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FEEDBACK COMPENSATION

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A popular feedback compensator is a rate sensor (tachometer feedback) ... Tachometer. Osman Parlaktuna Linear Control Systems Fall 2004. APPROACH 1 ... – PowerPoint PPT presentation

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Title: FEEDBACK COMPENSATION


1
FEEDBACK COMPENSATION
  • OSMAN PARLAKTUNA
  • OSMANGAZI UNIVERSITY
  • ESKISEHIR,TURKEY
  • www.ogu.edu.tr/oparlak

2
FEEDBACK COMPENSATION
R(s)
C(s)

K
K1
G1(s)
G2(s)
Minor loop
KfG(s)
Major loop
3
ADVANTAGES
  • Feedback compensation can yield faster responses.
  • It can be used in cases where cascade
    compensation cannot be used due to noise
    problems.
  • May not require additional amplification
  • A popular feedback compensator is a rate sensor
    (tachometer feedback).

4
TACHOMETER FEEDBACK

R(s)
G1(s)
K1
K
Tachometer
Kfs
5
APPROACH 1
By pushing K to the right past the summing
junction, pushing G2(s) to the left past the
pickoff point, and then adding the two feedback
paths, block diagram becomes
R(s)
C(s)
KK1G1(s)G2(s)
6
The open-loop transfer function
is G(s)H(s)K1G1(s)KfHc(s)KG2(s) Without
feedback, KfHc(s), the open-loop transfer
function is G(s)H(s)KK1G1(s)G2(s) Thus, the
effect of adding feedback is to replace the poles
and zeros of G2(s) with the poles and zeros of
KfHc(s)KG2(s) Hence this method is similar to
cascade compensation in that we add new poles and
zeros via H(s) to reshape the root locus to go
through the design point.
7
EXAMPLE
R(s)
C(s)
Design rate feedback compensation to reduce the
settling time by a factor of 4 while continuing
to operate the system with 20 overshoot.
First, checking the values for the uncompensated
system Plotting the root locus and determining
the 0.456 line intersection with the root locus
gives the dominant poles at -1.8?j3.5. The
settling time is 4/1.82.22sec and should be
reduced to 2.22/40.555sec
8
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9
R(s)

C(s)
1
Kfs
R(s)
C(s)
10
To achieve a fourfold decrease in settling time,
the real part of the pole must be increased by a
factor of 4 to 4(-1.8)-7.2. The imaginary part
is then, wd-7.2 tan(1800-62.870)14 Using the
compensated dominant pole position of -7.2j14,
the sum of the angles from the systems poles to
be -2770, then the angle condition of the root
locus requires the contribution from zero to be
970. The zero location is calculated as
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12
Since the third pole is not far away than the
dominant poles it cannot be neglected. Its
effect is seen in the step response
13
The gain at the desired point is found to be 259,
which is K1Kf. Since Kf is the reciprocal of the
compensator zero (1/5.48), then K11419.3 In
order to evaluate the steady-state error
characteristics, redraw the block diagram as
follows
C(s)
R(s)
14
APPROACH 2
Approach 2 allows us to use feedback compensation
to design a minor loops transient response
separately from the closed-loop system response.
The minor loop basically represents a
forward-path transfer function whose poles can be
adjusted with the minor-loop gain. Thus, rather
than reshaping the root locus with additional
poles and zeros, we can actually change the
plants poles through a gain adjustment. Finally,
the closed-loop poles are set by the loop gain.
15
MINOR-LOOP FEEDBACK COMPENSATION EXAMPLE
R(s)

C(s)
K
Kfs
Design minor-loop compensation to yield a damping
ratio of 0.8 for the minor loop and a damping
ratio of 0.6 for the closed-loop system.
16
The transfer function of the minor loop is
The real parts of the complex poles are constant
at -10. Using the damping ratio of 0.8 one can
determine the poles as -10 ?j7.5. The gain
Kf81.25 places the minor-loop poles in a
position to meet the specifications. These poles
together with the pole at the origin act as
open-loop poles that generate a root locus for
variations of the gain K.
17
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