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From: John Larkin on 4 Feb 2010 11:15
On Wed, 3 Feb 2010 21:38:46 -0800 (PST), Bret Cahill
<BretCahill(a)peoplepc.com> wrote:

>> > Obviously the same noise causing one signal to increase is causing the
>> > other to decrease.
>
>> False inference, there. �The 'sum of two signals' has twice the signal
>
>Depends on the 2 signals. That's generally not true.
>
>> but the noise is uncorrelated,
>
>In this case the sum of the two signals has the exact same shape and
>phase angle as each noise free signal.
>
>There's a very small chance this may suggest some algebraic solution
>where the noise can be somehow subtracted and lock in filtering can be
>avoided altogether.
>
>> so the sum of the two noises is
>> expected to be sqrt(2) times the individual noise values.
>
>The quotient of the two signals is the goal so any magnitude change
>from filtering will cancel out.
>
>> You are getting better signal/noise ratio in the sum than in either
>> of the two input terms, that's to be expected.
>
>It isn't merely better. The noise cancels out _altogether_ in the
>sum.
>
>The sum, however, isn't what is desired.
>
>The sum is only useful as a reference to clean up the two signals.
>
>It would be surprising in no one went down this path before on a
>similar situation.
>
>
>Bret Cahill
>
>
>

I have no idea what you're talking about, and I suspect you don't
either.

John

From: John Larkin on 4 Feb 2010 16:36
On Thu, 4 Feb 2010 08:45:07 -0800 (PST), Bret Cahill
<BretCahill(a)peoplepc.com> wrote:

>S1 = signal 1
>
>S2 = signal 2
>
>S1/S2 = desired output which would be a const. dc without noise.
>
>N = noise
>
>(S1 + N)/(S2 - N) = actual output.
>
>Noise is amplified in output.
>
>c = const.
>
>S1 + c(S2) is proportional to and has same phase angle as noise free
>S1 as well as noise free S2 but is completely impervious to noise and
>therefore can be used as a reference for lock in amp.


If S1 + c(S2) is as you say, then the final value of S1/S2 = K, where
K is any constant you like. So the output of your signal processing
box is a DC voltage, and you save a lot of money by eliminating input
connectors.

John

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