STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

[Omnibus]

Average dE/dt is mathematically different from taking the difference from end and starting point.

I am very pragmatic in this.
What do we want to know?
We want to know the different in energy consumption after a fixed time slot, so I use E=f(t) to calculate dE=f(t₁)-f(t₀).
Since we apply periodic input here in this case, fixed time slots should be used corresponding with the period time to avoid scattering of results.

What, in your view, is represented by average dE/dt?

That's the gist of the whole story and we have to discuss it thoroughly. I'm working now on the analytical aspect of it and will post it as soon as I get ready. Of course, the analytical solution will be more rigorous but the conclusion will be the same as we have it now with the numerical example (I don't think the discrepancy we see in the numerical example is due to floating point errors). Therefore, discussing the physics of the situation is the central point.

Speaking of physics, I'm not sure that even purely pragmatically (f(T) - f(0))/T provides the true energy change per unit time for the whole cycle. That would have been true only if the E-t relationship had been linear, which it isn't. Of course, we know for sure what the energy is at time T because we've done the integral. That's granted. Now, you do understand, however, that it's not what we're asking. What we need to know is the rate of energy change throughout the cycle and we cannot assume (and it obviously isn't) that the rate of energy change has been the same at every time t within the period [0, T]. That is, we cannot assume that said rate has been linear, as is the assumption when calculating it through (f(T) - f(0))/T. Therefore, in order to determine the real rate of energy change for the period we have to average the individual rates (which in general differ from each other) at every moment of time t.

[Omnibus]

@Omnibus,
I had a look at the definition of the "slope" function of Excel.
It seems to me your opinion of "slope" in Excel is different from it's definition.
As far as I understand your understanding is that "slope" represents sigma(dE/dt)/nr of rows.
Excel implementation is different.

Excel Definition:
Returns the slope of the linear regression line through data points in known_y's and known_x's.
Mathematical formula is depicted below.

Well, but that's how I understand it too. Isn't this the formula from the least squares method?

[Omnibus]

@gyulasun,

The last data were taken at 100kHz, as you required. Also, the scope has calibration terminal and I've calibrated the probes. Will do it again but please bear with me because at this point I'm on a different wavelength (no pun intended) -- will have to straighten out the theoretical aspect of this exercise. As for the calibration, I'd rather doubt the Keithley calibration because it's an older instrument. All in all, what needs to be done is to find out if I can trade in this scope for a higher class model which has 11 bit accuracy. This research has to take off from the hobby level, as it is now as far as apparatus is concerned, and get into a more definitive ground. That's why I'm looking for labs with appropriate technique and expertise to see what they will obtain with their more precise instruments. So far it has been a dead end. Anyone I contacted doesn't even want to hear about this.

[teslaalset]

Well, but that's how I understand it too. Isn't this the formula from the least squares method?

Yes, it is.
Maybe we should simplify this case a bit further.
Can you try an excel with just a resister directly to a sinus generator and skip the capacitor. Things should be a bit more handier to handle.

[Omnibus]

Yes, it is.
Maybe we should simplify this case a bit further.
Can you try an excel with just a resister directly to a sinus generator and skip the capacitor. Things should be a bit more handier to handle.

Yeah, I did that several posts back -- just the resistor. I'll do more on that but now I'm trying to figure out the theoretical part. Interesting, we don't even need to integrate, the only thing need is to calculate the IV and I^2R average values.