laplace

Thread Starter

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
5,447
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.
 

Thread Starter

hamza324

Joined Jul 10, 2011
33
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
 

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t_n_k

Joined Mar 6, 2009
5,447
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.:rolleyes:

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).
 

Thread Starter

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
5,447
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.
 
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