Unit step response and unit impulse response

[AlonsoMcLaren]

In my ODE course I learned that for a time invariant operator [tex] p(D)[/tex], if [tex] p(D)x=q(t)[/tex] then [tex] p(D)\dot{x}=\dot{q}(t)[/tex].

Then the professor "justified" that the derivative of the unit step response, v(t), is the unit impulse response, w(t), because the derivative of unit step function is the Dirac delta function.

However, w(t) and v(t) requires rest initial condition. There is no consideration of initial condition in the "justification" he gave.

Any thoughts?

 

[BackEMF]

I'm assuming you're disturbed by the fact that there's a jump discontinunity in the Dirac delta distribution* and Heaviside unit step? And you should be! Normal derivatives of normal functions do not exist at such points.

The answer is to extend to Generalised Functions and Distributions - you can't get an answer to your question with understanding this. Often you can ignore these issues in practical situations however.

*I haven't called this the Dirac delta function becuase it technically is not a function, it is a generalised function. But in general this technicality is dispensed with and most literature will actually refer to it as the Dirac delta function.