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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: 634 Thanks: 96 Math Focus: Electrical Engineering Applications Re: Inverse laplace transform Hi Rydog21, I believe that you need to use Euler's Formula: $e^{it}=\cos(t)+i\cdot \sin(t)$ 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: $\mathcal{L}^{-1}\left{\frac{5(-1+j)}{s-j500}+\frac{5(-1-j)}{s+j500}\right}$ Given that: $\mathcal{L}^{-1}\left{\frac{1}{s+\alpha}\right} \ \Leftrightarrow \ e^{-\alpha t} \$ we get: $\mathcal{L}^{-1}\left{\frac{5(-1+j)}{s-j500}+\frac{5(-1-j)}{s+j500}\right}=5(-1+j)e^{j500t}+5(-1-j)e^{-j500t}\$ [1] By Euler's formula, we can write $\ 5(-1+j) \$ as $\ 5\sqrt{2}\cdot e^{j\frac{3\pi}{4} \$ and we can write $\ 5(-1-j) \$ as $\ 5\sqrt{2}\cdot e^{-j\frac{3\pi}{4} \$ Substituting these into the right hand side of equation [1] we get: $5(-1+j)e^{j500t}+5(-1-j)e^{-j500t}=5\sqrt{2}\cdot e^{j\frac{3\pi}{4}}e^{j500t }+5\sqrt{2}\cdot e^{-j\frac{3\pi}{4}}e^{-j500t}$ $=5\sqrt{2}\cdot e^{j\left(500t + \frac{3\pi}{4}\right)}+5\sqrt{2}\cdot e^{-j\left(500t +\frac{3\pi}{4}\right)$ Using Euler's Formula again, this reduces to: $5\sqrt{2}\cdot \cos\left(500t+\frac{3\pi}{4}\right)+j\cdot 5\sqrt{2}\cdot \sin\left(500t+\frac{3\pi}{4}\right)+5\sqrt{2}\cdo t \cos\left(-500t-\frac{3\pi}{4}\right)+j\cdot 5\sqrt{2}\cdot \sin\left(-500t-\frac{3\pi}{4}\right)$ Since $\ \cos(x)=\cos(-x) \text{ and } \sin(x)=-\sin(-x) \$ the imaginary terms cancel and the real terms add giving: $\fbox{10\sqrt{2} \cdot \cos\left(500t+\frac{3\pi}{4}\right)}$ Note, of course, that $\ \frac{3\pi}{4} \$ could be replaced with $\ 135^o$
 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.

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