Smoothing A Quotient of Two |Sinwt| Signals

At any time t both numerator and denominator have the same phase angle
so that the quotient of the 2 time integrals should eliminate any
zeros in the denomenator and any trace of the original sin wave forms.

Smooting isn't necessary and only the amplitudes of the numerator and
denomenator are reflected in the quotient.

This would also work in analog as well as digital.


Bret Cahill

On Feb 14, 11:05&#4294967295;am, Bret Cahill <BretCah...@peoplepc.com> wrote:
> At any time t both numerator and denominator have the same phase angle
> so that the quotient of the 2 time integrals should eliminate any
> zeros in the denomenator and any trace of the original sin wave forms.
>
> Smooting isn't necessary and only the amplitudes of the numerator and
> denomenator are reflected in the quotient.
>
> This would also work in analog as well as digital.
>
> Bret Cahill

Is this a solution in search of a problem?

Bret Cahill wrote:
> At any time t both numerator and denominator have the same phase angle
> so that the quotient of the 2 time integrals should eliminate any
> zeros in the denomenator and any trace of the original sin wave forms.
> 
> Smooting isn't necessary and only the amplitudes of the numerator and
> denomenator are reflected in the quotient.
> 
> This would also work in analog as well as digital.

Context?

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

> > At any time t both numerator and denominator have the same phase angle
> > so that the quotient of the 2 time integrals should eliminate any
> > zeros in the denomenator and any trace of the original sin wave forms.
>
> > Smooting isn't necessary and only the amplitudes of the numerator and
> > denomenator are reflected in the quotient.
>
> > This would also work in analog as well as digital.
>
> > Bret Cahill
>
> Is this a solution in search of a problem?

The problem is an analog lock in amp takes time to smooth a single
sin^2(wt) [output after the signal is multiplied by the reference]
even if there is no noise whatsoever.

But if you are going to divide one signal into another signal that
except for amplitude and noise is identical in every way including
processing, then you shouldn't have to waste time smoothing each
signal before the division.

Just divide the raw [very wavy] signals as soon as enough integration
has taken place to overwhelm the noise,

Instead of waiting, say, 20 - 40 seconds to smooth each signal the
desired quotient would be smooth in as little as a couple seconds in
some situations.


Bret Cahill

The time to smooth a low SNR signal after PLL filtering isn't an issue
as it takes so long to integrate the noise out of the signal anyway.

But with a SNR of 5 - 20 or higher you can save a lot of time if you
can eliminate smoothing the sin^2 output resulting from multiplication
of the signal and the reference.

This is possible in a quotient situation where both the numerator and
denominator are PLL filtered with the same reference.

Similar low noise / quotient situations might be fairly common in
electronic filtering and it would be a little surprising if someone
hadn't already suggested eliminating the smoothing step.


Bret Cahill



> > > At any time t both numerator and denominator have the same phase angle
> > > so that the quotient of the 2 time integrals should eliminate any
> > > zeros in the denomenator and any trace of the original sin wave forms.
>
> > > Smooting isn't necessary and only the amplitudes of the numerator and
> > > denomenator are reflected in the quotient.
>
> > > This would also work in analog as well as digital.
>
> > > Bret Cahill
>
> > Is this a solution in search of a problem?
>
> The problem is an analog lock in amp takes time to smooth a single
> sin^2(wt) [output after the signal is multiplied by the reference]
> even if there is no noise whatsoever.
>
> But if you are going to divide one signal into another signal that
> except for amplitude and noise is identical in every way including
> processing, then you shouldn't have to waste time smoothing each
> signal before the division.
>
> Just divide the raw [very wavy] signals as soon as enough integration
> has taken place to overwhelm the noise,
>
> Instead of waiting, say, 20 - 40 seconds to smooth each signal the
> desired quotient would be smooth in as little as a couple seconds in
> some situations.
>
> Bret Cahill

On Feb 15, 5:05=A0am, Bret Cahill <BretCah...@peoplepc.com> wrote:
> At any time t both numerator and denominator have the same phase angle
> so that the quotient of the 2 time integrals should eliminate any
> zeros in the denomenator and any trace of the original sin wave forms.
>
> Smooting isn't necessary and only the amplitudes of the numerator and
> denomenator are reflected in the quotient.
>
> This would also work in analog as well as digital.
>
> Bret Cahill

0/0=3D??

> > At any time t both numerator and denominator have the same phase angle
> > so that the quotient of the 2 time integrals should eliminate any
> > zeros in the denomenator and any trace of the original sin wave forms.
>
> > Smooting isn't necessary and only the amplitudes of the numerator and
> > denomenator are reflected in the quotient.
>
> > This would also work in analog as well as digital.
>
> > Bret Cahill
>
> 0/0=??

For t > 0 the integral of the absolute value of the signal is
positive.

Starting the division after a little integration has taken place or
just after the signals pass 0 will eliminate any indeterminate
situation at t = 0.


Bret Cahill

The time to smooth a low SNR signal after PLL filtering isn't an issue
as it takes so long to integrate the noise out of the signal anyway.

But with a SNR of 5 - 20 or higher you can save a lot of time if you
can eliminate smoothing the sin^2 output resulting from multiplication
of the signal and the reference.

This is possible in a quotient situation where both the numerator and
denominator are PLL filtered with the same reference.

Similar low noise / quotient situations might be fairly common in
electronic filtering and it would be a little surprising if someone
hadn't already suggested eliminating the smoothing step.


Bret Cahill





> > At any time t both numerator and denominator have the same phase angle
> > so that the quotient of the 2 time integrals should eliminate any
> > zeros in the denomenator and any trace of the original sin wave forms.
>
> > Smooting isn't necessary and only the amplitudes of the numerator and
> > denomenator are reflected in the quotient.
>
> > This would also work in analog as well as digital.
>
> Context?
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
> =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF

Bret Cahill wrote:
>>> At any time t both numerator and denominator have the same phase angle
>>> so that the quotient of the 2 time integrals should eliminate any
>>> zeros in the denomenator and any trace of the original sin wave forms.
>>> Smooting isn't necessary and only the amplitudes of the numerator and
>>> denomenator are reflected in the quotient.
>>> This would also work in analog as well as digital.
>>> Bret Cahill
>> 0/0=??
> 
> For t > 0 the integral of the absolute value of the signal is
> positive.
> 
> Starting the division after a little integration has taken place or
> just after the signals pass 0 will eliminate any indeterminate
> situation at t = 0.

If the signal is noise free, you divide by zero at every crossing. Is it 
somewhat noise, then the pernicious divisions are close to the nominal 
zero crossing.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

> >>> At any time t both numerator and denominator have the same phase angle
> >>> so that the quotient of the 2 time integrals should eliminate any
> >>> zeros in the denomenator and any trace of the original sin wave forms.
> >>> Smooting isn't necessary and only the amplitudes of the numerator and
> >>> denomenator are reflected in the quotient.
> >>> This would also work in analog as well as digital.
> >>> Bret Cahill

> >> 0/0=??
>
> > For t > 0 the integral of the absolute value of the signal is
> > positive.
>
> > Starting the division after a little integration has taken place or
> > just after the signals pass 0 will eliminate any indeterminate
> > situation at t = 0.
>
> If the signal is noise free, you divide by zero at every crossing.

How can there be a t >0 crossing for the time integral of the absolute
value of the signal?

The noise is in there _before_ the absolute value or PLL filtering.


Bret Cahill