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
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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
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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
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PI Controller
s
K
KsK
s
KKsGc
2
11
21)(
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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
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Uncompensated system With lag compensation(root locus remains the same)
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Example
With damping ratio of 0.174, add lag Compensator to improve steady-state error by a factor of 10
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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
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3rd order approx. for lag compensator (= uncompensated system) makingSame transient response but 10 timesImprovement in ss response!!!
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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!!!
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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
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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
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Compensator: (s+3.006)
Might not have a pole-zero cancellation for compensated system
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PD Compensator
)(2
1212 K
KsKKsKGc
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Lead Compensation
Zeta2-zeta1=angular contribution
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Can put pairs of poles/zeros to get a desired θc
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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
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)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!!!
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