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April 6th, 2015, 02:09 PM  #1 
Senior Member Joined: Oct 2014 From: EU Posts: 224 Thanks: 26 Math Focus: Calculus  Confused on the solution of an ILT
Hi all ! I am trying to compute the Inverse Laplace Transform of the transfer function: $\displaystyle {(5\cdot 10^6 + 1.1s) \over (1.034\cdot10^{4})s^3 + 470.4s^2 + 3\cdot10^{6}s}$ and the poles are: $\displaystyle \begin{aligned} & s = 0\\ & s_1 = 4542936.50 = \alpha\\ & s_2 = 6382.98 = \beta \end{aligned}$ I wrote down the system of equations: $\displaystyle \begin{aligned} & {(5\cdot 10^6 + 1.1s) \over (1.034\cdot10^{4})s^3 + 470.4s^2 + 3\cdot10^{6}s}\\ & {(5\cdot 10^6 + 1.1s) \over s(s + \alpha)(s + \beta)} = {A \over s} + {B \over (s + \alpha)} + {C \over (s + \beta)}\\ & (5\cdot 10^6 + 1.1s) = A(s^2 + s(\alpha + \beta) + \alpha\beta) + B(s^2 + \beta s)+ C(s^2 + \alpha s)\\ \\ & 0 = A + B + C \\ & 1.1 = A(\alpha + \beta) + B\beta + C\alpha\\ & 5\cdot 10^6 = A(\alpha\beta) \end{aligned}$ And applying Cramer I obtain: A = 1.7243e04 B = 1.3440e10 C = 1.7243e04 But when I put these numbers into wolfram alpha as input: partial fractions ( 5000000 + 1.1s )/(3000000s + 470.44s^2 + 0.0001034s^3) The result is quiet different.. $\displaystyle {1.66667\over s}{1.66667 \over (6385.97+s)}+{(1.09963×10^6) \over (4.54332×10^6+s)}$ I don't understand where I am wrong. Thank you in advance. Last edited by szz; April 6th, 2015 at 02:19 PM. 
April 7th, 2015, 07:30 AM  #2 
Senior Member Joined: Oct 2014 From: EU Posts: 224 Thanks: 26 Math Focus: Calculus 
I solved.. My error was not considering the 1.034e4 coefficient.. 

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