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May 4th, 2014, 09:18 PM   #1
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Question Inverse Laplace Transform

Find the inverse Laplace transform of:

(3s + s*exp(-4s) +2*exp(-4s) +s^3)/((s^2)(s-1)(s+1))

- Suggest using partial fractions (This is where I had trouble).

Last edited by Bkalma; May 4th, 2014 at 09:25 PM.
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May 6th, 2014, 07:08 PM   #2
jks
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Hi Bkalma, and welcome to the forums. There may be a more elegant way to work this problem, but I took the simple approach of factoring out the $\displaystyle \large{e^{-4s}}$:

$\displaystyle \large{\frac{3s+s\cdot e^{-4s}+2\cdot e^{-4s}+s^3}{s^2(s-1)(s+1)}=\underbrace{\frac{s^3+3s}{s^2(s-1)(s+1)}}_{\text{term 1}}+\underbrace{\frac{(s+2)}{s^2(s-1)(s+1)}}_{\text{term 2}}\cdot e^{-4s}}$

The idea, as you can probably tell, is to use partial fractions on term 1 and term 2. Then for term 2, use the Laplace Transform pair:

$\displaystyle \large{f(t-t_0)\cdot u(t-t_0) \Leftrightarrow e^{-st_0}F(s)}$

to complete the solution.

I will work through the partial fraction solution for term 1 and leave term 2 and the final solution for you to complete (but if you get stuck, feel free to reply).

So for term 1:

$\displaystyle \large{\frac{s^3+3s}{s^2(s-1)(s+1)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s-1}+\frac{D}{s+1}}$

Multiply through by the denominator:

$\displaystyle \large{s^3+3s=A(s)(s-1)(s+1)+B(s-1)(s+1)+Cs^2(s+1)+Ds^2(s-1)}$ [1]

Letting s=0 in [1] we get:

$\displaystyle \large{0=B(-1)(1) \rightarrow B=0}$

Letting s=1 in [1] we get:

$\displaystyle \large{4=C(1)(2) \rightarrow C=2}$

Letting s=-1 in [1] we get:

$\displaystyle \large{-4=D(1)(-2) \rightarrow D=2}$

Since every variable except for A is known, let's just plug in s=2 in [1] to find A:

$\displaystyle \large{8+6=A(2)(1)(3)+0+2(2^2)(3)+2(2^2)(1) \rightarrow A=-3}$

So:

$\displaystyle \large{\frac{s^3+3s}{s^2(s-1)(s+1)}=-\frac{3}{s}+\frac{2}{s-1}+\frac{2}{s+1}}$

Are you able to finish the problem? Again, if you get stuck, feel free to reply.
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