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ECE 486 Assignment # 5
http://courses.ece.uiuc.edu/ece486/
Issued: February 27 Solutions Due: March 6, 2009
Problems:
12 The date is February 27, 2009 — a new (and worthless) compensator is born! The transfer function of the turbo-lag compensator (TLC) is expressed exactly as in the traditional lag compensator,
GTLC(s) =
s − z s − p
where again |z| > |p|. However, in the new TLC we choose the pole to be large, rather than small. Will this work? Let’s look at the example of February 27,
Gp(s) =
Km s(1 + τ s)
with Km = 100 and τ = 1/25. Recall that we want e∞ ≤ 0 .01 for a unit ramp input, and Mp < 0 .1.
(a) Draw the root locus plot obtained with proportional control, and verify that the proportional gain K 0 = 0.125 will give ζ =
2 /2. Find the resulting natural frequency ωn of the closed loop poles. (b) Choose p = −N ωn and z = 8p in your TLC, for a large value of N (say, N = 50 or 500). Sketch the resulting root locus 1 + KGTLCGp(s) = 0, computing any jω crossings. Where are the closed loop poles when K = K 0? Find the velocity error constant for your design - are the specifications met? (c) Can you think of any problems with the new compensator? Why is it that this approach never advocated in any textbook?
Solution: (a) This was worked out in lecture on Feb. 27. The poles are located at s = − 12. 5 ± 12. 5 j, which gives ωn = 12. 5
(b) Just as in consideration of the usual lag controller, this compensator does not have much affect on the locus for s ∼ ωn.
−500 −450 −400 −350 −300 −250 −200 −150 −100 −50 0
0
20
40
60
80
100
120
140
Real Axis
Imaginary Axis
Proportional control
Locus is almost unchanged for s in this region
TLC Compensator is not stabilizing for large K
(c) Although the locus is unchanged, the location of the poles is very different for a specific value of K: If s 0 is on the locus with K = K 0 then,
|Gc(s 0 )Gp(s 0 )| = 1/K 0 = 8.
For |s 0 | < 50 we have Gc(s 0 ) ≈ |z|/|p| = 8, giving
|Gp(s 0 )| ≈ 1
Hence the location of the poles are virtually identical to what would be obtained with proportional control using K 1 = K 0 |z|/|p| = 1. Conclusion: There is no value at all in placing a pole-zero pair far out in the LHP.
13 Consider the plant transfer function,
Gp(s) =
s
s^2 + 51s + 550
(a) Design a compensator U (s) = Gc(s)E(s) so that the dominant closed-loop poles yield,
ζ ≥ 0 .4; σ ≥ 7; kv ≥ 50sec−^1.
Use a combination of lead and lag compensators. Illustrate your design using root locus methods. (b) Obtain a step response using Matlab, and verify that the resulting overshoot and rise time specs are consistent with your closed loop poles.
Solution: (a) The plot on the left below shows the root locus obtained with PD compensation, and on the right the lead compensator obtained by adding an additional pole.
-5 -20 -15 -10 -5 0
0
5
10
15
20
Root Locus
Real Axis Real Axis
Imaginary Axis
-4 -50 -45 -40 -35 -30 -25 -20 -15 -10 -5 0
0
2
4
6
8
10
12
14
16
Root Locus Editor for Open Loop 1 (OL1) Rlocus for PD compensator that introduces a zero at -
Rlocus for lead compensator that introduces a zero at -20, and a pole at -
Pole location when K = 23600
The design is complete on introducing a lag compensator:
Real Axis
Imaginary Axis
Rlocus for lead compensator that introduces a zero at -20, and a pole at -140 and lag compensator that introduces zero at -1, pole at -0.05 to increase K_v by 20
Note that the introduction of the location of the closed loop poles -4^ -2 -50 -45 -40 -35 -30 -25 -20 -15 -10 -5 0 obtained from the lead design
0
2
4
6
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Root Locus Editor for Open Loop 1 (OL1)
