La place inverse

**mahm32** · April 25th 2010, 02:19 PM

F(s) = s+1/(s^2 + 2s + 10)^2 Find the La place inverse transform

**mr fantastic** · April 25th 2010, 05:38 PM

Originally Posted by mahm32

F(s) = s+1/(s^2 + 2s + 10)^2 Find the La place inverse transform

You have $F(s) = \frac{s + 1}{([s + 1]^2 + 3^2)^2} = G(s + 1)$ where $G(s) = \frac{s}{(s^2 + 3^2)^2}$ .

From a well known theorem: $LT^{-1}[G(s + 1) = e^t g(t)$ where $g(t) = LT^{-1}[G(s)]$ .

To get g(t), first note that $G(s) = \frac{dK}{ds}$ where $K(s) = - \frac{1}{2} \left(\frac{1}{s^2 + 3^2} \right)$ .

Now use the well known theorem that $LT^{-1}\left[\frac{dK}{ds}\right] = - t k(t)$ where $k(t) = LT^{-1}[K(s)]$ .

**mahm32** · April 26th 2010, 11:19 AM

Thanks a lot .

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