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

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

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

inverse laplace

\frac{e ^{1 - s}}{1 - s}

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

Solution steps

L^{- 1} {\frac{e ^{1 - s}}{1 - s}}

Apply inverse transform rule:

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

then

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

For

\frac{e ^{1 - s}}{1 - s} : a = 1, F (s) = \frac{e ^{- 1 s}}{- 1 s}

= e^{t} L^{- 1} {\frac{e ^{- 1 \cdot s}}{- 1 \cdot s}}

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

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

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