Al Clark wrote:
> 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.

What I'm saying is that there are two ways to make a windowed H 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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