inc341 design with root locus

INC 341 PT & BPINC 341 PT & BP

INC341Design with Root Locus

Lecture 9

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2 objectives for desired response

1. Improving transient responsePercent overshoot, damping ratio, settling

time, peak time

2. Improving steady-state errorSteady state error

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Gain adjustment

• Higher gain, smaller steady stead error, larger percent overshoot

• Reducing gain, smaller percent overshoot, higher steady state error

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Compensator• Allows us to meet transient and steady

state error.

• Composed of poles and zeros.

• Increased an order of the system.

• The system can be approx. to 2nd order using some techniques.

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Improving transient response

• Point A and B have the same damping ratio.

• Starting from point A, cannot reach a faster response at point B by adjusting K.

• Compensator is preferred.

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CascadeCompensator

FeedbackCompensator

The added compensator can change a pattern of root locus

Compensator configulations

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Types of compensator1. Active compensator

– PI, PD, PID use of active components, i.e., OP-AMP– Require power source– ss error converge to zero– Expensive

2. Passive compensator– Lag, Lead use of passive components, i.e., R L C– No need of power source– ss error nearly reaches zero– Less expensive


Improving steady-state errorPlacing a pole at the origin to increase system order; decreasing ss error as a result!!

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The pole at origin affects the transeint response adds a zero close to the pole to get an ideal integral compensator


Example

Damping ratio = 0.174 in both uncompensated and PI cases

Choose zero at -1

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• Draw root locus without compensator

• Draw a straight line of damping ratio

• Evaluate K from the intersection point

• From K, find the last pole (at -11.61)

• Calculate steady-state error

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Finding an intersection between damping ratio line and root locus

• Damping ratio line has an equation:

where a = real part, b = imaginary part of the intersection point,

• Summation of angle from open-loop poles and zeros to the point is 180 degrees

mab

18010

tan2

tan1

tan 111

a

b

a

b

a

b

))(tan(cos 1 m

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Arctan formula

AB

BABA

1tan)(tan)(tan 111

AB

BABA

1tan)(tan)(tan 111

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• Use the formula to get the real and imaginary part of the intersection point and get

• Magnitude of open loop system is 1

0.6936 -1.5893,a

3.9255- b

1321 ppp

K No open loop zero

53.164

1

1021 222222

bababa

K

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• Draw root locus with compensator (system order is up by 1--from 3rd to 4th)

• Needs complex poles corresponding to damping ratio of 0.174 (K=158.2)

• From K, find the 3rd and 4th poles (at -11.55 and -0.0902)

• Pole at -0.0902 can do phase cacellation with zero at -1 (3th order approx.)

• Compensated system and uncompensated system have similar transient response (closed loop poles and K are aprrox. The same)

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Comparason of step response of the 2 systems


PI Controller

s

K

KsK

s

KKsGc

2

11

21)(


Lag Compensator

•Build from passive elements•Improve ss error by a factor of Zc/Pc•To improve both transient and ss responses, put pole and zero close to the origin


Uncompensated system With lag compensation(root locus remains the same)


Example

With damping ratio of 0.174, add lag Compensator to improve steady-state error by a factor of 10


Step I: find an intersection of root locus and damping ratio line (-0.694+j3.926 with K=164.56)

Step II: find Kp = lim G(s) as s0 (Kp=8.228)

Step III: steady-state error = 1/(1+Kp)= 0.108

Step IV: want to decrease error down to 0.0108[Kp = (1 – 0.0108)/0.0108 = 91.593]

Step V: require a ratio of compensator zero to poleas 91.593/8.228 = 11.132

Step VI: choose a pole at 0.01, the corresponding Zero will be at 11.132*0.01 = 0.111

01.0

111.0

s

s


3rd order approx. for lag compensator (= uncompensated system) makingSame transient response but 10 timesImprovement in ss response!!!


If we choose a compensator pole at 0.001 (10 timescloser to the origin), we’ll get a compensator zero at 0.0111 (Kp=91.593)

001.0

0111.0

s

sNew compensator:

4th pole is at -0.01 (compared to -0.101) producing a longer transient response.

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SS response improvement conclusions

• Can be done either by PI controller (pole at origin) or lag compensator (pole closed to origin).

• Improving ss error without affecting the transient response.

• Next step is to improve the transient response itself.

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Improving Transient Response

• 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

cC zsG

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(a) Uncompensated system, (b) compensator zero at -2 (d) compensator zero at -3, (d) compensator zero at -4

Indicate settling time

Indicate peak time

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• Settling time & peak time: (b)<(c)<(d)<(a)• %OS: (b)=(c)=(d)=(a)• ss error: compensated systems has lower value than

uncompensated one cause improvement in transient response always yields an improvement in ss error

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Example

design a PD controller to yield 16% overshoot with a threefold reduction in settling time

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• Step I: calculate a corresponding damping ration (16% overshoot = 0.504 damping ratio)

• Step II: search along the damping ratio line for an odd multiple of 180 (at -1.205±j2.064) and corresponding K (43.35)

• Step III: find the 3rd pole (at -7.59) which is far away from the dominant poles 2nd order approx. works!!!


0)()2()2(

0))(06.22.1)(06.22.1(

02410

222223

23

bacsbacascas

csjsjs

Ksss

Kgainandpolethirdthegettosolve)(

10222

Kbac

ca

More details in step II and III

Characteristic equation:

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613.3107.1

44

sT

• Step IV: evaluate a desired settling time:

• Step V: get corresponding real and imagine number of the dominant poles

(-3.613 and -6.193)

sec107.13

320.3:systemdcompensate

sec320.3205.1

44:systemteduncompensa

s

ns

T

T

193.6))504.0(tan(cos613.3 1 d


Location of polesas desired is at-3.613±j6.192

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006.3

)607.95180tan(613.3

192.6

• Step VI: summation of angles at the desired pole location, -275.6, is not an odd multiple of 180 (not on the root locus) need to add a zero to make the sum of 180.

• Step VII: the angular contribution for the point to be on root locus is +275.6-180=95.6 put a zero to create the desired angle


Compensator: (s+3.006)

Might not have a pole-zero cancellation for compensated system


PD Compensator

)(2

1212 K

KsKKsKGc


Lead Compensation

Zeta2-zeta1=angular contribution


Can put pairs of poles/zeros to get a desired θc


Example

Design three lead compensators for the systemthat has 30% OS and will reduce settling time downby a factor of 2.

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sec972.3007.1

44

nsT

212.63K

• Step I: %OS = 30% equaivalent to damping ratio = 0.358, Ѳ= 69.02

• Step II: Search along the line to find a point that gives 180 degree (-1.007±j2.627)

• Step III: Find a corresponding K ( )• Step IV: calculate settling time of uncompensated

system

• Step V: twofold reduction in settling time (Ts=3.972/2 = 1.986), correspoding real and imaginary parts are:

014.2986.1

44

sT

253.5))358.0(tan(cos014.2 1 d

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• Step VI: let’s put a zero at -5 and find the net angle to the test point (-172.69)

• Step VII: need a pole at the location giving 7.31 degree to the test point.

96.42

31.7tan014.2

252.5

c

c

p

p


)96.42(

)5(rcompensatolead

s

s

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Note: check if the 2nd order approx. is valid for justify our estimates of percent overshoot and settling time– Search for 3rd and 4th closed-loop poles

(-43.8, -5.134)– -43.8 is more than 20 times the real part of

the dominant pole– -5.134 is close to the zero at -5

The approx. is then valid!!!

inc341 design with root locus

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