May 4th, 2014, 09:18 PM  #1 
Newbie Joined: May 2014 From: Australia Posts: 1 Thanks: 0  Inverse Laplace Transform
Find the inverse Laplace transform of: (3s + s*exp(4s) +2*exp(4s) +s^3)/((s^2)(s1)(s+1))  Suggest using partial fractions (This is where I had trouble). Last edited by Bkalma; May 4th, 2014 at 09:25 PM. 
May 6th, 2014, 07:08 PM  #2 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 626 Thanks: 90 Math Focus: Electrical Engineering Applications 
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 $\displaystyle \large{e^{4s}}$: $\displaystyle \large{\frac{3s+s\cdot e^{4s}+2\cdot e^{4s}+s^3}{s^2(s1)(s+1)}=\underbrace{\frac{s^3+3s}{s^2(s1)(s+1)}}_{\text{term 1}}+\underbrace{\frac{(s+2)}{s^2(s1)(s+1)}}_{\text{term 2}}\cdot e^{4s}}$ 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: $\displaystyle \large{f(tt_0)\cdot u(tt_0) \Leftrightarrow e^{st_0}F(s)}$ 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: $\displaystyle \large{\frac{s^3+3s}{s^2(s1)(s+1)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s1}+\frac{D}{s+1}}$ Multiply through by the denominator: $\displaystyle \large{s^3+3s=A(s)(s1)(s+1)+B(s1)(s+1)+Cs^2(s+1)+Ds^2(s1)}$ [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: $\displaystyle \large{\frac{s^3+3s}{s^2(s1)(s+1)}=\frac{3}{s}+\frac{2}{s1}+\frac{2}{s+1}}$ Are you able to finish the problem? Again, if you get stuck, feel free to reply. 

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inverse, laplace, transform 
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