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Homework Help: Bode Plot question

  1. Apr 10, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm doing a question that requires a Bode plot (magnitude and phase).

    3. The attempt at a solution
    I got a transfer function of
    [tex]H(\omega)=\frac{1+\frac{j*\omega}{3}}{1-0.1667(\omega)^{2}+0.4167j*\omega}[/tex]

    So there is a simple zero at 3 rad/s; complex pole at 2.45 rad/s; and no DC gain.
    I am able to plot the magnitude plot correctly, however my phase isn't exactly right.

    For phase I have essentially two contributions from the zero and the pole:
    From zero: 0.3<omega<30 rad/s with a slope of 45 degrees/decade
    From pole: 0.245<omega<24.5 rad/s with a slope of -90 degrees/decade

    So combining the two, I get (w==omega, i'm too lazy to write it out):
    0<w<0.245: 0 degrees,
    0.245<w<0.3: slope of -90 degrees/decade
    0.3<w<24.5: slope of -45 degrees/decade
    24.5<w<30: slope of 45 degrees/decade
    w>30: slope of 0

    However the solution my prof provided says:
    0<w<0.245: 0 degrees,
    0.245<w<0.3: slope of -90 degrees/decade
    0.3<w<24.5: slope of -45 degrees/decade
    w>24.5: slope of 0, flattens out at -90 degrees.

    Am I wrong? If so, why?
     
  2. jcsd
  3. Apr 11, 2010 #2

    berkeman

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    Staff: Mentor

    The zero is at -3 rad/s. Does that make a difference, or was that just a simple typo?
     
  4. Apr 11, 2010 #3
    Uh, neither. We're only dealing with positive frequencies (actually, absolute values, I suppose). We're calling them poles and zeros, but the values we're interested in is just the corner frequencies.
     
  5. Apr 11, 2010 #4

    berkeman

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    Staff: Mentor

    Sorry, then why do you say you have a zero at 3 rad/s?
     
  6. Apr 11, 2010 #5
    It's just the terminology used. Corner frequency is like absolute value of zero/pole.
     
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