What is inverse oflaplace (12e^{-s})/(s(10s+1)) ?

The answer to inverse oflaplace (12e^{-s})/(s(10s+1)) is \H(t-1)(12-12e^{-(t-1)/(10)})

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inverse oflaplace (12e^{-s})/(s(10s+1))

inverse laplace

\frac{12 e ^{- s}}{s ( 10 s + 1 )}

H (t - 1) (12 - 12 e^{- \frac{t - 1}{10}})

Solution steps

L^{- 1} {\frac{12 e ^{- s}}{s ( 10 s + 1 )}}

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 - 1) L^{- 1} {\frac{12}{s ( 10 s + 1 )}} (t - 1)

L^{- 1} {\frac{12}{s ( 10 s + 1 )}} : 12 H (t) - 12 e^{- \frac{t}{10}}

= H (t - 1) (12 H (t - 1) - 12 e^{- \frac{t - 1}{10}})

H (t - 1) = {01 t < 1 t \geq 1

= {0 12 - 12 e^{- \frac{t - 1}{10}} t < 1 t \geq 1

= H (t - 1) (12 - 12 e^{- \frac{t - 1}{10}})

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What is inverse oflaplace (12e^{-s})/(s(10s+1)) ?
The answer to inverse oflaplace (12e^{-s})/(s(10s+1)) is \H(t-1)(12-12e^{-(t-1)/(10)})