What is inverse oflaplace (e^{-s})/((s+1)^2+1) ?

The answer to inverse oflaplace (e^{-s})/((s+1)^2+1) is \H(t-1)e^{-t+1}sin(t-1)

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inverse oflaplace (e^{-s})/((s+1)^2+1)

inverse laplace

\frac{e ^{- s}}{( s + 1 ) ^{2} + 1}

H (t - 1) e^{- t + 1} sin (t - 1)

Solution steps

L^{- 1} {\frac{e ^{- s}}{( s + 1 ) ^{2} + 1}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

= H (t - 1) L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 1}} (t - 1)

L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 1}} : e^{- t} sin (t)

= H (t - 1) e^{- (t - 1)} sin (t - 1)

= H (t - 1) e^{- t + 1} sin (t - 1)

What is inverse oflaplace (e^{-s})/((s+1)^2+1) ?
The answer to inverse oflaplace (e^{-s})/((s+1)^2+1) is \H(t-1)e^{-t+1}sin(t-1)