Re: Odd length Hilbert FIR Implementation

Jerry Avins <jya@xxxxxxxx> wrote in news:Ot6dnYs9OrlK65nZRVn-hg@xxxxxxx:

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

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.
--------------------------------------------------------------------
Purveyors of Fine DSP Hardware and other Cool Stuff
Available at http://www.danvillesignal.com
.

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