Unit 4 The Performance of Second Order System

中華技術學院電子系講師 蔡樸生

Open Loop & Close Loop

Open Loop:

Close Loop:

E 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 2

3

2 4s s ( )U s ( )Y s

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

proportional control (Gain)

Homework3 : K=1, K=4 , K=8 , K=12 , K=36 Please explain the effect of P controller to the

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 Homework3

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 torque

is positive and rising rapidly. The large overshoot

and oscillations in the output because lack of damping. : The error signal is negative, the torque

is negative and slow down causes the direction of the

output to reverse and undershoot. : The torque is again positive, thus tending

to reduce the undershoot, the error amplitude is

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 interval

is too large ( 抑制 ) Decrease the amount of positive torque The retarding torque in the interval

is inadequate ( 增強 ) Increase the retarding torque

10 t t

1 2t t t

The Effect of PD Controller

: is negative; this will reduce the original torque due to alone.

: both and is negative; the negative retarding torque will be greater than that with only P controller.

: and have opposite signs. Thus the negative torque that originally contributes to t

he 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; x1=0;x2=0;dt=0.01;r=1;step=2000; kp=36;kd=6;pe=r-x1; for k=1:step t(k)=k*dt; e=r-x1; de=(e-pe)/dt; u=kp*e+kd*de; x1=x2*dt+x1; x2=(u-4*x2)*dt+x2; pos(k)=x1;vel(k)=x2;pe=e; end

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 pole at to the forward-path T.F. This means that the steady-state error of the

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; 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; for k=1:step t(k)=k*dt; e=r-x1; de=(e-pe)/dt; ie=ie+e*dt; u=kp*e+kd*de+ki*ie; x1=x2*dt+x1; x2=(u-2*x2-8*x1)*dt+x2; pos(k)=x1;vel(k)=x2;pe=e; end

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