Active Filter Derivation

  1. [​IMG]

    Can I get some help with this? I don't even know where to begin.
  2. jcsd
  3. 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. 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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