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Consider a signal $X$ filtered by a kernel $p$ with finite support $[t_0,t_1]$ and $\int_{t_0}^{t_1}p(t)\,\text{d}t = 1$, yielding the response function

$$\overline{X}(T) = \int_{t_0}^{t_1} X(T + t)\ p(t)\ \text{d}t$$

The response function for a step signal $X(t) = H(t)$ with $H$ being the Heaviside step function immediately gives $p(t)$:

$$\overline{X}(T) = \int_{t_0}^{t_1} H(T + t)\ p(t)\ \text{d}t = \int_{t_0+T}^{t_1+ T} H(t)\ p(t-T)\ \text{d}t \\= \int_{0}^{t_1+ T} p(t-T)\ \text{d}t = P(t_1) - P(-T)$$

Here $P$ is the anti-derivative of $p$ with $P(t) = \int_{-\infty}^t p(t')\ \text{d}t'$ so with $P(t_1) = 1$ we find $P(T) = 1 - \overline{X}(-T)$ and finally

$$p(t) = \overline{X'}(t)$$

I wonder if this finding has a specific name (that I can google for), and if it is worth being explicitly mentioned in textbooks (e.g. on signal processing). Or on the other hand, if it is too elementary or even trivial to be mentioned. (I have searched a bit and did not find it explicitly mentioned.)

Hans-Peter Stricker
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    This will probably get better answers at https://dsp.stackexchange.com/ – mhdadk Dec 28 '21 at 12:37
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    This is so basic that it's often just used automatically. See the derivation immediately below "intuition from probability" in my post at https://stats.stackexchange.com/a/43075/919 for one existing demonstration here on CV (which is completely general and rigorous, covering non-continuous distributions, too). – whuber Dec 28 '21 at 13:59
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    Answered and accepted [here](https://dsp.stackexchange.com/a/80837/4298) at dsp.SE. – Matt L. Jan 01 '22 at 14:28

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