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Given the input $$x(t)=u(t)$$ and the corresponding output signal measured as $$y(t)= 2 e^{-3t} u(t)$$ determine the impulse response $h(t)$.

This what have done so far: $$ h(t)= \mathscr{L}^{-1} \left\{ \frac{Y(s)}{X(s)} \right\} = \frac{2/(s+3)}{1/s} = \frac{2s}{s+3} $$.

I need to find the Laplace inverse of this, I can't figure out the approach.

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Your approach is correct. Rewrite $H(s)$ as

$$H(s)=\frac{2s}{s+3}=2\frac{s+3-3}{s+3}=2-\frac{6}{s+3}\tag{1}$$

and use basic Laplace transform identities to obtain $h(t)$ from $(1)$.

Note that you don't need to use the Laplace transform. A time domain approach as suggested in oxuf's answer is even more straightforward.

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When feeding a filter with a unit step, you are actually integrating the impulse response of the filter. So

$$ h(t) = \frac{dy(t)}{dt} = -6e^{-3t}\cdot u(t) + 2e^{-3t}\cdot \delta(t) = -6e^{-3t}\cdot u(t) + 2 \delta(t) $$

is its impulse response.

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    $\begingroup$ It's true that the step response is the integrated impulse response, but your expression for $h(t)$ is not the derivative of the given step response. $\endgroup$ – Matt L. Jun 26 '17 at 16:17
  • $\begingroup$ Opss, you are right, I forgot the chain rule. Please, check now. $\endgroup$ – oxuf Jun 26 '17 at 16:27
  • $\begingroup$ That looks right. $\endgroup$ – Matt L. Jun 26 '17 at 16:31

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