thanks rstofer, in the meantime I analyze d / dt (as soon as I archive this I read well your intervention 674 for the integrated)
I'm starting to understand something, and I also learned what slew rate is in op.amp datasheets
If I look at your test with the Rigol, I see that if you enable the diff function, the display shows you the value that is 1MV / s, and everything makes sense.
Now look at my attached image: usual compensation signal, as you can see the square wave values are 3.12v and a rise time of 984ns.
As you taught me, the d / dt should be: 3.12v / 0.000000984s = 3170731V / s
As you can see, the only data he shows me related to d / dt are the vertical divs (100kv / s) and the position on the display.
What good can this view be for me, if it doesn't even tell me the d / dt value? or maybe I'm missing something?
In your Rigol it makes sense, you enable the diff and it indicates the V / s relative to the ramp rise time of the signal ... but in my Siglent what do I do with this d / dt graph?
I thought your rise time on that compensation signal was a few microseconds, nowhere near nanoseconds. Are you sure?
Notice how your dv/dt trace is broad at the bottom leading and trailing edge. Assuming there is any basis in reality, that tells me that the leading and trailing edges of the source are rounded, the derivative is not yet very high. This would be consistent with a slower rise time. I would expect to see a spike just one pixel wide for a very fast rise time because the derivative dV/dt gets very high when dt approaches zero. In fact, that's what this is all about. dt - how fast is something happening.
In the Integral post, I talk about taking the integral of sin(t) and getting -cos(t). Here, with the derivative function, if you plug in sin(t), you should get cos(t) so you should see a couple of traces like my integral image except the cos(t) will be rightside up. It will start at a high positive value and swing down to a matching low value.
I didn't get a useful display on my Rigol because the sine wave coming in isn't a pure sine, it is a synthesized sine. As a result, there is some noise as it changes steps and the derivative function sees those steps as a change in voltage over time and posts it as a derivative. Mathematically, this is correct and it is also the reason that analog computing doesn't use differentiators, they emphasize the noise.
Even though analog computing is used to solve (or model) differential equations, they don't use differentiators. Lord Kelvin came up with the answer and his genius shows up in (1) and (2) in this paper. Integrate until the derivatives disappear! Then close the feedback loop. This is so much easier than trying to come up with a solution using a slide rule! How come they don't teach this stuff in college?
http://chalkdustmagazine.com/features/analogue-computing-fun-differential-equations/Using the diff function on a scope is one place where I wish I had a 'pure' sine wave.