Design with Root Locus

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Design with Root Locus

• Lecture 9

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Objectives for desired response

• Improving transient response
• Percent overshoot, damping ratio,
• settling time,
• peak time
• Improving steady-state error
• Steady 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.
• We have pole at A on the root locus, but we want
  response like at B.
• Compensator is preferred.

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Compensator configulations
Cascade Compensator
Feedback Compensator
The added compensator can change a pattern of
root locus
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compensator
Method of implementing compensator 1.
Proportional control systems feed the error
forward to the plant. 2. Integral control
systems feed the integral of the error to the
plant. 3. Derivative control systems feed the
derivative of the error to the plant.
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Types of compensator

• Active compensator
• PI, PD, PID use of active components, i.e.,
  OP-AMP
• Require power source
• ss error converge to zero
• Expensive
• 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 error (PI)
Placing a pole at the origin to increase system
order decreasing ss error as a result!!
(a) Original system without compensation (b) Add
a pole at the origin but angular contribution at
point A is no longer 180
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Improving Steady-State Error (PI)
Also add a zero close to the pole at the origin.
As angular contribution of the compensator zero
and pole cancels out, point A is still on the
root locus, and the system type has
been increased.
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Improving Steady-State Error (PI)Example
Choose zero at -1
Damping ratio 0.174 in both uncompensated and
PI cases
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Improving Steady-State Error (PI)
gtgt z1 gtgt nconv(1 3 2,1 10) gtgt
systf(z,n)
gtgt sgrid(0.174,2,10) gtgt k prlocfind(sys)
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Improving Steady-State Error (PI)
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Improving Steady-State Error (PI)
As shown in the figure, the step response of the
PI compensated system approaches unity in the
steady-state, while the uncompensated system
response approaches 1-0.108 0.892. The
simulation shows that it takes 18 seconds for the
compensated system to reach and stay within 2
of the final value of unity, while the
uncompensated system takes about 6 seconds to
settle to within 2 of its final value of 0.892.
This is because there is no pole-zero cancelation
and the pole not canceled is very close to the
origin.
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Improving Steady-State Error (PI)
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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

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Arctan formula
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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

No open loop zero
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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 (K158.2)
• 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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PI Controller
A compensator with a pole at the origin and a
zero close to the pole is called an ideal
integral compensator, or PI controller
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Lag Compensator
Ideal integral compensation pole is in the
origin, requires active network (costly). Real
(passive) integral compensation pole is close to
origin (not in the origin), cheaper.
(a) Type 1 uncompensated system (b) Type 1
compensated system
Type is not increased. What about steady-state
error
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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 (-0.694j3.926 with
K164.56) Step II find Kp lim G(s) as s?0
(Kp8.228) Step III steady-state error
1/(1Kp) 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 pole as 91.593/8.228 11.132 Step VI
choose a pole at 0.01, the corresponding Zero
will be at 11.1320.01 0.111
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3rd order approx. for lag compensator (
uncompensated system) ? making Same transient
response but 10 times Improvement in ss
response!!!
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If we choose a compensator pole at 0.001 (10
times closer to the origin), well get a
compensator zero at 0.0111 (Kp91.593)
New 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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PD controller Improving transient response
System above controlled by a pure gain (P
controller) in the forward path has its root
locus going through point A for some value of
gain K. Our goal is to speed up the response at
A to that at B, while keeping the percent
overshoot unchanged. The above root locus with
a P controller cannot go through point B (sum of
angles from the open-loop finite poles and zeros
to point B is not an odd multiple of 180?). A
solution is to add a (nonzero) zero to the
forward path (e.g., PD controller).
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PD controller Improving transient response
Transfer function of the PD controller Gc(s)
K2 s K1 K2(sK1/K2) K(szc) introduces a
zero at -zc Into the forward path. Effect of
the added zero The added zero will contribute to
make the sum of angles from the open-loop finite
poles and zeros to the desired point (point B) be
an odd multiple of 180?. Note an added zero has
the effect of pushing the root locus to the left
while an added pole has the effect of pushing it
to the right. The new root locus can meet the
specific transient response (with shorter
settling time) by going through point B for some
value of gain K.
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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

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(a) Uncompensated system, (b) compensator zero at
-2 (d) compensator zero at -3, (d) compensator
zero at -4
Indicate peak time
Indicate settling time
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• Settling time peak time (b)lt(c)lt(d)lt(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 -1.205j2.064) 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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More details in step II and III Characteristic
equation
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• Step IV evaluate a desired settling time
• Step V get corresponding real and imagine number
  of the dominant poles
• (-3.613 and -6.193)

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Location of poles as desired is at -3.613j6.192
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• 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 275.6-18095.6 ? put a
  zero to create the desired angle

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Compensator (s3.006)
Might not have a pole-zero cancellation for
compensated system
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PD Compensator
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Lead Compensation
A PD controller can be approximated with a lead
compensator, which is implemented with a passive
network. If the lead compensator pole is
farther from the imaginary axis than the
compensator zero, the angular contribution of
the compensator is still positive and thus
approximates an equivalent single zero. The
advantages of a passive lead compensator over an
active PD controller are that (1) no additional
power supplies are required and (2) noise due to
differentiation is reduced.
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Lead Compensation

• The concept behind lead compensation the
  difference between 180? and the sum of the angles
  from the uncompensated systems poles and zeros
  to the design point (desired pole location) must
  be the angular contribution required of the
  compensator.
• That is,
• q2 -q1 -q3 -q4 q5 (2k1)180?
• where q2 -q1 qc is the angular contribution of
  the lead compensator.

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The angular contribution qc can be determined
from the rays originating from the desired
closed-loop pole and terminating at the
compensator pole and zero. These rays can be
rotated about the desired closed-loop pole and
thus different pairs of compensator pole and zero
can be used to meet the transient response
requirement. Different possible lead
compensators differences are in the values of
the static error constants, the static gain, the
difficulty in justifying a second-order
approximation when the design is complete, and
the ensuing transient response. For design we
arbitrarily select either a lead compensator pole
and zero and find the angular contribution at the
design point of this pole or zero with the
uncompensated systems open-loop poles and zeros.
The difference between this angle and 180? is the
required contribution of the remaining
compensator pole or zero.
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Example
Design three lead compensators for the
system that has 30 OS and will reduce settling
time down by a factor of 2.
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• Step I OS 30 equaivalent to damping ratio
  0.358, ? 69.02
• Step II Search along the line to find a point
  that gives 180 degree (-1.007j2.627)
• Step III Find a corresponding K (
  )
• Step IV calculate settling time of uncompensated
  system
• Step V twofold reduction in settling time
  (Ts3.972/2 1.986), correspoding real and
  imaginary parts are

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• Step VI lets 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.

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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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