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Question No. 11 EEE | GATE 2019

The inverse Laplace transform of $ H(s)=\frac{s+3}{s^2+2s+1} $ for $ t\geq0 $ is


Answer : (C) $ 2te^{-t}+e^{-t} $


Solution of Question No 11 of GATE 2019 EEE Paper

Given, $ \mathrm H(\mathrm s)=\frac{\mathrm s+3}{\mathrm s^2+2\mathrm s+1}\mathrm{for}\;\mathrm t\geq0 $

$ \begin{array}{l}\Rightarrow\;\;\;\;\mathrm L^{-1}\left[\mathrm H(\mathrm s)\right]=\mathrm L^{-1}\left[\frac{\mathrm s+3}{\mathrm s^2+2\mathrm s+1}\right]=\mathrm L^{-1}\left[\frac{\mathrm s+3}{\left(\mathrm s+1\right)^2}\right]\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\mathrm L^{-1}\left[\frac{\mathrm s+1+2}{\left(\mathrm s+1\right)^2}\right]=\mathrm L^{-1}\left[\frac1{\mathrm s+1}+\frac2{\left(\mathrm s+1\right)^2}\right]\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\mathrm L^{-1}\left[\frac1{\mathrm s+1}\right]+2\mathrm L^{-1}\left[\frac1{\left(\mathrm s+1\right)^2}\right]=\mathrm e^{-1}\left(1\right)+2\mathrm e^{-\mathrm t}\mathrm t\\\Rightarrow\;\;\;\;\mathrm L^{-1}\left[\mathrm H(\mathrm s)\right]=2\mathrm{te}^{-\mathrm t}+\mathrm e^{-\mathrm t}\end{array} $

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