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Bode Plot

  1. Aug 29, 2015 #1
    1. The problem statement, all variables and given/known data

    [itex] \frac{0.5}{1 - \frac{3145j}{w}} [/itex], sketch the bode plot

    2. Relevant equations


    3. The attempt at a solution

    Moving the jw to the bottom
    [itex] \frac{0.5}{1+ \frac{3145}{jw}} [/itex]

    Let s = jw

    [itex] \frac{0.5}{1+ \frac{3145}{s}} [/itex]

    [itex] = \frac{0.5}{\frac{s+3145}{s}} [/itex]

    [itex] = (0.5) \frac{s}{s+3145} [/itex]

    [itex] = (0.5)(s) \frac{1}{3145(\frac{s}{3145}+1)} [/itex]

    [itex] = (\frac{0.5}{3145})(s) (\frac{1}{(\frac{s}{3145}+1)}) [/itex]

    So are those the three separate bode functions that I have to sketch ?
     
  2. jcsd
  3. Aug 29, 2015 #2

    LvW

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    I would start somewhat earlier (3rd line from the bottom formula): Two separate functions to be combined:
    1.) Differentiating function : H1=0.5*s : Straight line with a slope of +20dB/dec crossing the f-axis at w=2)
    2.) First order lowpass H2=1/(s+3145) : Horizontal line (1/3145) below the pole frequency ; -20dB/dec above the pole frequency.
     
  4. Aug 29, 2015 #3

    rude man

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    Homework Helper
    Gold Member

    What? It's just one function: F(s) = ks/(Ts+1)
    where you can easily identify k and T, right?
    You did just fine, your last expression is in the most convenient form for drawing the Bode plot.
    There are two segments, one corresponding to the zero at s=0 and one to the pole at s = -1/T.
     
  5. Aug 30, 2015 #4
    For number 2, I was taught that it has to be in the form 1/(s+1), which is why I took the 3145 out
     
  6. Aug 30, 2015 #5
    Yes, It's one function but I meant that I would draw those three separate functions and then add them together

    So the first one would be a constant line 20log(0.5/3145), the second one, s, would be a differentiation with +20 db/dec slope and the last one would be zero gain from 0 to 3145 rad/s and then -20db/dec slope
     
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