j )

=1

=1

L 14

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Lesson 14

1. Extension:364


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2. Extension:

Design of C (s) in the OL/FD

design of C (s) in the CL/TD

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1. Design inner (fast) loop without consideringouter loop. Main objective: speed

2. Design outer (slow) loop with inner loop active

(closed). Main objective: accuracy

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Case 1: output-feedback design367


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output-feedback controller (aggressive!)

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Case 2: cascaded-feedback design, part 1369


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inner feedback (fast) controller

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Case 2: cascaded-feedback design, part 2

outer transfer function with inner loop closed

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outer feedback (accurate) controller

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Bridge between classical and modern methods

Works in the time domain and closed loop

MUST be accompanied by a post-design frequency

domain check

Basic Idea

Plot poles of CLOSED-LOOP system as a functionof STATIC loop gain

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=> poles = solution of

Main assumption: L(s)asymptotically stable

(if L(s)unstable see Section 9.3)

Poles closed loop dened by rational equation:

Pro Memoria:378


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Poles closed loop dened by rational equation:

Characteristic polynomial:

Poles

Main Rules1. The root locus is a set of curves in the complex plane2. The curves start ( k =0 ) at the poles of the OL system

3. For k approaching infinity, m of the curves approachthe m finite zeros of the OL system

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4. The n-m=r other curves go to infinity approachingstraight asymptotes

5. The center of these asymptotes is at a + j 0 6. The angle of the asymptotes is i 7. A point in the complex plane is part of the root locus

iff it satisfies the phase condition

Center of asymptotes ( r straight lines)

Rules for asymptotes380


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Angle of asymptotes

i = 1, , n m > 0

L ( s ) =b (s )a (s )

=s + 1

(s + 2) s 2 + 2 s + 3( )

Root locus = curve of solutions of

Example:381


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Root locus curve of solutions of

Why is this a one-dimensional curve?

Can we choose any point in the complex plane to be

on the root locus?

Re

Im

a ( s ) = 0

b ( s ) = 0

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Re

1=

2

=

3=

1 =

a =1

[ ]

Number of asymptotes =

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Phase condition (most useful!)A point z is part of the root locus iff:

zeros of L(s) poles of L(s)

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i = 1,0,1,

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L(

s)

=b (s )a (s )

=b

m (s 1) (s 2 )

(s 1) (s 2 ) (s 3 )

arg L (s ){ }= arg b (s ){ } arg a (s ){ }

=

args

1{ }+

[ ]

args

1{ }+

[ ]

Re

Im

Illustration: z

z- 2 z- 1

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poles 1, 2zero

( + ) =

1

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Plant:

Specs:

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1

2

3

Controller:

spec 1 OK

spec 2 and 3 to be satised by appropriatechoice of the poles of the CL system

2

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T ( s ) =

0

s 2 + 2 0 s + 02 where (Sect. 10.3)

= ( ) = log(0.02)

2 + log(0.02) 2 0.78

0 = 0 ( , t 90 ) = 0.14 + 0.4 [ ]2

1.3 2.2 rad / s

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C 1 (s) = k p = 1 C 2 (s) = k p (s + ) = s +

C 1 (s) = k p = 1

C (s)1

open-loop zero

open-loop pole

closed-loop pole

Im

2

4

6

k = 0

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desired pole

p p

Re-5 0

-6

-4

-2

0

n-m=2

=0.5(-1-3)=-2

C (s)2

open-loop zero

open-loop pole

desired pole

closed-loop pole

Im

-2

0

2

4

6

( + ) =

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Re-10 -5 0

-6

-4

Outlook: Better control system design methods, in

particular for MIMO and nonlinear systems

Realization aspects (SW and HW, sensors,

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Realization aspects (SW and HW, sensors,actuators, , COST)

System modeling and experimental validation

Applications

State Feedback

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Reset (Integrator) Windup

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Gain Scheduling

( how to cope with real-life problems )

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400

Hardware I


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401

Hardware II

slides fall 10

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