<magoldfish@gmail.com> wrote in message 
news:1140115903.996500.291670@g43g2000cwa.googlegroups.com...
>
> fizteh89 wrote:
>> You don't get any meaningful results by just computing analytic signal
>> and estimating instantaneous frequncy.
> I agree with this statement for complex signals like speech, but what
> about for simple fm signals/chirps?  Shouldn't the IF give me the
> modulating signal?  This is the kind of information I am seeking.
>>


Yes, ideal FM demodulation is the process of finding the instantaneous 
frequency. Practically it is not usually done by Hilbert transforms, but it 
can be. I've done it before. It is easier to use a pair of all pass filters 
whose phase response differ by 90 degrees. But once your signal has been 
either analytically extended via Hilbert transforms or split into orthogonal 
signals, the instantaneous calculation is the same. Namely  (I dQ - Q dI) / 
(I^2 + Q^2) or one of its suitable approximations.


> Has anyone actually used the Hilbert transform to estimate the IF of an
> AM-FM signal + noise?  What do you practically do with the negative
> frequencies it returns?
>

The interpretation depends on your application. Typically with FM, the 
carrier is modulated to frequencies above and below nominal. So in this case 
negative just means the instantaneous frequency is below your nominal 
carrier.

Quite often in DSP based radios the signal is mixed down to nearly baseband 
before sampling. I.e., let's say you have an RF signal whose bandwidth is 30 
kHz, the radio then can be made to mix the channel's center frequency to say 
25kHz and then the sampling may be done at 100kHz. So spectrally your signal 
now occupies 25+-15 = 10 to 40 kHz. In a case like this the instantaneous 
frequency varies about 25 kHz. Certainly with noise, your calculation may 
end up with spikes and such. But these can be easily filtered since the 
transmitted signal likely had a slew rate limit filter going into the 
modulator. Sometimes a median filter works well here. If you mixed the 
signal all of the way down to base band, you may encounter difficulties with 
the 90 degree phase splitting. Notice in the above scenario, I allowed for 
10 kHz guard bands on both sides of the signal. Since in this case the 
discrete Hilbert transform doesn't have to function below 10kHz (0.1 of the 
sampling rate) it is actually realizable for low cost.  Some radios do the 
phase splitting in hardware and use two samplers - others use a single 
sampler and use a DSP to split the signal. There are trade offs with each 
method. I.e., amplitude matching overall and while in compression, phase 
errors, and computational overhead.

IHTH,

Clay S. Turner

Stan Pawlukiewicz <spam@spam.mitre.org> writes:

> People who live in Hartford don't spell that way! ;)

Damn Yankees! :-)

-- 
"And he sees the vision splendid 
of the sunlit plains extended
And at night the wondrous glory of the everlasting stars."

 

in article 1140116271.124235.289870@o13g2000cwo.googlegroups.com,
magoldfish@gmail.com at magoldfish@gmail.com wrote on 02/16/2006 13:57:

>> if you visualize that analytic signal as a function of time, it might look
>> like a sorta cork-screw wrapping around the t-axis.  the rate that it wraps
>> around is the instantaneous frequency.  if the instantaneous frequency
>> changes signs, that means that the parametric curve that represents the
>> analytic signal has stopped wrapping in one direction or sense (say,
>> counter-clockwise) and has begun wrapping in the other direction (now
>> clockwise).
>> 
>> with only words that's about the only way i can illustrate it.

> Thanks-- makes perfect sense.  Guess I missed the phasor
> interpretation.
> 
> Are phase-direction reversals usually signifcant in terms of signal
> properties (e.g., change-points in the signal), or just artifacts of
> the algorithm?  Does it make sense to ignore the direction of the
> phasor rotation, and just look at the absolute value of the IF?

i dunno.  i guess it depends on whatever it is that you're doing.  some
applications of the phase vocoder sorta ignore phase and just have it
continue on from the phase of an earlier frame.  but in using the HT to
identify the instantaneous frequency is not the same.

usually this analytic signal thing is used for signals that are "bandpass"
signals.  the analytic signal:

    a(t) = x(t) + j*Hilbert{x(t)} = r(t) * exp(j*theta(t))

where 

    r(t) = |a(t)| >= 0

    theta(t) = arg{a(t)} + k(t)*2*pi

where k(t) always takes on integer values and is used to "unwrap" theta(t)
from its principal value so as to minimize discontinuities. the
instantaneous frequency f(t) (in Hz) is

    f(t) = 1/(2*pi) * (d/dt)theta(t) = 1/(2*pi) * theta'(t)

now, in communications, sometimes a linear trend in theta(t) can be noted:

    f(t) = 1/(2*pi) * theta'(t) = f0 + f_delta(t)

where

    f_delta(t) = f(t) - f0           (read "f sub delta")

and
                   t
    theta(t) = integral{ 2*pi*f(u) du}
                 -inf
                               t
             = 2*pi*f0*t + integral{2*pi*f_delta(u) du}
                             -inf

             = 2*pi*f0*t + theta_delta(t)    (read "theta sub delta")

where

    f_delta(t) = 1/(2*pi) * (d/dt)theta_delta(t)

and then we see that

    a(t) = r(t)*exp(j*theta(t))

         = (r(t)*exp(j*theta_delta(t))) * exp(j*2*pi*f0*t)

so it looks like that the complex "baseband" signal,

            r(t)*exp(j*theta_delta(t))

which may not have symmetrical spectrum about f=0 (because it is not
necessarily real) but, if f0 is properly chosen (by least square fitting a
strait line to theta(t)), it should have spectrum approximately equally in
quantity on both sides of DC (f=0).  this baseband signal is getting
modulated up (or the spectrum "bumped up") to be centered around f0 instead
of 0.

anyway, in this case i would never expect theta(t) to turn around and wrap
around the t-axis in the other direction.  it is heavily biased by f0 to
wrap only in the positive direction or sense.  but the "baseband" signal,
r(t)*exp(j*theta_delta(t)), could go either way.  if the instantaneous
frequency of x(t) is less than f0, then the instantaneous frequency of the
baseband signal is negative.

i dunno what to do with this info.  but it's there to possibly help you
interpret data.


-- 

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

> if you visualize that analytic signal as a function of time, it might look
> like a sorta cork-screw wrapping around the t-axis.  the rate that it wraps
> around is the instantaneous frequency.  if the instantaneous frequency
> changes signs, that means that the parametric curve that represents the
> analytic signal has stopped wrapping in one direction or sense (say,
> counter-clockwise) and has begun wrapping in the other direction (now
> clockwise).
>
> with only words that's about the only way i can illustrate it.
Thanks-- makes perfect sense.  Guess I missed the phasor
interpretation.

Are phase-direction reversals usually signifcant in terms of signal
properties (e.g., change-points in the signal), or just artifacts of
the algorithm?  Does it make sense to ignore the direction of the
phasor rotation, and just look at the absolute value of the IF?

Marcus

For a zero-mean FM signal I expect it to return the modulating
function.

Marcus

fizteh89 wrote:
> You don't get any meaningful results by just computing analytic signal
> and estimating instantaneous frequncy.
I agree with this statement for complex signals like speech, but what
about for simple fm signals/chirps?  Shouldn't the IF give me the
modulating signal?  This is the kind of information I am seeking.

> The whole methodology is called Hilbert-Huang Transform and it can be
Thanks for the pointer.  Already used the HHT-- seems very nice-- but
for the simple problem I am considering I think it is overkill: the
signal I am analyzing is not the sum of many AM-FM components, merely a
single AM-FM component.  Again, I expected the direct application of
the Hilbert transform + IA/IF estimation to be effective.

Has anyone actually used the Hilbert transform to estimate the IF of an
AM-FM signal + noise?  What do you practically do with the negative
frequencies it returns?

Peter K. wrote:
> "dbell" <dbell@niitek.com> writes:
> 
> 
>>What do you expect the instantaneous frequency of a speech signal to be
>>indicative of?
> 
> 
> Coloured noise?

People who live in Hartford don't spell that way! ;)
> 

dbell wrote:

> I can think of easier ways to generate colored noise.

Me too. :-)

I can think of easier ways to generate colored noise.

Dirk

Peter K. wrote:
> "dbell" <dbell@niitek.com> writes:
>
> > What do you expect the instantaneous frequency of a speech signal to be
> > indicative of?
>
> Coloured noise?
>
> --
> "And he sees the vision splendid
> of the sunlit plains extended
> And at night the wondrous glory of the everlasting stars."

You don't get any meaningful results by just computing analytic signal
and estimating instantaneous frequncy.

To get something meaningful you might have to do empirical mode
decomposition first, then compute analytic signal and instantaneous
frequency of each mode and finally arrive at the so called Hilbert
spectrum.

The whole methodology is called Hilbert-Huang Transform and it can be
found at
http://www.fuentek.com/technologies/hht.htm