2nd order filter transfer function normalization

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The discussion focuses on the normalization of the transfer function for a second-order lowpass filter as described in a Texas Instruments guide. The original equations presented include a coefficient form and a normalized version, leading to confusion regarding the second term's coefficient. The contributor identifies a potential misprint in the equations, suggesting that the second term should include FSF instead of being omitted. They clarify that the correct form should reflect the relationship between the coefficients and the cutoff frequency. Understanding that ω_c equals √a_0 is also emphasized as a key point.
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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: s^2 + a_1s + a_0

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

Substituting s = j2\pi f, \omega_c = 2\pi f_c, a_1 = \frac{1}{Q}, \sqrt{a_0} = FSFP(f) = -(\frac{f}{FSF*f_c})^2 + \frac{1}{Q}\frac{jf}{FSF*fc} + 1

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 \sqrt{a_0} = FSF?
 
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I made an error in the LaTeX of the first equation, I've corrected it. :redface:
 
I think I see the problem, it must be a misprint on their part. The equation should be:

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

Also remembering that \omega_c = \sqrt{a_0} helps.
 
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