Find inverse Laplace transform of:
G(s)= S / (S+1) (S+1+j1)^2 (S+1-j1)^2
Solve step by step and briefly.
L−1{ss2+4s+5}=L−1{s+2(s+2)2+1−2⋅1(s+2)2+1}Use the linearity property of Inverse Laplace TransformFor functions f(s), g(s) and constants a, b:L−1{a⋅f(s)+b⋅g(s)}=a⋅L−1{f(s)}+b⋅L−1{g(s)}=e−2tcos(t)−2e−2tsin(t)L^{-1}\left\{\frac{s}{s^2+4s+5}\right\}\\ =L^{-1}\left\{\frac{s+2}{\left(s+2\right)^2+1}-2\cdot \frac{1}{\left(s+2\right)^2+1}\right\}\\ Use\:the\:linearity\:property\:of\:Inverse\:Laplace\:Transform\\ \mathrm{For\:functions\:}f\left(s\right),\:g\left(s\right)\mathrm{\:and\:constants\:}a,\:b:\quad L^{-1}\left\{a\cdot f\left(s\right)+b\cdot g\left(s\right)\right\}=a\cdot L^{-1}\left\{f\left(s\right)\right\}+b\cdot L^{-1}\left\{g\left(s\right)\right\}\\ =e^{-2t}\cos \left(t\right)-2e^{-2t}\sin \left(t\right)L−1{s2+4s+5s}=L−1{(s+2)2+1s+2−2⋅(s+2)2+11}UsethelinearitypropertyofInverseLaplaceTransformForfunctionsf(s),g(s)andconstantsa,b:L−1{a⋅f(s)+b⋅g(s)}=a⋅L−1{f(s)}+b⋅L−1{g(s)}=e−2tcos(t)−2e−2tsin(t)
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