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How to find $\mathcal{L^{-1}} \left[ \frac{F(s)}{s+a} \right]$where $F(s)$ is the Laplace transform of $f(t).$

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    The title is seriously misleading.2012-10-15

3 Answers 3

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Use laplace transform properties : $ G(s)=\frac{F(s)}{s+a} $ $ \mathcal{L^{-1}}[G(s)]=g(t) $ $ \mathcal{L} \left[ e^{at}g(t) \right] = G(s-a)=\frac{F(s-a)}{s} $ $ \frac{1}{s} \triangleq \int_0^t $ $ e^{at}g(t)=\int_0^t e^{a \tau }f( \tau ) d \tau $ $ g(t)=e^{-at} \int_0^t e^{a \tau }f( \tau ) d \tau $ or we can write : $ g(t)= \int_0^t e^{-a(t- \tau) }f( \tau ) d \tau $

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    Nice approach +1.2012-12-22
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Hint:

You know that $\mathcal{L}(f*g)=F(s)G(s)$ so $\mathcal{L^{-1}}\big(F(s)G(s)\big)=f*g$ wherein $f*g=\int_0^tf(\kappa)g(t-\kappa)d\kappa$.

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    You're doing well for MSE!2013-02-12
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If you know about convolution, this is just a piece of cake.

$\mathcal L^{-1}\left\{\dfrac{F(s)}{s+a}\right\}$

$=\mathcal L^{-1}\left\{\dfrac{1}{s+a}\right\}*\mathcal L^{-1}\{F(s)\}$

$=e^{-at}*f(t)$

$=\int_0^te^{-a(t-\tau)}f(\tau)~d\tau$

$=e^{-at}\int_0^te^{a\tau}f(\tau)~d\tau$