May 4th, 2014, 10:18 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 10:25 PM. 
May 6th, 2014, 08:08 PM  #2 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 634 Thanks: 96 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. 

Tags 
inverse, laplace, transform 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Inverse laplace transform  Rydog21  Calculus  2  October 15th, 2013 10:35 AM 
inverse laplace transform  capea  Complex Analysis  5  August 3rd, 2013 03:30 PM 
Laplace tranform and inverse of Laplace transform  Deiota  Calculus  1  April 28th, 2013 11:28 AM 
inverse laplace transform  defunktlemon  Real Analysis  1  April 14th, 2012 03:19 PM 
Inverse Laplace Transform  timh  Applied Math  1  June 7th, 2010 07:44 PM 