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Lead compensator for double integrator

  1. May 2, 2015 #1
    I'm trying to design a lead compensator for a double integrator plant, which is plant = tf([1],[1 0 0])

    I started with the angle condition. So, poles contribute (+) angle and zeros contribute (-) angle and that sums to 180

    Desired pole location is 4.4 +- 4.4j

    So, 45 + 45 = 90 degrees comes from the poles at the origin, assume that zero of the lead compensator is placed at -5, then atand(4.4/1) = 77.19 contribution from zero of the lead compensator.

    45 + 45 - 77.19 + Lp = 180

    Where Lp is the pole of the lead compensator. Lp should contribute +167.19 degrees, so tand(167.19) = -0.2274

    At this point I wouldn't expect to get a negative number. Anyways, I proceeded and found the location of the Lp as 4.4/0.2274 = 19.3492, so Lp should be located at -19.3492 - 4.4 = 23.75

    So the resulting Lead compensator is lead = tf([1 5],[1 23.75])

    When I close the loop with unity feedback closedLoop = feedback(plant*lead,1) and check the gain with sisotool(closedLoop), I see that my desired pole location is not reached. at -4.4 I got -+6.84i

    What am I doing wrong ?
     
  2. jcsd
  3. May 2, 2015 #2
    It sums to ##r180^\circ,r = \pm1,\pm3,\pm5,\dots## if the test point is on the root locus.

    Should be ##s = -4.4 \pm j4.4##?

    ##\arg(s) = \pm135^\circ##, and ##\arg(s - z) \approx \pm82.2^\circ,z = -5##.

    If ##s = -4.4 + j4.4##, then the angle sum is ##135^\circ + 135^\circ - 82.2^\circ = 187.8^\circ##. You want to remove those ##7.8^\circ##, but that's not possible with a single pole (it adds to the angle sum).

    The problem is your zero. You want it to subtract more than ##90^\circ##, so the angle sum dips below ##180^\circ##, and the pole makes up the difference, i.e. you should have ##z > -4.4##.
     
  4. May 2, 2015 #3
    Thank you so much. I was sure that I was missing something.
     
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