What is inverse oflaplace 20+(10)/s*1/(8s+1) ?

The answer to inverse oflaplace 20+(10)/s*1/(8s+1) is 20\delta(t)+10\H(t)-10e^{-t/8}

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inverse oflaplace 20+(10)/s*1/(8s+1)

inverse laplace

20 + \frac{10}{s} \cdot \frac{1}{8 s + 1}

20 δ (t) + 10 H (t) - 10 e^{- \frac{t}{8}}

Solution steps

L^{- 1} {20 + \frac{10}{s} \cdot \frac{1}{8 s + 1}}

Expand

20 + \frac{10}{s} \frac{1}{8 s + 1} : 20 + \frac{10}{8 s ^{2} + s}

= L^{- 1} {20 + \frac{10}{8 s ^{2} + s}}

Take the partial fraction of

\frac{10}{8 s ^{2} + s} : \frac{10}{s} - \frac{80}{8 s + 1}

= L^{- 1} {20 + \frac{10}{s} - \frac{80}{8 s + 1}}

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} {20} + L^{- 1} {\frac{10}{s}} - L^{- 1} {\frac{80}{8 s + 1}}

L^{- 1} {20} : 20 δ (t)

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

L^{- 1} {\frac{80}{8 s + 1}} : 10 e^{- \frac{t}{8}}

= 20 δ (t) + 10 H (t) - 10 e^{- \frac{t}{8}}

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What is inverse oflaplace 20+(10)/s*1/(8s+1) ?
The answer to inverse oflaplace 20+(10)/s*1/(8s+1) is 20\delta(t)+10\H(t)-10e^{-t/8}