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

> 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 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

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.

BTW, better was determined strictly by plotting results for each case
from my filter program (Momentum Data Systems - QEDesign). The frequency
range of interest is near Fs/4, which is the center frequency of the
hilbert bandpass filter. I used a Kaiser Window in all cases.

Al

--
Al Clark
Danville Signal Processing, Inc.
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