[SOLVED] Chebyshev type I band-pass frequency scaling

[bubulescu]

Hello

For a low-pass prototype Chebyshev type I filter, if the ripple in the pass-band is Rp, then the corner frequency will be @-Rp, but if the ripple needs to be less than the attenuation (e.g. Rp=-1 A=-2), one would use this formula:

\[ \omega_{sc} = \frac{\omega_p}{cosh \left(\frac{1}{N} acosh \left( \frac{\epsilon_p}{\epsilon_r} \right) \right)} \]

and, for a high-pass transformation, there will be \[\omega_p \] multiplied by [...], instead of division.


The way I'm using the general transfer function for band-pass is:

\[ H(s) = \frac{a_{1_1} s}{s^2 + b_{1_1} s + b_{0_1} } \cdot \frac{a_{1_2} s}{s^2 + b_{1_2} s + b_{0_2} ) \cdot ... \]

(and, for the band-stop, they will have s^2+a0), adapted to fit the bandwidth requirements for being greater than, or less than 2 (normalized). As you can see I cannot apply the frequency scaling like for the low- or high-pass filters.


So, what I'd like to know is how to use frequency scaling for band-pass/band-stop filters.


Thank you in advance,
Vlad

 

[bubulescu]

This is either a curse, or luck, and it's not the first time... After posting I realized that, the way I am using the transfer function is derived from the state-space direct formula which made it possible to find the poles (,zeroes) and use their wp=1 normalized value, resulting in (using idt() as an integral):

\[ H(s)=idt \left( a_1 v_{in}-b_1 v_{out}+ idt \left( a_0 v_{in} - b_0 v_{out} \right) \cdot \omega_p \cdot \omega_{sc} \right) \cdot \omega_p \cdot \omega_{sc} + a_2 v_{in}\]

therefore w_sc was directly used in the final formula, but the correct scaling comes from altering the poles(,zeroes).