The answer is simple: it has absolutely nothing to do with the instrument, but the signal. Perform the measurement correctly (consistently -- measuring the step response
with a step!) and you will observe the rated spec of the instrument.
Quite frankly, Dave is most likely following the same rules of thumb as anyone else. I expect he hasn't, for example, worked out the impulse response and dispersion relations of typical filter types, and derived the rules independently and accurately -- he's much too lazy (= good) of an engineer to do that! Moreover, given his lack of enthusiasm for higher math(s), he probably doesn't know how to work that out (or very quickly), anyway. (No offense, if you're reading, Dave -- I hope I'm not underestimating your knowledge level here.)
And to be fair, I don't know either, but, I'm not afraid of working a little calculus to figure things out, and if I were charged with finding it, I feel confident I could.
I guess if you don't have a solid footing in filter and signals theory, and Fourier analysis, there isn't much I can quickly tell you that will convey this, other than to continue to assert that I am right (that the signals are different, and incomparable by the same rule of thumb, which depends upon gross assumptions). Are you incapable of taking that in confidence?
As for physical analogies, we can work with that. Acceleration to velocity is a single pole filter (namely, an ideal integrator), and acceleration with drag is a single pole filter at some frequency (where the cutoff frequency is given by mass and drag coefficient). Suppose we label acceleration as input, and velocity as output.
The pulse is not simply a "rocket going up", but a proper impulse, such as a rocket engine which burns out much faster than the drag time constant, or, say, the blast wave from an explosion. The rocket instantaneously accelerates (it's able to, because this is a single pole system; aside from the few scopes with deflection plates tied directly to the input transmission line, this is not representative of scopes, though), then gradually slows down at a rate determined by that time constant.
The risetime is therefore zero, because it can accelerate instantly from a stop, limited only by the rate of the impulse itself. FWHA is proportional to one fall time.
Whereas a step applies variable acceleration, so as to achieve a certain steady-state value (velocity, in this case). If we apply a constant acceleration, it will eventually balance against the drag force, so the total acceleration goes to zero, and the velocity stabilizes to a constant. The 10-90% risetime is something like 2.2 time constants (assuming the usual exponential single pole response).
If it were a two pole filter of some type (like critical damping), the impulse response won't rise instantly, and the result will be a hump that's going up and down completely on its own, with no force from the impulse as it goes. Whereas the step has to continue dragging through the whole range, like a tsunami coming in.
Oh, it may also help just to realize the relationship between ideal step and impulse responses: impulse is the derivative of step. The entire rising edge of the step is the entire width of the impulse. If the 'corners' are symmetrical (such as a Gaussian passband has), 10-90% points on the step correspond to 10-10% points on the impulse! (Or something like that.) The 10-90% points on the impulse are only the beginning of the toe of the entire step.
Tim
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