What is inverse oflaplace e^{-2s}(1/(4s)) ?

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

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inverse oflaplace e^{-2s}(1/(4s))

inverse laplace

e^{- 2 s} (\frac{1}{4 s})

\frac{1}{4} H (t - 2)

Solution steps

L^{- 1} {e^{- 2 s} (\frac{1}{4 s})}

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

L^{- 1} {\frac{1}{4 s}} : \frac{1}{4} H (t)

= H (t - 2) \frac{1}{4} H (t - 2)

= \frac{1}{4} H (t - 2)

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x \to 0 lim (\frac{( x )}{x})

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