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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 = 1.7243e-04 B = 1.3440e-10 C = -1.7243e-04 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 1.034e-4 coefficient..  Tags confused, ilt, solution Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Josh Algebra 2 November 13th, 2013 02:38 PM najaa Calculus 3 April 13th, 2012 03:41 AM Valar30 Advanced Statistics 2 September 14th, 2011 12:02 PM TreeTruffle Algebra 2 March 27th, 2010 02:22 AM mikeportnoy Applied Math 1 December 31st, 1969 04:00 PM

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