I think we are missing the meaning behind the impulse response here and whether or not we can create very narrow pulses to imitate the ideal impulse.
Whether or not something is an impulse does not really depend on the shape of the waveform but what its duration is with respect to the system it is being appllied. Let me explain.
Let's say you have a simple RC filter and you want to find it impulse response. To do this, you have to apply an impulse voltage at the input and measure its response (the impluse response) across the capacitor.
But how narrow should you make the impulse width? Well, let's say the time constant of the RC circuit is 1 second (RC=1). Then, if you make the width of the pulse to be \[1\mu s\], then for all practical purposes, that pulse will feel like an impulse to the RC circuit.
Now the other thing to consider is the \[weight\] of the impulse, that is the are under the pulse itself. If you applied a voltage of 5V with \[1 \mu s\] duration, then the weight of the impulse is \[5 \mu V \cdot s\]. Note that the units of V-s, is actually flux, which may also be expressed in Webers (Wb).
Here's the most important idea. As long as you apply a pulse that is much, much less than the time constant of the circuit and the applied waveform has the same area, the response will be the same. What I am trying to say is that it is the area of the curve that is important, not necessarily the shape.
Try this with a circuit simulator. Apply a square pulse, triangular pulse, anything that you like. Just make sure the are is the same and the time duration is small. Look at the output of your RC circuit -- amazing!
Best regards,
v_c