The inverse Laplace transform of H(s)=\frac{s+3}{s^{2}+2 s+1} for t \geq 0 is

Correct Answer is : 2te−t+e−t

GATE Electrical Engineering (EE) 2019 Solved Paper

Question : 15 of 65

Marks: +1, -0

The inverse Laplace transform of

H(s)=\;{s+3}/{s^{2}+2 s+1}

for

t ≥ 0

Solution:

H(s)=\;{s+3}/{(s+1)^{2}}=\;{s}/{(s+1)^{2}}+\;{3}/{(s+1)^{2}}

{t}^{ {n}} {e}^{--2 t} {u}( {t}) ↔ \;{ {n} !}/{( {s}+ {a})^{ {n}+1}}

{n}=1 ; {a}=1

{te}^{-t} {u}( {t}) ↔ \;{1}/{( {s}+1)^{2}}

.......(1)

3 {t} {e}^{- {t}} {u}( {t}) ↔ \;{3}/{( {s}+1)^{2}}

....(2)
Apply Time Differentiation property

\;{ {d}}/{ {dt}}[t e^{-t}] ⟷ \;{ {s}}/{( {s}+1)^{2}}

{e}^{-t}- {te}^{-t} ↔ \;{ {s}}/{( {s}+1)^{2}}

.....(3)

{e}^{-t}+2 {t} {e}^{-t} ↔ \;{ {s}+3}/{ {s}^{2}+2 {s}+1}

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