Rectification Quotient Sync Filter

Rectify both the signal and the ref along with their noise components _then_ divide.

Since there are no humps from multiplying two in sync signals as with lock in, only the noise needs to be smoothed. The time constant could be lower, much much lower at low noise levels.

At high noise levels the time constant would increase but it would it would always remain below lock in.

Rectification quotient sync is superior to lock in for every application.

Bret Cahill

"Fun is the father of invention."

-- Bret Cahill

Reply to
Bret Cahill
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Loonier and loonier. There's long-established mathematics that you seem to know nothing about. You may as well try to violate conservation of energy.

John

Reply to
John Larkin

oh? so what about friction?

Reply to
Ace0f_5pades

Someone will need to show it.

I just finished trying to do that at the local pool.

Assume:

The pool temp is 93 degrees F.

The delta T through the skin is 3 F.

The Reynolds number for a swimmer at 2.6 ft/sec is 10^6.

The HX coefficient of a swimmer in water is 500 watt/m^2 - C.

Swimming 2.6 ft/sec requires 200 watts mechanical energy.

If human metabolism is 30% efficient what is my surface area?

Bret Cahill

Reply to
Bret Cahill

Rectification doesn't do much to solve the unbounded quotient problem. Instead of +/- infinity you get + infinity.

An alert math person brought this to my attention.

Bret Cahill

"Be alert. America needs more lerts."

Reply to
Bret Cahill

Start here:

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John

Reply to
John Larkin

uld

It doesn't seem too long but I don't think I need it now.

Basic strategies to keep time constant(s) down might be interesting.

Dividing low noise AC signals will work if they are first either rectified or multiplied by a ref and then smoothed just enough to keep the denominator a certain distance that increases with noise above zero. After that one is divied by the other so only the noise needs to be smoothed, not the humps from the wave forms.

The two low time coinstants would be much lower than the one big one if the waves are smoothed in one step after rectification.

This has probably been done before and there might even be a name for it.

Bret Cahill

Reply to
Bret Cahill

You obviously don't need to know anything to shoot off your mouth! I recommend staying that way as you'll use less Prozac.

Reply to
Benj

sounds like an envelope detector.

that'll not have the narrow-band response of the lock-in,

Reply to
Jasen Betts

There's no modulation, just one signal being divided by another signal that happens to be in sync. The amplitudes only change ~ 1% over 4 -

6 cycles.

Both signals could be rectified and smoothed and then divided for the output but the output needs to be 99.5% smooth in just 4 - 6 cycles so smoothing each signal all at once _before_ the division probably won't work.

To save time both signals could be smoothed just enough before division to keep the denominator above zero.

After that it will be possible divide the slightly modified signals for a relatively smooth output.

If a second smoothing is necessary both time constants should be less than smoothing all at once before the division.

Noise isn't directly an issue because the noise is only 10% of the signal and already has a short enough time const..

True. My problem isn't noise but time. I only get 6 cycles max.

Bret Cahill

Reply to
Bret Cahill

You should get out more.

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John

Reply to
John Larkin

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Someone else also suggested matched filters a few weeks ago. That may be the way to go.

Plan "A" is to get away with as few sensors and wires as possible.

Still it's good to know there is a plan "B" or "C."

Bret Cahill

Reply to
Bret Cahill

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My problem isn't too much noise but too little time. =A0I only get 6 cycles max.

Some EE or electronics poster, maybe from here, posted on sci.mech.engr that he had invented a heat sink that relied on the high heat transfer due to condensation and evaporation of a fluid in a tube. Someone told him that heat pipes had been on the market and studied in depth for decades.

He was delighted.

Now we can easily guess several things:

  1. dividing 2 in sync signals is done all the time.

  1. the time to smooth a rectified AC to a DC output is frequently an issue.

  2. someone decades ago probably tried to lower the time constant in the situation above by using a low time constant prefilter to keep the denominator a small amount above zero and then letting the division of the two signals do most of the heavy lifting as far as smoothing the quotient to DC in a short time.

Bret Cahill

Bret Cahill

Bret Cahill

Reply to
Bret Cahill

I've designed and studied a lot of electronics, and I've never seen it done. Got any references?

We use things called "filters", about which there's a lot of history and math.

Doubt it.

John

Reply to
John Larkin

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Now that you mention it, there was never anything similar to my particular application/solution, at least not in that field. Judging from that alone maybe it shouldn't be too astounding that there was never any motivation for the circuit before now.

But nothing on quotients of 2 signals . . .

They have all these circuits as complicated as the Palo Verde nuke plant so it is hard to believe that something as simple as this situation, a DC output that consists of the quotient of two AC signals, hasn't appeared before now.

I'm not very interested in patenting a "filter" or time const. lowering strategy that only has one application.

Bret Cahill

Reply to
Bret Cahill

No, usually one samples the signals, say X and Y, and does a least-squares fit to determine Y=3D aX + b and on finding b approximately equal to zero, one has a good value for the ratio, 'a'. Division isn't a necessary step at all. Accumulating (in this case, integrating) values like X, Y, X**2, X*Y, and doing some arithmetic ON THE SUMS.

It's less than 'an issue', it's a failure to define the terms. AC measurement by making a DC conversion only works if you have some kind of average in mind (AC is a vector, DC is a scalar; they aren't ever 'equivalent'). It's not clear that the best 'm' value in the linear equation is related to any averaging after division.

Collect the data, do a least-squares fit, extract 'm' value. There's no particular need for any 'time constant' or prefilter, no concern with denominators.

Reply to
whit3rd

Oh, division of analog signals has been done for 60 years at least. But not in the way you suggest.

John

Reply to
John Larkin

No, usually one samples the signals, say X and Y, and does a least-squares fit to determine Y= aX + b

=================================== If Y = a.sin(b.X+c) can you still do a least-squares fit to the curve?

and on finding b approximately equal to zero, one has a good value for the ratio, 'a'. Division isn't a necessary step at all. Accumulating (in this case, integrating) values like X, Y, X**2, X*Y, and doing some arithmetic ON THE SUMS.

It's less than 'an issue', it's a failure to define the terms. AC measurement by making a DC conversion only works if you have some kind of average in mind (AC is a vector, DC is a scalar;

============================================ You have that back arse-wards, boy.

Reply to
Androcles

Yes, of course, but that isn't a line-fit, and simple linear least- squares formulae for a line are built into calculators. Nonlinear fit with three variable parameters (a, b, c) is usually done by computer search followed by iterative improvements.

We know the temperature and composition of the Sun because of multiparameter fits to sunlight's spectrum. Helium was discovered by folk who found that improving the fit required a new (hitherto unknown) element.

Reply to
whit3rd

Yes, of course, but that isn't a line-fit, and simple linear least- squares formulae for a line are built into calculators. Nonlinear fit with three variable parameters (a, b, c) is usually done by computer search followed by iterative improvements. ============================================ Who uses a calculator?

============================================= We know the temperature ============================================ What does that have to do with DC being a vector?

Reply to
Androcles

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