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

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

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

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

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

The Response of ( )y t

Homework2 : The effect of damping ratio

(1) 0 undamped

(2) 0 1 underdamped

(3) 1 critically damped

(4) 1 overdamped

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

0 5 10 15 200

0.2

0.4

0.6

0.8

1

1.2

1.4

Solution of HW3

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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