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If the step response of a system is given as $$c(t)=(10e^{-5t}-2e^{-10t}+1)u(t)$$ then its impulse response is _______?

My Approach:
we know: $$h(t)= \frac {d}{dt} c(t)$$ Process(1): $$h(t)= \{ 10e^{-5t}(-5) - 2e^{-10t}(-10)+0 \} u(t)$$ $$\implies h(t)= \{ -50e^{-5t} + 20e^{-10t} \} u(t)$$ Process(2): $$h(t)= \{ 10e^{-5t}(-5) - 2e^{-10t}(-10)+ 0 \} u(t) + (10e^{-5t}-2e^{-10t}+1) \delta (t)$$ $$\implies h(t)= \{ -50e^{-5t} + 20e^{-10t} \} u(t) + 9 \delta (t)$$

so which process is correct? please help...

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HINT:

For $t\neq 0$ both solutions are identical and correct. So you only need to consider $t=0$ to find out which of the two solutions is correct. Does the step response jump at $t=0$ or is it continuous? What would a jump at $t=0$ imply for the impulse response? Alternatively, you should try to compute the step responses corresponding to the two impulse responses you obtained. Only one of them will be identical to the given step response.

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  • $\begingroup$ Hi Matt and user9198116. I understand the first approach. Could you shed light on the second one. I don't follow what's being done there. Thanks. $\endgroup$ – mark leeds Mar 14 '18 at 8:17
  • $\begingroup$ @markleeds: Take a look at this answer. You need to use the product rule taking into account that the derivative of the step $u(t)$ is the Dirac delta impulse $\delta(t)$. $\endgroup$ – Matt L. Mar 14 '18 at 8:46

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