Re: Odd length Hilbert FIR Implementation

Al Clark wrote:

> Jerry Avins <jya@xxxxxxxx> wrote in
> news:BOOdnfXlxoWte57ZnZ2dnUVZ_sadnZ2d@xxxxxxx:
>
>> I sent this earlier, but I don't see it, so here's a repeat.
>>
>> Al Clark wrote:
>>
>>
>>> As many of you know, Hilbert pairs are often constructed by using an
>>> odd- length FIR filter with antisymmetrical coefficients. The I part
>>> is
>>
>>
>> taken from the middle of the delay line and the Q from the output of
>> the filter.
>>
>>> If you use a Parks/McClellan method for the filter, you will have
>>
>>
>> coefficients for each tap of the filter.
>>
>>> If you use a window approach instead, The coefficients are 0 for
>>
>>
>> every other value, so in principle, you need about 1/2 the MACs since
>> half the MACs are 0.
>>
>>> Assuming a Window approach:
>>>
>>> If the length of the filter is 4i-1, that is 3,7,11,..... there will
>>
>>
>> be 2i nonzero coefficients and 2i-1 zero coefficients. The sequence is
>> w0,0,w2,0,......w(4i-2)
>>
>>> If the length of the filter is 4i+1, that is 5,9,13,..... there will
>>
>>
>> be 2i nonzero coefficients and 2i+1 zero coefficients. The sequence is
>> 0, w1, 0, w3,....0
>>
>>> There is the same number of non zero coefficients for 4i-1 and 4i+1
>>
>>
>> length filters.
>>
>>> I calculated both N=11 and N=13 hilbert filters using a Kaiser
>>
>>
>> Window. The N=13 had a flatter passband even though the number of non
>> zero coefficients are the same. Since the end points of the N=13
>> coefficents are 0, I could truncate the filter to make it smaller. In
>> essense, I now have a N=11 filter.
>>
>>> I guess this doesn't make too much sense to me. The coefficients are
>>
>>
>> different, but I am surprised that the filter actually looks much
>> better.
>>
>>> What am I missing?
>>
>>
>>
>> Filters using windows without a pedestal (Hann, as opposed to Hamming,
>> e.g.) have zero coefficients at the ends even if all the unwindowed
>> coefficients are non zero, because the window value is zero there. For
>> such filters, compute the window for n + 2, where n is the number of
>> taps. The longer window gives better results. Computing the entire
>> filter for n + 2 and relying on the window of n + 2 elements to
>> shorten the result to n may be better yet; I haven't tried it.
>>
>> Jerry
>
>
>
> Since the 7,11,15,set have non zero values at the endpoints, I assume that this must have been done for all the filters. I used a Kaiser window using QED1000.


What I'm saying is that there are two ways to make a windowed filter with 2n-1 coefficients (including any zeros). Both use a window with 2n+1 points (including the zeros at the ends). One starts with 2n-1 points, the other with 2n+1 points. They are clearly not the same. Just as clearly, the same applies to even numbers of taps. A Kaiser window of n+2 points produces a filter of n points whether the unwindowed filter has n points or n+2.

I lost track of the question. :-)

Jerry
--
Engineering is the art of making what you want from things you can get.
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.

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