Jerry Avins <[email protected]> wrote in news:[email protected]:

> Al Clark wrote:
>
> ...
>
>> The basic question is why does the hilbert filter with length 4i+1,
>> for example 9 or 13 have better performance than a fir filter of
>> length 4i-1, for example 7 or 11, even though the number of non zero
>> coefficients is the same for the 4i+1 & 4i-1 filters. In the 4i+1
>> case, the first and last coefficients are zero, which allows the
>> filter size to be reduced to a 4i-1 length.
>
> The 4n+1 filter has two more points. It's just that the end two being
> zero allows you the efficiency of not including them in the
> computation. Try it with a low-pass. The same thing happens but it's
> harder to see in the plots. Unless the window includes a pedestal,
> windowed FIRs are always better with a window that's wider than the
> raw coefficients by two.
>
> ...
>
> Jerry

Sometimes a remez exchange FIR is desirable for hilbert transformers
since you may be able to reduce ripple over a wider range of frequencies.
The catch is that every coefficient will be non zero. The window
techniques reduce your computation load by about a factor of 2, if you
take advantage of the trivial multiplies (by ignoring them). This makes
it interesting to compare a fir filter with twice the length using
windowing techniques versus a remez exchange method.

It seems to me that the 4n+1 tap filter converted to a 4n-1 tap filter is
a little trick since it my give you a little better performance for no
added cost.

I still can't say I understand everything that is happening, but I guess
it might be added to our little bag of hilbert transformer tricks.

--
Al Clark
Danville Signal Processing, Inc.
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