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Drawing the step response of a second order system

  1. Oct 30, 2012 #1
    1. The problem statement, all variables and given/known data
    I have to draw the step response of the following two systems.
    G1 = (4+3s)/(s^2+4s+4)
    G2 = 3/(s^2+4s+4)

    So I started to draw the step response of the second system first. It has to be in the funky standard form:
    [itex]
    \frac{ω2}{s2 + 2ζωs + ω2}
    [/itex]

    EDIT:
    Seems like the above doesn't work..

    2)/(s2 + 2ζωs + ω2)
    Problem is that 4 =/= 3 ;) So i'm not sure how to go about fixing that..

    Furthermore, I have NO idea how to even do the first system, G1.

    In the solutions they just wrote the following:
    G1 has a slight overshoot due to the zero at -4/3
    4/(s^2+4s+4) + 3s/(s^2+4s+4)

    That is exactly what they wrote. Now I'm curious as to WHY there is an overshoot just because the zero is -4/3? What is the definition for that?

    So how would I go about solving this problem?

    Thanks!
     
  2. jcsd
  3. Oct 31, 2012 #2
    What about multiplying by a constant?

    s F(s) <-> df/dt

    The question wants you to think about what happens as you change the location of the zero (better to write the zero as 4s+a to have the correct form immediately). Something special happens at (4s+8) but what you should get out of it is what happens if it's close to the poles, close but not too close, and what happens if it's far away.
     
    Last edited: Oct 31, 2012
  4. Nov 1, 2012 #3

    LCKurtz

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    Don't use the superscript and subscript keys in a tex expression. TeX also has built in capability for Greek letters. Do it like this:$$
    \frac{\omega^2}{s^2 + 2\zeta\omega s + \omega^2}
    $$Right click on that expression to see the TeX.
     
  5. Nov 4, 2012 #4

    rude man

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    The formal way to get step response is to multiply your G(s) by 1/s, then do a partial fraction expansion of the resulting G(s)/s and inverse-transform each term.

    Better yet, get a good table of transforms.

    You might have noticed that s^2 + 4s + 4 = (s+2)^2.

    Also remember that if f(t) → F(s), then
    df/dt → sF(s) - f(0+) but for transfer functions initial conditions are ignored by definition.
    Also, ∫f(t')dt' from 0 to t → F(s)/s.

    I would not spend too much time wondering why the zero creates overshoot. Probably a snow-job by your prof.
     
    Last edited: Nov 4, 2012
  6. Nov 5, 2012 #5
    The step response for a system in standard second order form can be determined from the coefficients. The damping ratio is particularly important.

    Not so. The unexpected bump in the step response is easily seen by noticing the total response is the sum of the usual step response plus some fraction of the derivative of the step response. The derivative comes from s*F(s) and its origin is the zero. The relative size of the derivative term depends on how close the zero is to the poles of the second order system. A special situation occurs when the zero is equal to one of the poles, in which case you get pole/zero cancellation and a first order system will result.

    A lot of system design is focussed on where the poles are and ignore where the zeroes are, which you can't always do. In control system design, a condition might be to have the zeroes far enough in the left side of the plane while designing the dominant poles so that the response you are trying to get is not changed by the zeroes.
     
    Last edited: Nov 5, 2012
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