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Chapter 5 Design using Transformation Technique – Classical Method. 0. time. Transient response. D( s ). G( s ). +. -. unity feedback. Steady-state response. Rule of Thumb. ex). Design by Emulation. Design specifications: Overshoot to a step input less than 16%. - PowerPoint PPT Presentation
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Robotics Research Labo-ratory
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Chapter 5
Design using Transformation Technique – Classical Method
Robotics Research Labo-ratory
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2
2 2( )
2n
n n
ωH s
s ζω s ω
σ
jω
nω dω
σ 0θ
2
cos
1n n d
θ ζ
s ζω jω ζ σ jω
Robotics Research Labo-ratory
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( )y t
1.0 9
pTrT
.0 1
1%
sT
time
( ) 1 cos sinσtd d
d
σy t e ω t ω t
ω
Transient response
Robotics Research Labo-ratory
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2
2
/ 1
2
1.8 rise time
4.6 4.6 settling time
1
0 <1 %OS
0.6(1 )
PM (phase margin)100
1 2
rn
sn
pd n
πζ ζp
p
BW n
Tω
Tζω σ
π πT
ω ω ζ
M e ζ
ζ M
PMζ
ω ω ζ
4 24 4 2ζ ζ
Robotics Research Labo-ratory
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Steady-state response
G(s)D(s)( )R s ( )Y s( )E s
unity feedback
+ -
0 0
( ) 1 1 or ( ) ( )
( ) 1 ( ) ( ) 1 ( ) ( )
1lim ( ) lim ( )
1 ( ) ( )ss s se sE
E sE s R s
R s D s G s D s G
s s R sD s s
s
G
Robotics Research Labo-ratory
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0
0
If the system is stable,
1Type 0 , lim ( ) ( ) for step input
1
1Type 1 , lim ( ) ( ) for ramp input
1Type 2 ,
ss p sp
ss v sv
ssa
e k D s G sk
e k sD s G sk
ek
2
0lim ( ) ( ) for parabolic inputa s
k s D s G s
1 2 3
1 2 3
Remark: Forward transfer function, not the closed-loop transfer function
( )( )( ) ( ) ( )
( )( )( )
If 0, the system type is zero.
If 1, the system ty
n
s z s z s zD s G s
s p s p s p
n
n
s
pe is one.
Robotics Research Labo-ratory
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Rule of Thumb
A reasonable choice of is one that results in at least
4 samples in the rise time.
For better and smoother control result, more than 10
samples in the rise ti
T
me.
1.8 1.8 1.8
10 10 2 35
should be 35 times faster than the natural frequency .
s sn
r
s n
ω ωω
T T π
ω ω
ex)
Robotics Research Labo-ratory
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1.83 / sec
For 0.125sec, 50 / sec
For 0.25 sec, 25 / sec
For 0.5sec, 12.5 / sec
For 1.0 sec, 6.28 / sec
n
s
s
s
s
ω rad
T ω rad
T ω rad
T ω rad
T ω rad
Robotics Research Labo-ratory
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Design by Emulation
) Design of Antenna Angle-Tracker Servo Controller
in p.215 of Frankin's
1 ( ) System type ?
(10 1)
ex
G ss s
Design specifications:
1. Overshoot to a step input less than 16%.2. Settling time to 1% to be less than 10sec. 3. Tracking error to a ramp input of slope 0.01 rad/sec to be less than 0.01 rad.4. Sampling time to give at least 10 samples in a rise time.
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Transient response
i) %OS 0.16 0.5
4.6 ii) 0.46s
ζ
σ T
-0.46 0
forbidden region
cos 0.5 60oθ θ
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0 0
Steady-state response
iii) Type 1 system
1 lim ( ) ( ) lim ( )
(10 1)
0.01 1
0.01
10 1 ( ) - one of many solutions
1
v s s
vss
k sD s G s sD ss s
Rk
e
sD s
s
( ) ( )
( ) ,( ) ( )
( ))( )
,
. .. .
2
Since 1
1 1
1 3 31 1( 2 2 2 2 Since 1
1 8 1 8 from , 0 18 0 2
10
n
rn n
D s G sH s
D s G s
H ss s s j s j
ω
T Tω ω
Robotics Research Labo-ratory
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1
1
( ) 9.15( 0.9802) 9.15(1 0.9802 )( )
( ) 0.8187 1 0.8187
( 0.8187) ( ) 9.15( 0.9802) ( )
Forward
( 1) 0.8187 ( ) 9.15( ( 1) 0.9802 ( ))
U z z zD z
E z z z
z U z z E z
u k u k e k e k
0.11
1
z 1 0
Pole-Zero Mapping
0.9802 iv) ( )
0.8187
lim ( ) lim ( ) 1
0.9802 ( ) 9.15 where 0.2
0.8187
T
T
s
z z z e zD z K K K
z p zz e
D z D s
zD z T
z
Robotics Research Labo-ratory
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1 sTe
s
r(t) u(k) c(t)e(t) e(k)
T=0.2
.
. .
0 8187 1
9 15 0 982 1
u k u k
e k e k
1
10 1s s
10 1
1
s
s
1
10 1s s R E C
( ) . ( ) . ( ( ) . ( ))
Backward
0 8187 1 9 15 0 9802 1u k u k e k e k
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2
ex) Antenna angle-tracker servo with slow sampling in . 220
1 0.1 ( )
( 0.1)
0.99340.00199
( 1)( 0.9802)
1sec
0.9802 ( ) 9.15
0.8187
1 0.( )
0.2sec
p
zG z
z s s
z
z z
zD
z
z
T
T
z
zG z
z
z 2
1 0.96720.0484
( 1)( 0.9048)( 0.1)
0.9048 ( ) 6.64
0.3679
z
z zs s
zD z
z
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10-1
100
101
10-2
10-1
100
101
Bode plot of the continuous design for the antenna control
Mag
nitu
de
10-1
100
101
-180
-160
-140
-120
-100
-80
Pha
se, d
egre
es
Phase Margin of continuous-time system 51.8PM
cpω
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0.2 4.5 47.3
1 23 28.8
T φ PM
T φ PM
Remark: ( ) sin2
sample and hold (sec)2
sample and hold ( ) 2
where is the crossover frequency.
e
time delay
phase
x) In this exampl
de
2
lay
e
ho
cp
cp
ωT ωG jω
T
Tω rad
ω
T
, 0.1 sec and 0.8 / sec 45.8 / sec2
ocp
Tω rad
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T=0.2 T=1
0 2 4 6 8 10 12 14 16 18 20-0.5
0
0.5
1
1.5
2
OU
TPU
T,
Y
and
CO
NTR
OL,
U
/10
TIME (SEC)
Fig. 7.7 Step response of the 1-Hz controller
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Direct Design by Root Locus
in the z-plane The method for continuous-time systems can be extended w/o modifi-
cation The effects of the system gain and/or sampling period can be investi-
gated
Performance
-1 10 Rez
0ζ
Imz
damping ( %OS )
2
( )
1
d
d
nn
σ jω TsT
jω TσT
jω T ζζω T
z e e
e e
e e
1ζ
0.8ζ
0.6ζ
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s-plane1ζ ζ
0 -1 1
Overshoot
1σ2σ
1djω1dω
1σ Te
Settling time s-plane z-plane
z-plane
1djω1dω
2σ Te
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0.5
ex) From the given specification for transient response
0.5, 1 / sec, 0.2sec
For 0.9048
For damping ratio ( ) 0.5
For
n
Ts
p
ζ ω rad T
T r e
M ζ
2 ( ) 1 0.75
p p d nd
πT T ω ω ζ
ω
-1 1
nζω TσTr e e 1dω
1dω
Robotics Research Labo-ratory
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1
Consider the steady-state error
( ) ( )
1 ( ) ( )
( ) where ( ) 1
Remark: System type of discrete-time control system
the number of open-loop poles at 1
1 ( ) (
( 1)
)N
R zE z
D z G z
G sG z z
s
z
B z
zG z
A
z
( )
where B(z)/A(z) contains neither a pole nor zero at 1
z
z
Robotics Research Labo-ratory
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1
For a unit step input
1( )
11 ( ) ( )
1( ) lim( 1)
11 ( )
1 1
1 (1) (1) 1
Type 0 finite error
Type 1 zero
z
p
zE z
z D z G z
ze z
z D z G z
D G k
Position error constant
Robotics Research Labo-ratory
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2
21
1 1
For a unit ramp input
1( )
1 ( ) ( )( 1)
1( ) lim( 1)
1 ( ) ( )( 1)
lim lim( 1)(1 ( ) ( )) ( 1) ( ) ( )
1
z
z z
v
TzE z
D z G zz
Tze z
D z G zz
Tz Tz
z D z G z z D z G z
k
Velocity error constant
2
21
2
21
Similarly,
( ) lim( 1) (1 ( ) ( ))
1 lim
( 1) ( ) ( )
z
za
Tze
z D z G z
Tz
kz D z G z
Acceleration error constant
Robotics Research Labo-ratory
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1 2
1 2
the overall transfer function
Assume
( )( ) ( ) ( )
( )( ) (
Remark: ( Truxal's Rule )
( )( ) :
( )
and ( ) results
)
from a Type 1 sy
For unit
stem of (
step
).
, st
n
n
z z z z z zH z K
z p z
Y z
p z p
G
H zR z
H zz
eady-state error must be zero
( ) ( ) ( ) ( ) 0
(1) 1 --- (1)
( ) ( ) 1 Remark: ( ), ( ) ( ) ( ), and
( ) ( ) 1 ( )
E z R z H z R z
H
Y z E zH z Y z G z E z
R z R z G z
Robotics Research Labo-ratory
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2
21
1
For a ramp input
( ) ( )(1 ( ))
(1 ( ))( 1)
1( ) lim( 1) (1 ( ))
( 1)
1 1 ( ) lim --- (2)
1
We can use L'Hopital's rule.
zv
zv
E z R z H z
TzH z
z
Tze z H z
kz
H z
Tk z
Robotics Research Labo-ratory
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1 1
1
1
1
1 ( / )(1 ( )) ( )lim lim
( / )( 1)
1Note : ln ( ) ( ) ( ) since (1) 1
( )
1lim ln ( )
( ) lim ln
( )
lim ln(
z zv
zv
i
zi
iz
d dz H z dH z
Tk d dz z dz
d d dH z H z H z H
dz H z dz dz
dH z
Tk dz
z zdK
dz z p
dz z
dz
1
1 1
) ln( ) ln
1 1 lim ( ) ( )
1 1 1
1 1
i
i
n n
i i i
ii i
v i
z
z p K
d dz z z p
z z dz z p
Tk p z
dz
Robotics Research Labo-ratory
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Imz
Rez-1 0 1
pole
zero
pole 0
ze ro 1vk
Large overshoot Poor dynamic response
Small steady-state error against Good transient response
Errors are decreased
1 1
1 1 1
1 1
n n
i iv i iTk p z
Robotics Research Labo-ratory
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0.5
Characteristic polynomial
0.9672 1 0.0484 0
( 1)( 0.9048)
Design specifications :
1 , 10sec 0.5 , 1sec
4.6 0.5 0.61
v s
s
zK
z z
k T ζ T
σ r eT
0.9672( ) 0.0484 where 1 sec
( 1)( 0.9048)
( ) ( proportional controller - static gain)
( )( )
1 ( )
zG z T
z z
D z K
KG zH z
KG z
ex) Antenna system in p. 228 of Franklin’s
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-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0.1p/T
0.2p/T
0.3p/T
0.4p/T0.5p/T
0.6p/T
0.7p/T
0.8p/T
0.9p/T
p/T
0.1p/T
0.2p/T
0.3p/T
0.4p/T0.5p/T
0.6p/T
0.7p/T
0.8p/T
0.9p/T
p/T
0.10.2
0.30.40.50.60.70.80.9
Discrete root locus with and without compensation
Real Axis
Imag
inar
y A
xis
1 2 1
1 2 1
without compensator ( )
( 1.0, 0.9048, 0.9672)
0.9048with ( ) 6.64 , 1sec (emulation)
0.3679 ( 1.0, 0.3697, 0.9672)
D Z K
p p z
zD z T
zp p z
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The system goes unstable at 19 where 0.92.
That means there is no value of gain that meets
the steady-state specification.
"Trial and error" based controller design
vK k
Imz
Rez-1 10 0.61
ζ
r
Robotics Research Labo-ratory
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0.9048 ( ) 6.64 1sec (design by emulation)
0.3679
Direct design using root locus (trial and error)
0.8) ( ) 6
0.05
0.88) ( ) 13
0.5
0.8) ( ) 9 (hidden ocillation)
0.8
)
zD z T
z
zi D z
z
zii D z
z
ziii D z
z
iv
0.88 ( ) 13 (delay)
( 0.5)
zD z
z z
( . ~ )Dynamic compensation 228 235 pp
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-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0.1p/T
0.2p/T
0.3p/T
0.4p/T0.5p/T
0.6p/T
0.7p/T
0.8p/T
0.9p/T
p/T
0.1p/T
0.2p/T
0.3p/T
0.4p/T0.5p/T
0.6p/T
0.7p/T
0.8p/T
0.9p/T
p/T
0.10.20.30.40.50.60.70.80.9
Root locus for antenna design
Real Axis
Imag
inar
y A
xis
1 2 3 0 1
0.8 ( ) 6
0.05 ( 1.0, 0.9048, 0.05, 0.8, 0.9672)
zD z
zp p p z z
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0 2 4 6 8 10 12 14 16 18 20-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
OU
TP
UT
, Y
an
d C
ON
TR
OL,
U
/10
TIME (SEC)
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-1 -0.5 0 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0.1p/T
0.2p/T
0.3p/T
0.4p/T0.5p/T
0.6p/T
0.7p/T
0.8p/T
0.9p/T
p/T
0.1p/T
0.2p/T
0.3p/T
0.4p/T0.5p/T
0.6p/T
0.7p/T
0.8p/T
0.9p/T
p/T
0.10.2
0.30.40.50.60.70.80.9
Root locus for compensated Antenna Design
Real Axis
Imag
inar
y A
xis
1 2 3 0 1
0.88 ( ) 13
0.5 ( 1.0, 0.9048, 0.5, 0.88, 0.9672)
zD z
zp p p z z
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0 2 4 6 8 10 12 14 16 18 20-1.5
-1
-0.5
0
0.5
1
1.5
OU
TP
UT,
Y a
nd C
ON
TR
OL,
U/1
0
TIME (SEC)
Step Response of Compensated Antenna
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-1 -0.5 0 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
10.160.340.50.64
0.76
0.86
0.94
0.985
0.160.340.50.640.76
0.86
0.94
0.985
0.20.40.60.811.21.4
Root locus for Compensated Antenna Design
Real Axis
Imag
inar
y A
xis
1 2 3 0 1
0.8 ( ) 9
0.8 ( 1.0, 0.9048, 0.8, 0.8, 0.9672)
zD z
zp p p z z
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0 2 4 6 8 10 12 14 16 18 20-1
-0.5
0
0.5
1
1.5
OU
TP
UT,
Y
an
d
CO
NT
RO
L,
U/1
0
TIME (SEC)
Step response of compensated Antenna Design
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-1 -0.5 0 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
10.160.340.50.64
0.76
0.86
0.94
0.985
0.160.340.50.640.76
0.86
0.94
0.985
0.2
0.4
0.6
0.8
1
1.2
0.2
0.4
0.6
0.8
1
1.2
Root locus for compensated Antenna Design
Real Axis
Imag
inar
y A
xis
1 2 3 4 0 1
0.88 ( ) 13
( 0.5)
( 1.0, 0.9048, 0.5, 0, 0.88, 0.9672)
zD z
z
p p p p z z
z
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0 2 4 6 8 10 12 14 16 18 20-1
-0.5
0
0.5
1
1.5
2
OU
TP
UT,
Y a
nd C
ON
TR
OL,
U/1
0
TIME (SEC)
Step response for compensated antenna Design
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Frequency Response Methods1. The gain/ phase curve can be easily plotted by hand.
2. The frequency response can be measured experimentally.
3. The dynamic response specification can be easily interpreted in terms of gain/ phase margin.
4. The system error constants and can be read directly from the low frequency asymptote of the gain plot.
5. The correction to the gain/phase curves can be quickly computed.
6. The effect of pole/ zero gain changes of a compensator can be easily deter-mined.
Note : 1, 5, 6 above are less true for discrete frequency response design
using z - transform.
vkpk
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Nyquist Stability Criterion Continuous case
zeros of the closed-loop characteristic equation, n(s) + d(s) = poles of the closed-loop system, n(s)+d(s)
( ) ( ) ( )
( ) ( ) ( ) ( ) 1+
( )
( )1
n sKD s G s
H sn sKD s G ss
d s
d
( )
( )( )
(
( )( ) ( ) ( ))
n s n s d
n s
n sd s
d ssd s
known
open-loop system
closed-loop system
characteristic equation
Robotics Research Labo-ratory
Z (unknown) = # of unstable zeros (same direction) of 1 + K D(s) G(s) ( or # of unstable poles of H(s) )
P (known) = # of unstable poles (opposite direction) of 1 + K D(s) G(s) ( or # of unstable poles of KD(s)G(s))
N(known after mapping) = # of encirclement (same direction) of the origin of 1+KD(s)G(s) ( or -1 of KD(s)G(s) )
Z must be zero for stability
43
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unstable poles
0 -1
1+KD(s)G(s)-plane KD(s)G(s)-plane
Z – P =N
or Z = P + N
S-plane
-1/K
D(s)G(s)-plane
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Discrete case ( The ideas are identical )
Unstable region of the z-plane is the outside of the unit circle
Consider the encirclement of the stable region.
N = { # of stable zeros } - { # of stable poles}
= { n – Z } – { n – P }
= P – Z Z = P – N
In summary,
1. Determine the number, P, of unstable poles of KDG.
2. Plot KD(z)G(z) for the unit circle, and .
3. Set N equal to the net number of CCW encirclements of the point
-1 on the plot
4. Compute Z = P – N. This system is stable iff Z = 0.
jωTz e 0 2ωT π
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1( ) , 2
( 1)
1.135 ( 0.523)( )
( 1) ( 0.135)
1
G s ZOH at Ts s
zG z
z z
K
ex) p. 241 (Franklin’s)
The unit feedback discrete system with the plant transfer
function with sampling rate ½ Hz and zero-order hold
0, # of CCW encirclements of the point 1 0
0
P N
Z
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-2 -1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1 Nyquist plot from Example 1 using contour
Real Axis
Imag
inar
y A
xis
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Design Spec. in the Frequency Domain
Gain Margin (GM) : The factor by which the gain can be increased before the system to go unstable
Phase Margin (PM): A measure of how much additional phase lag
or time delay can be tolerated in the loop before
instability results.ex) p. 243
2
2
1( ) with ZOH at 0.2 sec
( 1)
( 3.38) ( 0.242)( ) 0.0012
( 1) ( 0.8187)
1
G s Ts s
z zG z
z z
K
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Bode plot Nyquist plot
GM=1.8, PM=18
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Sensitivity Function
• Stability robustness – stability
1( ) ( )
1 ( ) ( )
1( ) ( ) ( ) ( )
1 ( ) ( )
sensitivity function
E s R sD s G s
E jω R jω S jω R jωD jω G jω
• Tracking error – performance
Remarks:
( ) ( )i) ( ) complementary sensitivity function
1 ( ) ( )
ii) ( ) 1 ( )
D jω G jωT jω
D jω G jω
T jω S jω
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-1PM
1
GM
1VGM
Re
Im
max
1min(1 ) the distance of the closest point from -1
| |
1 1 1
1
DGS
S S
VGM S
SVGM
S
VGM : vector gain margin
( )1
1 DGS
( ) ( )ω ωD j G j plane
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Performance
1
1
1
| | | || | where is an error bound| |
| | 1
| |Define ( ) performance frequency function
| | | | 11
| | 1| |
b b
b
b
E S R e eR
Se
RW ω e
S W
W DGS
1
Remark:
At every frequency, the point on the Nyquist plot lies
outside the disk of the center -1, radius ( ) .
DG
W jω
-1
1| |W
DG
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1
If the loop gain is large | | 1
1 1 | | where
| | 1
| | | |
DG
S SDG DG
W DG
1
) 0.005, 1 below 100
0 2 200
1 = | | 2000.005
b
b
ex e R Hz
ω πf π
R We
DG
200
2001( )W ω
ωThe higher the magnitude curve at low frequency,
the lower the steady-state errors
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Robust Stability
0 2
0 2
( ) ( )[1 ( ) ( )]
multiplicative uncertainty
( ) ( ) ( ) ( )
additive uncertainty
G jω G jω w ω jω
G jω G jω w ω jω
0
2
2 2
2
where ( ) : nominal transfer function
( ) : magnitude
( ) : phase
| ( ) | 1
( ) ( ) upper bound
( ) : robust stability frequen
G s
w ω
jω
jω
w ω W ω
W ω
cy function
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02 2
0 0
2
0
2 2
0 0
1
( )
1
1
1
sB
ωKG s
s s sB
ω ω
s
ωK
s s sB
ω ω
2 ( )
2
0
2
0 0
1
s
ωw ω
s sB
ω ω
ex) p. 250
small for low frequencies
and large for high frequencies
Robotics Research Labo-ratory
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10-1
100
101
102
10-2
10-1
100
101
102
103
Fig.7.27 Plot of typical plant uncertainty
10-1
100
101
10-2
10-1
100
101
102
Fig.7.26 Model uncertainty for disk read/write head assembly
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2 2 2
2 2
Suff & Nec. Cond.
1 not to be zero any
1 1
Tw w W
T w T W
0
0
0
0 2
Define ( ) 1 ( )
complementary sensitivity function1
1Assume that the nominal system is stable, 1 0
1 0 (for stability robustness)
1 [1 ]
T jω S jω
DG
DG
DGS
DG
DG w
00 2
0
0 2
0
(1 ) 1 01
(1 )(1 ) 0
DGDG w
DG
DG Tw
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0 0
0
2
since is small for high frequencies.
1
T DG DG
DGW
200 ω
Bode plot
0DG
1
robust performance
W
2
1
robust stability
W
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2
2 0
0
2 0 0
Remark:
1
1 1
1
W T
W DGω
DG
W DG DG
2
Remark:
At every frequency, the critical point, -1, lies
outside the disk of the center , radius ( ) ( ) ( ) .DG W jω D jω G jω
Im
Re
-1
0DG
| 2 0 |W DG
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Re
Robust Performance
Im
-1
0DG
| 2 0 |W DG
1W
1
2
Remark:
For each frequency , construct two closed disks;
one with center -1, radius radius ( ) ;
the other with center , radius ( ) ( ) ( ) .
Then robust performance holds iff for each
ω
W jω
DG W jω D jω G jω
ω, these two disks are disjoint.
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Bode plot
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Remark:
1 2 bilinear rule 1 22 1
1
sT
T sz e z
T s
zω
T z
There exists big difference in the Bode diagram at the high frequency.
Types of Compensator:
phase-lead ( high pass ~ PD) transient response
phase-lag ( low pass ~ PI ) steady-state response
PID : a special case of a phase lead-lag compensator
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100
Type 0 system
20 log at the low-frequency magnitude in Bode log-magni
Transient response
Steady-state response
Remark:
p
PMς
K
tude plot
Type 1 system
is the intersection (extension of the initial -20 dB/decde) with frequency
axis in Bode log-magnitude plot
Ty
vK
pe 2 system
is the intersection (extension of the initial -40 dB/decde) with frequency
axis in Bode log-magnitude plot aK
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Design Procedure in Frequency
1.
2. Bode plot
3.
4.
1 2 bilinear transformation 1 2
T νz
T ν
( )G shold
( )G z ( )G ν
( )G ν ( )G jων jω
( ) ,
( ).
Assuming that the low frequency gain of is unity
design
D ω
D ω
2 1
1
zω
T z
( ) ( )D ω D z
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1
ex) . 230- (Ogata's)
1 10( )
10
1 10 10 ( ) (1 )
10 ( 10)
0.6321
0.3679
Ts
Ts
p
eG s
s s
eG z z
s s s s
z
z z
1 sTe
s
10
10s Tδ
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1 ( / 2) 1 0.05
1 ( / 2) 1 0.05
0.6321 0.6321(1- 0.05 )( )
1 0.05 0.6321 0.068400.36791 0.05
1 0.059.241
9.241
T v vz
T v v
vG v
v vv
v
v
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0
1
ex) .236- (Ogata's)
1 ( )
( 1)
Design Specifications:
phase margin: 50 gain margin: at least 10db
velocity error constant : 2sec
sampling peri
Ts
v
p
e KG s
s s s
K
2
od 0.2
1 ( )
( 1)
( 0.9356) 0.01873
( 1)( 0.8187)
(0.08173 0.01752)
1.8187 0.8187
Ts
T
e KG z
s s s
K z
z z
K z
z z
z
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T v vz
T v v
vK
vG v
v v
v vv v
KK v v
v vv v
2
2
2
1 ( / 2) 1 0.1
1 ( / 2) 1 0.1
1 0.10.08173( ) 0.01752
1 0.1( )
1 0.1 1 0.1( ) 1.8187( ) 0.81871 0.1 1 0.1
(1 )(1 )( 0.000333 0.09633 0.9966) 300 10
( 1)0.9969
How to determ Kine the gain ?
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0
2
2
lim ( ) ( ) 2
2(1 )(1 )2( 0.000333 0.09633 0.9966) 300 10( )
( 1)0.9969
From Bode plot, and GM 14.5dB
PM should be properly
PM 3
adj
0
usted.
What type of compensator is neede
o
v DvK vG v G v K
v vv v
G vv vv v
d?
1 ( ) , 0 1 (pha se-lead compensator)
1D
τvG v α
ατv
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72
D
D
τvG v
ατv
φ τν ατν
d
τvG v
αv
τ αα α
φαα
τ
φ τ ατ
dν τν ατν
τ
α
v
p
1 1
max
1 1a
2
m
2
x
1( )
1
tan tan
1 ( ) 1 ( )
( and ) (1)
( from t
1
1 1tan
1 break
sin12
h
frequencies from ( )1
e req
Review ( .628 ιn Νιse's):
D
D
DG
jjτω αG jω
jατω j α
α
G jω G j
j
ω ω
ωα
ma
maxmax
m
max max
x
m
x
x
a
a
uired phase) (2)
11
1( )
1 1
(magnitude at the peak of phase curve) (3)
where .
Solving (2), we can obtain
( )
1(
.
( )
U
at
)
sing τ (3) and (1), we can determine .
Robotics Research Labo-ratory
73
o
max ma
1
x
1max
max
max
How to design ( )?
1 sin 28 0.361
1
120 log 4.425 dB
0.361
At 1.7, ( ) 4.425 dB
11.7 0.9790
1 1 0.( ) ( )
1
1 1tan sin
121
( )
1
D
D
D
D
α αφ
αα
G jωα
v
G v
αα
α
v G jω
ττ α
τvG
τ α
v G vατv
9790.
1 0.3534
v
v
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75
D
D
D
vG v
vz
zG zz
zz
z
z zG z G z
z z z
make it a difference equatio
1 0.9790( )
1 0.35341
1 0.9790(10 )1( )1
1 0.3534(10 )1
2.3798 1.9387
0.5589
2.3798 1.9387 0.03746( 0.9356)( ) (
n!
)0.5589 ( 1)( 0.8187)
z z
z z z
C z z z
R z z z z
z z
z z j z j
2
3 2
2
3 2
0.0891 0.0108 0.0679
2.377 1.8352 0.4576
( ) 0.0891 0.0108 0.0679
( ) 2.2855 1.8460 0.5255
0.0891( 0.9357)( 0.8145)
( 0.8126)( 0.7379 0.3196)( 0.7379 0.3196)
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77
Design by Direct Method (Analytic Method)
Ts
n nn
n nn
eG z G s
s
C z D z G zH z
R z D z G z
H zD z
G z H z
b z b z bH z
a z a z a
10 1
10 1
1( ) ( )
( ) ( ) ( )( )
( ) 1 ( ) ( )
1 ( )( )
( ) 1 ( )
where ( )
z
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78
n nn
n
nn
a z a z aH z
z
a a z a z
10 1
10 1
( )
Direct Design Method of Ragazzini
i) (transient) determine characteristic equation according to the specification
denominator of ( )
ii) (steady-state) determine coefficient of numerator of ( )
The system sh
H z
H z
1
ould be causal.
(1) 1 for Type 1 system
1 |z
v
H
dHT
dz K
Deadbeat Controller (unique characteristic in digital controller)
– minimum settling time with zero steady-state error
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79
Digital PID Controller
0
1 ( )( ) ( ) ( ) ( )
1( ) 1 ( )
1
where 1 is a low-pass filter , 3 10
(0) ( ) ( ) (2 )( ) [ ( ) [
2 2
(( 1) ) ( ) ( ) (( 1)]
2
t
di
d
di
d
i
d
de tu t K e t e t d t T
T dt
T sU s K E s
T sT sN
T s NN
T e e T e T e Tu kT K e kT
T
e k T e kT e kT e k TT
)]
trapezoidal backward T
Positional form
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80
1 1
1
1
1 1
1
Define
(( 1) ) ( ) ( ), (0) 0
2(( 1) ) ( )
( )2
(( 1) ) ( )
2
( )
1 ( ) ( ) (0)
1 1
1( ) ( ) ( )
2
h h
k k
h
k
h
k
e k T e kTf kT f
e k T e kTf kT
e k T e kT
f kt
F zF z f
z z
zF z f kt E z
z
z
z
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10
0
1 Show that ( ) ( )
1
( ) ( ), 0,1,2,
(0) (0)
(1) (0) (1)
( ) (0) (1) (2) ( )
( ) ( 1)
k
h
k
h
Z x h X zz
y k x h k
y x
y x x
y k x x x x k
y k y k
1
1
10
( )
( ) ( ) ( )
1 ( ) ( )
1
1 ( ) ( ) ( ) ( )
1
k
h
x k
Y z z Y z X z
Y z X zz
Z x h Z y k Y z X zz
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1
10
1 ( 1)
0
1Show that ( ) ( ) ( ) where 1 1
1
( ) ( ) ( ) ( 1) ( )
( ) ( ) ( 1) ( )
Noting that
( ) ( ) ( ) (0) (
k ih
h i h
k
h i
i k
k
k
Z x h X z x h z i kz
y k x h x i x i x k
X z x i z x i z x k z
X z Z x k x k z x x
1 2
1
0
1
1
1 10
1) (2)
( ) ( ) ( )
Since
( ) ( 1) ( ), , 1, 2,
( ) ( ) ( )
1 1 ( ) ( ) ( ) ( ) ( )
1 1
ih
h
k ih
h i h
z x z
X z X z x h z
y k y k x k k i i i
Y z z Y z X z
Z x h Y z X z X z x h zz z
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83
( )
,
11
1
11
11
1( ) ( ) ( ) 1 ( )
2 1
(1 ) ( )1
2 2
( )( ) (1 )
( ) 1
d
i
Ip D
Ip
i
dI D
i
ID p D
TT zU z K E z E z z E z
T z T
KK K z E z
z
KKTK K K
T
KTKTK K
T T
KU zG z K K z
E z z
112 1i i
T T
T T z
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84
( ) ( ) (( 1) )
[ ( ) (( 1) ] ( )
+ [ ( ) 2 (( 1) (( 2) ]
If ( ) ( ) ( )
( ) [ ( ) (( 1) ) ( ) (( 1) ]
+ [ ( ) (
p I
D
p
I
U kT U kT U k T
K e kT e k T K e kT
K e kT e k T e k T
e kT r kT c kT
U kT K r kT r k T c kT c k T
K r kT c k
If ( ) is a constant reference input
)]
+ [ ( ) ( ) ]
( ) [ ( ) (( 1) )]
[ ( ) ( 1)]
[ ( )
(or step change)
2 (( 1) (( 2
D
p
I
D
T
K r kT c kT
U kT K c kT c k T
K r kT c k
K c
r k
kT k k
T
c T c
) ]
Incremental form or velocity form
T
Velocity form
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85
( ) (( ) ) ( )
( ) ( ) ( ) ( )
( ( ) ( ))
( ) ( )
( )( ) ( ) ( ) ( )
)
(
1 1
1 2
11
1
1 1
1 2
11
p
I
D
p I D
U kT U k T U kT
z U z K z C z
K R z C z
K z z C z
C zU z K C z K K z C z
z
R z
velocity form
A drawback of velocity form is that it can not be used for P- or PD-controllers. In these cases the controller will not be unable to keep the reference value.
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Remarks: Integral action in PID Controller (Astrom’s)
Integrator windup :
A controller with integral action combined with an actuator that becomes saturated can give some undesirable effects.
If the controller error is so large that the integrator saturates the actuator, the feedback path will be broken, because the actuator will remain satu-rated even if the process output changes.
When the error is finally reduced, the integral may be so large that it takes considerable time until the integral assumes a normal value again.
Antiwindup:
i) Stop updating the integral when the actuator is saturated.
ii) Or an extra feedback path is provided by measuring the actuator output and forming error signal as the difference between the actuator output and the controller output and feeding this error back to the integrator through a proper gain(tracking-time constant).
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Unsaturated
1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101 106 111 116 121 126 131 136 1410
0.5
1
1.5
2
2.5
3
r(t)
u(t)
y(t) [Unsaturated]
Robotics Research Labo-ratory
Saturated
1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101 106 111 116 121 126 131 136 1410
0.5
1
1.5
2
2.5
3
r(t)
uc(t)
u(t)
y(t) [Saturated]
Robotics Research Labo-ratory
Anti-windup
1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101 106 111 116 121 126 131 136 1410
0.5
1
1.5
2
2.5
3
r(t)
uc(t)
u(t)
y(t) [Anti-windup]
Robotics Research Labo-ratory
Comparison (u)
1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101 106 111 116 121 126 131 136 1410
0.5
1
1.5
2
2.5
3
3.5
u(t)
uc(t)
u(t)
uc(t)
u(t)
Unsaturated
Saturated
Anti-Windup
Robotics Research Labo-ratory
Comparison (y)
1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101 106 111 116 121 126 131 136 1410
0.2
0.4
0.6
0.8
1
1.2
1.4
y(t) [Unsaturated]
y(t) [Saturated]
y(t) [Anti-windup]
r(t)
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93
Bumpless transfer:
Practically all PID controllers can run in two modes: manual and automatic.
When there are changes of modes and parameters, it is essential to avoid switching transients.
When the system is in manual mode, the controller produces a control sig-nal that may be different from the manually generated control signal.
It is necessary to make sure that the value of the integrator is correct at the time of switching. This is called bumpless transfer.
Bumpless transfer is easy to obtain for a controller in incremental form (ve-locity form) because the switching only influence the increments , there will not be any large transients. Initialization is not necessary when the opera-tion is switched from automatic to manual.
Recommended