# Getting different impulse responses by solving with Laplace transform and plugging delta-dirac

If I know that the output of a system $$y(t) = 2e^{-3t} \cdot u(t)$$ and it's specified that the input is $$x(t) = u(t)$$

I was trying to find the impulse response h(t).

So I solved it using Laplace transform and I got the output $$h(t) = 2\delta(t) - [6 e^{-3t} \cdot u(t)]$$ by doing the following: $$y(t) = x(t) * h(t)$$ therefore $$Y(s) = X(s) \cdot H(s)$$ and by doing the laplace transform to y(t) and x(t) I optain: $$\frac{2}{s+3} = \frac{1}{s} \cdot H(s)$$ $$H(s) = \frac{2s}{s + 3}$$ $$H(s) = \frac{2(s + 3) - 6}{s + 3}$$ $$H(s) = 2 + \frac {-6}{s + 3}$$

therefore, by using the inverse Laplace transform, I got: $$h(t) = 2\delta(t) - [6e^{-3t} \cdot u(t)]$$

Until now I think I understand it. From now on, I'd like you to correct me if I'm wrong at anything.

I know that h(t), the impulse response is the output if the input is the impulse function delta.

so I did the following:
I substituted x(t) in place of u(t) in the output equation, as x(t) = u(t) $$y(t) = 2e^{-3t} \cdot x(t)$$ and then I applied delta as the input to get the impulse response.

$$h(t) = 2e^{-3t} \cdot \delta(t)$$ which can be reduced to $$h(t) = 2\delta(t)$$

I don't know what I've done wrong in that process to obtain a wrong answer in the second way.

Also, I'm pretty sure that the first impulse response is the right one because I've solved it in another way and obtained the same result (getting the step response and then differentiating it to obtain the impulse response.). It just didn't work out when I try to plug in delta.

Also, I apologize if the tags are not very accurate, but I just don't know how to categorize the question.

• This step is wrong $y(t)=2e−3t⋅x(t)$ it should be $y(t)=(h*x)(t)$. Convolution $\neq$ multiplication. – jDAQ Dec 10 '19 at 17:06
• Well, I know that convolution is not the same as multiplication. but all I did is that I substituted $u(t)$ and put $x(t)$ in its place, because $x(t) = u(t)$. – Tortellini Teusday Dec 10 '19 at 17:11
• Ok, but you have to looks at the subtleties, you cannot substitute $u(t)$ by $x(t)$ as you did. $y(t)=(h*x)(t)$ where $h(t) = 2\delta(t) - [6 e^{-3t} \cdot u(t)]$, so $y(t)$ would be... $y(t)=h(t)$, that is the whole point of impulse responses, right? – jDAQ Dec 10 '19 at 17:17

The step where you substituted $$\x(t)\$$ for $$\u(t)\$$ was incorrect. It's like saying "$$\x(t) = 1\$$, so I can substitute $$\x(t)\$$ for $$\1\$$ everywhere in my expression".
Observing that $$\h(t) \star u(t) = \int_0^t h(\tau) d\tau\$$ and then taking the derivative is correct, however.