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Active Filter Derivation

  1. Apr 29, 2011 #1

    Can I get some help with this? I don't even know where to begin.
  2. jcsd
  3. Apr 29, 2011 #2
    Bandpass filters usually employ both inductors and capictors.

    The inductor cuts off the high frequencies, while the capacitor cuts off the low frequencies, ideal for a bandpass filter.
  4. Apr 29, 2011 #3
    Whoops, I just noticed that the the answer is given in the figure. I can see they did not use a capacitor!

    I managed to derive the transfer function for this circuit and found the following,

    [tex]T(jw) = \frac{-z}{(jwL + z)(jwL + R)}[/tex]

    Now we simply need to select our Z such that the requirement for resonant frequency is obatined.

    If we select,

    [tex]z = R_{z}[/tex]

    We rearrange the transfer function,

    [tex]\frac{-\frac{1}{R}}{(1 + \frac{jwL}{R_{z}})(1 + \frac{jwL}{R})}[/tex]

    This becomes,

    [tex]\frac{-\frac{1}{R}}{(1 + \frac{jw}{w_{L}})(1 + \frac{jw}{w_{H}})}[/tex]

    Now we know what wL and wH are we can solve for fL and fH since,

    [tex]f = 2\pi w[/tex]

    Then our desired resonant frequency,

    [tex]f_{r} = \sqrt{f_{L} \cdot f_{H}}[/tex]

    Simply solve for, [tex]R_{z}[/tex].

    You should find that, [tex]R_{z} = \frac{100\pi^{2}L^{2}}{R}[/tex]
    Last edited: Apr 29, 2011
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