Elements of Feedback Control

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Prototype 2nd order system
target
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Settling time
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Effects of additional zeros
Suppose we originally have
i.e. step response
Now introduce a zero at s -z
The new step response
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Effects

• Increased speed,
• Larger overshoot,
• Might increase ts

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When z lt 0, the zero s -z is gt 0, is in the
right half plane. Such a zero is called a
nonminimum phase zero. A system with nonminimum
phase zeros is called a nonminimum phase system.
Nonminimum phase zero should be avoided in
design. i.e. Do not introduce such a zero in
your controller.
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Effects of additional pole
Suppose, instead of a zero, we introduce a pole
at s -p, i.e.
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L.P.F. has smoothing effect, or averaging effect
Effects

• Slower,
• Reduced overshoot,
• May increase or decrease ts

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Root locus

• A technique enabling you to see how close-loop
  poles would vary if a parameter in the system is
  varied
• Can be used to design controllers or tuning
  controller parameters so as to move the dominant
  poles into the desired region

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• Recall step response specs are directly related
  to pole locations
• Let p-sjwd
• ts proportional to 1/s
• Mp determined by exp(-ps/wd)
• tr proportional to 1/p
• It would be really nice if we can
• Predict how the poles move when we tweak a system
  parameter
• Systematically drive the poles to the desired
  region corresponding to desired step response
  specs

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Root Locus
k s(sa)
y
e
r
Example

-
Two parameters k and a. would like to know how
they affect poles
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The root locus technique

• Obtain closed-loop TF and char eq d(s) 0
• Rearrange terms in d(s) by collecting those
  proportional to parameter of interest, and those
  not then divide eq by terms not proportional to
  para. to get
• this is called the root locus equation
• Roots of n1(s) are called open-loop zeros, mark
  them with o in s-plane roots of d1(s) are
  called open-loop poles, mark them with x in
  s-plane

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1. The o and x marks falling on the real axis
   divide the real axis into several segments. If a
   segment has an odd total number of o and/or x
   marks to its right, then n1(s)/d1(s) evaluated on
   this segment will be negative real, and there is
   possible k to make the root locus equation hold.
   So this segment is part of the root locus. High
   light it. If a segment has an even total number
   of marks, then its not part of root locus. For
   the high lighted segments, mark out going arrows
   near a pole, and incoming arrow near a zero.

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• Let npolesorder of system, mzeros. One root
  locus branch comes out of each pole, so there are
  a total of n branches. M branches goes to the m
  finite zeros, leaving n-m branches going to
  infinity along some asymptotes. The asymptotes
  have angles (p 2lp)/(n-m). The asymptotes
  intersect on the real axis at

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• Imaginary axis crossing
• Go back to original char eq d(s)0
• Use Routh criteria special case 1
• Find k value to make a whole row 0
• The roots of the auxiliary equation are on jw
  axis, give oscillation frequency, are the jw axis
  crossing points of the root locus
• When two branches meet and split, you have
  breakaway points. They are double roots. d(s)0
  and d(s) 0 also. Use this to solve for s and k.
• Use matlab command to get additional details of
  root locus
• Let num n1(s)s coeff vector
• Let den d1(s)s coeef vector
• rlocus(num,den) draws locus for the root locus
  equation
• Should be able to do first 7 steps by hand.