unit 4 the performance of second order system
DESCRIPTION
Unit 4 The Performance of Second Order System. 中華技術學院電子系 副教授 蔡樸生 副教授 林盈灝. Open Loop & Close Loop. Open Loop :. Close Loop :. The Performance of Second Order System. The Response of Second Order System. Homework 1. 1. Steady State Error :. 2. Overshoot :. 0.163. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/1.jpg)
Unit 4 The Performance of Second Order System
中華技術學院電子系中華技術學院電子系副教授 蔡樸生副教授 蔡樸生副教授 林盈灝副教授 林盈灝
![Page 2: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/2.jpg)
Open Loop & Close Loop
Open Loop:
Close Loop:
R Y
E YR
( )( ) ( )
1 ( ) ( )
G sY s R s
G s H s
( ) ( ) ( )Y s G s R s
2
( 2 )n
n
w
s s w
2
2 22n
n n
w
s w s w
![Page 3: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/3.jpg)
The Performance of Second Order System
2:
1p
n
The Peak Time Tw
2: exp( )
1pThe Overshoot M
4 1.8
: , :s rn n
The Settling Time T Rise Time Tw w
2
2 2:
2n
n n
wSecond Order System
s w w
,: :nw natural frequency damping ratio
![Page 4: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/4.jpg)
The Response of Second Order System
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
Time(s)
y(t)
ssepM
pt
![Page 5: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/5.jpg)
Homework 1 42
32 ss
)(sU )(sY
1. Steady State Error :
2 0
3( ) ( ) lim ( ) 0.75
( 2 4) sY s y t s Y s
s s s
2. Overshoot : 0.163
3. The Peak Time : 1.8138 s
4. The Rise Time : 0.9 s
5. The Setting Time : 4 s [Hint] : max
![Page 6: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/6.jpg)
The Response of ( )y t
![Page 7: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/7.jpg)
Homework2 : The effect of damping ratio
(1) 0 undamped
(2) 0 1 underdamped
(3) 1 critically damped
(4) 1 overdamped
![Page 8: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/8.jpg)
P Controller This type of control action is formally known as This type of control action is formally known as
proportional control (Gain)proportional control (Gain)
Homework3Homework3 : K=1, K=4 , K=8 , K=12 , K=36 : K=1, K=4 , K=8 , K=12 , K=36 Please explain the effect of P controller to thePlease explain the effect of P controller to the
second order system second order system
1
( 4)s s E YR
K
![Page 9: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/9.jpg)
0 5 10 15 200
0.2
0.4
0.6
0.8
1
1.2
1.4
Solution of HW3
![Page 10: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/10.jpg)
PD Controller
2
( 2 )n
n
w
s s wPK
DK s
( )R s
( )E s( )U s
( )Y s
( )c P DG s K K s ( )
( ) ( )P D
de tu t K e t K
dt
2 ( )( )( ) ( ) ( )
( ) ( 2 )n P D
c pn
w K K sY sG s G s G s
E s s s w
![Page 11: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/11.jpg)
0 1 2 3 4 5 60
0.5
1
1.5
0 1 2 3 4 5 6-0.5
0
0.5
1
0 1 2 3 4 5 6-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
( )y t
( )e t
( )de t
dt
1t 2t 3t 4t
![Page 12: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/12.jpg)
The Performance of P Controller
: : The error signal is positive, the torqueThe error signal is positive, the torque
is positive and rising rapidly. The large overshootis positive and rising rapidly. The large overshoot
and oscillations in the output because lack of damping.and oscillations in the output because lack of damping. : : The error signal is negative, the torqueThe error signal is negative, the torque
is negative and slow down causes the direction of theis negative and slow down causes the direction of the
output to reverse and undershoot.output to reverse and undershoot. : The torque is again positive, thus tending: The torque is again positive, thus tending
to reduce the undershoot, the error amplitude isto reduce the undershoot, the error amplitude is
reduced with each oscillations.reduced with each oscillations.
10 t t
1 3t t t
3 4t t t
![Page 13: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/13.jpg)
The contributing factors to the high overshoot
The positive correcting torque in the intervalThe positive correcting torque in the interval
is too large is too large (( 抑制抑制 )) Decrease the amount of positive torqueDecrease the amount of positive torque The retarding torque in the intervalThe retarding torque in the interval
is inadequate is inadequate (( 增強增強 )) Increase the retarding torqueIncrease the retarding torque
10 t t
1 2t t t
![Page 14: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/14.jpg)
The Effect of PD Controller
: : is negative; this will reduce this negative; this will reduce the original torque due to alone.e original torque due to alone.
: both and is negative; th: both and is negative; the negative retarding torque will be greater thae negative retarding torque will be greater than that with only P controller.n that with only P controller.
: and have opposite signs.: and have opposite signs. Thus the negative torque that originally contributes to Thus the negative torque that originally contributes to
the undershoot is reduced also.the undershoot is reduced also.
10 t t ( )de tdt
( )e t
1 2t t t ( )e t
2 3t t t
( )de tdt
( )e t ( )de tdt
![Page 15: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/15.jpg)
Homework 4
1
( 4)s s PK
DK s
( )R s
( )E s( )U s
( )Y s
.
Design the PD Controller such that the
response of the original system is optimal
![Page 16: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/16.jpg)
Solution of PD Controller clear;clear; x1=0;x2=0;dt=0.01;r=1;step=2000;x1=0;x2=0;dt=0.01;r=1;step=2000; kp=36;kd=6;kp=36;kd=6;pe=r-x1pe=r-x1;; for k=1:stepfor k=1:step t(k)=k*dt;t(k)=k*dt; e=r-x1;e=r-x1; de=(e-pe)/dtde=(e-pe)/dt;; u=kp*e+kd*de;u=kp*e+kd*de; x1=x2*dt+x1;x1=x2*dt+x1; x2=(u-4*x2)*dt+x2;x2=(u-4*x2)*dt+x2; pos(k)=x1;vel(k)=x2;pos(k)=x1;vel(k)=x2;pe=epe=e;; endend
![Page 17: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/17.jpg)
PI Controller
2
( 2 )n
n
w
s s wPK
IK
s
( )R s
( )E s( )U s
( )Y s
( ) Ic P
KG s K
s
( ) ( ) ( )P Iu t K e t K e t dt 2
2
( )( )( ) ( ) ( )
( ) ( 2 )n P I
c pn
w K s KY sG s G s G s
E s s s w
![Page 18: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/18.jpg)
HW5 : The Effect of PI Controller
Adds a zero at to the forward-path T.F.Adds a zero at to the forward-path T.F. Adds a pole at to the forward-path T.F.Adds a pole at to the forward-path T.F. This means that the steady-state error of the This means that the steady-state error of the
original system is improved by one order. original system is improved by one order.
I
P
Ks
K
0s
2
k
s as b ( )R s ( )E s ( )Y s
2 2
( ) 1 1, ( )
( ) 1 ( ) ( ) ( ) ( )
E s k kE s
R s G s H s s as k b s s as k b
a=2,b=8,k=1
![Page 19: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/19.jpg)
Program of PID Controller clear;clear; x1=0;x2=0;dt=0.01;r=1;step=2000;x1=0;x2=0;dt=0.01;r=1;step=2000; kp=1;kd=6;ki=0.1;pe=r-x1;ie=(r-x1)*dt;kp=1;kd=6;ki=0.1;pe=r-x1;ie=(r-x1)*dt; for k=1:stepfor k=1:step t(k)=k*dt;t(k)=k*dt; e=r-x1;e=r-x1; de=(e-pe)/dt;de=(e-pe)/dt; ie=ie+e*dt;ie=ie+e*dt; u=kp*e+kd*de+ki*ie;u=kp*e+kd*de+ki*ie; x1=x2*dt+x1;x1=x2*dt+x1; x2=(u-2*x2-8*x1)*dt+x2;x2=(u-2*x2-8*x1)*dt+x2; pos(k)=x1;vel(k)=x2;pe=e;pos(k)=x1;vel(k)=x2;pe=e; endend
![Page 20: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/20.jpg)
PID Controller
2
( 2 )n
n
w
s s wPK
IK
s
( )R s
( )E s
( )U s ( )Y s
( ) Ic P D
KG s K K s
s
( )( ) ( ) ( )P I D
de tu t K e t K e t dt K
dt
DK s
![Page 21: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/21.jpg)
Homework6 針對以下系統 , 憑藉經驗值調諧 PID 三個參數值 , 使得系統響應為最佳化
2
1
2 8s s PK
IK
s
( )E s
( )U s( )Y s
DK s
( )R s
1 , ,P ss sK e t OS
cp cpK T Critical Stable
![Page 22: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/22.jpg)
Homework7 : Ziegler-Nichols Tuning
1( ) (1 )I P
c P D P d P P di i
K KG s K K s K T s K K T s
s T s T s
Step 1 : Let until the occur of critical stableStep 2 : Optimal Parameter Tuning
, 0,i d PT T K
P
PI
PID
PKiT dT
0.5 cpK
0.45 cpK
0.6 cpK
1.2cpT
0.5 cpT 0.125 cpT
0
0
![Page 23: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/23.jpg)
Homework8: Pendulum System
1
2
3
4
x x
x x
x
x
x
uM
1 2
2 3
3 4
4 3
1
1
x x
mx g x u
M Mx x
M mx g x u
M l M l
0(0) 0.2 , (0) 0.2x m
2
0.2
0.6
M Kg
m kg
l m
![Page 24: Unit 4 The Performance of Second Order System](https://reader035.vdocuments.site/reader035/viewer/2022070402/568137e2550346895d9f873b/html5/thumbnails/24.jpg)
Feedback Controller Design
1 1
2 2
3 3
4 4
1
1 2
2 3
4
0 1 0 0 0
10 0 0
0 0 0 1 0
10 0 0
1 0 0 0
0 0 1 0
x xmgx xM M u
x x
x xM m
M l M l
x
y x
y x
x
[ 487 227 944 187]K State Feedback Controller