INC342 Lecture 4: Frequency Response Method Dr. Benjamas - - PowerPoint PPT Presentation

INC342

Lecture 4: Frequency Response Method

• Dr. Benjamas Panomruttanarug

Benjamas.pan@kmutt.ac.th

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Improving Transient Response Using Compensator

• Objective is to

– Decrease settling time – Get a response with a desired %OS (damping ratio)

• Techniques can be used:

– PD controller (ideal derivative compensation) – Lead compensator

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Ideal Derivative Compensator

• So called PD controller
• Compensator adds a zero to the system

at –Zc to keep a damping ratio constant with a faster response

c C

z s G  

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Example

design a PD controller to yield 16%

• vershoot with a threefold reduction in

settling time

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16% overshoot = 0.504 damping ratio

Location of poles as desired is at

• 3.613±j6.192

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PD Compensator

) (

2 1 2 1 2

K K s K K s K Gc    

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Improving Both Steady‐State Error and Transient Response

• PI, Lag improve steady‐state error
• PD, Lead improve transient response
• PID, Lead‐lag improve both

(PID = Proportional plus Intergal plus Derivative controller)

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PID Controller

s K K s K K s K s s K K s K s K s K K s Gc ) ( ) (

3 2 3 1 2 3 2 3 2 1 3 2 1

        

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PID controller design

• 1. Evaluate the performance of the uncompensated

system

• 2. Design PD controller to meet transient response

specifications

• 3. Simulate and Test, redesign if necessary
• 4. Design PI controller to get required steady‐state

error

• 5. Find K constant of PID
• 6. Simulate and Test, redesign if necessary

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Example

Design PID controller so that the system can operate with a peak time that is 2/3 of uncompensated system, at 20% OS, and steady-state error of 0 for a step input

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sec 297 . 57 . 10      

d p

T

Step 1

• %OS = 20%  damping ratio = 0.456

 Ѳ = 62.87

• Search along ther line to find a point of 180

degree (‐5.415±j10.57)

• Find a correspoding K=121.51
• Then find the peak time

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Step 2

• Decrease peak time by a factor of 2/3  get

imaginary point of a compensator pole:

• To keep a damping ratio constant, real part of the

pole will be at

• The compensator poles will be at ‐8.13±j15.867

867 . 15 ) 297 . )( 3 / 2 (      

p d

T

13 . 8 ) 87 . 62 tan(  

 d

 

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92 . 55 ) 37 . 18 tan( 13 . 8 87 . 15   

c c

z z



ดPD controller is (s+55.92)

• Sum of the angles from uncompensated poles and

zeros to the test point (‐8.13±j15.867) is ‐198.37

• The contribution angle for the compensator zero is

then 180‐198.371 = 18.37

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Step 3

• Simulate the PD compensated system to see if

it reduces peak time and improves ss error

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Step 4

• design PI compensator (one pole at origin and

a zero near origin; at ‐0.5 in this example)

• Find a new point along the 0.456 damping

ratio line (‐7.516±j14.67), with an associate gain of 4.6

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s s s GPI 5 . ) (  

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s s s s s s s s s K s GPID ) 96 . 27 42 . 56 ( 6 . 4 ) 5 . )( 92 . 55 ( 6 . 4 ) 5 . )( 92 . 55 ( ) (

2

        

K1 = 259.5, K2 = 128.6, K3 = 4.6

Step 5

• Evaluate K1, K2, K3 of PID controller
• Compare to

s s K K s K s Gc

2 3 2 1

) (   

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Step 6

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Physical System Realization

R1 R2 C Vi(s) Vo(s) 2 2 1

1 ( ) s R C R C s R s          PI Compensator

R1 R2 C Vi(s) Vo(s)

2 1

1 ( )

c

G s R C s R C          PD Compensator

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R1 R2 C2 Vi(s) Vo(s) C1

2 1 1 2 2 1 1 2

1 ( )

c

R C R C G s R C s R C s                        PID Compensator

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