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2nd order filter transfer function normalization

  1. Feb 3, 2010 #1
    I'm looking at a guide by Texas Instruments on active filter design. In it are the following equations for a second order lowpass filter, verbatim:

    The coefficient form of the denominator: [tex]s^2 + a_1s + a_0[/tex]

    Normalized: [tex]P(s) = (\frac{s}{\sqrt{a_0}*\omega_c})^2 + \frac{a_1s}{a_0*\omega_c} + 1[/tex]

    Substituting [tex] s = j2\pi f, \omega_c = 2\pi f_c, a_1 = \frac{1}{Q}, \sqrt{a_0} = FSF [/tex]

    [tex]P(f) = -(\frac{f}{FSF*f_c})^2 + \frac{1}{Q}\frac{jf}{FSF*fc} + 1 [/tex]

    Maybe I'm missing something obvious here, but why is it that it is not FSF^2 in the second term of the last equation, if [tex]\sqrt{a_0} = FSF[/tex]?
    Last edited: Feb 4, 2010
  2. jcsd
  3. Feb 4, 2010 #2
    I made an error in the LaTeX of the first equation, I've corrected it. :redface:
  4. Feb 5, 2010 #3
    I think I see the problem, it must be a misprint on their part. The equation should be:

    [tex]P(s) = (\frac{s}{\sqrt{a_0}*\omega_c})^2 + \frac{a_1s}{\sqrt{a_0}*\omega_c} + 1[/tex]

    Also remembering that [tex] \omega_c = \sqrt{a_0} [/tex] helps.
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