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Answer to Question #261614 in Electrical Engineering for Mafizur Alam

Question #261614

Find the Inverse Laplace transform of X(s)= -s^2-2s+1/(s+2)(s+3)^2. ROC: Re{s}>-2

1
2021-11-06T01:37:09-0400

"L^{-1}\\left\\{\\frac{-s^2-2s+1}{\\left(s+2\\right)\\left(s+3\\right)^2}\\right\\}\\\\\n=L^{-1}\\left\\{\\frac{1}{s+2}-\\frac{2}{s+3}+\\frac{2}{\\left(s+3\\right)^2}\\right\\}\\\\\n\\mathrm{Use\\:the\\:linearity\\:property\\:of\\:Inverse\\:Laplace\\:Transform:}\\\\\n\\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\\}\\\\\n=L^{-1}\\left\\{\\frac{1}{s+2}\\right\\}-L^{-1}\\left\\{\\frac{2}{s+3}\\right\\}+L^{-1}\\left\\{\\frac{2}{\\left(s+3\\right)^2}\\right\\}\\\\\n=e^{-2t}-2e^{-3t}+e^{-3t}\\cdot \\:2t"

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