Bilinear Transformation in Audio EQ Cookbook

Asked Feb 06 '23 at 17:03

Active Feb 06 '23 at 18:23

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In the Audio EQ CookBook there is $$\omega_0 = 2\pi\cdot\frac{f_0}{f_s}$$so frequency warping will be $$\omega_r = \frac{2}{T}\tan(\frac{\omega_0}{2})$$

Then $$s \longleftarrow \frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}} = \frac{\omega_r}{\tan\frac{\omega_0}{2}}\frac{1-z^{-1}}{1+z^{-1}}$$ But in the in the CookBook, we're given: $$s \longleftarrow \frac{1}{\tan\frac{\omega_0}{2}}\frac{1-z^{-1}}{1+z^{-1}}$$

What does $\omega_0 = 2\pi\cdot\frac{f_0}{f_s}$ exactly mean?

edited Feb 06 '23 at 18:23

Jdip

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asked Feb 06 '23 at 17:03

arlen

Robert Bristow-Johnson is a major contributor on this website and can probably answer your question. As far as $\omega_0$, it is the normalized frequency in radians/sample. See this Dan Boschen answer on the subject – Jdip Feb 06 '23 at 18:27
@RobertBristow-Johnson (I think that alerts Robert to the question, not sure) – Dan Boschen Feb 06 '23 at 22:27
@DanBoschen Users are only notified of @ if they comment on the same post – OverLordGoldDragon Feb 07 '23 at 11:20
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@OverLordGoldDragon Got it. Well I doubt he would miss the title “Audio EQ Cookbook”. – Dan Boschen Feb 07 '23 at 11:34

Bilinear Transformation in Audio EQ Cookbook

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