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dklein
02-26-2005, 11:25 AM
Suppose I have an ARMA filter of order N with known A and B coefficients.
Does there exist another ARMA filter of order N whose impulse respons
(and output) is the Hilbert transform of the original filter? If so, wha
is the expression relating the two sets of coefficients?

Thanks!






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Andor
02-26-2005, 02:08 PM
You can just multiply the two transfer functions, ie. multiply each AR
coefficient with j.

Andor
02-26-2005, 02:10 PM
Sorry. Multiply each AR coefficient with -j, or the MA coefficients
with j.

02-26-2005, 02:15 PM
Andor <[email protected]> wrote:
> Sorry. Multiply each AR coefficient with -j, or the MA coefficients
> with j.

Which of course isn't true :)

Try your transformation on only MA (FIR) process and check hilbert transform
orthogonality property.

Andor
02-26-2005, 02:27 PM
Yeah, that was silly. It's one of my s-z domain confusions. scuse me!

djklein
02-28-2005, 07:29 PM
I don't completely understand your reply. What do you mean by "your
transformation" and what do you mean by "check hilbert transform
orthogonality"?

I understand the operations required in the Fourier domain, and I was
hoping this would transfer to a simple operation on ARMA coefficients
(e.g., a rotation, as the multiplication by would imply).

Thanks,
David

[email protected] wrote:
> Andor <[email protected]> wrote:
> > Sorry. Multiply each AR coefficient with -j, or the MA coefficients
> > with j.
>
> Which of course isn't true :)
>
> Try your transformation on only MA (FIR) process and check hilbert
transform
> orthogonality property.

03-01-2005, 09:38 AM
djklein <[email protected]> wrote:
>
> I don't completely understand your reply. What do you mean by "your
> transformation"

Andor's proposition

> and what do you mean by "check hilbert transform
> orthogonality"?

Original signal and Hilbert transformed are orthogonal.

> I understand the operations required in the Fourier domain, and I was

Frequency domain isn't necessary but often used. Why? Because it's simple
method but has drawbacks (errors in time domain).

Hilbert transform is defined as time domain integral.

> hoping this would transfer to a simple operation on ARMA coefficients
> (e.g., a rotation, as the multiplication by would imply).

No, doing in frequency domain you will transform a "transfer function" of
arma filter. After IFFT you get an impulse response - where are your new
coefficients??? You must fit new arma to new impulse response.

Ask google about "IIR Hilbert" or "90 phase allpass" - in second case you
will connect serially (convolute) original arma and allpass. Real hilbert
transform solutions are always approximations, they works only in band
limited case if you arma transfer function doesn't fall to zero at both 0
and nyquist frequencies you are in trouble.