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 October 13th, 2013, 09:19 AM #1 Newbie   Joined: Oct 2013 Posts: 13 Thanks: 0 Inverse laplace transform Hi, I am trying to do the inverse Laplace transform for this equation I arrived at with partial fractions. Two of the terms have complex numbers in them and I see that they are simplified to Cos and Sin terms from their exponential form (the process is quoted as 'Eulers identity'). Could somebody explain how this process is carried out? L^-1 : 5*(-1+j)/(s+j500) and 5*(-1-j)/(s+j500) The inverse of these I have as (-1+j)e^(j500*t) and (-1-j)e^(-j500*t) respectively (I hope that is correct!) Where my total equation is equal to these two terms and one other non-complex exponential term added together (it is for current in a resistor inductor circuit with a sinusoidal source voltage). Many thanks, Ryan October 13th, 2013, 06:23 PM #2 Senior Member   Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications Re: Inverse laplace transform Hi Rydog21, I believe that you need to use Euler's Formula: I also think that there is a mis-statement in your problem. The reason is that the denominators are both the same but the signs of the exponents in the Inverse Laplace Transform are opposite. I will assume that you meant to find: Given that: we get:  By Euler's formula, we can write as and we can write as Substituting these into the right hand side of equation  we get: Using Euler's Formula again, this reduces to: Since the imaginary terms cancel and the real terms add giving: Note, of course, that could be replaced with October 15th, 2013, 10:35 AM #3 Newbie   Joined: Oct 2013 Posts: 13 Thanks: 0 Re: Inverse laplace transform Thanks for that, I understand now. You're the man. Tags inverse, laplace, transform ### eulers identity inverse laplace

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