# impulse response

#### hamza324

Joined Jul 10, 2011
33
i want to know the impulse response to the following

4y''(t) = 2x(t) - x'(t)

I am able to solve this problem without the x'(t) on the right hand side but i am not sure how to deal with x'(t). The way i was told to this is to integrate it through "0 negative " to " 0 positive" thanks

#### tgotwalt1158

Joined Feb 28, 2011
110
I find a clue from my old differential equation recollection, might help you!

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#### hamza324

Joined Jul 10, 2011
33
thanks.this will help..

#### t_n_k

Joined Mar 6, 2009
5,455
Another approach might be using the Laplace Transform

$image=http://forum.allaboutcircuits.com/mimetex.cgi?4s%5E2Y%28s%29=2X%28s%29-sX%28s%29&hash=fd7d97071141d195f392f2aee04b8477$

or

$image=http://forum.allaboutcircuits.com/mimetex.cgi?H%28s%29=%5Cfrac%7BY%28s%29%7D%7BX%28s%29%7D=%28-%5Cfrac%7B1%7D%7B4%7D%29%5Cfrac%7B%28s-2%29%7D%7Bs%5E2%7D&hash=3ccfd8febb55e5288bb9ffe7e2b73e5a$

For which the impulse response is the inverse transform of H(s), giving the output y(t) as ....[for t>=0]

$$y(t)=\frac{t}{2}-\frac{1}{4}$$

Which agrees with tgotwalt1158's solution.

#### hamza324

Joined Jul 10, 2011
33

#### hamza324

Joined Jul 10, 2011
33
i solved another question for impulse response
y'' + 9 y= -6x'

after doing laplace i get

Y = -6s/(s^2 + 9)

and so y(t) =-6 cos3t .

is it correct..??

#### t_n_k

Joined Mar 6, 2009
5,455
That's correct