Reply by Bret Cahill September 23, 20112011-09-23
> >> So there are more than two nodes! I need a picture. > > > See the three "---" lines between Vs(t) and L1 and L and �Vn(t)? > > > Ground -- Vs(t) --- L1 --- L --- Vn(t) -- Ground > > > ? �? �? > > > Each one is generally at a different voltage. > > > . . . > > >>>>> If Vn(t) is significant and in the same band as Vs(t) then the noise > >>>>> from Vn(t) can be filtered by calculating Vs(t) as a noise free > >>>>> reference: > > >>>>> Vs(t) = Vm(t) + L1(di/dt) = reference > > >>>>> For phase sensitive rectification, > > >>>>> Integral [Vm(t) * (Vm(t) + L1(di/dt))] / Integral [(di/dt) * (Vm(t) + > >>>>> L1(di/dt))] => � � �L > > >>>> How do you measure or compute di/dt? > > >>> Analog or digital? > > >> Either way works. > > > Is there any reason to discuss something that has already been > > invented? > > > Do you have any comments or questions on the reference, > > > Vm(t) + L1(di/dt) > > > for filtering the circuit, > > > Ground -- Vs(t) -- L1 -- L -- Vn(t) -- Ground > > OK. So Vs(t) and Vn(t) are voltage generators with one end grounded. > Presumably, Vs(t) can have a very high SNR.
Between 3 - 20.
> Moreover, there are four > nodes, and L and L! form an inductive voltage divider. Why is noise a > problem? Are the inductances very small?
Small compared to what? In this case L = ~ 4 L1
> Why not simply short out Vn(t)?
Supposing that isn't possible?
> > to determine the unknown inductance L, > > > Integral [Vm(t) * (Vm(t) + L1(di/dt))] / Integral [(di/dt) * (Vm(t) + > > L1(di/dt))] � => � � L > > There is no need to differentiate. The ratio of the voltages across the > inductors is the ratio of the inductances.
Supposing the voltage across L1 is unknown or very expensive and/or inaccurate? Bret Cahill
Reply by Bret Cahill September 23, 20112011-09-23
The noise free reference,

Vm(t) + L1(di/dt)

for filtering the circuit,

Ground -- Vs(t) -- L1 -- L -- Vn(t) -- Ground

is used to determine the unknown inductance L,

Integral [Vm(t) * (Vm(t) + L1(di/dt))] / Integral [(di/dt) * (Vm(t) +
L1(di/dt))]   =>    L


> > Phase sensitive rectification.
> Oh cripes and here I thought we were talking about adaptive filters.
Well? Could any adaptive filter be made to work with the reference above?
> But, I've noticed you mentioned matched filters along the way .. I > wasn't "getting it".
> I wouldn't generally associate the two directly. �Maybe a good Master's > thesis topic:
> "The Relationship Between Adaptive Filters and Matched Filters"
What about, "Various Categorizations of Reference Based Filters"
> But, somehow I think the answer is trivial .. according to my notion of > what those two things are:
> - A Matched Filter is one that is matched to a KNOWN signal and outputs > the best SNR in the presence of white Gaussian noise.
Try match filtering a noisy signal using the reference above, Vm(t) + L1(di/dt) to determine L and compare it with PSR. Try both filters using the same signal, same noise and the same reference. You can do everything on Excel. To save time construct the reference in the frequency domain.
> - An Adaptive Filter is one that attempts to either remove noise (by > some definition) in the form of an ANC or ALE - one with a noise > reference and one without in their simplest forms. �AND is capable of > changing in dynamic conditions of signal and noise.
> I suppose if the signal is stable and the noise is white Gaussian then > an ALE may tend to the matched filter. �But I don't' think I know that > for sure. �And, an ANC will simply shut off and not do anything with > that kind of noise. > > But "Phase Senssitive Rectification"? �Where did that come from in this > context?
The question is if an accurate determination of L could be accomplished using another reference with _any_ filter. So far the answer seems to be "no." Bret Cahill
Reply by Fred Marshall September 23, 20112011-09-23
On 9/22/2011 7:02 PM, Bret Cahill wrote:
> Phase sensitive rectification.
Oh cripes and here I thought we were talking about adaptive filters. But, I've noticed you mentioned matched filters along the way .. I wasn't "getting it". I wouldn't generally associate the two directly. Maybe a good Master's thesis topic: "The Relationship Between Adaptive Filters and Matched Filters" But, somehow I think the answer is trivial .. according to my notion of what those two things are: - A Matched Filter is one that is matched to a KNOWN signal and outputs the best SNR in the presence of white Gaussian noise. - An Adaptive Filter is one that attempts to either remove noise (by some definition) in the form of an ANC or ALE - one with a noise reference and one without in their simplest forms. AND is capable of changing in dynamic conditions of signal and noise. I suppose if the signal is stable and the noise is white Gaussian then an ALE may tend to the matched filter. But I don't' think I know that for sure. And, an ANC will simply shut off and not do anything with that kind of noise. But "Phase Senssitive Rectification"? Where did that come from in this context? Fred
Reply by Bret Cahill September 22, 20112011-09-22
> > Do you have any comments or questions on the reference, > > > Vm(t) + L1(di/dt) > > > for filtering the circuit, > > > Ground -- Vs(t) -- L1 -- L -- Vn(t) -- Ground > > > to determine the unknown inductance L, > > > Integral [Vm(t) * (Vm(t) + L1(di/dt))] / Integral [(di/dt) * (Vm(t) + > > L1(di/dt))] � => � � L > > I think I understand the circuit diagram now. �Thank you. > > Next, where did this expression of integrals come from? �Its derivation > doesn't just jump out at me.
Phase sensitive rectification. Multiply the noisy signal by the noise free reference and to then integrate or otherwise low pass filter. The clean part of the signal correlates with the reference and increases with the integration time. The noise part of the signal doesn't correlate with the reference eventually disappears as a percentage of the integral value. PSR both Vm(t) and di/dt over the same time t and then take the quotient to get the unknown inductor L. If no one has ever derived a reference like this before now it is understandable since it's so easy to get Vs(t) and, for that matter, inductance. Bret Cahill
Reply by Jerry Avins September 22, 20112011-09-22
On 9/22/2011 7:30 PM, Bret Cahill wrote:
>> So there are more than two nodes! I need a picture. > > See the three "---" lines between Vs(t) and L1 and L and Vn(t)? > > Ground -- Vs(t) --- L1 --- L --- Vn(t) -- Ground > > ? ? ? > > Each one is generally at a different voltage. > > > . . . > > >>>>> If Vn(t) is significant and in the same band as Vs(t) then the noise >>>>> from Vn(t) can be filtered by calculating Vs(t) as a noise free >>>>> reference: >> >>>>> Vs(t) = Vm(t) + L1(di/dt) = reference >> >>>>> For phase sensitive rectification, >> >>>>> Integral [Vm(t) * (Vm(t) + L1(di/dt))] / Integral [(di/dt) * (Vm(t) + >>>>> L1(di/dt))] => L >> >>>> How do you measure or compute di/dt? >> >>> Analog or digital? >> >> Either way works. > > Is there any reason to discuss something that has already been > invented? > > Do you have any comments or questions on the reference, > > Vm(t) + L1(di/dt) > > for filtering the circuit, > > Ground -- Vs(t) -- L1 -- L -- Vn(t) -- Ground
OK. So Vs(t) and Vn(t) are voltage generators with one end grounded. Presumably, Vs(t) can have a very high SNR. Moreover, there are four nodes, and L and L! form an inductive voltage divider. Why is noise a problem? Are the inductances very small? Why not simply short out Vn(t)?
> to determine the unknown inductance L, > > Integral [Vm(t) * (Vm(t) + L1(di/dt))] / Integral [(di/dt) * (Vm(t) + > L1(di/dt))] => L
There is no need to differentiate. The ratio of the voltages across the inductors is the ratio of the inductances. Jerry -- Engineering is the art of making what you want from things you can get.
Reply by Fred Marshall September 22, 20112011-09-22
On 9/22/2011 4:30 PM, Bret Cahill wrote:
> Do you have any comments or questions on the reference, > > Vm(t) + L1(di/dt) > > for filtering the circuit, > > Ground -- Vs(t) -- L1 -- L -- Vn(t) -- Ground > > to determine the unknown inductance L, > > Integral [Vm(t) * (Vm(t) + L1(di/dt))] / Integral [(di/dt) * (Vm(t) + > L1(di/dt))] => L
I think I understand the circuit diagram now. Thank you. Next, where did this expression of integrals come from? Its derivation doesn't just jump out at me. Fred
Reply by Bret Cahill September 22, 20112011-09-22
> So there are more than two nodes! I need a picture.
See the three "---" lines between Vs(t) and L1 and L and Vn(t)? Ground -- Vs(t) --- L1 --- L --- Vn(t) -- Ground ? ? ? Each one is generally at a different voltage. . . .
> >>> If Vn(t) is significant and in the same band as Vs(t) then the noise > >>> from Vn(t) can be filtered by calculating Vs(t) as a noise free > >>> reference: > > >>> Vs(t) = Vm(t) + L1(di/dt) = reference > > >>> For phase sensitive rectification, > > >>> Integral [Vm(t) * (Vm(t) + L1(di/dt))] / Integral [(di/dt) * (Vm(t) + > >>> L1(di/dt))] => � �L > > >> How do you measure or compute di/dt? > > > Analog or digital? > > Either way works.
Is there any reason to discuss something that has already been invented? Do you have any comments or questions on the reference, Vm(t) + L1(di/dt) for filtering the circuit, Ground -- Vs(t) -- L1 -- L -- Vn(t) -- Ground to determine the unknown inductance L, Integral [Vm(t) * (Vm(t) + L1(di/dt))] / Integral [(di/dt) * (Vm(t) + L1(di/dt))] => L ? ? ? Bret Cahill
Reply by Jerry Avins September 22, 20112011-09-22
On 9/22/2011 5:16 PM, Bret Cahill wrote:
>>> The circuit is a simple loop: >> >>> Ground -- Vs(t) -- L1 -- L -- Vn(t) -- Ground >> >> I read that as ground -- L1 -- L -- ground, > > Feel free to start another thread if you want a circuit w/o voltage or > current sources. > > The circuit on this thread is: > > Ground -- Vs(t) -- L1 -- L -- Vn(t) -- Ground > >> with Vs(t) and Vn(t) >> referenced to ground. Where are their other ends connected? > >>> Vs(t) is the unknown clean signal. > >>> Vn(t) is unknown uncorrelated noise. > >> Not correlated to what? > > You get 3 guesses and the 1st 2 don't count. > >>> L(1) is the known inductor > >> Whet use does it have? > > 1. Some circuits have to have it. > > 2. Without it then Vm(t) would = Vs(t) which may be more expensive > and less accurate to measure than with L1. > > 3. When Vm(t) = Vs(t) then Vm(t) does not need to be filtered and it > can be used a reference to filter di/dt but this isn't as interesting > as with L1 in the circuit. > >>> L is the unknown inductor to be determined. > >>> Vm(t) is the voltage measured at the node between L1 -- L and ground. >>> (Not shown) > >> Is Vm(t) the same as Vn(t)? > > Not as long as L1 is between the 2 voltages.
Between which two voltages? For a circuit with only two nodes (that one of them is ground doesn't matter) there is only one voltage to measure.
>> If not, where is it in your scheme? > > Vm(t) is measured between ground and the node between the two > inductors
Agreed. That is the only place a voltage can be measured.
>> i is the current in the loop. >> >>> If you know >> >>> 1. the voltmeter voltage Vm(t) measured between ground and the node >>> between the inductors. >> >> Since your circuit is a loop with two nodes (one of them ground), there >> is only one place to measure any voltage. > > You think the entire circuit is at 1 voltage?
Yes. Don't you? I said I didn't know what you were talking about. A picture would help.
>> The same voltage is across >> both L1 an L. > > Do you mean the same voltage _drop_? The voltage drop over each > inductors will generally be different.
Is there a difference between a voltage and a voltage drop?
>> What causes it? > > It's plugged into something.
So there are more than two nodes! I need a picture.
>>> 2. the current i through the loop >> >>> 3. the noise, Vn(t) = 0 >> >>> then it's easy to determine L: >> >>> L = Vm(t)/(di/dt) >> >>> (except near crossings) >> >>> If Vn(t) is significant and in the same band as Vs(t) then the noise >>> from Vn(t) can be filtered by calculating Vs(t) as a noise free >>> reference: >> >>> Vs(t) = Vm(t) + L1(di/dt) = reference >> >>> For phase sensitive rectification, >> >>> Integral [Vm(t) * (Vm(t) + L1(di/dt))] / Integral [(di/dt) * (Vm(t) + >>> L1(di/dt))] => L >> >> How do you measure or compute di/dt? > > Analog or digital?
Either way works. I asked you. Jerry -- Engineering is the art of making what you want from things you can get.
Reply by Bret Cahill September 22, 20112011-09-22
> > The circuit is a simple loop: > > > Ground -- Vs(t) -- L1 -- L -- Vn(t) -- Ground > > I read that as ground -- L1 -- L -- ground,
Feel free to start another thread if you want a circuit w/o voltage or current sources. The circuit on this thread is: Ground -- Vs(t) -- L1 -- L -- Vn(t) -- Ground
> with Vs(t) and Vn(t) > referenced to ground. Where are their other ends connected?
> > Vs(t) is the unknown clean signal.
> > Vn(t) is unknown uncorrelated noise.
> Not correlated to what?
You get 3 guesses and the 1st 2 don't count.
> > L(1) is the known inductor
> Whet use does it have?
1. Some circuits have to have it. 2. Without it then Vm(t) would = Vs(t) which may be more expensive and less accurate to measure than with L1. 3. When Vm(t) = Vs(t) then Vm(t) does not need to be filtered and it can be used a reference to filter di/dt but this isn't as interesting as with L1 in the circuit.
> > L is the unknown inductor to be determined.
> > Vm(t) is the voltage measured at the node between L1 -- L and ground. > > (Not shown)
> Is Vm(t) the same as Vn(t)?
Not as long as L1 is between the 2 voltages.
> If not, where is it in your scheme?
Vm(t) is measured between ground and the node between the two inductors
> > i is the current in the loop. > > > If you know > > > 1. �the voltmeter voltage Vm(t) measured between ground and the node > > between the inductors. > > Since your circuit is a loop with two nodes (one of them ground), there > is only one place to measure any voltage.
You think the entire circuit is at 1 voltage?
> The same voltage is across > both L1 an L.
Do you mean the same voltage _drop_? The voltage drop over each inductors will generally be different.
> What causes it?
It's plugged into something.
> > 2. �the current i through the loop > > > 3. �the noise, Vn(t) = 0 > > > then it's easy to determine L: > > > L = Vm(t)/(di/dt) > > > (except near crossings) > > > If Vn(t) is significant and in the same band as Vs(t) then the noise > > from Vn(t) can be filtered by calculating Vs(t) as a noise free > > reference: > > > Vs(t) = Vm(t) + L1(di/dt) = reference > > > For phase sensitive rectification, > > > Integral [Vm(t) * (Vm(t) + L1(di/dt))] / Integral [(di/dt) * (Vm(t) + > > L1(di/dt))] => �L > > How do you measure or compute di/dt?
Analog or digital? Bret Cahill
Reply by Jerry Avins September 22, 20112011-09-22
On 9/22/2011 1:05 AM, Bret Cahill wrote:
> The circuit is a simple loop: > > Ground -- Vs(t) -- L1 -- L -- Vn(t) -- Ground
I read that as ground -- L1 -- L -- ground, with Vs(t) and Vn(t) referenced to ground. Where are their other ends connected?
> Vs(t) is the unknown clean signal. > > Vn(t) is unknown uncorrelated noise.
Not correlated to what?
> L(1) is the known inductor
Whet use does it have?
> L is the unknown inductor to be determined. > > Vm(t) is the voltage measured at the node between L1 -- L and ground. > (Not shown)
Is Vm(t) the same as Vn(t)? If not, where is it in your scheme?
> i is the current in the loop. > > If you know > > 1. the voltmeter voltage Vm(t) measured between ground and the node > between the inductors.
Since your circuit is a loop with two nodes (one of them ground), there is only one place to measure any voltage. The same voltage is across both L1 an L. What causes it?
> 2. the current i through the loop > > 3. the noise, Vn(t) = 0 > > then it's easy to determine L: > > L = Vm(t)/(di/dt) > > (except near crossings) > > If Vn(t) is significant and in the same band as Vs(t) then the noise > from Vn(t) can be filtered by calculating Vs(t) as a noise free > reference: > > Vs(t) = Vm(t) + L1(di/dt) = reference > > For phase sensitive rectification, > > Integral [Vm(t) * (Vm(t) + L1(di/dt))] / Integral [(di/dt) * (Vm(t) + > L1(di/dt))] => L
How do you measure or compute di/dt? I wouldn't presume to tell you that you don't know what you're talking about. I can say with confidence that I don't know what you're talking about. Jerry -- Engineering is the art of making what you want from things you can get.