I rearranged the ideas on the integral mathematical function of the oscilloscope, I divide into two posts for the usual confusion speech on the attached images! I can say I have two doubts, one in this post and one in the next!
In this first attached image, I have rearranged the three phases of the integral relative to the compensation square wave signal; not, I used the signal in AV coupling for the ramp and the descent, so I could also have the signal negative and therefore the ramp down: I instead used the channel in dc coupling to be able to have the zero volt flat integral
I isolated the three phases using the range gate.
if I understood correctly, when the integral ramp is rising it means that the signal is emitting positive voltage; when the ramp is down, it means that the signal is emitting negative voltage; when the ramp is horizontal, at that moment the signal is at rest = zero volts.
It is obvious that if the signal were AC coupling, the integral part would not be flat as it would tansit only for an instant on zero volts ..
I only have a doubt about the unit of measurement: if we look at the rising ramp, we see that the wave area of the integral measures 3V vertically and 500uS horizontally: so we have 3 divs of 1V * 500uS: I see that the scale is indicated in xxxuV * S. Wouldn't it be more accurate to indicate xxxV * uS? (my doubt would be: u is indicated at the time, not the voltage right?)