laplace

Discussion in 'Homework Help' started by hamza324, Sep 28, 2011.

Jul 10, 2011
33
1
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

2. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
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.

Jul 10, 2011
33
1

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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4. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
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).

Jul 10, 2011
33
1
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..

6. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
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: Sep 28, 2011
hamza324 likes this.