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Nyquist plot, damping factor and phase margin

  1. Feb 6, 2014 #1
    1. The problem statement, all variables and given/known data

    A Nyquist (polar) plot of a standard second-order system is shown below (drawn to scale).
    attachment.php?attachmentid=66371&stc=1&d=1391738703.jpg
    Suppose a unit-step function is applied as the input to this system. Determine the peak percentage overshoot expected in the system output.

    2. Relevant equations



    3. The attempt at a solution

    My idea was to trace backwards from the ω→+∞ at the origin until I reach a vector from the origin to a point on the curve with magnitude one. This ω will correspond to a gain cross over frequency and provide me with the phase margin.

    I could then link this value of the phase margin back to zeta through the following equation,

    [tex]\text{P.M.} = \gamma = tan^{-1}\left( \frac{2\zeta}{\sqrt{-2\zeta^{2}+\sqrt{1+4\zeta^{2}}}}\right)}[/tex]

    The first point I could find that would give me a vector of magnitude one resides at (0.6, -0.8) yielding a phase margin of 126.87°. Unfortunately this PM yields a negative value for zeta.

    Any idea where I went wrong or an easier way to solve the problem?
     

    Attached Files:

  2. jcsd
  3. Feb 10, 2014 #2
    Bump, still looking for help!
     
  4. Feb 10, 2014 #3

    AlephZero

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    It might help if you defined what your textbook calls a "standard" 2nd order system.

    Looking carefully at the plot, it doesn't go exactly through (0.6, -0.8), but it does go exactly through (1, 0), (0, 0) and (0, -0.8), and the slope at (1,0) is vertical. Is that enough data to fix the transfer function, for your "standard" system?
     
  5. Feb 10, 2014 #4
    A standard 2nd order system would be in the form,

    [tex]G(s) = \frac{\omega_{n}^{2}}{s^{2}+2\zeta \omega_{n} s + \omega_{n}^2}[/tex]

    The only information I can extract out of those three points is that,

    [tex]G(j\omega_{n}) = \frac{1}{j2 \zeta} = -j0.8[/tex]

    [tex]|G(j\omega_{n}| = \frac{1}{2 \zeta} = 0.8[/tex]

    [tex]\Rightarrow \zeta = 0.625[/tex]

    So,

    [tex]\text{OS%} = 100 e^{\frac{-\pi \zeta}{\sqrt{1 - \zeta^{2}}}} = 8.083 \text{%}[/tex]
    attachment.php?attachmentid=66491&stc=1&d=1392080153.jpg

    Does the slope at the point (1,0) somehow link back to ζ as well?
     

    Attached Files:

    Last edited: Feb 10, 2014
  6. Feb 10, 2014 #5

    rude man

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    The problem I have with this is that you speak of a standard 2nd order transfer function, which is well defined: F(s) = (1/(s2 + 2ζwns + wn2), which I assume is the closed-loop transfer function.

    However, a Nyquist plot is that of the open loop, not the closed loop, and the open loop is not unique to the closed loop.

    One candidate however is F(s) = G/(1+G) which is G/(1+GH) with H=1 (unity fedback).

    The closed loop is also written as F(s) = 1/(s+s1)(s+s2) where s1 and s2 are complex-conjugate pole pairs:
    s1 = a +jb
    s2 = a -jb
    where a and b are functions of ζ and wn.

    Solve for G(a,b) from G/(1+G) = 1/(s+a+jb)(s+a-jb) with s = jw.

    You can now generate a "template" Nyquist plot of G(jw) with unknown a and b. Then pick values for a and b so as to generate your given plot. This may not be as bad as it seems if you pick your w point judiciously. Knowing a and b you can find wn and ζ . Finally you have the entire closed loop transfer function and can input a step function and solve for the overshoot.

    This seems a really horrible problem that way, there is probably a short-cut of some kind equating open-loop Nyquist plots with the corresponding closed-loop transfer functions, but I don't know it. I'll try to research this a bit further.
     
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