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Integrals of the form $$\int_{a}^bf(t)\delta^{(n)}(t-a)dt\tag{1}$$ with $\delta^{(n)}(t)$ being the $n^{th}$ generalized derivative of the Dirac delta impulse, are undefined. What is well-defined are integrals of the form $$\int_{a^+}^bf(t)\delta^{(n)}(t-a)dt\tag{2}$$ or $$\int_{a^-}^bf(t)\delta^{(n)}(t-a)dt\tag{3}$$ In $(2)$, the Dirac impulse (or its ...

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Just using your initial formula, you get $$Y = (-1)^2 \cdot \frac{\partial^2 }{\partial t^2}\left [ 3t^3+9 \right ]_{t=-1} =$$ $$\left [ 18 \cdot t \right ]_{t=-1}=-18$$ EDIT: per Matt's answer: this is only correct if the integration interval explicitly includes $-1$. If it's explicitly excluded, the answer would be $0$, it it's undefined, the answer is ...

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The answer (How to calculate the impulse response of an RC circuit using time-domain method) provides a direct time-domain solution of an RC circuit for the impulse reponse $h(t)$. Now this new answer modifies it to solve for the step-response $s(t)$ instead and then computes the impulse response according to : $$h(t) = s(t)'$$ The differential equation ...

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A simple way to derive the differential equation from the impulse response is to transform the latter to the frequency domain, rewrite the input/output relation, and then transform the resulting equation back to the time domain. The Laplace transform of the given impulse response $h(t)$ is \begin{align} H(s) &=\frac{A}{\tau}\frac{1}{\left(s+\frac{1}{\...

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