How do I find the uncertainty in a timestep for an oscilloscope

Thread Starter

GreyPoupon24

Joined Sep 14, 2022
6
I am trying to calculate the energy delivered by a short electrical pulse. I am using an oscilloscope to capture the waveform and then exporting it to a CSV file. My scope took measurements at a rate of once every 8 microseconds which means that each time time value could potentially vary by plus or minus 4 microseconds. The manual of my scope gives no mention of uncertainties whatsoever, but it says that the maximum capture rate that can be used is once very 1 microsecond. My question is whether my uncertainty should be taken as half of the time increment I am using (± 4 microseconds) or half of the smallest possible time increment the scope can capture (± 0.5 microseconds).

Here comes the real issue, to calculate energy, we must square voltage to get power, and add up power values multiplied by how long each power value lasted for (energy=timestep*SUM(powers)). We are dealing with the error of the oscilloscope's time resolution. To get the final uncertainty we must add up the relative uncertainties of the timestep and uncertainty of the sum of powers. If I am to say that the error in the time resolution is ±4 microseconds, then I necessarily must have a relative error of at least 4μs/8μs=0.5 which is tremendous. That does not seem right, so which method of estimating uncertainty is correct here?

Thank you all
 

tautech

Joined Oct 8, 2019
292
I am trying to calculate the energy delivered by a short electrical pulse.
This alone is the core issue and is done by measuring the pulse duration and voltage and current magnitudes.
You can sometimes use a current shunt however unless you can reference to one side (ground loop issues) but the best solution is to obtain a scope current probe.
Then with current and voltage amplitude and pulse duration measurements the total pulse energy can be easily calculated.
 

WBahn

Joined Mar 31, 2012
27,395
I am trying to calculate the energy delivered by a short electrical pulse. I am using an oscilloscope to capture the waveform and then exporting it to a CSV file. My scope took measurements at a rate of once every 8 microseconds which means that each time time value could potentially vary by plus or minus 4 microseconds. The manual of my scope gives no mention of uncertainties whatsoever, but it says that the maximum capture rate that can be used is once very 1 microsecond. My question is whether my uncertainty should be taken as half of the time increment I am using (± 4 microseconds) or half of the smallest possible time increment the scope can capture (± 0.5 microseconds).

Here comes the real issue, to calculate energy, we must square voltage to get power, and add up power values multiplied by how long each power value lasted for (energy=timestep*SUM(powers)). We are dealing with the error of the oscilloscope's time resolution. To get the final uncertainty we must add up the relative uncertainties of the timestep and uncertainty of the sum of powers. If I am to say that the error in the time resolution is ±4 microseconds, then I necessarily must have a relative error of at least 4μs/8μs=0.5 which is tremendous. That does not seem right, so which method of estimating uncertainty is correct here?

Thank you all
First off, squaring the voltage does NOT give you power. You can infer power from it only if you know the reactance of what that voltage is appearing across.

The error in the timing of the samples is not going to be anywhere near half of the period between samples. What you need to know is the nature of the jitter in the sampling rate.

How long are these pulses? In particular, how may samples are you getting per pulse. If it's only a few, then that is where you are going to run into big uncertainties because of the uncertainty in when the pulse started and when it ended.

Let's say that you have data showing five samples during the pulse. Let's further say that the pulse amplitude is a firm, fixed 10 V and that the scope's sampling clock is jitter-free, meaning that the samples are taken exactly once every 8 µs. How much "energy" (per your method) was delivered by the pulse?

What was it if the pulse started just after the last sample that showed no pulse (before the pulse data) and ended just before the first sample that showed no pulse afterward?

What was it if the pulse started just before the first sample that showed the presence of a pulse and ended just after the last one that did?

Both of these situations would give you the same data set.
 

MrChips

Joined Oct 2, 2009
27,122
We could be of better assistance if we knew the make and model of the oscilloscope.

If we were to assume that the digital sampling oscilloscope is using an 8-bit ADC, then 1 in 256 represents a resolution of about 0.4%. I would expect that the sampling jitter would be better than 0.4%. The reason I say this is that any jitter in the sampling time would degrade the voltage measurement.

Let us make an assumption that the jitter is 0.5%. The error at 8μs is ±40ns.
 

tautech

Joined Oct 8, 2019
292
We could be of better assistance if we knew the make and model of the oscilloscope.

If we were to assume that the digital sampling oscilloscope is using an 8-bit ADC, then 1 in 256 represents a resolution of about 0.4%. I would expect that the sampling jitter would be better than 0.4%. The reason I say this is that any jitter in the sampling time would degrade the voltage measurement.

Let us make an assumption that the jitter is 0.5%. The error at 8μs is ±40ns.
TBH trigger jitter contribution to measurement uncertainty at a few MHz won't contribute much error when a reasonable DSO without a 10M reference has a trigger jitter spec of <100 ps
Certainly not enough to impact on accuracy to loose any sleep about.
 

Thread Starter

GreyPoupon24

Joined Sep 14, 2022
6
I'm confused.
What determines the 8µs period for the oscilloscope?
Is that the same as the electrical pulse period, or is that different?
I am sorry for any confusion, 8µs is the time between the datapoints in the CSV file of my scope capture. I have one voltage at 0µs then one at 8µs then one at 16µs etc. The actual pulse is around 200µs.
 

Thread Starter

GreyPoupon24

Joined Sep 14, 2022
6
We could be of better assistance if we knew the make and model of the oscilloscope.

If we were to assume that the digital sampling oscilloscope is using an 8-bit ADC, then 1 in 256 represents a resolution of about 0.4%. I would expect that the sampling jitter would be better than 0.4%. The reason I say this is that any jitter in the sampling time would degrade the voltage measurement.

Let us make an assumption that the jitter is 0.5%. The error at 8μs is ±40ns.
We could be of better assistance if we knew the make and model of the oscilloscope.

If we were to assume that the digital sampling oscilloscope is using an 8-bit ADC, then 1 in 256 represents a resolution of about 0.4%. I would expect that the sampling jitter would be better than 0.4%. The reason I say this is that any jitter in the sampling time would degrade the voltage measurement.

Let us make an assumption that the jitter is 0.5%. The error at 8μs is ±40ns.
Thanks for your help, the scope is a SainSmart DS212. I couldn't find mention of the jitter or time uncertainty in the manual. Amazon says it has 1Mhz analog bandwidth, whatever that means. Does this mean that it is capable of sampling a million times a second?
 

Thread Starter

GreyPoupon24

Joined Sep 14, 2022
6
This alone is the core issue and is done by measuring the pulse duration and voltage and current magnitudes.
You can sometimes use a current shunt however unless you can reference to one side (ground loop issues) but the best solution is to obtain a scope current probe.
Then with current and voltage amplitude and pulse duration measurements the total pulse energy can be easily calculated.
Sadly, I do not have a scope current probe. I am stuck basing this off of voltage measurements so I will have to make the best of it.
 

Thread Starter

GreyPoupon24

Joined Sep 14, 2022
6
First off, squaring the voltage does NOT give you power. You can infer power from it only if you know the reactance of what that voltage is appearing across.

The error in the timing of the samples is not going to be anywhere near half of the period between samples. What you need to know is the nature of the jitter in the sampling rate.

How long are these pulses? In particular, how may samples are you getting per pulse. If it's only a few, then that is where you are going to run into big uncertainties because of the uncertainty in when the pulse started and when it ended.

Let's say that you have data showing five samples during the pulse. Let's further say that the pulse amplitude is a firm, fixed 10 V and that the scope's sampling clock is jitter-free, meaning that the samples are taken exactly once every 8 µs. How much "energy" (per your method) was delivered by the pulse?

What was it if the pulse started just after the last sample that showed no pulse (before the pulse data) and ended just before the first sample that showed no pulse afterward?

What was it if the pulse started just before the first sample that showed the presence of a pulse and ended just after the last one that did?

Both of these situations would give you the same data set.
Jitter! That's the magic word, I need to know the jitter. I knew the plus minus 50% was extremely unreasonable, it was an absolute highest upper bound I came up with because the user manual didn't provide me with info on uncertainties. Amazon says the scope has "1Mhz analog bandwidth" if that means anything to you.
 

MrChips

Joined Oct 2, 2009
27,122
SainSmart DS212 oscilloscopes are based on STM32 ARM MCU which typically use an 8MHz quartz crystal oscillator.

Typical instability is about 5ppm. If we were to assume a worse case instability of 100ppm this is ±0.01% or ±800ps at 8μs.
 

WBahn

Joined Mar 31, 2012
27,395
I am sorry for any confusion, 8µs is the time between the datapoints in the CSV file of my scope capture. I have one voltage at 0µs then one at 8µs then one at 16µs etc. The actual pulse is around 200µs.
If you have N samples during the pulse, the pulse width, W, will be somewhere in the range:

T(N-1) < W < T(N+1)

where T is the sampling period.

Written another way:

W = T·N ± T

which is a fractional uncertain of ±(1/N).

For a 200 µs pulse and a sampling period of 8 µs, you would expect to see N = 25, giving your pulse width an uncertainty of ±4%, which for a nominally rectangular pulse, is also the minimum uncertainty in your energy estimate.

The jitter in the interior samples is really neither here nor there -- since you only have a signal value at the sampling points, you don't have any basis to claim that knowing the exact sampling times would yield any better result than just assuming uniform sampling. It's only the first and last samples that could affect the results and the uncertainly of when the samples were actually taken is completely overshadowed by the uncertainly in when the pulse actually started and stopped.
 

Thread Starter

GreyPoupon24

Joined Sep 14, 2022
6
If you have N samples during the pulse, the pulse width, W, will be somewhere in the range:

T(N-1) < W < T(N+1)

where T is the sampling period.

Written another way:

W = T·N ± T

which is a fractional uncertain of ±(1/N).

For a 200 µs pulse and a sampling period of 8 µs, you would expect to see N = 25, giving your pulse width an uncertainty of ±4%, which for a nominally rectangular pulse, is also the minimum uncertainty in your energy estimate.

The jitter in the interior samples is really neither here nor there -- since you only have a signal value at the sampling points, you don't have any basis to claim that knowing the exact sampling times would yield any better result than just assuming uniform sampling. It's only the first and last samples that could affect the results and the uncertainly of when the samples were actually taken is completely overshadowed by the uncertainly in when the pulse actually started and stopped.
Oh! Oh! I see the confusion, I see the confusion. This pulse is not even remotely rectangular, I will share an example of one of my scope's captures right now(this one is a different pulse but similar shape and same order of magnitude period and sampling frequency).

1663233208555.png
 

rindonejj

Joined Oct 20, 2022
1
I'm confused.
What determines the 8µs period for the oscilloscope?
Is that the same as the electrical pulse period, or is that different?
The intervals are set by the sweep generator. Basic scope is a display, a vertical amp and a horizontal generator. Using the scope at best might be a 1 to 3% device. Using the variable Vernier adjust better accuracies are available. Capturing events on O'scopes can be difficult. 1 usecond seems incredibly fast. Until you start getting spike of over 100 VAC @ picosecond duration. Why use a scope? Time interval counters can accurately do that. In fact some decent scope can too.
 
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