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Pole-zero analysis

  1. Oct 22, 2011 #1
    a network has the equation s2 + (3+6K1)s + 6K2 = 0
    it has to be stable and cannot decay faster than e-3t

    show that the network must meet the following criteria:
    K2 > 0
    |K1| < 1/2
    K2 > 3K1

    My attempt at the solution:
    solving the quadratic formula for s, i get
    [-(3+6K1)2 + √(3+6K1)2 - 4(6K2)] / 2

    i believe the first two are true because the coefficients cannot be < 0
    i also believe that ω should be less than 3. i am having trouble with the last condition, K2 > 3K1. where does this come from??
     
    Last edited: Oct 22, 2011
  2. jcsd
  3. Oct 22, 2011 #2

    rude man

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    Hint #1: Better check your solution to the quadrature equation first.

    Hint #2: you will get either a complex-conjugate pole pair or two real poles. You need to examine the pole locations for both cases for stability and ensure that no real pole exists < -3.

    Hint # 3: k2 > 0 always required.
    k1 > -1/2 is imposed by stability requirement
    k1 < +1/2 is imposed by real pole location > -3
    k2 > 3k1 is imposed by real pole location > -3.

    So I think you can see that your initial observations need to be re-examined.
    This is not an easy exercise!
     
  4. Oct 22, 2011 #3
    wow, that's a lot more involved than i thought
    the solution to the quad should be [-(3+6K1) + √((3+6K1)2 - 4(6K2))] / 2
     
  5. Oct 22, 2011 #4
    i'm a little confused by "ensure that no real pole exists < -3"
    i thought that was determined by the imaginary part (in the square root) and that it would be positive 3?
     
  6. Oct 22, 2011 #5

    rude man

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    No. That requirement stems from the need for the output to drop no faster than exp(-3t).

    The quantity inside the square root sign may be either + or -. If it's negative you get a complex-conjugate pole pair, so only one real part, which is the "b" in your quadratic solution (did you fix it yet?). If the quantity inside the square root sign positive you get two real poles (poles on the real axis).

    It's possible that the problem intended for the solution to be limited to the two-real-pole case. That's because their statement "no faster than exp(-3t)" might preclude an oscillatory solution, which you get with a complex-conjugate pole pair. It's a question of semantics, and you should raise that issue with your TA or whoever. I interpreted the requirement as meaning the envelope of the output is restricted to no faster than exp(-3t). If you allow a complex-conjugate solution your output will look like exp(-at)*sin(wt) with a being the real part of the c/c pole pair and +/-w the imaginary parts.
     
  7. Oct 22, 2011 #6

    rude man

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    Well, you're right, but you have to come up with the restrictions on k1 and k2 as imposed by the sadist who gave you the problem. :-)
     
  8. Oct 22, 2011 #7

    rude man

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    Actually, looking back on the solution, I see that all the conditions placed on the two-real-pole solution (positive number inside the square root sign) also apply to the cc pair. So just limit yourself to + values inside the square root sign & you'll get it all.
     
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