Reading actual values of MDO4104C oscilloscope reference level

Is it possible to read actual values of reference level computed by oscilloscope for rise time measurement?
For example, reference levels are set to default so low=10% and high=90%. Once waveform is captured, how to read actual voltage values?

Hi,

This functionality is not given directly for the reference levels set to percentage mode (although it is available for absolute levels), but there are two options:

1) If you are using the scope manually, you can turn on markers for the rise time measurement and you can then measure the values from the graticule

2) If controlling the instrument remotely, you can calculate the values by getting the high and low values from the scope and then calculating the reference values yourself from the known percentages.

" PERCent specifies that the reference levels are calculated as a percent relative to HIGH and LOW. The percentages are defined using the MEASUrement:REFLevel:PERCent commands "

Hi,

Actually, I am looking for an answer about reference level used for computing when captured waveform has noise like in attached picture. If method parameter is set to auto, then oscilloscope pick MinMax or Histogram method. I suspect that the rise time measurement can be affected by noise so would like to know actual values of reference level.
 

Hi,

For a waveform with such high amplitude peaks, you will certainly want to use the histogram mode.

From page 118 of the user manual, on the Low measurement:

" This value is used as 0% whenever high reference, mid reference, or low reference values are needed, such as in fall time or rise time measurements. Calculate using either the min/max or histogram method. The min/max method uses the minimum value found. The histogram method uses the most common value found below the midpoint. This value is measured over the entire waveform or gated region. "

So, your reference level in histogram mode will be 10% above the most common value below the midpoint (this should almost completely eliminate the effect of noise you see on your waveform).

If you wish to know what that exact value is, then I will refer you back to 2) of my first answer.