What is inverse oflaplace (e^{-pis})/(s^2+4) ?

The answer to inverse oflaplace (e^{-pis})/(s^2+4) is \H(t-pi) 1/2 sin(2(t-pi))

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inverse oflaplace (e^{-pis})/(s^2+4)

inverse laplace

\frac{e ^{- π s}}{s ^{2} + 4}

H (t - π) \frac{1}{2} sin (2 (t - π))

Solution steps

L^{- 1} {\frac{e ^{- π s}}{s ^{2} + 4}}

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 - π) L^{- 1} {\frac{1}{s ^{2} + 4}} (t - π)

L^{- 1} {\frac{1}{s ^{2} + 4}} : \frac{1}{2} sin (2 t)

= H (t - π) \frac{1}{2} sin (2 (t - π))

What is inverse oflaplace (e^{-pis})/(s^2+4) ?
The answer to inverse oflaplace (e^{-pis})/(s^2+4) is \H(t-pi) 1/2 sin(2(t-pi))