> Ahmed,
>
> If I'm interpreting [schartzcommtechniques] (section 3-9, around page
> 157) correctly, the threshold extension offered by FMFB techniques (FM
> with FeedBack, i.e., using a PLL for FM demodulation) is only available
> when the input signal is relatively wideband FM (beta >> 1). Since beta
> = 1 in the BTSC SAP channel, this technique is not applicable here and
> a plain, old FM discriminator is about as good as you're gonna' get.
>
> Would you agree with this? If not, why not?
>
> Is there another technique for extending the threshold that works
> with narrowband FM?
>
> --Randy

Yes Randy, you're right, the threshold effect is more of a problem at
higher modulation indices (Beta) and similarly the threshold extension
that a PLL provides is greater for a larger modulation index. Looking
at "Principles of Communication Systems" by Taub and Schilling, 2nd
Ed, 1986, the figure on pg 427 shows that the threshold using a
limiter discriminator is at around 25 dB for Beta=12 while it is at
17dB for Beta=3. And the PLL provides a threshold extension of 3dB and
2.5dB for Beta=12 and 3 respectively.

So not only is the threshold lower for a smaller modulation index, but
the threshold extension due to a PLL is also less. For your case, the
threshold is probably somewhere around 10-12dB CNR. That said, I would
guess that you could get some (not a lot) threshold extension using a
PLL, so it really depends on how important it is for you to operate at
lower SNRs around threshold. Probably, it would be best for you to
first implement the limiter discriminator and then, if you feel it is
worth it for you, you could look into implementing a PLL.

The other technique for threshold extension that I've come across is
Click noise elimination and it is also usually applied to larger
modulation indices. A good paper on it is "On FM Threshold Extension
by Click Noise Elimination" by M. Polacek et. al, IEEE Trans on Comm,
1988.

Feel free to share your results with us when you get the LD working!

On Apr 6, 4:03 pm, "dbell" <bellda2...@cox.net> wrote:
> On Apr 6, 3:48 pm, "robert bristow-johnson"
>
>
>
>
>
>
>
> <r...@audioimagination.com> wrote:
> > On Apr 6, 1:51 pm, "dbell" <bellda2...@cox.net> wrote:
>
> > > On Apr 6, 1:26 pm, "robert bristow-johnson" wrote:
>
> > ...
>
> > > > but it does (or *might*) require a division.  if you are confident
> > > > that the denominator
>
> > > >      I[n]*I[n-1] + Q[n]*Q[n-1]
>
> > > > is close enough to
>
> > > >     I[n]*I[n] + Q[n]*Q[n]
>
> > > > and that is constant, you can skip the division.  even if not, you can
> > > > compute the *new* denominator, multiply it by the previous reciprocal
> > > > and use the difference of that from 1 in a sorta servo feedback loop
> > > > to skip the division.  also the arctan() might be able to be tossed if
> > > > you know that the change in angle is small for each sample.  then you
> > > > get
>
> > > >     omega[n] = Constant*( Q[n]*I[n-1] - I[n]*Q[n-1] )
>
> > > > and that's about as simple as it can get.
>
> > ...
>
> > > omega[n] = arg{ I[n] + j*Q[n] }  -  arg{ I[n-1] + j*Q[n-1] }
> > >                 (adjusted for wrap)
> > >    = arg{ (I[n] + j*Q[n]) * conjugate( I[n-1] + j*Q[n-1] ) }
> > >                  with no division
> > >    = arg{ (I[n] + j*Q[n]) * (I[n-1] - j*Q[n-1]) }
> > >                  with no division
>
> > well, not yet anyway.
>
> > so assuming omega[n] doesn't get so big, then
>
> > omega[n] = arg{ (I[n] + j*Q[n]) * (I[n-1] - j*Q[n-1]) }
>
> >          = arg{ I[n]*I[n-1] + Q[n]*Q[n-1]
> >                  + j*(I[n-1]*Q[n] - I[n]*Q[n-1] ) }
>
> > now, Dirk, how're you gonna do that arg{} ?
>
> > r b-j- Hide quoted text -
>
> > - Show quoted text -
>
> Depends.
>
> Not someway that depends on unity magnitude going into arg of course.

i dunno.  might not be such a bad assumption.

> IIRC coordic algorithm would do, but I would need to verify that.
>

maybe you'll have to do the arg{} the old-fashioned way by dividing
the imag part with the real part and arctanning.  it might never be
that big, going into the arg and maybe you can just call it the unity
gain function.  maybe not.

r b-j

"Ahmed" <ahsenahmed@hotmail.com> writes:
> [...]
> It is shown to have performance comparable to the Limiter-
> Discriminator receiver. So even though you would not have to worry
> about phase wrapping issues, the performance will suffer below
> threshold due to click noise.

Ahmed,

If I'm interpreting [schartzcommtechniques] (section 3-9, around page
157) correctly, the threshold extension offered by FMFB techniques (FM
with FeedBack, i.e., using a PLL for FM demodulation) is only available
when the input signal is relatively wideband FM (beta >> 1). Since beta
= 1 in the BTSC SAP channel, this technique is not applicable here and
a plain, old FM discriminator is about as good as you're gonna' get.

Would you agree with this? If not, why not?

Is there another technique for extending the threshold that works
with narrowband FM?

--Randy


@book{schwartzcommtechniques,
  title = "Communication Systems and Techniques",
  author = "Mischa Schwartz and William R. Bennett and Seymour Stein",
  publisher = "IEEE Press (reissue, originally New York: McGraw-Hill 1966)",
  year = "1996"}

-- 
%  Randy Yates                  % "She tells me that she likes me very much,
%% Fuquay-Varina, NC            %     but when I try to touch, she makes it
%%% 919-577-9882                %                            all too clear."
%%%% <yates@ieee.org>           %        'Yours Truly, 2095', *Time*, ELO  
http://home.earthlink.net/~yatescr

Ahmed wrote:

>>FYI, I'm really leaning toward Vladimir's solution (Q'I - I'Q / (I^2 + Q^2))

> This technique is named the "cross-correlator receiver" in some
> papers.

What a fancy name for the simple thing.

> It is shown to have performance comparable to the Limiter-
> Discriminator receiver.

It is obvious. This is just one of many different ways of implementing 
the old good discriminator. As long as the underlying math is the same, 
all methods are equivalent.

> So even though you would not have to worry
> about phase wrapping issues, the performance will suffer below
> threshold due to click noise.

To be more precise, it has to do with the suboptimal approach to the 
task of the frequency estimation in the presence of noise. All of the 
discriminators do the frequency measurement in the bandwidth of the 
input signal, and then average the result to the bandwidth of the 
modulating signal.


Vladimir Vassilevsky

DSP and Mixed Signal Design Consultant

http://www.abvolt.com

On Apr 6, 3:48 pm, "robert bristow-johnson"
<r...@audioimagination.com> wrote:
> On Apr 6, 1:51 pm, "dbell" <bellda2...@cox.net> wrote:
>
>
>
>
>
> > On Apr 6, 1:26 pm, "robert bristow-johnson" wrote:
>
> ...
>
> > > but it does (or *might*) require a division.  if you are confident
> > > that the denominator
>
> > >      I[n]*I[n-1] + Q[n]*Q[n-1]
>
> > > is close enough to
>
> > >     I[n]*I[n] + Q[n]*Q[n]
>
> > > and that is constant, you can skip the division.  even if not, you can
> > > compute the *new* denominator, multiply it by the previous reciprocal
> > > and use the difference of that from 1 in a sorta servo feedback loop
> > > to skip the division.  also the arctan() might be able to be tossed if
> > > you know that the change in angle is small for each sample.  then you
> > > get
>
> > >     omega[n] = Constant*( Q[n]*I[n-1] - I[n]*Q[n-1] )
>
> > > and that's about as simple as it can get.
>
> ...
>
> > omega[n] = arg{ I[n] + j*Q[n] }  -  arg{ I[n-1] + j*Q[n-1] }
> >                 (adjusted for wrap)
> >    = arg{ (I[n] + j*Q[n]) * conjugate( I[n-1] + j*Q[n-1] ) }
> >                  with no division
> >    = arg{ (I[n] + j*Q[n]) * (I[n-1] - j*Q[n-1]) }
> >                  with no division
>
> well, not yet anyway.
>
> so assuming omega[n] doesn't get so big, then
>
> omega[n] = arg{ (I[n] + j*Q[n]) * (I[n-1] - j*Q[n-1]) }
>
>          = arg{ I[n]*I[n-1] + Q[n]*Q[n-1]
>                  + j*(I[n-1]*Q[n] - I[n]*Q[n-1] ) }
>
> now, Dirk, how're you gonna do that arg{} ?
>
> r b-j- Hide quoted text -
>
> - Show quoted text -

Depends.

Not someway that depends on unity magnitude going into arg of course.
IIRC coordic algorithm would do, but I would need to verify that.

Dirk

On Apr 6, 1:51 pm, "dbell" <bellda2...@cox.net> wrote:
> On Apr 6, 1:26 pm, "robert bristow-johnson" wrote:
>
>
...
>
> > but it does (or *might*) require a division.  if you are confident
> > that the denominator
>
> >      I[n]*I[n-1] + Q[n]*Q[n-1]
>
> > is close enough to
>
> >     I[n]*I[n] + Q[n]*Q[n]
>
> > and that is constant, you can skip the division.  even if not, you can
> > compute the *new* denominator, multiply it by the previous reciprocal
> > and use the difference of that from 1 in a sorta servo feedback loop
> > to skip the division.  also the arctan() might be able to be tossed if
> > you know that the change in angle is small for each sample.  then you
> > get
>
> >     omega[n] = Constant*( Q[n]*I[n-1] - I[n]*Q[n-1] )
>
> > and that's about as simple as it can get.
>
...
>
> omega[n] = arg{ I[n] + j*Q[n] }  -  arg{ I[n-1] + j*Q[n-1] }
>                 (adjusted for wrap)
>    = arg{ (I[n] + j*Q[n]) * conjugate( I[n-1] + j*Q[n-1] ) }
>                  with no division
>    = arg{ (I[n] + j*Q[n]) * (I[n-1] - j*Q[n-1]) }
>                  with no division

well, not yet anyway.


so assuming omega[n] doesn't get so big, then

omega[n] = arg{ (I[n] + j*Q[n]) * (I[n-1] - j*Q[n-1]) }

         = arg{ I[n]*I[n-1] + Q[n]*Q[n-1]
                 + j*(I[n-1]*Q[n] - I[n]*Q[n-1] ) }

now, Dirk, how're you gonna do that arg{} ?

r b-j

On Apr 6, 1:26 pm, "robert bristow-johnson"
<r...@audioimagination.com> wrote:
> On Apr 6, 10:00 am, Grant Griffin <nos...@yahoo.com> wrote:
>
> > Vladimir Vassilevsky wrote:
> > >>> The straightforward way would be not to use the phase
> > >>> representation. Work with the complex derivatives directly:
>
> > >>> IdQ/dt - QdI/dt
> > >>> ---------------
> > >>> I^2  + Q^2
>
> > >> I guess the key advantage is that you don't have to implement an
> > >> arctan function?
>
> > > Atan is not that big of a deal.
>
> especially when you know that the argument to it is small.
>
> > > What is bad is that you have to do the
> > > costly division no matter if you are using atan() or not. What is good
> > > is that with I' and Q' there is no phase wrapping problem.
>
> > I guess I've always just used delta(atan(Q/I)) for this.  Even if the
> > data representation doesn't automatically handle the phase wrapping,
> > handling it specially is less computation than doing all the arithmetic
> > above.
>
> well, if-then statements (needed for explicit phase unwrapping) are
> sorta costly and ugly in DSP code, me thinks.  this is why i liked:
>
>     omega[n] = arg{ I[n] + j*Q[n] }  -  arg{ I[n-1] + j*Q[n-1] }
>
>              = arg{  (I[n] + j*Q[n]) / (I[n-1] + j*Q[n-1]) }
>
>              = arctan(  (Q[n]*I[n-1] - I[n]*Q[n-1])
>                              / (I[n]*I[n-1] + Q[n]*Q[n-1])  )
>
> but it does (or *might*) require a division.  if you are confident
> that the denominator
>
>      I[n]*I[n-1] + Q[n]*Q[n-1]
>
> is close enough to
>
>     I[n]*I[n] + Q[n]*Q[n]
>
> and that is constant, you can skip the division.  even if not, you can
> compute the *new* denominator, multiply it by the previous reciprocal
> and use the difference of that from 1 in a sorta servo feedback loop
> to skip the division.  also the arctan() might be able to be tossed if
> you know that the change in angle is small for each sample.  then you
> get
>
>     omega[n] = Constant*( Q[n]*I[n-1] - I[n]*Q[n-1] )
>
> and that's about as simple as it can get.
>
> > Coincidentally, one of the very first DSP designs I ever did, circa
> > 1985, was a hardware implementation of the above, using
> > now-seemingly-gigantic DIP parts such as a look-up EPROM, adder, etc.
>
> > =g2
> > _____________________________________________________________________
>
> > Grant R. Griffin
> > Publisher of dspGuru                          http://www.dspguru.com
> > Iowegian International Corporation            http://www.iowegian.com
> > Seehttp://www.iowegian.com/img/contact.giffore-mail address
>
> it's nice to see this sig again.
>
> r b-j

Hi RBJ,

omega[n] = arg{ I[n] + j*Q[n] }  -  arg{ I[n-1] + j*Q[n-1] }
(adjusted for wrap)
               = arg{ (I[n] + j*Q[n]) * conjugate( I[n-1] +
j*Q[n-1] ) } with no division
               = arg{ (I[n] + j*Q[n]) * (I[n-1] - j*Q[n-1]) } with no
division

Dirk

"Ahmed" <ahsenahmed@hotmail.com> writes:

>> FYI, I'm really leaning toward Vladimir's solution (Q'I - I'Q / (I^2 + Q^2))
>> since
>>
>>   1) it doesn't require an arctan approximation
>>
>>   2) it is "perfect"
>>
>>   3) a division has to be done somewhere one way or the other
>>
>>   4) the differentiator is not a big deal
>>
>>   5) I can easily remove the systematic DC error due to the uncentered
>>   spectrum.
>>
>> --Randy
>
> Randy,
>
> This technique is named the "cross-correlator receiver" in some
> papers. It is posted in many discussions on FM demod, but references
> are usually not provided. One of the earliest papers to propose it is:
>
> "An FM Detector for Low S/N" by John Park, IEEE Trans on Comm, 1970
>
> and the analytical performance derivation has been done in:
>
> "Performance of the Cross-Correlator Receiver for Binary Digital
> Frequency Modulation" by Kevin Farrell et. al, IEEE Trans on Comm,
> 1997.

Thanks very much, Ahmed! 

> It is shown to have performance comparable to the Limiter-
> Discriminator receiver. So even though you would not have to worry
> about phase wrapping issues, the performance will suffer below
> threshold due to click noise.

Yes, I was afraid of that. 

> By the way, what modulation index are you using?

beta = 1. This is the SAP on a BTSC signal. So it's neither
narrow-band nor wideband, according to [schwartzcomm].

--Randy


@book{schwartzcomm,
  title = "Information Transmission, Modulation, and Noise",
  author = "{Mischa~Schwartz}",
  publisher = "McGraw-Hill",
  edition = "fourth",
  year = "1990"}
-- 
%  Randy Yates                  % "Rollin' and riding and slippin' and
%% Fuquay-Varina, NC            %  sliding, it's magic."
%%% 919-577-9882                %  
%%%% <yates@ieee.org>           % 'Living' Thing', *A New World Record*, ELO
http://home.earthlink.net/~yatescr

On Apr 6, 10:00 am, Grant Griffin <nos...@yahoo.com> wrote:
> Vladimir Vassilevsky wrote:
> >>> The straightforward way would be not to use the phase
> >>> representation. Work with the complex derivatives directly:
>
> >>> IdQ/dt - QdI/dt
> >>> ---------------
> >>> I^2  + Q^2
>
> >> I guess the key advantage is that you don't have to implement an
> >> arctan function?
>
> > Atan is not that big of a deal.

especially when you know that the argument to it is small.

> > What is bad is that you have to do the
> > costly division no matter if you are using atan() or not. What is good
> > is that with I' and Q' there is no phase wrapping problem.
>
> I guess I've always just used delta(atan(Q/I)) for this.  Even if the
> data representation doesn't automatically handle the phase wrapping,
> handling it specially is less computation than doing all the arithmetic
> above.

well, if-then statements (needed for explicit phase unwrapping) are
sorta costly and ugly in DSP code, me thinks.  this is why i liked:

    omega[n] = arg{ I[n] + j*Q[n] }  -  arg{ I[n-1] + j*Q[n-1] }

             = arg{  (I[n] + j*Q[n]) / (I[n-1] + j*Q[n-1]) }

             = arctan(  (Q[n]*I[n-1] - I[n]*Q[n-1])
                             / (I[n]*I[n-1] + Q[n]*Q[n-1])  )

but it does (or *might*) require a division.  if you are confident
that the denominator

     I[n]*I[n-1] + Q[n]*Q[n-1]

is close enough to

    I[n]*I[n] + Q[n]*Q[n]

and that is constant, you can skip the division.  even if not, you can
compute the *new* denominator, multiply it by the previous reciprocal
and use the difference of that from 1 in a sorta servo feedback loop
to skip the division.  also the arctan() might be able to be tossed if
you know that the change in angle is small for each sample.  then you
get

    omega[n] = Constant*( Q[n]*I[n-1] - I[n]*Q[n-1] )

and that's about as simple as it can get.

> Coincidentally, one of the very first DSP designs I ever did, circa
> 1985, was a hardware implementation of the above, using
> now-seemingly-gigantic DIP parts such as a look-up EPROM, adder, etc.
>
> =g2
> _____________________________________________________________________
>
> Grant R. Griffin
> Publisher of dspGuru                          http://www.dspguru.com
> Iowegian International Corporation            http://www.iowegian.com
> Seehttp://www.iowegian.com/img/contact.giffor e-mail address

it's nice to see this sig again.

r b-j

> FYI, I'm really leaning toward Vladimir's solution (Q'I - I'Q / (I^2 + Q^2))
> since
>
>   1) it doesn't require an arctan approximation
>
>   2) it is "perfect"
>
>   3) a division has to be done somewhere one way or the other
>
>   4) the differentiator is not a big deal
>
>   5) I can easily remove the systematic DC error due to the uncentered
>   spectrum.
>
> --Randy

Randy,

This technique is named the "cross-correlator receiver" in some
papers. It is posted in many discussions on FM demod, but references
are usually not provided. One of the earliest papers to propose it is:

"An FM Detector for Low S/N" by John Park, IEEE Trans on Comm, 1970

and the analytical performance derivation has been done in:

"Performance of the Cross-Correlator Receiver for Binary Digital
Frequency Modulation" by Kevin Farrell et. al, IEEE Trans on Comm,
1997.

It is shown to have performance comparable to the Limiter-
Discriminator receiver. So even though you would not have to worry
about phase wrapping issues, the performance will suffer below
threshold due to click noise.

By the way, what modulation index are you using?