# laplace

#### hamza324

Joined Jul 10, 2011
33
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

#### t_n_k

Joined Mar 6, 2009
5,455
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.

#### hamza324

Joined Jul 10, 2011
33

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

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#### t_n_k

Joined Mar 6, 2009
5,455
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).

#### hamza324

Joined Jul 10, 2011
33
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..

#### t_n_k

Joined Mar 6, 2009
5,455
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.

Last edited:
• hamza324