mesb374 system modeling and analysis pid controller design

MESB374 System Modeling and Analysis

PID Controller Design

PID Controller

• Structure of Controller

• Effects of Proportional, Integral and Derivative Actions

• Design of PID Controllers

GC (s)

H(s)

+

Reference Input

R(s)

Error

E(s)++ Output

Y(s)

Disturbance

D(s)

Plant

G(s)Control Input

U(s)Gf (s)

Proportional plus Integral plus Derivative (PID) Control

0

( ) ( )t

CP CI CDu t K e t K e d K e t

21 2( ) CDCD CP CI

c

U s K s z s zK s K s KG s

E s s s

PID Controller

In time domain

In s-domain

1( )CP CI CDU s K K K s E s

s

1 2

1 2

CP CD

CI CD

K K z z

K K z z

current information passed information (prediction of) future information

• Derivative Action (KCD s): Provides added damping to the closed-loop system; reduces overshoot and oscillation in step response; tends to slow down the closed-loop system response. Dominant during initial transient, due to the effect of differentiation.

Effect of P-I-D Actions

• Integral Action (KCI /s): Eliminates steady-state error to step inputs; tends to destabilize the closed-loop system; has averaging effect. Dominant during steady-state by producing an accumulation of steady-state error to increase control effort.

• Proportional Action (KCP): Introduces immediate action due to error; improves system response time. Has similar control authority for both transient and steady-state.

Effect of PID Actions•Ex: For a given process

4 1

2 1 0.5 1R s

Y ss s s

4

(2 1)(0.5 1)s s Reference input

R(s) +

+ Output

Y(s)

Dis

turb

ance

D(s

)

•Unit step reference response •Unit step disturbance response

4 1

2 1 0.5 1D s

Y ss s s

Step Disturbance Response

Time (sec)

Am

plit

ud

e

0 2 4 6 8 10 120

0.5

1

1.5

2

2.5

3

3.5

4Step Reference Response

Time (sec)

Am

plit

ud

e

0 2 4 6 8 10 120

0.5

1

1.5

2

2.5

3

3.5

4

Effect of P Action•Objective: Design a system that has zero steady state error for step inputs with %OS<10% and Ts (2%)<6 [sec]

4

(2 1)(0.5 1)s s GC (s)+

Reference Input

R(s)

Error

E(s)++ Output

Y(s)

Disturbance

D(s)

Plant

Control Input

U(s)

1

0.05 1s

(A.) Let’s try Proportional control: GC(s)=KCP

CLTF:

42 1 0.5 1

4 11 12 1 0.5 1 0.05 1

CPC P

YRC P

CP

KG s G s s s

G sG s G s H s K

s s s

42 1 0.5 1

4 11 12 1 0.5 1 0.05 1

4 0.05 1

2 1 0.5 1 0.05 1 4

PYD

C PCP

CP

G s s sG s

G s G s H s Ks s s

s

s s s K

Effect of P Action(A.1) Design P Control using Root Locus:

1 2

2 2

1 2

4 1 1.125 2.55 0.05 0

7.1414 0,

14.1 0.25

CP

CP CP

K j

K K

CLCE:

4 11

2 1 0.5 1 0.05 1

14

10.05 0.5 2 20

CP

N s

CP

D s

Ks s s

Ks s s

-15

-10

Real Axis

-25 -20 -15 -10 -5 0 5

-5

0

5

10

15

Img. Axis

-7.5

(2 1) [rad]kP Z

kN N

3

5

3

0 7.5i i

P Z

p z

N N

2

1 2

( )0, 3 45 51 0

( )

1.235, 13.765

d D ss s

ds N s

s s

Not valid

Stability: 0<K CP <14.1

-0.667

53.76

Effect of P Action(A.2) Check for Steady State Error:

4

1 11 4

CPss ss

rCP

Ke r y

K

Stability: 0<K CP <14.1

Unit Step Reference Response

Unit Step Disturbance Response

0

1 4lim 0

1 4ss YD YDs

CP

y sG s Gs K

Larger gain results in smaller steady-state error.

Larger gain results in stronger attenuation of disturbance.

Do NOT forget

Think about the change of overshoot when gain increases

Effect of PI Actions

4

(2 1)(0.5 1)s s GC (s)+

Reference Input

R(s) Error

E(s)++ Output

Y(s)

Disturbance

D(s)

Plant

Control Input

U(s)

1

0.05 1s

(B.) Add integral action (PI control):

CLTF:

42 1 0.5 1

4 11 12 1 0.5 1 0.05 1

42 1 0.5 1

4 11 12 1 0.5 1 0.05 1

4 0.05 1

2 1 0.5 1 0.05 1 4

PICP

C PYR

PIC PCP

PYD

PIC PCP

CP PI

s zK

G s G s s s sG s

s zG s G s H s Ks s s s

G s s sG s

s zG s G s H s Ks s s s

s s

s s s s K s z

Effect of PI Actions(B.1) Design PI Control using Root Locus:

CLCE:

41 0

0.05 0.5 2 20

PI

N s

CP

D s

s z

Ks s s s

Real Axis

-25 -20 -15 -2 -1 0 1

5

Img. Axis

-15

-10

-5

0

10

15

(2 1) [rad]kP Z

kN N

3

5

3

0 7.3i i

P Z

p z

N N

2

1 2

( )0, 3 44 40 0

( )

0.9737, 13.6929

d D ss s

ds N s

s s

1 2

2 2

1 2

2 0.55 1 0.025 0

6.3246 0,

11 0

CP

CP CP

K j

K K

Not valid

Stability: 0<K CP <11

Choose zPI = -0.5:

-7.3

Effect of PI Actions(B.2) Check for Steady State Error:

1 0 1 0ss ss YRr

e r y G

Unit Step Reference Response

Unit Step Disturbance Response

0

1lim 0 0ss YD YDs

y sG s Gs

By using Integral action. steady state error is eliminated.

By using Integral action, the effect of a constant disturbance can also be eliminated.

Has transient performance been improved? Not much

Effect of PID Actions

(C.) Add derivative action (PID control):

CLTF:

1 2

1 2

1 2

1 2

42 1 0.5 1

4 111

2 1 0.5 1 0.05 1

42 1 0.5 1

4 111

2 1 0.5 1 0.05 1

4 0.05 1

2 1 0.5 1 0.05 1 4

CDC P

YRC P

CD

PYD

C PCD

CD

s z s zK

G s G s s s sG s

s z s zG s G s H sK

s s s s

G s s sG s

s z s zG s G s H sK

s s s s

s s

s s s s K s z s z

1 21C CP CI CD CD

s z s zG s K K K s K

s s

4

(2 1)(0.5 1)s s GC (s)+

Reference Input

R(s) Error

E(s)++ Output

Y(s)

Disturbance

D(s)

Plant

Control Input

U(s)

1

0.05 1s

Effect of PID Actions(C.1) Design PI Control using Root Locus:

CLCE:

1 2

41 0

0.05 0.5 20 20N s

CD

D s

s z s z

Ks s s s

Real Axis

-20 -10 -15 -2 -1 0 1

5

Img. Axis

-15

-10

-5

0

10

15

(2 1) [rad]kP Z

kN N

23

2

0 9.95i i

P Z

p z

N N

3 2

1 2 3

( )0, 4 56.6 184.8 168 0

( )

9.92, 2.60, 1.62

d D ss s s

ds N s

s s s

Stability: 0<K CD

Choose z1 = -0.5, z 2=-2.1

-9.95

-2.6

-2.1

-9.92 -1.62

Effect of PID Actions(C.2) Check for Steady State Error:

1 0 1 0ss ss YRr

e r y G

Unit Step Reference Response

Unit Step Disturbance Response

0

1lim 0 0ss YD YDs

y sG s Gs

By using Integral action. steady state error is eliminated.

By using Integral action, the effect of a constant disturbance can also be eliminated.

Has transient performance been improved? Yes

PID Controller Design via Root Locus

Design ProcedureStep1: Select the position of the two zeros such that the root locus will intersect with the desired performance region.Step 2: Pick the controller gain KC such that CL poles are in the performance regionStep 3: Find the corresponding PID gain using the above formula.

2

1 21 2

1 2

,CD C

CDCD CP CIC C CP C

CI C

K KN s K s z s zK s K s K

G s K K K z zD s s s

K K z z

Pole-Zero structure of PID Controller:

PID Controller adds one open-loop pole at origin and two open-loop zeros, z1 and z2. These two open-loop zeros could be either real or complex conjugate pair.

Design of PID ControllerEx: ( Motion Control of Hydraulic Cylinders )

M

qqININ

Recall the example of the flow control of a hydraulic cylinder that takes into account the capacitance effect of the pressure chamber. The plant transfer function is:

2

2 2

( )( )

( ) 2p p

IN p p p

CV sG s

Q s s s

VV CC

BB

AA

where 6 rad/sec, 0.1, 0.2.p p pC

Output

V(s)

2

2 22p p

p p p

C

s s

GC (s)+

Reference Velocity

R(s)

Error

E(s)Plant

Control Input

QIN (s)

Dis

turb

ance

D

(s)

++

Design of PID Controller

1 2

2 22

22 2 2

0.2 6 0.2 6( )

2 1.2 61.885 18.755 1.885 18.755

P p

IN p p p

p p

CV sG s

Q s s s s ss j s j

We would like to design a controller such that the closed loop system is better damped (smaller OS%)

CLTF:

( ) ,

with ( ) , and ( )

C P CCL

P C C P C

C PC

C P

V s K N s N sG s

R s D s D s K N s N s

N s N sG s G s

D s D s

Output

V(s)

2

2 22p p

p p p

C

s s

GC (s)+

Reference Velocity

R(s)

Error

E(s)Plant

Control Input

QIN (s)

Dis

turb

ance

D

(s)

++

Design of PID Controller

2

1 2

1 2

0.2 61 0C

s z s zK

s s p s p

2

1 21 2

1 2

,CD C

CDCD CP CIC C CP C

CI C

K KN s K s z s zK s K s K

G s K K K z zD s s s

K K z z

arctan

ln %X

Closed-loop Characteristic Equation

PID controller design

Real Axis

-30 -25 -20 -15 -10 -5 0

-5

0

5

10

15

Img. Axis

20

-15

-10

-20

Can a PID controller be designed to satisfy the transient design specifications (smaller overshoot and faster settling time) ?

t TTS S

S

(2%)4 4

Transient performance region