slides fall 10
TRANSCRIPT
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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
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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
-
-
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(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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165
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The Tent-Pole Theorem
s 2
166
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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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171
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Poles:
Parameters:
172
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174
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(linearity, table 6.2, and damping law)
Y (s)
175
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(linearity, table 6.2, and damping law)
Y (s)
176
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177
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178
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179
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180
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QC
181
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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
182
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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
183
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Result:184
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185
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< k < +
| k | case no finite zero
| k | 0 case worst finite zero
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m
l
187
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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 )
190
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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)
191
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(linearity, table 6.2, and damping law)
Y (s)
192
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k =193
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194
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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
197
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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
199
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(s )
astab
Explanation
200
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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( )[ ]
202
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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
205
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(s )
astab
Nyquist Diagram, example206
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frequency is animplicit parameter
2 2
207
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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; !
209
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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) !
210
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211
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212
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213
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214
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1
s2
+ s + 1
1s ( s + 3)
s + 1
s2
215
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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
217
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219
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220
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221
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( s ) =1
z
s + 1
222
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( s ) =
zs + 1
1
223
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224
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Lesson 9
225
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226
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227
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Physical interpretation (for a mechanical system, similar for allother areas of system dynamics)
228
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229
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Remark: small damping added in all subsequent plots/calculationsto avoid numerical problem; this has no inuence on the results
230
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231
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232
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233
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234
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Nominal model described by TF
True plant described by unknown TF
235
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Set of possible TF
Uncertainty generator
Uncertainty bound
Quick Check:
100 1 mm
236
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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?
240
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loop gain
Definitions242
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sensitivity
complementarysensitivity
return diference
243
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If plant and/or controller unstable then CL
244
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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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246
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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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250
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252
-
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return difference
253
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Re
Im-1
254
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255
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log
||
256
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0
258
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0
259
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260
-
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261
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| L( j )|
262
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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:
265
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T (s)= k n(s)(s)
ImPoles of the CL system for varying k :
266
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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
268
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Time delays are similar to NMP zeros.
e-sT 1/T
269
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k p
e -sT
270
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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:
274
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Maximum-Modulus Theorem:
Robst Stability Theorem:
Therefore:
Interpretation:
Both NMP zeros and unstable poles:275
-
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276
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277
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Lesson 11
Good Design278
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Why Not?100 dB
|L(j )|
279
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log( )
log( )
-100 dB
-45
Arg{L(j )}
If:
280
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Then:
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282
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r(t ), = h(t )
uP
yrC
-e
283
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Rise time t 90Max overshoot
time
CL step response
284
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Step1: CL TD -> CL FD 285
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Step2: CL FD -> OL FD
=>
286
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L( j ) =
02
j ( j + 2 0 )
287
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cross-over frequency
phase margin
insert
288
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289
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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
292
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|P(j)|
| L(j)|
293
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= 0.53
294
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Frequency-Domain Closed -Loop Specs 295
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Definition:
Im
L(j )
S max
1/ S max
296
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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
300
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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 !!
301
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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
302
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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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304
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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
306
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c
0.03
t 90 1.7 /0.03 56 s
Arg{ P ( j
c)} = 110
(finally )
307
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(finally )
First approach: choose a suitable controller structure and optimize the parameter of that controller.
308
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90-95% of all controllers are PI(D) controllers
309
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l ( )
dB
Why is PI(D) so common?310
-
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0
|P ( j )|
log( )
-20 dB/dec
311
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|L(j )|
c
312
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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
315
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Ratio of time constants
must be small (below 0.3)
-
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Interpretation
317
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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
320
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kp*| P( j )|
|P(j)|
321
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Arg{ P( j )}
|P( j )|
P(s)
d
yC(s)
-
322
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323
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Limits of Ziegler-Nichols Approach
> 0.3
324
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Iterate
Objective: improve shape of loop gain
Badly damped resonances, NMP zeros,
325
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Numerical solution:
y(t)
326
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t
y(t)
ZN
LS
1 2
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328
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Lesson 13
329
-
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Lead
Lag
Lead
330
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Lag
331
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332
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Recap: How to Design a PID-Controller?
Ziegler-Nichols
333
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Parametric loop shaping
Lead/Lag loop shaping
{k p ,T i,T d , }
Plant:
334
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Specs:
In a second step, a 2nd-order Lead is added
335
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C 2(s)=C l2(s).C(s)
l2
with
336
-
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There is much more in RT II:
337
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and in the subsequent lessons
338
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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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340
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Dangerous!
Rule #1: never cancel unstable poles or non-
341
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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
342
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Desired loop gain
Plant343
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Resulting controller
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345
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346
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348
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349
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Why not? Example:
Case 2:
350
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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?
351
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2 (not restricting)
First iteration
Choose suchthat cross-over frequency is and 0.2
352
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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
353
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354
-
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355
-
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P(s)
d
yC(s)
-
356
-
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trade-off robustness versus performance
| 1 + L ( j ) |
357
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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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-
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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 )|
360
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Therefore to reach more than+180 phase shift
361
-
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to make controller realizable
first high-pass than low-pass bevior
1
L( j ) C ( j )
1
362
-
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=1P( j )
=1
=1
L 14
363
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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