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# What is the impulse response?

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Hello,
I have a question regarding Network Analysis.
Suppose we are given that the response to a step input at t=0 to a network is given as say
i(t) = -2exp(-t) + 4exp(-3t)

Then what is the impule response of the network and how can it be calculated??
Any suggestions are welcome.

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That means that step input is :

| 1 , t>0
u(t)= |
| 0 , t<0

u(t)
|
1 |--------------------------
|___________________ t
0

Right?

INPUT : x(t)= u(t)
OUTPUT : y(t)= impule response (unkown?)

The relation between them is y(t)=h(t)*x(t)

but h(t) can't be calculated easily right away... so you work at the frequency domain, where:

INPUT : x(s)= L [u(t)] -> x(s)= 1/s (where L is Laplace tranform)
OUTPUT : y(s)=H(s)*x(s)

So IF you know (or can calculate) H(s) of your system then you can find y(s) becuase x(s)=1/s. Finally making reverse Laplace transform (L^-1) of y(s) you can find y(t) which is the IMPULSE RESPONSE:

y(t) = (L^-1)[y(s)]

Do you know H(s) or can u calculate it for your system?

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Well as I wrote here i(t) (given by me) is the response to u(t) therefore

i(t) = u(t)*h(t) where h(t) is the impulse response, so we can calculate h(t) in the frequency domain as:

H(s) = sI(s)

but this creates a confusion since if the response of u(t) is i(t) the response to D(t) (A unit impulse = du(t)/dt ) should be di(t)/dt which becomes in the s domain as:

sI(s) - i(0+) hence we get 2 different answers.

Stretching this to the basics we have Laplace of u(t) = 1/s so the Laplace for a unit impulse D(t) should be (from the differentiation rule) = s.(1/s) - u(0+) = 1 - 1 = 0. But its simply taken s.(1/s) = 1.
Why is this? Any help would be appreciated.

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