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Bode Plot - Calculating ωgc and ωpc analytically

  1. Dec 12, 2011 #1
    I am learning to draw Bode Plots. I am able to figure out ωgc, ωpc, Phase Margin & Gain Margin graphically from the Bode Plot. But I was wondering if there is a way to calculate ωgc and ωpc mathematically with some formulas - how do I do this?
     
    Last edited: Dec 12, 2011
  2. jcsd
  3. Dec 12, 2011 #2
    There is the Laplace transform that converts a time domain equation to an "s" domain equation, where

    [tex]s = i\omega[/tex]
    or
    [tex]s = \sigma + i\omega[/tex]

    In this case, the time domain equation might be some differential equation that defines the behavior of a filter.
     
  4. Dec 12, 2011 #3
    I calculated ωgc analytically & compared it to the one I got from a Bode plot.

    When Gain is rather low (say 2), then the calculated ωgc varies a lot from the one obtained from an asymptotic Bode Plot.

    For eg.

    let's take
    G(s)H(s) = 80/(s)(s+2)(s+20)

    = 2/(s)(1 + 0.5s)(1 + 0.05s).

    In this case the ωgc is very close to where the approximation error happens for the first cornering frequency (2 rad/s - corresponding to (1 + 0.5s)).

    My calculated ωgc = 1.57 rad/s.
    The one on the graph (where the Magnitude plot intersects 0) is around 2 rad/s.

    On a semilog paper the horizontal distance in the 2 lines (ω = 1.57 & ω = 2) is rather big.

    Is this a known issue?
     
  5. Dec 13, 2011 #4
    I use Bode plot for years to design all sort of closed loop control systems and I consider myself pretty good at taming them. I never get into the s-plane stuff. Bode plot and the ωgc is almost two different thing, you draw your Bode plot from knowing the pole and zero frequency of the system, not the other way around. When you use Bode Plot, you try to avoid all the calculation and use graph to design the system. If you want to do calculation, don't use Bode Plot.

    When you design a closed loop system, you measure the system poles and zeros and design the amplifier circuit with poles and zeros to get single pole cross over with enough phase margin. Identifying the system poles and zeros are the most difficult part.

    When you identify all the poles and zeros of the system, your job is over 90% done, the rest is just defining which part belong to the forward gain and which part be the reverse feedback and draw the Bode plot and add your own gain, poles and zeros to get the single pole cross over with phase margin!!!
     
    Last edited: Dec 13, 2011
  6. Dec 13, 2011 #5
    i remember calictlating phase cross over frequency by equating argument of open loop transfer function to -180 degrees and gain cross over frequency by equating magnitude to 1. . .
     
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