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I'm trying to understand impulse and unit step responses, but I'm not sure if my understanding is correct. In general, are the following statements true:

a) The impulse response for a system $x(t) \mapsto y(t)$ is $y(t)$ when $x(t) = \delta(t)$ where $\delta$ is the Dirac delta, and

b) The unit step response for a system $x(t) \mapsto y(t)$ is $y(t)$ when $x(t) = u(t)$ where $u$ is the unit step (Heaviside step function)

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    $\begingroup$ That is correct. $\endgroup$ – Brian Sep 22 '15 at 14:12
  • $\begingroup$ Wow, I'm almost shocked - all the resources I consulted made this sound so much more complicated than it actually is. $\endgroup$ – user2398029 Sep 22 '15 at 15:22
  • $\begingroup$ I'm not sure this is correct (it may be just that I'm misunderstanding the notation). If the input to an LTI system is $x(t)$, then the output is $y(t)=x(t) \ast h(t)$, where $h(t)$ is the system's impulse response. To me this is not the same as statement a) in the question. $\endgroup$ – MBaz Sep 22 '15 at 15:39
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In response to MBaz's comment, if we define the impulse response (edit) for Linear Time Invariant systems as

"the function $h(t)$ such that the mapping $x(t) \mapsto y(t)$ is computed as $y(t) = x(t) * h(t)$"

then notice that if $x(t) = \delta(t)$ you get $y(t) = \delta(t) * h(t) = h(t)$.

Thus louism's statement a) and MBaz's comment agree.

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  • $\begingroup$ Yes, as long as it's made clear that this is only applicable to LTI systems (this was not specified in the question). $\endgroup$ – MBaz Sep 22 '15 at 16:48
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    $\begingroup$ Agreed. And to make it explicit, louism's definition in the original post is the general definition of "impulse response" (even for non-LTI systems). The impulse response of a system is generally only of interest for LTI systems because for LTI systems, the output of the system can be written as a convolution of the input and the impulse response. This last statement is your definition. $\endgroup$ – Austin A. Sep 22 '15 at 17:04

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