# Write down the qualitative form of the inverse Laplace transform of the following function. For each question first write down the poles of the function , X(s) a) X(s)=frac{s+1}{(s+2)(s^2+2s+2)(s^2+4)} b) X(s)=frac{1}{(2s^2+8s+20)(s^2+2s+2)(s+8)} c) X(s)=frac{1}{s^2(s^2+2s+5)(s+3)}

Write down the qualitative form of the inverse Laplace transform of the following function. For each question first write down the poles of the function , X(s)
a) $X\left(s\right)=\frac{s+1}{\left(s+2\right)\left({s}^{2}+2s+2\right)\left({s}^{2}+4\right)}$
b) $X\left(s\right)=\frac{1}{\left(2{s}^{2}+8s+20\right)\left({s}^{2}+2s+2\right)\left(s+8\right)}$
c) $X\left(s\right)=\frac{1}{{s}^{2}\left({s}^{2}+2s+5\right)\left(s+3\right)}$
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The given equation of Laplace transformation is:
$X\left(s\right)=\frac{s+1}{\left(s+2\right)\left({s}^{2}+2s+2\right)\left({s}^{2}+4\right)}$
Poles are the zeros of the denominator the equation of Laplace transformation
$⇒\left(s+2\right)\left({s}^{2}+2s+2\right)\left({s}^{2}+4\right)=0$
$⇒s+2=0,{s}^{2}+2s+2=0,{s}^{2}+4=0$
So , the poles are : $s=-2,-1+-i,±2i$
To find the inverse Laplace transformation, use the partial fraction
$X\left(s\right)=-\frac{1}{16\left(s+2\right)}+\frac{-7s-2}{80\left({s}^{2}+4\right)}+\frac{3s+4}{20\left({s}^{2}+2s+2\right)}$
${L}^{-1}\left[X\left(s\right)\right]={L}^{-1}\left[-\frac{1}{16\left(s+2\right)}\right]+{L}^{-1}\left[\frac{-7s-2}{80\left({s}^{2}+4\right)}\right]+{L}^{-1}\left[\frac{3s+4}{20\left({s}^{2}+2s+2\right)}\right]$
Using the property of inverse Laplace transformation, we get
$=\frac{-1}{16}{L}^{-1}\left[\frac{1}{\left(s+2\right)}\right]+\frac{-7}{80}{L}^{-}1\left[\frac{s}{\left({s}^{2}+4\right)}\right]-\frac{1}{80}{L}^{-}1\left[\frac{2}{\left({s}^{2}+4\right)}\right]+\frac{3}{20}{L}^{-}1\left[\frac{s}{\left(s+1{\right)}^{2}+1}\right]+\frac{1}{20}{L}^{-}1\left[\frac{4}{\left(s+1{\right)}^{2}+1}\right]$
Using the formula of inverse Laplace transformation, we get
${L}^{-1}\left[\frac{1}{\left(s-a\right)}\right]={e}^{at}$
${L}^{-1}\left[\frac{s}{{s}^{2}+{a}^{2}}\right]=\mathrm{cos}\left(at\right)$
${L}^{-1}\left[\frac{1}{{s}^{2}+{a}^{2}}\right]=\frac{1}{a\mathrm{sin}\left(at\right)}$
${L}^{-1}\left[\frac{A\left(s-\lambda \right)+B}{\left(s-\lambda {\right)}^{2}+{\mu }^{2}}\right]={e}^{\lambda t}\left(A\mathrm{cos}\left(\mu t\right)+B\mathrm{sin}\left(\mu t\right)\right)$
Part(b)
The given equation is:
$X\left(s\right)=\frac{1}{\left(2{s}^{2}+8s+20\right)\left({s}^{2}+2s+2\right)\left(s+8\right)}$
So to find the poles set denominator $=0$
$\left(2{s}^{2}+8s+20\right)\left({s}^{2}+2s+2\right)\left(s+8\right)=0$ poles are:
$s=-8,-4±2i,-1±i$
To find the inverse Laplace transformation use the partial fraction, we get
$X\left(s\right)=\frac{3s+20}{600\left({s}^{2}+8s+20\right)}+\frac{-9s+4}{1500\left({s}^{2}+2s+2\right)}+\frac{1}{1000\left(s+8\right)}$
Using the inverse Laplace properties and formula, we get