1. Find the Laplace transform of e^(-2t) sinηπt please help. thx
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Originally Posted by ghazala 1. Find the Laplace transform of e^(-2t) sinηπt please help. thx Where are you stuck? Theorem: $\displaystyle LT\left[ e^{at} f(t) \right] = F(s - a)$ where $\displaystyle F(s) = LT[f(t)]$ In your case a = -2 and $\displaystyle f(t) = \sin (nt)$. You should be able to look up a table of Laplace transforms to get F(s) .....
yup thts wht i did.. i got this answer: (ηπ + 2)/ (ηπ + 2)^2 + s^2 i think i made a mistake smwhere :S
Originally Posted by ghazala yup thts wht i did.. i got this answer: (ηπ + 2)/ (ηπ + 2)^2 + s^2 i think i made a mistake smwhere :S Laplace transform of $\displaystyle f(t) = \sin (nt)$ is $\displaystyle F(s) = \frac{n}{s^2 + n^2} \, ....$
thank you
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