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bitrex
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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: [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]?
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]?
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