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

> 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.

--
Al Clark
Danville Signal Processing, Inc.
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