The given equation of Laplace transformation is:

\(X(s)=\frac{s+1}{(s+2)(s^2+2s+2)(s^2+4)}\)

Poles are the zeros of the denominator the equation of Laplace transformation

\(\Rightarrow (s+2)(s^2+2s+2)(s^2+4)=0\)

\(\Rightarrow s+2=0, s^2+2s+2=0 , s^2+4=0\)

So , the poles are : \(s=-2, -1+-i , \pm2i\)

To find the inverse Laplace transformation, use the partial fraction

\(X(s)=-\frac{1}{16(s+2)}+\frac{-7s-2}{80(s^2+4)}+\frac{3s+4}{20(s^2+2s+2)}\)

\(L^{-1}\left[X(s)\right]=L^{-1}\left[-\frac{1}{16(s+2)}\right]+L^{-1}\left[\frac{-7s-2}{80(s^2+4)}\right]+L^{-1}\left[\frac{3s+4}{20(s^2+2s+2)}\right]\)

Using the property of inverse Laplace transformation, we get

\(=\frac{-1}{16}L^{-1}\left[\frac{1}{(s+2)}\right]+\frac{-7}{80}L^-1\left[\frac {s}{(s^2+4)}\right]-\frac{1}{80}L^-1\left[\frac{2}{(s^2+4)}\right]+\frac{3}{20}L^-1\left[\frac {s}{(s+1)^2+1}\right]+\frac{1}{20}L^-1\left[\frac{4}{(s+1)^2+1}\right]\)

Using the formula of inverse Laplace transformation, we get

\(L^{-1}[\frac{1}{(s-a)}]=e^{at}\)

\(L^{-1}[\frac{s}{s^2+a^2}]=\cos(at)\)

\(L^{-1}[\frac{1}{s^2+a^2}]=\frac{1}{a \sin(at)}\)

\(L^{-1}[\frac{A(s-\lambda)+B}{(s-\lambda)^2+\mu^2}]=e^{\lambda t}(A\cos (\mu t)+B\sin (\mu t))\)

Part(b)

The given equation is:

\(X(s)=\frac{1}{(2s^2+8s+20)(s^2+2s+2)(s+8)}\)

So to find the poles set denominator =0

\((2s^2+8s+20)(s^2+2s+2)(s+8)=0\) poles are:

\(s=-8,-4\pm2i,-1\pmi\)

To find the inverse Laplace transformation use the partial fraction, we get

\(X(s)=\frac{3s+20}{600(s^2+8s+20)}+\frac{-9s+4}{1500(s^2+2s+2)}+\frac{1}{1000(s+8)}\)

Using the inverse Laplace properties and formula, we get

\(L^{-1}\left[X(s)\right]=L^{-1}\left[\frac{3s+20}{600(s^2+8s+20)}\right]+L^{-1}\left[\frac{-9s+4}{1500(s^2+2s+2)}\right]+L^{-1}\left[\frac{1}{1000(s+8)}\right]\)

\(L^{-1}[X(s)]=\frac{1}{200}L^{-1}\left[\frac{s+4}{(s+4)^2+4}\right]+\frac{1}{75}L^{-1}\left[\frac{1}{(s+4)^2+4}\right]+L^{-1}\left[\frac{-9s+4}{1500((s+1)^2+1)}\right]+L^-1\left[\frac{1}{1000(s+8)}\right]\)

Therefore, the required Laplace transformation is:

\(L^{-1}[X(s)]=\frac{1}{200}e^{-4t}\cos(2t)+\frac{1}{150}e^{-4t}\sin(2t)-\frac{3}{500}e^{-t}\cos(t)+\frac{13}{1500}e^{-t}\sin(t)+\frac{1}{1000}e^{-8t}\)

\(X(s)=\frac{s+1}{(s+2)(s^2+2s+2)(s^2+4)}\)

Poles are the zeros of the denominator the equation of Laplace transformation

\(\Rightarrow (s+2)(s^2+2s+2)(s^2+4)=0\)

\(\Rightarrow s+2=0, s^2+2s+2=0 , s^2+4=0\)

So , the poles are : \(s=-2, -1+-i , \pm2i\)

To find the inverse Laplace transformation, use the partial fraction

\(X(s)=-\frac{1}{16(s+2)}+\frac{-7s-2}{80(s^2+4)}+\frac{3s+4}{20(s^2+2s+2)}\)

\(L^{-1}\left[X(s)\right]=L^{-1}\left[-\frac{1}{16(s+2)}\right]+L^{-1}\left[\frac{-7s-2}{80(s^2+4)}\right]+L^{-1}\left[\frac{3s+4}{20(s^2+2s+2)}\right]\)

Using the property of inverse Laplace transformation, we get

\(=\frac{-1}{16}L^{-1}\left[\frac{1}{(s+2)}\right]+\frac{-7}{80}L^-1\left[\frac {s}{(s^2+4)}\right]-\frac{1}{80}L^-1\left[\frac{2}{(s^2+4)}\right]+\frac{3}{20}L^-1\left[\frac {s}{(s+1)^2+1}\right]+\frac{1}{20}L^-1\left[\frac{4}{(s+1)^2+1}\right]\)

Using the formula of inverse Laplace transformation, we get

\(L^{-1}[\frac{1}{(s-a)}]=e^{at}\)

\(L^{-1}[\frac{s}{s^2+a^2}]=\cos(at)\)

\(L^{-1}[\frac{1}{s^2+a^2}]=\frac{1}{a \sin(at)}\)

\(L^{-1}[\frac{A(s-\lambda)+B}{(s-\lambda)^2+\mu^2}]=e^{\lambda t}(A\cos (\mu t)+B\sin (\mu t))\)

Part(b)

The given equation is:

\(X(s)=\frac{1}{(2s^2+8s+20)(s^2+2s+2)(s+8)}\)

So to find the poles set denominator =0

\((2s^2+8s+20)(s^2+2s+2)(s+8)=0\) poles are:

\(s=-8,-4\pm2i,-1\pmi\)

To find the inverse Laplace transformation use the partial fraction, we get

\(X(s)=\frac{3s+20}{600(s^2+8s+20)}+\frac{-9s+4}{1500(s^2+2s+2)}+\frac{1}{1000(s+8)}\)

Using the inverse Laplace properties and formula, we get

\(L^{-1}\left[X(s)\right]=L^{-1}\left[\frac{3s+20}{600(s^2+8s+20)}\right]+L^{-1}\left[\frac{-9s+4}{1500(s^2+2s+2)}\right]+L^{-1}\left[\frac{1}{1000(s+8)}\right]\)

\(L^{-1}[X(s)]=\frac{1}{200}L^{-1}\left[\frac{s+4}{(s+4)^2+4}\right]+\frac{1}{75}L^{-1}\left[\frac{1}{(s+4)^2+4}\right]+L^{-1}\left[\frac{-9s+4}{1500((s+1)^2+1)}\right]+L^-1\left[\frac{1}{1000(s+8)}\right]\)

Therefore, the required Laplace transformation is:

\(L^{-1}[X(s)]=\frac{1}{200}e^{-4t}\cos(2t)+\frac{1}{150}e^{-4t}\sin(2t)-\frac{3}{500}e^{-t}\cos(t)+\frac{13}{1500}e^{-t}\sin(t)+\frac{1}{1000}e^{-8t}\)