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Popular >

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

Solution

inverse laplace

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

Solution

H (t - 2) (- 1 + e^{t - 2})

Solution steps

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

Apply inverse transform rule:

if

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}{s ( s - 1 )}} (t - 2)

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

= H (t - 2) (- H (t - 2) + e^{t - 2})

H (t - 2) = {01 t < 2 t \geq 2

= {0 - 1 + e^{t - 2} t < 2 t \geq 2

= H (t - 2) (- 1 + e^{t - 2})

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Frequently Asked Questions (FAQ)

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