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I'm attempting to better understand the relationship between step responses, impulse responses, and convolutions. Say that I have a system where if I apply a constant input, my output decays from a certain value to an equilibrium value. This gives me the step response of a system that I believe to be LTI based on the system's differential equation.

PDE:

$$\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}-\frac{u}{r^2}=\frac{1}{H_A k}\frac{\partial u}{\partial t}+\frac{1}{H_A k}\frac{r}{2}\frac{\partial \epsilon}{\partial t} $$

ODE in S Domain

$$\frac{\partial ^2 \bar{u}}{\partial \hat{r}^2}+ \frac{1}{\hat{r}}\frac{\partial \bar{u}}{\partial \hat{r}}-\frac{\bar{u}}{\hat{r}^2}-s\bar{u}=\frac{s\bar{\epsilon}\hat{r}}{2} $$

My specific step response is fairly complex and I went more in-depth into this specific system in another post, but a simpler version that still captures my issue could be given as:

$$ Y(t) = 1+e^{-0.1t} $$ enter image description here

In an LTI System, it's my understanding that the derivative of the step response yields the impulse response, which can then be used to obtain the output of the system for any arbitrary input through convolution.

$$ h(t) = \frac{dY(t)}{dt} = -0.1e^{-0.1t} $$

enter image description here

So now, if I have this impulse response, which is negative, and I attempt to convolve this with a unit step function to recover my original Y(t), I get something that is both the wrong shape and the wrong sign. Can anyone help point out my mistake?

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You're forgetting the discontinuity at $t=0$. Your step response is actually

$$Y(t)=(1+e^{-0.1t})u(t)\tag{1}$$

where $u(t)$ is the unit step function. Differentiating $(1)$ gives

$$h(t)=-0.1e^{-0.1t}u(t)+(1+e^{-0.1t})\delta(t)=-0.1e^{-0.1t}u(t)+2\delta(t)\tag{2}$$

where $\delta(t)$ is the Dirac delta impulse. Convolving $(2)$ with a unit step gives the original step response:

$$\begin{align}\int_{-\infty}^{\infty}h(\tau)u(t-\tau)d\tau&=-0.1u(t)\int_{0}^{t}e^{-0.1\tau}d\tau+2\int_{-\infty}^{t}\delta(\tau)d\tau\\ &=u(t)(e^{-0.1t}-1)+2u(t)\\&=(1+e^{-0.1t})u(t)\end{align}$$

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  • $\begingroup$ Thank you Matt, this is very helpful. Gave me a much clearer idea of how my system can be described by the impulse response. One quick question though: In (2), you simplify $$(1+e^{-0.1t})\delta(t)$$ to $$2\delta(t)$$ It's not immediately obvious to me how you did that simplification. Is it just that the delta is only "active" at t=0, and at t=0 the exponential is 1, so it's equivalent to the delta function itself? $\endgroup$ – Dillon Brown Sep 15 '17 at 12:37
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    $\begingroup$ @DillonBrown: yes, for any $f(t)$ that is continuous at $t=0$ you have $f(t)\delta(t)=f(0)\delta(t)$. $\endgroup$ – Matt L. Sep 15 '17 at 14:38

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