Question

Find the inverse Laplace transform of $G(s) = \dfrac{s}{s^{2}+2s+2}; \sigma >-1$.

Answer 1

Use the shifting property of Laplace transform to find the inverse Laplace transform.
If $\mathcal{L}(F(t)) = f(s)$, then by shifting property, $\mathcal{L}(e^{at}F(t)) = f(s-a)$.
For example, $$\mathcal{L}(e^{3t}\cos 2t) = \dfrac{s-3}{(s-3)^{2}+4}.

Complete step-by-step answer:
Given function is [G(s) = \dfrac{s}{s^{2}+2s+2}; \sigma >-1$. Simplify it into two terms as,$\begin{align*}\dfrac{s}{s^{2}+2s+2} &= \dfrac{s}{s^{2}+2s+1+1}\\ &= \dfrac{s}{(s+1)^{2}+1}\\ &= \dfrac{s+1-1}{(s+1)^{2}+1}\\ &= \dfrac{s+1}{(s+1)^{2}+1}-\dfrac{1}{(s+1)^{2}+1}\end{align*}$Now apply the inverse Laplace on both sides as,$\begin{align*}\mathcal{L}^{-1}(G(s)) &= \mathcal{L}^{-1}\dfrac{s+1}{(s+1)^{2}+1}-\dfrac{1}{(s+1)^{2}+1} \\ g(t) &= \mathcal{L}^{-1}\dfrac{s+1}{(s+1)^{2}+1} -\mathcal{L}^{-1}\dfrac{1}{(s+1)^{2}+1}\end{align*}$Now,$\mathcal{L}(\sin at) = \dfrac{a}{s^{2}+a^{2}}$and$\mathcal{L}(\cos at) = \dfrac{s}{s^{2}+a^{2}}$. Using the shifting property,$\mathcal{L}(e^{at}\cos bt) = \dfrac{s-a}{(s-a)^{2}+b^{2}}$and$\mathcal{L}(e^{at}\sin bt) = \dfrac{-a}{(s-a)^{2}+b^{2}}$, the inverse Laplace transform is given as,$\begin{align*}g(t) &= \mathcal{L}^{-1}\dfrac{s+1}{(s+1)^{2}+1} -\mathcal{L}^{-1}\dfrac{1}{(s+1)^{2}+1}\\ &= e^{-t}\cos t-e^{-t}\sin t\\ &= e^{-t}(\cos t-\sin t)\end{align*}$$

Note: Laplace transform satisfy the linear property, i.e.,
$\mathcal{L}{f(t)+g(t)} = \mathcal{L}{f(t)}+ \mathcal{L}{g(t)}$
Laplace transform is an integral transform that converts a function of a real variable to a function of a complex variable.