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

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

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

inverse laplace

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

H (t) - H (t - 1)

Solution steps

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

Expand

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

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

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

= H (t) - H (t - 1)

\frac{d}{d x} (\frac{e ^{x}}{cos ( x )})

x^{3} - 4 x + lo g_{e} (x)

\frac{\partial}{\partial x} (- 3 x y^{2} + 2 y cos (x))

\frac{\partial}{\partial y} (2 (y + 1) x^{2})

\int - \frac{1}{2} x^{3} e^{- x} d x

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