is the cheby2 filter actually maximally flat anyway? i mean, yes - it is flat, but as far as i know, maximum flatness was not a design criterion in this case. the design criterion was actually to be equiripple in the stopband - that the magnitude is flat in the passband seems to be a byproduct in this case.urosh wrote: ↑Sun Oct 10, 2021 3:10 amBut, two pass Cheb2 is not as flat. We would need filter designed in such way that two such filters in series give maximally flat response, have no idea how simple/hard/doable that is.
but be that as it may - it may indeed be interesting to look at that question for a butterworth-like design. if M(w) denotes the magnitude response, the butterworth design requires that as many derivatives as possible should vanish at w=0. ...now we would apply the same requirement to M^2(w) instead. using the product rule, we would get M^2'(w) = M' * M + M * M' = 2*M*M' ...and now we need to require that to vanish at w=0....next: M^2'' = ...blablabla ...must vanish, too...etc. hmmm...am i on the right track? might indeed be useful to work that out.
or wait - no - in the case of the butterworth design, it is actually already required that the derivatives of M^2 should vanish. that seems to translate to the condition that M itself is also maximally flat. so maybe the same then also holds for (M^2)^2 = M^4? i don't know, but it seems plausible at first glance. squaring or taking the square-root doesn't seem to destroy the condition. but could also be a brainfart. i did not really think this through yet. edit: but then - no - linkwitz-riley is actually already well known as butterworth-squared filter and it is not maximally flat (because then it would be butterworth). ...maybe we need to design some sort of butterworth-sqrt filter. in the sense of: when the magnitude-squared response is squared, the result should be maximally flat
i'm not sure, if this is even possible. the polynomial in the denominator of the magnitude-squared funtion of an n-th order butterworth filters is: B_n(x) = 1 + x^(2*n). now we would look for a polynomial A_n(x) such that (A_n(x))^2 = B_n(x). but that would mean that B_n(x) must be factored into A_n(x)*A_n(x) which would require B_n(x) to have roots of even multiplicity. but for our given butterworth polynomial B_n(x), all roots are simple - so it can't be factored this way...take all with a grain of salt....these are just some quick ideas