Sketching Bode Plots (Which one is correct?)

Discussion in 'Homework Help' started by jegues, Jan 22, 2011.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    Hello all,

    I've got a question with regards to sketching Bode plots.

    I'm supposed to sketch the bode plot of my transfer function,

    T(s) = \frac{-390000 - 0.585s}{164368 + 0.019302s}

    by hand.

    EDIT: Whoops! I think I mixed my zeroes and poles up!

    I should have a zero at,

    \omega_{z} = \frac{390000}{0.585}

    and one pole at,

    \omega_{p} = \frac{164368}{0.019302}

    When we compute these, it seems as though the zero occurs at a high frequency than the pole, so the bode plot would resemble a low pass filter, but when I modeled the circuit in Multisim it gives me a graph like the following,
    [​IMG]

    I attached the original schematic of the circuit I'm working with as well incase someone can tell by inspection whether the circuit will act as a low, or high pass filter.

    There is a possibility I may have made an error in my calculator computing the values of the transfer function, but I'd like to be sure before I redo all my calculations. (There's alot, and it's really messy)

    EDIT: I just redid my calculations again and I arrive at the same result. Am I misunderstanding how to plot the bode plot correctly?

    Now that I've got my zeros and poles sorted out, my plot should match this one!

    What do you guys think?
     
    Last edited: Jan 22, 2011
  2. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
    718
    I kept trying to scratch the specks off my screen before I realized it was your graphic. :D

    Can you size them a bit smaller?
     
  3. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    In a Bode plot, when a pole occurs, the graph goes up by 20dB/decade. When a zero occurs, the graph goes down by 20dB/decade.

    As a result, the graph will first rise when it meets the pole, and then level down when it reaches the zero.

    Your assumption that it would be a low-pass filter is wrong, in my opinion.

    <edit> This post is flowed. Read below for amends.
     
    Last edited: Jan 22, 2011
  4. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    I think it is just the opposite. A pole sends the slope of the magnitude down while a zero sends the slope of the magnitude up. In the equations given above, the zero occurs before the pole, so the resultant graph appears correct.
     
  5. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    You guys are correct. I don't know how I remembered this the wrong way.

    I admit my mistake and confirm that a zero sends the plot up and a pole sends it down. Sorry for any misunderstanding.
     
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