Digital FM Demodulation - Filtering

Matt Boytim wrote:
> 
> Well, if for example a bandlimited i(t)+jq(t) has a trajectory through
> the origin, phi(t) will instantaneously flip 180 degrees which would
> have infinite bandwidth.

I don't agree. In order for phi(t) to instantaneously flip, so must
i(t) and or q(t). 

> But, in general, the FM modulation of a bandlimited signal produces a
> signal which has infinite bandwidth, 

I agree with this, but this doesn't make your point. Your point was
that phi(t) is infinite-bandwidth even when z(t) = I(t) + i*Q(t) is
not. 

> which we typically approximate as
> finite bandwidth.  The truncation of bandwidth is equivalent to adding
> modulation components that cancel the truncated components; these
> components extend to infinite frequency.  

That's a very interesting way to look at it. Could be true. Got anything 
a little more concrete?

> Does this help?

A little. Thanks Matt.

> 
> Matt
> 
> Randy Yates <yates@ieee.org> wrote in message news:<3F6FD798.9CB600BD@ieee.org>...
> > Matt Boytim wrote:
> > >
> > > Even if the signal
> > > i(t)+j*q(t) is bandlimited to nyquist, the signal phi(t)=arg(i+jq) is
> > > not (except in trivial cases)
> >
> > Hey Matt,
> >
> > Really? I can't off-the-cuff think of why this may or may not be. Are
> > you sure? Do you have a pointer? This (FM demodulation) is actually
> > of significant interest to me as well.


-- 
%  Randy Yates                  % "...the answer lies within your soul
%% Fuquay-Varina, NC            %       'cause no one knows which side 
%%% 919-577-9882                %                   the coin will fall."
%%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*, ELO   
http://home.earthlink.net/~yatescr

Randy Yates <yates@ieee.org> wrote in message news:<3F73B451.83543DF7@ieee.org>...
> Matt Boytim wrote:
> > 
> > Well, if for example a bandlimited i(t)+jq(t) has a trajectory through
> > the origin, phi(t) will instantaneously flip 180 degrees which would
> > have infinite bandwidth.
> 
> I don't agree. In order for phi(t) to instantaneously flip, so must
> i(t) and or q(t). 

A bandlimited signal is allowed to pass through zero, but the flip in
phi is a discontinuity.

A complex carrier exp(j*w0*t)=cos(w0*t)+j*sin(w0*t) is clearly
bandlimited, then phi(t)=w0*t (also bandlimited if taken for all t). 
Now, add a DC offset 1+exp(j*w0*t)=1+cos(w0*t)+j*sin(w0*t) (so that it
rotates about 1 instead of the origin), still clearly bandlimited but
phi(t) is now periodic with a 180 degree discontinuity every cycle so
obviously not bandlimited.

Matt

> 
> > But, in general, the FM modulation of a bandlimited signal produces a
> > signal which has infinite bandwidth, 
> 
> I agree with this, but this doesn't make your point. Your point was
> that phi(t) is infinite-bandwidth even when z(t) = I(t) + i*Q(t) is
> not. 
> 
> > which we typically approximate as
> > finite bandwidth.  The truncation of bandwidth is equivalent to adding
> > modulation components that cancel the truncated components; these
> > components extend to infinite frequency.  
> 
> That's a very interesting way to look at it. Could be true. Got anything 
> a little more concrete?
> 
> > Does this help?
> 
> A little. Thanks Matt.
> 
> > 
> > Matt
> > 
> > Randy Yates <yates@ieee.org> wrote in message news:<3F6FD798.9CB600BD@ieee.org>...
> > > Matt Boytim wrote:
> > > >
> > > > Even if the signal
> > > > i(t)+j*q(t) is bandlimited to nyquist, the signal phi(t)=arg(i+jq) is
> > > > not (except in trivial cases)
> > >
> > > Hey Matt,
> > >
> > > Really? I can't off-the-cuff think of why this may or may not be. Are
> > > you sure? Do you have a pointer? This (FM demodulation) is actually
> > > of significant interest to me as well.

Matt Boytim wrote:
> 
> Randy Yates <yates@ieee.org> wrote in message news:<3F73B451.83543DF7@ieee.org>...
> > Matt Boytim wrote:
> > >
> > > Well, if for example a bandlimited i(t)+jq(t) has a trajectory through
> > > the origin, phi(t) will instantaneously flip 180 degrees which would
> > > have infinite bandwidth.
> >
> > I don't agree. In order for phi(t) to instantaneously flip, so must
> > i(t) and or q(t).
> 
> A bandlimited signal is allowed to pass through zero, but the flip in
> phi is a discontinuity.

Matt (... sigh ...),

I must've had the biggest stupid hat on I've had for years when I responded
to this. I might as well have said 2+2=5. Yes, of course you are right in
this respect. I have embarassed myself here. 

> A complex carrier exp(j*w0*t)=cos(w0*t)+j*sin(w0*t) is clearly
> bandlimited, then phi(t)=w0*t (also bandlimited if taken for all t).
> Now, add a DC offset 1+exp(j*w0*t)=1+cos(w0*t)+j*sin(w0*t) (so that it
> rotates about 1 instead of the origin), still clearly bandlimited but
> phi(t) is now periodic with a 180 degree discontinuity every cycle so
> obviously not bandlimited.

Well, sorry, but I have to disagree here. The phase angle of a 
trajectory that is a circle centered at (1, 0) instead of (0, 0)
with radius 1 would be between +/- A (A near 45 degrees), right? 

I think the point you were making could be made with just the simple
complex exponential (simple complex??) e^{j*w*t}. Its angle will jump
from -180 to +180 periodically using plain arctan. The issue here
is phase unwrapping. With phase unwrapping, this type of discontinuity
goes away. And it's not a bandwidth issue either - you could sample
at a google Hz and still have this problem.

Nevertheless, I do see your original point and now I agree. However, 
there are some interesting things going on that I haven't gotten my 
mind (hopefully the one without the stupid hat) around here. For 
example, most FM detectors AGC the signal before operating on it, 
so the signal would never go through the origin. 
-- 
%  Randy Yates                  % "...the answer lies within your soul
%% Fuquay-Varina, NC            %       'cause no one knows which side 
%%% 919-577-9882                %                   the coin will fall."
%%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*, ELO   
http://home.earthlink.net/~yatescr

Randy Yates wrote:
> 
> Matt Boytim wrote:
> >
> > Randy Yates <yates@ieee.org> wrote in message news:<3F73B451.83543DF7@ieee.org>...
> > > Matt Boytim wrote:
> > > >
> > > > Well, if for example a bandlimited i(t)+jq(t) has a trajectory through
> > > > the origin, phi(t) will instantaneously flip 180 degrees which would
> > > > have infinite bandwidth.
> > >
> > > I don't agree. In order for phi(t) to instantaneously flip, so must
> > > i(t) and or q(t).
> >
> > A bandlimited signal is allowed to pass through zero, but the flip in
> > phi is a discontinuity.
> 
> Matt (... sigh ...),
> 
> I must've had the biggest stupid hat on I've had for years when I responded
> to this. I might as well have said 2+2=5. Yes, of course you are right in
> this respect. I have embarassed myself here.
> 
> > A complex carrier exp(j*w0*t)=cos(w0*t)+j*sin(w0*t) is clearly
> > bandlimited, then phi(t)=w0*t (also bandlimited if taken for all t).
> > Now, add a DC offset 1+exp(j*w0*t)=1+cos(w0*t)+j*sin(w0*t) (so that it
> > rotates about 1 instead of the origin), still clearly bandlimited but
> > phi(t) is now periodic with a 180 degree discontinuity every cycle so
> > obviously not bandlimited.
> 
> Well, sorry, but I have to disagree here. The phase angle of a
> trajectory that is a circle centered at (1, 0) instead of (0, 0)
> with radius 1 would be between +/- A (A near 45 degrees), right?

If you zoom in very very very close to the origin, the imaginary axis
and the cirlce are indistinguisable, hence the angle is +/- 90 degrees.

> 
> I think the point you were making could be made with just the simple
> complex exponential (simple complex??) e^{j*w*t}. Its angle will jump
> from -180 to +180 periodically using plain arctan. The issue here
> is phase unwrapping. With phase unwrapping, this type of discontinuity
> goes away. And it's not a bandwidth issue either - you could sample
> at a google Hz and still have this problem.

Your simplified example is not the same.  Atan is defined to return the
principal value and there is a point of discontinuity but these jumps
are 360 degrees, not 180 degrees, and although the simple atan may have
jumps of 360 degrees the true phase of a complex tone rotating about the
origin will not have the 360 jumps.  In my example, the tone rotating
about the point 1+j0 goes through the origin; as it approaches the
origin the angle approaches 90 degrees, just after it passes through the
origin the phase is -90 degrees (or 270 degrees, if it truly went
through the origin you really don't know what to call it on the other
side), hence the 180 degree jump or discontinuity (there's a 180 degree
jump if we call it -90 or 270).  If the amplitude of the complex tone is
just less than one (still rotating about 1+j0) then the situation is
similar except the jump from 90 degrees to -90 degrees is sharp but
continuous and unambiguous.  If the amplitue is just greater than one
then the jump is from 90 degrees to 270 degrees and is unambiguous (not
-90 degrees).  These are the conditions around FM threshold.  (Actually,
since true phase runs from -infinity to +infinity (not -180 to +180) the
angles in my examples can have any multiple of 360 degrees added to
them.)

> 
> Nevertheless, I do see your original point and now I agree. However,
> there are some interesting things going on that I haven't gotten my
> mind (hopefully the one without the stupid hat) around here. For
> example, most FM detectors AGC the signal before operating on it,
> so the signal would never go through the origin.

AGC'ing doesn't really help because zero times any finite gain is still
zero.  Detectors that convert FM to AM need a normalized signal which
can be obtained using an AGC or (in continuous time) a limiter/filter. 
There may be other reasons for an AGC in an FM detector, I'm not sure.

An FM signal itself (at complex baseband) cannot go through the origin -
it would take noise or interference for this to happen.  But I was using
this only as an example to demonstrate that a signal which is
bandlimited in its rectangular form is not necessarily bandlimited in
its polar form.

Matt

> --
> %  Randy Yates                  % "...the answer lies within your soul
> %% Fuquay-Varina, NC            %       'cause no one knows which side
> %%% 919-577-9882                %                   the coin will fall."
> %%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*, ELO
> http://home.earthlink.net/~yatescr

-- 
matt@boytim.org

Matt Boytim wrote:
> 
> Randy Yates wrote:
> >
> > Matt Boytim wrote:
> > >
> > > Randy Yates <yates@ieee.org> wrote in message news:<3F73B451.83543DF7@ieee.org>...
> > > > Matt Boytim wrote:
> > > > >
> > > > > Well, if for example a bandlimited i(t)+jq(t) has a trajectory through
> > > > > the origin, phi(t) will instantaneously flip 180 degrees which would
> > > > > have infinite bandwidth.
> > > >
> > > > I don't agree. In order for phi(t) to instantaneously flip, so must
> > > > i(t) and or q(t).
> > >
> > > A bandlimited signal is allowed to pass through zero, but the flip in
> > > phi is a discontinuity.
> >
> > Matt (... sigh ...),
> >
> > I must've had the biggest stupid hat on I've had for years when I responded
> > to this. I might as well have said 2+2=5. Yes, of course you are right in
> > this respect. I have embarassed myself here.
> >
> > > A complex carrier exp(j*w0*t)=cos(w0*t)+j*sin(w0*t) is clearly
> > > bandlimited, then phi(t)=w0*t (also bandlimited if taken for all t).
> > > Now, add a DC offset 1+exp(j*w0*t)=1+cos(w0*t)+j*sin(w0*t) (so that it
> > > rotates about 1 instead of the origin), still clearly bandlimited but
> > > phi(t) is now periodic with a 180 degree discontinuity every cycle so
> > > obviously not bandlimited.
> >
> > Well, sorry, but I have to disagree here. The phase angle of a
> > trajectory that is a circle centered at (1, 0) instead of (0, 0)
> > with radius 1 would be between +/- A (A near 45 degrees), right?
> 
> If you zoom in very very very close to the origin, the imaginary axis
> and the cirlce are indistinguisable, hence the angle is +/- 90 degrees.

Once again I stand corrected. You are right. I was thinking of a circle
at (1, 0) with a radius < 1; when the circle butts right up against the
origin, you get this discontinuity problem. 

This is a good example - thank you for it. See the end of this
message for another example based on your "through the origin" idea
that is even simpler. 

> > I think the point you were making could be made with just the simple
> > complex exponential (simple complex??) e^{j*w*t}. Its angle will jump
> > from -180 to +180 periodically using plain arctan. The issue here
> > is phase unwrapping. With phase unwrapping, this type of discontinuity
> > goes away. And it's not a bandwidth issue either - you could sample
> > at a google Hz and still have this problem.
> 
> Your simplified example is not the same. 

Yeah, I see that now. No need to explain in detail, but thanks for the
attempt to be clear.

> > Nevertheless, I do see your original point and now I agree. However,
> > there are some interesting things going on that I haven't gotten my
> > mind (hopefully the one without the stupid hat) around here. For
> > example, most FM detectors AGC the signal before operating on it,
> > so the signal would never go through the origin.
> 
> AGC'ing doesn't really help because zero times any finite gain is still
> zero.

How often is the input going to be precisely zero? It seems like you
can't treat this like a one-dimensional signal, which is frequently
going through zero.

> Detectors that convert FM to AM need a normalized signal which
> can be obtained using an AGC or (in continuous time) a limiter/filter.
> There may be other reasons for an AGC in an FM detector, I'm not sure.

There can be good reasons to AGC even in a digital discriminator, e.g., if the
arctan() function is performed via a lookup table.

> An FM signal itself (at complex baseband) cannot go through the origin -
> it would take noise or interference for this to happen.  But I was using
> this only as an example to demonstrate that a signal which is
> bandlimited in its rectangular form is not necessarily bandlimited in
> its polar form.

Yes. Point well-made. 

Here's another example of a signal that is clearly bandlimited but
produces an infinite phase spectrum:

  z(t) = sin(2*pi*f*t).

Clearly this signal is bandlimited to f Hz. The IQ diagram would be a 
horizontal line, but as the point moved through the origin, the phase 
will instantaneously shift. Pretty basic at one level, but from the
phase (phase in the IQ plane, that is) POV, shows how a simple signal
can result in a discontinuous phase and therefore an infinite phase
spectrum. 

It may be of practical significance to the original poster that 
the phase problem caused by wrapping can be overcome by a) sampling
the signal at a sufficient rate that the *intended* (transmitted) 
phase change never exceeds some maximum delta, 0 < delta <= 180, and
b) unwrapping the result.
-- 
%  Randy Yates                  % "...the answer lies within your soul
%% Fuquay-Varina, NC            %       'cause no one knows which side 
%%% 919-577-9882                %                   the coin will fall."
%%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*, ELO   
http://home.earthlink.net/~yatescr

Randy Yates wrote:

> There can be good reasons to AGC even in a digital discriminator, e.g., if the
> arctan() function is performed via a lookup table.

CORDIC has problems for very small magnitudes as well.

> Here's another example of a signal that is clearly bandlimited but
> produces an infinite phase spectrum:
> 
>   z(t) = sin(2*pi*f*t).
> 
> Clearly this signal is bandlimited to f Hz. The IQ diagram would be a
> horizontal line, but as the point moved through the origin, the phase
> will instantaneously shift. Pretty basic at one level, but from the
> phase (phase in the IQ plane, that is) POV, shows how a simple signal
> can result in a discontinuous phase and therefore an infinite phase
> spectrum.

Yes, the complex angle of any real signal is 0 or 180 degrees.

> 
> It may be of practical significance to the original poster that
> the phase problem caused by wrapping can be overcome by a) sampling
> the signal at a sufficient rate that the *intended* (transmitted)
> phase change never exceeds some maximum delta, 0 < delta <= 180, and
> b) unwrapping the result.

As our examples show, there really is no sample rate that is 'fast
enough' unless you know that you are always well above threshold.  But
while the angle (arctan) has discontinuities, the delta angle is well
behaved, so if you are interested in FM then the derivative can be
computed without being oversampled relative to the modulated bandwidth
and without an arbitrary delta.

The continuous derivative can be approximated with the first difference,
or any reasonable approximation to the continuous derivative can be
factored into a first difference and a 'correction' filter.  For a
discrete time signal, the delta angle (first difference) is always
bounded by +/- 180 degrees (to be precise, you have your choice to call
this [-180.180) or (-180,180] depending on whether you represent -fs/2
or +fs/2 but that's another thread).  Atan returns the principal value;
all you need is a first difference that returns the principal value as
well and you're in (just a modulo operation which is free in fixed point
if you choose the right representaion for angles which happens to be the
obvious one).  For true angle (PM) yo can actually differentiate the
phase in this way and then re-integrate to get rid of the
discontinuities (extended range CORDIC).  There are other applications
for the bounded first difference.

Matt