Re: FIR Hilber Transformer

I. R. Khan wrote:

Jerry Avins wrote:

throws the communication out of order.
Top post is confusing because it

I. R. Khan wrote:

But odd symmetric coefficients do not necessarily make a HT. A
differentiator also has odd symmetric coefficients.

Clay wrote:

By simple inspection you can see if the phase repsonse matches that of
a Hilbert transformer. The coefs need to have odd symmetry. This is
necessary in order to have a 90 degree phase shift.


I assume you want the HT to occupy the widest bandwidth between DC and
Fs/2. Number the coefficients c[n] according to their position from the
center, negative for one side, positive. For an odd number of
coefficients, c[0] is zero, and so are all other even-numbered ones. The
odd numberer coefficients, multiplied by their indices, will be a
constant.


Are you talking about windows based design?

Yes. Before windowing.

This does not seem true for
other designs. For example, for the following HT
{-1/16, 0, -9/16, 0, 9/16, 0, 1/16}

Did you plot the magnitude response? What is the useful bandwidth?
Compare it to [-6 0 -10 0 -30 0 +30 0 +10 0 +6]

I found a very nice chapter on HTs in Rick's book, and got the
information that I needed. Thanks for your reply.

That book is good for a lot of things; you're welcome.

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

  • Re: FIR Hilber Transformer
    ... Clay wrote: ... a Hilbert transformer. ... The coefs need to have odd symmetry. ... Number the coefficients caccording to their position from the ...
    (comp.dsp)
  • Re: FIR Hilber Transformer
    ... The coefs need to have odd symmetry. ... Number the coefficients caccording to their position from the ... Are you talking about windows based design? ...
    (comp.dsp)