INC341 Design Using Graphical Tool (continue)

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INC341Design Using Graphical Tool(continue)

• Lecture 10

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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
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PID controller design

• Evaluate the performance of the uncompensated
  system
• Design PD controller to meet transient response
  specifications
• Simulate and Test, redesign if necessary
• Design PI controller to get required steady-state
  error
• Find K constant of PID
• 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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Step 1

• OS 20 ? damping ratio 0.456
• ? ? 62.87
• Search along ther line to find a point of 180
  degree (-5.415j10.57)
• Find a correspoding K121.51
• 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 -8.13j15.867

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• Sum of the angles from uncompensated poles and
  zeros to the test point (-8.13j15.867) is
• -198.37
• The contribution angle for the compensator zero
  is then 180-198.371 18.37

?PD controller is (s55.92)
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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 (-7.516j14.67), with an associate gain of
  4.6

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Step 5

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

K1 259.5, K2 128.6, K3 4.6
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Step 6
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Lead-Lag Compensator Design

• Same procedures as in designing PID
• Begin with designing lead compensator to get the
  desired transient response
• design lag compensator to improve steady-state
  error

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Example
Design lead-lag compensator so that the system
can operate with 20 OS, twofold reduction in
settling time, and tenfold improvement in
steady-state error for a ramp input
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Step 1

• OS 20 ? damping ratio 0.456
• ? ? 62.87
• Search along ther line to find a point of 180
  degree (-1.794j3.501)
• Find a correspoding K192.1
• Then find the settling time

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Step 2

• Decrease settling time by a factor of 2 ? get a
  real part of a compensator pole
• To keep a damping ratio constant, imaginary part
  of the pole will be at
• The compensator poles will be at -3.588j7.003

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• Select the compensator zero at -6 to coincide
  with the open-loop pole
• Sum of the angles from uncompensated poles and
  zeros to the test point (-3.588j7.003) is
• -164.65
• The contribution angle for the compensator zero
  is then 180-164.65 15.35

Lead compensator is
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Then find a new K at the design point (K1977)
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Step 3

• Simulate the lead compensated system

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Step 4

• Originally the uncompensated system has the
  transfer function

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• After adding the lead compensator, the system has
  changed to
• Static error constant, Kv, is then 6.794 (lead
  compensator has improved ss error by a factor of
  6.794/3.2012.122)
• So the lag compensator must be designed to
  improve ss error by a factor of 10/2.1224.713

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Step 5

• Pick a pole at 0.01, then the associated zero
  will be at 0.04713
• Lag-lead compensator
• Lag-lead compensatated open loop system

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Step 6
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Conclusions
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Feedback Compensation
Put a compensator in the feedback path
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Tachometer
Popular feedback compensator, rate sensor
Tachometer generates a voltage output
proportional to input rational speed
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rate feedback
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Example
Design a feedback compensator to decrease
settling time by a factor of 4 and keep a
constant OS of 20
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Step 1
OS 20 ? damping ratio 0.456 ? ?
62.87 Search along the daping ratio line to get
a summation of angle of 180 degrees at
-1.809j3.531
Find the corresponding K from the magnitude rule
settling time
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Step 2
Reduce the Settling time by a factor of 4
A new location of poles is at -7.236j14.123
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At dominant pole -7.236j14.123, KG(s)H(s) has a
net angle of -277.33 ? needs an additional
angle from zero of 277.33-180 97.33
Find the corresponding K to the pole at
-7.236j14.123 using the magnitude rule
K 256.819
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Feedback block is 0.185(s5.42)
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Physical System Realization
PI Compensator
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Lag Compensator
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PD Compensator
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Lead Compensator
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PID Compensator