I hope someone can help improve my understanding ... I have a need to figure uncertainty for the Ratio function of a digital multimeter. I have been doing this for other measurement parameters for years, but have never needed to do it for ratio before ... it's been one of those specifications that gets sort of glanced over with a muttered "someday I'll get to that". Well, "someday" is here.

The measurement is a DC Volt ratio. The DMM manual says the ratio is (Input / Reference). OK so far.

The Accuracy specification is +/-(Input error + Reference error).


Input error is the "total error for the input signal measurement function" - DC V in this case, and let's assume an unknown voltage very close to 100 mV. For the DMM I am using (HP 3458A option 2) this works out to about 1.58 microvolts. OK so far.
Reference error is "1.5 x total error for the range of the reference DC input" - and let's say that the reference is 100.0 mV DC. This works out to 2.37 microvolts. OK so far.
Adding the two together gives a sum of 3.95 micro-somethings.
Let's assume that applying the voltages correctly results in a ratio display of +0.999874 which means that the unknown voltage is 99.9874 mV.
Here's the question:

The accuracy specification for the RATIO function says to ADD these two microvolt values.

Does the resulting number remain as microvolts? (Logically it should.) If so, then exactly how does it get applied as an uncertainty to the RATIO?
Or, is the resulting number somehow magically transmuted into a ratio that applies in a plus/minus fashion to the ratio displayed on the DMM? If so, then what is the explanation and proof?
In either case, what is the uncertainty of (a) the ratio, and (b) the unknown voltage?
(Or, is it something else entirely and am I so far out that it will take the rest of the millenium for Voyager to get to me?) :)

I would appreciate any reasonable-sounding instruction on this topic because I am sure I'm not the only one who has been avioding this for as long as possible ... Thanks.

I could understand the specs if they dealt with a relative error, like ppm or percent error, rather than absolute error. For example, if the reference voltage was accurate to +/- 10 ppm and the input voltage was good to +/- 6 ppm, then the ratio of the two should be good to +/- 16 ppm. That one I could give a quick proof for.

(By extension, if you can prove that adding does work for the relative error, then it is a pretty good bet that it doesn't work for absolute error.)

Tim Folkerts

Hi Graeme,

I don't claim any profound knowledge of the instrumentation involved, but I can suggest a way to calculate the uncertainty of a ratio or quotient.

From calculus, if f(x)=u(x)/v(x), then delta f=f ' (x)*delta x, where f ' the derivative of the quotient, or (vu'-uv')/v**2,

In your problem, v=ref=100mv, u=input =(for all practical purposes 100mv),
u'~1.58uv/100mv, v'~2.37uv/100mv, so the value of the derivative is 0.79 x
10(-6) [dimensionless]. I take it that the delta x should be the sum of the input error and the reference error voltage, or (1.58uV+2.37uV) or 3.95uV.

The total uncertainty would be ridiculously small, about 3.2 x 10 (-12).

I'm not sure if the result is reasonable, or if this mathematical model is the appropriate for the physical phenomena being modeled. I just wanted to offer the idea to possibly stimulate others.

Good luck

Jerry Haddock, P.E.

OK, here are two other proofs, followed by some specific results.

1) For the hardcore calculus folks:
Let u = input, v= reference, f = ratio = u/v

The "differential" of f is

df = (df/du) du + (df/dv) dv
= 1/v du - u/v^2 dv
(where the italic d represents the partial deriviative

If we divide both sides by f = u/v, we get

df/f = du/u - dv/v

Since we are looking for the "worst case" combination in the preceding equation, we should take the absolute values for the two terms, or

df/f = du/u + dv/v


2) The laid-back algebraic approach:
I'll do this for multiplication because it saves a couple of steps, but the result is the same for division. (You could call v = 1/reference voltage if you like).

f = uv is the best estimate for the product of the two term.

F = (u +du)(v +dv) is the worst it could be.
= uv + udv + vdu + dudv

We can drop the final term (du dv) because both of these numbers are much smaller than u or v, so their product will be much, much smaller than either uv, udv or vdu.

The uncertainty is df = F-f = uv + udv + vdu - uv = udv + vdu

The relative uncertainty will be (exactly as before, I might add)
df/f = udv/uv + vdu/uv = dv/v + du/u

Interpretation

du/u = 1.58uV / 100mV = 0.0000158 = 0.00158% = 15.8 ppm is the relative uncertainty in the input

dv/v = 2.37uV / 100mV = 0.0000237 = 0.00237% = 23.7 ppm is the relative uncertainty in the reference.

df/f = 15.8 ppm + 23.7 ppm ~ 40 ppm is the relative uncertainty in the reading

If your meter reads f = +0.999874, then the absolute uncertainty is +0.999874*(0.000040) = 0.000040


Tim F

P.S. You'd never guess I had to teach this recently, would you? :)

Nice job Tim!

Jerry Haddock, P.E.

Jerry and Tim --

Thank you both for your explanations! :thanks:

I will spend some time first converting everything to standard notation (it's too bad that a text-based system like this can't support an equation editor ... ) and then studying and applying it, and writing down the results so I can explain it to someone else -- the best way to learn something.

Thanks again! :thanx:

I hope someone can help improve my understanding ... I have a need to figure uncertainty for the Ratio function of a digital multimeter. I have been doing this for other measurement parameters for years, but have never needed to do it for ratio before ... it's been one of those specifications that gets sort of glanced over with a muttered "someday I'll get to that". Well, "someday" is here.

The measurement is a DC Volt ratio. The DMM manual says the ratio is (Input / Reference). OK so far.

The Accuracy specification is +/-(Input error + Reference error).


Input error is the "total error for the input signal measurement function" - DC V in this case, and let's assume an unknown voltage very close to 100 mV. For the DMM I am using (HP 3458A option 2) this works out to about 1.58 microvolts. OK so far.
Reference error is "1.5 x total error for the range of the reference DC input" - and let's say that the reference is 100.0 mV DC. This works out to 2.37 microvolts. OK so far.
Adding the two together gives a sum of 3.95 micro-somethings.
Let's assume that applying the voltages correctly results in a ratio display of +0.999874 which means that the unknown voltage is 99.9874 mV.
Here's the question:

The accuracy specification for the RATIO function says to ADD these two microvolt values.

Does the resulting number remain as microvolts? (Logically it should.) If so, then exactly how does it get applied as an uncertainty to the RATIO?
Or, is the resulting number somehow magically transmuted into a ratio that applies in a plus/minus fashion to the ratio displayed on the DMM? If so, then what is the explanation and proof?
In either case, what is the uncertainty of (a) the ratio, and (b) the unknown voltage?
(Or, is it something else entirely and am I so far out that it will take the rest of the millenium for Voyager to get to me?) :)

I would appreciate any reasonable-sounding instruction on this topic because I am sure I'm not the only one who has been avioding this for as long as possible ... Thanks.




Whoa! I just looked it up, and if you are using 100mV input, and 100mV reference, then it gets much simpler! You were using the uncertainty of cross-ranged inputs (like your input is 100mV, but you are using a 10V reference cell, so your front panel is using the 100mV range, and your rear panel is using the 10V range).

Basically, you are on the 100mV range on both the front and rear panels, which means you are doing a "Transfer Accuracy" measurement. Your uncertainty becomes: 0.5 ppm of reading + 0.5 ppm of range. Therefore, since you are applying a basically full scale reading, your uncertainty due to the 3458A Opt 2 is 1 ppm. Easy math, my personal favorite!

Ryan

Basically, you are on the 100mV range on both the front and rear panels, which means you are doing a "Transfer Accuracy" measurement. Your uncertainty becomes: 0.5 ppm of reading + 0.5 ppm of range. Therefore, since you are applying a basically full scale reading, your uncertainty due to the 3458A Opt 2 is 1 ppm. Easy math, my personal favorite!

Ryan,

Thanks! :thanks: I will keep that in mind when everything is on the same range, and even try to make it so whenever possible. After all, I like easy math as well. I must have had a brain malfunction and overlooked the transfer stuff.

But from what you said I assume that if the two are on DIFFERENT ranges then I am likely to have to analyze things as suggested by Tim and Jerry. (As an example, the reference might be 9.999987 V and the unknown might be something very close to 1.018123 V. Then I would have to worry about the 10 V range and the 1 V range and break out the TI-68 for some heavy math ...) Is that correct?

Graeme

Whoa! I just looked it up, and if you are using 100mV input, and 100mV reference, then it gets much simpler! ... Therefore, since you are applying a basically full scale reading, your uncertainty due to the 3458A Opt 2 is 1 ppm. Easy math, my personal favorite!

Ryan

Why did you have to wait until after :bonk: I had done the derivation! :)

Tim F

Ryan,

Thanks! :thanks: I will keep that in mind when everything is on the same range, and even try to make it so whenever possible. After all, I like easy math as well. I must have had a brain malfunction and overlooked the transfer stuff.

But from what you said I assume that if the two are on DIFFERENT ranges then I am likely to have to analyze things as suggested by Tim and Jerry. (As an example, the reference might be 9.999987 V and the unknown might be something very close to 1.018123 V. Then I would have to worry about the 10 V range and the 1 V range and break out the TI-68 for some heavy math ...) Is that correct?

Graeme

Tim's math was very good, but you made things more difficult than you needed to Graeme, even for inter-range ratios. I'll work it out for this example.

Unknown input = 1.018123 VDC, and will use the 1 V Range
Reference input = 9.999987 VDC, and will use the 10 V Range
Device = 3458A Opt 2
Specifications for this example = 1 Year (omitting the 2 ppm error due to "factory traceability", and assuming that a math null has been performed)
All figures are assumed to be k=2

The Accuracy specification is ±(Input error + Reference error).

Input error = "total error for the input signal measurement function" = 4 ppmReading + 0.3 ppmRange = 4.072 µV + 0.3 µV = 4.372 µV

Reference error = "1.5 x total error for the range of the reference DC input" = 1.5 x 0.05 ppmRange¹ = 1.5 x 0.5 µV = 0.75 µV

Note ¹ - The "ppm of Reading" is omitted in this calculation because the source of this error is the reference error, and you are not using the internal reference. I will have to add (RSS) the uncertainty of your 9.999987 V source later (I'll make one up).

3458A Opt 2 Accuracy = ± (4.372 µV + 0.75 µV) = ± 5.122 µV

Now I need to add the uncertainty due to your reference source, which I will call ± 1 ppm for this example.

Reference Source Uncertainty = 9.999987 VDC * 1 ppm = 10 µV
Reference Source Contribution to final uncertainty = Ref Source Uncertainty x Ratio(Input:Reference) = 10 µV x 1/10 = 1 µV

Combine your uncertainties = 2 * SQRT((5.122 / 2)² + (1 / 2)²) µV = 2 * SQRT(6.559 + 0.25) µV = 5.219 µV

Therefore, your result is 1.018123 VDC ± 5.2 µV

Oh, except you probably used cables, so you have some thermal EMFs to deal with, and a bit of noise because of those danged flourescent lights, and...


Ryan