Reply by Cedron ●December 3, 20152015-12-03
>On Wed, 02 Dec 2015 07:17:41 -0800, benjamin.couillard wrote:
>

[...snip...]

>> 
>> there are some similarities with my problem at hand, however the

author

>> know what kind of signal he's looking for, which is not my case

(except

>> that I know that frequency content can be in the same order as the
>> carrier frequency).
>> 
>> Regards
>


This will cause trouble with Tim's solution, even with Tim's assumption
being true.  See below.


>Check that the maximum value of B * y(t) is much smaller than 1 in your 
>high frequency case.
>
>When B * y(t) is small, regardless of bandwidth, then you can

approximate

>
>cos(2*pi*f*t + B*y(t))
>
>as
>
>cos(2*pi*f*t) + B*y(t) * sin(2*pi*f*t)
>
>-- this is just a simple first-order Taylor's series expansion of the cos



>term.  

After you have done the angle addition formula for cosines.  I think you
meant:

cos(2*pi*f*t)*cos(B*y(t)) - sin(2*pi*f*t)*sin(B*y(t))

Which then approximates, with Tim's assumption, to:

cos(2*pi*f*t) - B*y(t) * sin(2*pi*f*t)


>
>-- 
>
>Tim Wescott
>Wescott Design Services
>http://www.wescottdesign.com


Finding the value of y(t) is going to be very difficult in the
neighborhood of sin(2*pi*f*t) = 0.  This corresponds to where the peaks
and troughs are for cos(2*pi*f*t).  Since the signal is near horizontal
there, it makes sense that horizontal resolution will be difficult. 
Therefore, y(t) can be best determined when the cos signal is doing a zero
crossing and the sin signal has its maximum magnitudes.  If y(t) were
slowly varying in respect to the carrier signal, you could calculate the
values at the signal's zero crossings and interpolate for the peak and
trough values.  If y(t) is varying as much as the carrier, then the
interpolations are not likely to be accurate.

Ced 


---------------------------------------
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Reply by Cedron ●December 3, 20152015-12-03
>Tim Wescott <seemywebsite@myfooter.really> wrote:
>> On Wed, 02 Dec 2015 07:17:41 -0800, benjamin.couillard wrote:
> 
>>> say that we have a signal of this kind
> 
>>> x(t) = A*cos(2*pi*f*t + B*y(t));


[...snip...]


>
>As far as I know, there is more study on FM than PM, though they
>are related. 
> 


[...snip...]

Please note that the equation can be rewritten as:

x(t) = A*cos(g(t)*t)

where 

g(t) = 2*pi*f + B*y(t)/t

Thus, any frequency modulation could be interpreted as phase modulation,
and vice versa.

So, not only are they related, but they are indistinguishable
mathematically.  Ultimately, it turns into an equation of the form:

x(t) = A*cos( c0 + c1 * t + [higher order terms of t] )

Ced


---------------------------------------
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Reply by Tim Wescott ●December 2, 20152015-12-02
On Wed, 02 Dec 2015 18:48:00 +0000, glen herrmannsfeldt wrote:


> Tim Wescott <seemywebsite@myfooter.really> wrote:
>> On Wed, 02 Dec 2015 07:17:41 -0800, benjamin.couillard wrote:
>  
>>> say that we have a signal of this kind
>  
>>> x(t) = A*cos(2*pi*f*t + B*y(t));
>  
>>> say that f is 1 kHz.
>  
>>> If y(t) has a low frequency content compared to the carrier frequency,
>>> I find that it easy to extract the B*y(t) signal either using
>>> quadrature demodulation or the Hilbert transform.
>  
>>> However, in the case where y(t) has a high bandwidth, for example a
>>> wideband signal with frequency content up to 1 kHz, a lot of high
>>> frequency sidebands are generated, therefore the quadrature
>>> demodulation does not work to extract the B*y(t) signal.
> 
> As far as I know, there is more study on FM than PM, though they are
> related.
>  
>>> Do you have ideas on to extract the B*y(t) signal in this situation?
>>> Keep in mind that frequency "f" cannot be changed in that situation.
>  
> (snip)
> 
>> Check that the maximum value of B * y(t) is much smaller than 1 in your
>> high frequency case.
>  
>> When B * y(t) is small, regardless of bandwidth, then you can
>> approximate
>  
>> cos(2*pi*f*t + B*y(t))
>  
>> as
>  
>> cos(2*pi*f*t) + B*y(t) * sin(2*pi*f*t)
>  
>> -- this is just a simple first-order Taylor's series expansion of the
>> cos term.  Consequently, you can treat low-deviation phase modulation
>> as double-sideband modulation with quadrature sidebands.
> 
> OK, FM radio (which isn't PM) has about a 100MHz carrier, and a
> deviation of 75kHz, so the ratio is about 0.075.
> 
> The modulation can include audio (below 19kHz for stereo),
> the stereo subcarrier at 38kHz, with +/- 19kHz sidebands,
> and SCA at 67kHz.
> 
> Is that small enough?
>  
>> If the magnitude of B*y(t) grows with bandwidth enough that it's
>> maximum values start spending a lot of time approaching or exceeding 1
>> (if, for instance, you're widening a low-pass filter without depressing
>> its DC gain) then the approximation no longer holds, and the essential
>> nonlinearity of the modulation takes hold and makes lots of sidebands.
> 
> In theory, you have infinite sidebands even at very low frequency,
> but in most cases they fall off fast enough not to worry about.
> 
> FM radio fits all that in a 200kHz (+/- 100kHz) band.


If you do the math it's really the phase deviation that causes the 
nonlinearity, not the frequency deviation.

If you concentrate all the audio band into a 1kHz tone with a 75kHz 
deviation then the phase varies by +/- 75 radians, which is considerably 
more than 1, or 2*pi, or any other measure you may pick to claim that it 
is "small".

Even claiming that 1kHz is much too small and choosing 10kHz as the 
centroid of the audio band leaves you with a phase deviation of 7.5 
radians, which is still way above linear.

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Reply by glen herrmannsfeldt ●December 2, 20152015-12-02
Tim Wescott <seemywebsite@myfooter.really> wrote:

> On Wed, 02 Dec 2015 07:17:41 -0800, benjamin.couillard wrote:

 

>> say that we have a signal of this kind

 

>> x(t) = A*cos(2*pi*f*t + B*y(t));

 

>> say that f is 1 kHz.

 

>> If y(t) has a low frequency content compared to the carrier frequency, I
>> find that it easy to extract the B*y(t) signal either using quadrature
>> demodulation or the Hilbert transform.

 

>> However, in the case where y(t) has a high bandwidth, for example a
>> wideband signal with frequency content up to 1 kHz, a lot of high
>> frequency sidebands are generated, therefore the quadrature demodulation
>> does not work to extract the B*y(t) signal.


As far as I know, there is more study on FM than PM, though they
are related. 
 

>> Do you have ideas on to extract the B*y(t) signal in this situation?
>> Keep in mind that frequency "f" cannot be changed in that situation.

 
(snip)


> Check that the maximum value of B * y(t) is much smaller 
> than 1 in your high frequency case.

 

> When B * y(t) is small, regardless of bandwidth, then you can approximate

 

> cos(2*pi*f*t + B*y(t))

 

> as

 

> cos(2*pi*f*t) + B*y(t) * sin(2*pi*f*t)

 

> -- this is just a simple first-order Taylor's series expansion of the cos 
> term.  Consequently, you can treat low-deviation phase modulation as 
> double-sideband modulation with quadrature sidebands.


OK, FM radio (which isn't PM) has about a 100MHz carrier, and 
a deviation of 75kHz, so the ratio is about 0.075. 

The modulation can include audio (below 19kHz for stereo),
the stereo subcarrier at 38kHz, with +/- 19kHz sidebands,
and SCA at 67kHz. 

Is that small enough?
 

> If the magnitude of B*y(t) grows with bandwidth enough that it's maximum 
> values start spending a lot of time approaching or exceeding 1 (if, for 
> instance, you're widening a low-pass filter without depressing its DC 
> gain) then the approximation no longer holds, and the essential 
> nonlinearity of the modulation takes hold and makes lots of sidebands.


In theory, you have infinite sidebands even at very low frequency,
but in most cases they fall off fast enough not to worry about.

FM radio fits all that in a 200kHz (+/- 100kHz) band.

-- glen
Reply by Tim Wescott ●December 2, 20152015-12-02
On Wed, 02 Dec 2015 07:17:41 -0800, benjamin.couillard wrote:


> Hi everyone,
> 
> say that we have a signal of this kind
> 
> x(t) = A*cos(2*pi*f*t + B*y(t));
> 
> say that f is 1 kHz.
> 
> If y(t) has a low frequency content compared to the carrier frequency, I
> find that it easy to extract the B*y(t) signal either using quadrature
> demodulation or the Hilbert transform.
> 
> However, in the case where y(t) has a high bandwidth, for example a
> wideband signal with frequency content up to 1 kHz, a lot of high
> frequency sidebands are generated, therefore the quadrature demodulation
> does not work to extract the B*y(t) signal.
> 
> Do you have ideas on to extract the B*y(t) signal in this situation?
> Keep in mind that frequency "f" cannot be changed in that situation.
> 
> I found this reference
> http://www.acoustics.asn.au/conference_proceedings/AAS2009/papers/

p83.pdf

> 
> there are some similarities with my problem at hand, however the author
> know what kind of signal he's looking for, which is not my case (except
> that I know that frequency content can be in the same order as the
> carrier frequency).
> 
> Regards


Check that the maximum value of B * y(t) is much smaller than 1 in your 
high frequency case.

When B * y(t) is small, regardless of bandwidth, then you can approximate

cos(2*pi*f*t + B*y(t))

as

cos(2*pi*f*t) + B*y(t) * sin(2*pi*f*t)

-- this is just a simple first-order Taylor's series expansion of the cos 
term.  Consequently, you can treat low-deviation phase modulation as 
double-sideband modulation with quadrature sidebands.

If the magnitude of B*y(t) grows with bandwidth enough that it's maximum 
values start spending a lot of time approaching or exceeding 1 (if, for 
instance, you're widening a low-pass filter without depressing its DC 
gain) then the approximation no longer holds, and the essential 
nonlinearity of the modulation takes hold and makes lots of sidebands.

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Reply by ●December 2, 20152015-12-02
> 
> there are some similarities with my problem at hand, however the author know what kind of signal he's looking for, which is not my case (except that I know that frequency content can be in the same order as the carrier frequency).
> 
> Regards


If the modulation bandwidth is wide compared to the carrier frequency, some of the modulation sidebands will extend below 0 Hz to negative frequencies.

I think you can sort this out if you have a complex  rather than real signal.

I'm not sure what that means in the context of your application, but it probably means two signals of some sort.

Mark
Reply by ●December 2, 20152015-12-02
Hi everyone,

say that we have a signal of this kind

x(t) = A*cos(2*pi*f*t + B*y(t));

say that f is 1 kHz.

If y(t) has a low frequency content compared to the carrier frequency, I find that it easy to extract the B*y(t) signal either using quadrature demodulation or the Hilbert transform.

However, in the case where y(t) has a high bandwidth, for example a wideband signal with frequency content up to 1 kHz, a lot of high frequency sidebands are generated, therefore the quadrature demodulation does not work to extract the B*y(t) signal. 

Do you have ideas on to extract the B*y(t) signal in this situation? Keep in mind that frequency "f" cannot be changed in that situation.

I found this reference http://www.acoustics.asn.au/conference_proceedings/AAS2009/papers/p83.pdf

there are some similarities with my problem at hand, however the author know what kind of signal he's looking for, which is not my case (except that I know that frequency content can be in the same order as the carrier frequency).

Regards