PID_slides.pdf

8/10/2019 PID_slides.pdf

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PID Controls

Issues on Control Design


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Res onse S ecs Review

MAAE 4500: Feedback Control 2


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jtan

tann n

s

x djn

2

r

d

np

d

n

p 21e

s cos cosnn

(5%); (2%)



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Movin Poles1,2 ds j

( ) cosd

y t Ae t

const.



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1,2

( ) cos

d

t

d

s j

y t Ae t

const.



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21

pM e

const.



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Effect of

Pole(s)

Location(s)

Time

Response



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Effect of

Pole(s)

Location(s)

Time

Response



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Effect of

Pole(s)

Location(s)

Time

Response



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Effect of

Pole s

Location(s)

on System

Response



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11p d

i

K T sT s

Plant

1

d

Plant

pK

1

i



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Ziegler-Nichols Rules

Industrial applications: engineers can tunethe response characteristics of a plant on site

Based on experimental step responses

Useful when a good mathematical modeldoes not exist (or it is too complicated)



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r e e e e -

If the plant does not involve integrator(s) nor dominant

comp ex-con uga e po es, e un s ep response can e

approximated by a S-shape curve as follows (first-orderres onse lus a time dela :

K

( )c t ( )u t

1

( ) LsC s Ke

ttangent to inflection point

t


TL


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First Method for Tuning the Controller

controllerP

pK

iT

dT

T

PI

L 0

T L

PID

0.9L 0.3

1.2L 2L 2



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o e a or e , e rs e o g ves or e

compensator TF:

1 1( ) 1 1.2 12 2

c p d

i

T LG s K T s sT s L Ls

21

0.6

s

LT

s

one ole at the ori in and one double zero at s = 1L

Q. Not good for type I systems. Why?



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Ste 1: Desi n a ro ortional controller T = 0

Ti=) and increase the gain until marginalabsolute stability is reached (pure imaginary. , p= cr

The method does not apply if the condition does not

exist for a finite gain

Step 2: Determine the period of the correspondingundamped oscillations, Pcr (P=2/)

Step 3: Use the next table to tune the controller Fine-tune analyzing the time response



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Second Method for Tuning the Controller

controllerP

pK

iT

dT

K

PI

2

cr 0

P

PID

.cr

1.2

cr

P

. cr 2

cr

. cr



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o e a or e , e econ e o g ves or

the compensator TF:

1 2( ) 1 0.6 1 0.125c p d cr cr

i cr

G s K T s K P sT s P s

2

4

cr

s

P

. cr cr s

one ole at the ori in and one double zero at s = 4Pcr



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Example 1 (PID, Ziegler-Nichols):

11

p dK T s

1

1 5s s s

( )C s( )R s

excluded. Use the second method and set Td= 0, Ti=:

3

( ) 1 5p

pR s s s s K

2 6

30

ps K

K

3 26 5p

ps s s K

0

6

p

s

s K

cr



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using the critical value of Kinto the closed-loop

characteristic equation:

3 26 5 30 0s s s From the table:

0.6 18K K

3 2

2

6 5 30 0j j j

0.5 1.405

0.125 0.3512

i crT P

T P

2 22.81

crP

24

2.81s

2

. .

6.322 1.423

c s

s


s


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Unit step response

too large overshoot

2

4 3 2

( ) 6.322 18 12.811

( ) 1 6 11.322 18 12.811

c

c

G GC s s s

R s G G s s s s



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26.322 1.423s

cs

Tr modif onl the controller ain. Draw root-locus lot:

2 1 C s.K

s 1 5s s s

B s

2

1.423( ) K sB sKG s G s H s

( ) 1 5R s s s s


.


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Root-locus

1 = 4.1328

6.322 = 0.3737 1.6213

s

K s

Overshoot is first

insensitive to K,

4 = 1.1197s

increase!

Lines of constant overshoot, for:= . =cons



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For an overshoot of 20%:

21

20.2 ln 0.2 1.61 0.456

1p

M e

ne w ave o c ange e ou e zeroes pos on: s

found by try and error that by moving the zeroes to0.6:

2

0.6sK

s

1

1 5s s s

( )C s( )R s

( )B s

2

0.6( ) K sB s

2( ) 1 5c

R s s s s

the next root-locus lot is roduced:



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Root-locus

must be acceptable

Mp < 20%Double pole

at s=0

Single pole at

s=0.6

2

0.6s 1 ( )C s( )R s

s=5

s

1 5s s s

( )B s



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Unit step response

, a ou

2

2

.30

1 5( )

( ) 1 0.6 1

c

c

s

s s s sG GC s

R s G G s

2

1 5

30 36 10.8

s s s s

s s

.s s s s



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Example 2 (PID, Nise): Given the system below,

design a PID controller so that the system operates

with 2/3 of the uncompensated peak time at 20%

-

21 3K K s

s 3 6 10s s s

Step 1: Evaluate performance of uncompensatedsystem operating at 20% overshoot (= 0.456).

a n e roo - ocus p o an raw eovershoot line to determine the closed-loop dominantpoles:



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Step 2: Design a PD

compensa or o meethe transient response

specs (determine the

zero location and the

PD controller gain).

parts of the dominant

pole must be:0.297

10.569p

d

t

.

2 3 0.297

8.13

d

d


an .


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,compensator. The

contribution from the

8.13 15.867

8

3 6 10s j

s

s s s

compensator zero must

be:

.

. .

From the Figure:

15.867

tan 18.3718.13

55.909

cz

z

1( ) 55.909G s K s



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Step 3: Design a PI

2 1

55.909 0.5 8( ) ( ) ( )

3 6 10

K s s sG s G s G s

s s s s

compensa or o meethe steady-state error

(determine the zero

location arbitrarily near

the origin [e.g.0.5]

0.5cs z s

the origin):

2s s

Re-draw the root locus

to determine the finalgain and the static

dis lacement error and

2 10

lim (0) (0) (0)

0

ps

ss

K G G G

e

(type 1)


the steady-state error)


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Step 4 Check the specs:

uncompensated PD PID

Dominant poles 5.415 10.569j 8.13 15.867j 7.519 14.676j

Gain, K121.510 5.339 4.604

Damping ratio, 0.456 0.456 0.456

Natural freq., n 11.875 17.829 16.49

Max. overshoot,Mp0.2 0.2 0.2

Peak time, tp 0.267 0.198

(=2/3* 0.267)

0.214

(~4/5 * 0.267)

Static dis l. error K

Steady-state error, ess 0.156 0.070 0


If not satisfactory redesign the PI controller


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Unit step response

PID compensated

Compensation yields the original system

faster and with no steady-state error; the

peak time is still a little large for the specs

.



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2K 8K s ( )C s( )R s

1 3s 3 6 10s s s

2 1

4.604 55.909 0.5( ) ( )

s sG s G s

s

2 12

259.74.604 56.409 27.954

128.7

4.604

Ks s

Ks

K

(One pole at the origin, one zero at 0.5 and one zero at


.


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Summary of Controllers and Their Usage



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Documents

PID_slides.pdf