I am trying to derive the coefficients used for a IIR implementation for the lowpass portion of a SVF filter. I've seen finished expressions for the coefficients (smith (p.8) and victor), their discretizized transfer function looks something like (not super important for my question)
$$ H(z) = \frac{k_f^2 \, z^{-1}}{1-(2-k_f(k_f+k_q))z^{-1} + (1-k_f k_q)z^{-2}} \, , $$ with $k_q = \frac{1}{Q}$ and (next is central to my question) $k_f = 2 \, \sin (\pi f_c T) $ .
I would like to derive these parameters myself, which I have attempted, but I am having trouble with the parameters involving the cutoff frequency $f_c$!
This is my starting point, (from analysing SVF circuit), I get the following transfer function for the lowpass portion,
$$ H(s) = \frac{\Omega_0^2}{s^2 + \frac{\Omega_0}{Q}s + \Omega_0^2} \, . $$
Next step is performing the discretization. I did this using the bilinear transform ($z = e^{sT}$), using from what I gather the Tustin/trapezoidal approximation ($s = \frac{1}{T} \log(z) \approx \frac{2}{T} \frac{z-1}{z+1})$.
Prewarping
This transform requires prewarping and was done as follows,
$$ \Omega_0 = \tfrac{2}{T} \tan \left( \tfrac{\omega_0}{2} \right) \,, $$
where $\Omega_0 = 2 \pi f_c$ and $f_c$ is my wanted cutoff frequency.
Now, already I can see that my derivation is going to result in a different expression than what the two links above are getting. Their frequency parameters shows up in expressions that look like $ f = 2 \sin(\pi f_c T) $
My Question
Basically, where is the $\sin()$ term coming from?
Could it be that they use Euler forward (or backward) as an approximation?
$$ s = \frac{1}{T} \log(z) \approx \frac{z - 1}{T} $$
And then done some prewarping for that case? I have looked around the web and can not find any mention of prewarping done for euler backward/froward. Is that usually done?
That would give me $$ \begin{split} s = j \Omega &= \frac{z-1}{T} \\ j \Omega &= \frac{e^{j \omega}-1}{T} \\ j \Omega &= \frac{e^{j \omega/2}-e^{-j \omega/2}}{T e^{-j \omega/2}} \\ j \Omega &= \frac{j 2 \sin(\omega/2)}{T e^{-j \omega /2}} \\ \Omega &= \frac{2 \sin(\omega/2)}{T e^{-j \omega /2}} \\ \end{split} $$
Not sure what I am missing.
--
Thank you very much for any help regarding this!