Is this legit?

Not to be confused with a Bissell filter, which are widely used in Vacuum cleaners. ;-)

Reply to
Michael A. Terrell
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the finer differences between filters - you are confined to Bessel, Butterw orth and Chebyshev.

^^ for... oops

Reply to
George Herold

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There are "transitional Gaussian" filters that have good time-domain response and better frequency response than a Bessel. They behave like a Bessel until some number of dB down, then drop off fast.

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Reply to
John Larkin

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Oo, will have to retract that (should have started from the start of the thread. Of course it could be steeper than a butterworth - the butterworth parameters are chosen to be maximally flat in the pass band, the "L" filter doesn't have peaks and troughs but it does sage a little.

Reply to
David Eather

GROAN!

--
"For a successful technology, reality must take precedence  
over public relations, for nature cannot be fooled." 
                                       (Richard Feynman)
Reply to
Fred Abse

I think you missed the point of the original post. Leave out elliptic filters since we don't want transmission zeroes.

The point was at infinity, N poles is N poles, so dammit, the falloff rate should be the same for all classes of filters. I think at infinity, that is true. But for practical purposes, you aren't at infinity. The poles are not all at the same frequency, so as you look at the response with increasing frequency, you are getting further away from DIFFERENT poles.

Reply to
miso

You need to be specific as to what a "direct high order solution" is. If we are talking ladder filters, you have less component sensitivity than cascaded biquads. But in continuous time active filters, ladders always have more components than a cascade of biquads simply because you need the inverted phase signal.

If I may quote Dan Senderovich, "It is hard to move a ladder."

Reply to
miso

You guys know better than to leave an opening! ;-)

Reply to
Michael A. Terrell

The shape of the Chebychev's ripple says it's a 4th order. Given that the Butterworth has 60 dB loss at a normalized frequency of 10, it's only a 3rd order.

Reply to
PeteD

less than 60

Reply to
PeteD

ev.svg

... . Regarding your last question "Is there any point..." . Apparently so. A good example is the Linkwitz-Riley crossover network. It 's not "optimal" in any classic sense. Mr Riley, while fooling around with filters, discovered that by cascading two 2nd order Butterworth filters, he obtained a 4th order filter, later dubbed the "Linkwitz Riley" filter. He then discovered that a L-R lowpass combined with a L-R Highpass results in a loudspeaker crossover network with near ideal characteristics: The two f ilters combine to produce an exact flat voltage response throughout the ent ire spectrum. The Highpass and Lowpass outputs remain exactly 360 degrees out of phase (ie, in phase) throughout the entire frequency spectrum. This phase response eliminates the "Lobing" problem in loudspeaker design. A f ine example of tinkering that produced useful results. . Regarding the Legendre filter. It has been proven to be the monotonic filt er that is optimum in the sense that it has maximum attenuation in the stop band. Its response is steeper than the Butterworth (and all other filters that are monotonic in the passband and stopband) at frequencies close to t he characteristic frequency, but it eventuall attains a slope of 6n per oct ave, where n is the order of the filter.

Reply to
Jon Lark

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