DSPRelated.com
Forums

Hilbert Transform multiplication rule

Started by Scott Seidman August 1, 2006
Bedrosian's derivation of the Hilbert product rule,

H[f(x)g(x)]=f(x)H[g(x)], where g is often the high frequency carrier 
cos(w_0t+P), stipulates that either F(w) and G(w) don't overlap, or f(x) 
and g(x) are both analytic.

A few questions:

A) does non-overlapping mean that F(w) and G(w) don't overlap, or that 
F(w)*G(w) goes to zero at some point in between the two spectra (like a 
Nyquist-type arrangement?

B) What happens when they overlap?  Is the result of a Hilbert-based 
demodulation technique for overlapping F(w) and G(w) a simple aliasing?

C) (this one might be tough)  The original Bedrosian paper refers to a 
more general form than the r(t)cos(w_0t +P) type modulation, which is
r(t)cos(w_0t+P(t)), where the phase offset is a function of time (derived 
by Kelly).  Bedrosian points out that the Hilbert product rule for a 
carrier with constant phase offset requires non-overlapping signals or 
two analytic signals, while the less specific form requires no spectral 
restriction.  This suggests that the less specific form is not subject to 
aliasing.  Is this true?  What does this mean practically?  Does it mean 
that if you have access to the actual carrier, instead of doing a sort of 
carrier recovery, that you don't need to worry about aliasing?  Does it 
mean something else entirely?  I have the Kelly J Soc Indust App Math 
paper, but its a little dense, and I'm hoping that some facile person 
could help me find a shortcut through this.

BTW, my motivation is that I'm trying to analyze a physiological systems 
output w/ envolope detection, where the output really seems to be an AM 
signal.  The problem is that the carrier frequencies are very low 
(<10Hz), and I have no way of knowing for a fact that the message signal 
is non overlapping (though most of the energy must be nonoverlapping).  
This isn't a huge worry, but I'd like to have some more grip on how this 
impacts the methodology.

Thanks in advance,
-- 
Scott
Reverse name to reply