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Inverse Laplace Transform
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[QUOTE="roam, post: 5467678, member: 120460"] Thank you very much for this hint. I expanded the fraction and got: $$\frac{1}{8} \left( \frac{1}{s+2} + \frac{j}{4(s+2j)} - \frac{j}{4(s-2j)} \right)$$ now that we reduced the expression into a sum of standard forms, the only transform applicable in this case is ##\frac{1}{s-a} \iff e^{at}.## Therefore we obtain $$f(t) = \frac{1}{8} \left( e^{-2t} + \frac{j}{4} \left( e^{-2jt} + e^{2jt} \right) \right).$$ Is this correct? When you say you end up with a trigonometric function, do you mean that I should express the exponentials in the last equation as sines and cosines? [/QUOTE]
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Inverse Laplace Transform
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