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Math Help - Fastest roll-off of a filter

  1. #1
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    Fastest roll-off of a filter

    I was wondering: can a filter be optimised to have maximum roll off. Filters are usually specified as:
    |H(\omega)|^2 = \frac{1}{1 + \alpha(\omega)} where \alpha is a polynomial function. I was wondering if it was possible to maximise the steepness of the roll-off by specifying the passband and stopband ripples.

    What is usually done is:
    |H(\omega)|^2 = H(\omega)H(-\omega) = \frac{1}{1 + \alpha(\omega)}
    This gives you the polls and zeros. Then you'll need to do some optimisation of some kind.
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  2. #2
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    Re: Fastest roll-off of a filter

    Optimum "L" filter - Wikipedia, the free encyclopedia particular the two references in the bibliography.

    Maximum roll off usually means undesired characteristics somewhere else, such as too much passband ripple or phase distortion.
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