i just want to verify the answer i am getting is correct or not.

-y'' + 3y' = 3x + 5x''

now

h(t) = - 18 e^(-3t) - 5


using some other method i got
h(t) = - 16 e^(-3t) - 3

can some bdy help me out plzz.
thnx

 

Using Laplace

\(-y^{''}+3y^'=3x+5x^{''}\)

Transforming

\(-s^2Y(s)+3sY(s)=3X(s)+5s^2X(s)\)

\(\frac{Y(s)}{X(s)}=\frac{5s^2+3}{-s^2+3s}\)

\(\frac{Y(s)}{X(s)}=\frac{-5s^2-3}{s(s-3)}\)

\(\frac{Y(s)}{X(s)}=\frac{-5s}{s-3}-\frac{3}{s(s-3)}\)

Inverse Transform for impulse response h(t)

\(h(t)=-5(\delta(t)+3e^{3t})+(1-e^{3t})\)

\(h(t)=-5\delta(t)+1-16e^{3t}\)

which is an unusual result in that the function increases unbounded with time. Are you sure of the original equation?

Clearly you now have three different results.

 

Hi i got your answer.....

but my professor told me slightly different procedure....

i am getting the answer above except the impulse function in the asnwer..

i.e h(t) = - 16 e^(3t) + 1 but missing that impulse thing..

i have attached the file...plz give it a look

 

i am getting the answer above except the impulse function in the asnwer..

i.e h(t) = - 16 e^(3t) + 1 but missing that impulse thing..

Yes - blink and you'll miss it.

One would have to ensure that the boundary conditions are met. At t=0 your answer is h(0)=0. What will you see in practice? I suspect an impulse output y(t) at t=0 is likely, due to the presence of the second derivative of the input x(t).

 

i did the changed at h(0) at t=0 but answer isn't matching..

but i understand your point so i think i would have to add this impulse at the end once i get the h(t) = - 16 e^(3t) + 1 expression.....am i right..

 

By definition, the delta impulse function only exists at t=0. Elsewhere it is zero so the function reduces to

h(t) = - 16 e^(3t) + 1

This is still an increasing value function with time given the positive exponential index.

 

thanks a lot for your help...