What is inverse oflaplace 1/(s(2s^2+2s+1)) ?

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

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inverse oflaplace 1/(s(2s^2+2s+1))

inverse laplace

\frac{1}{s ( 2 s ^{2} + 2 s + 1 )}

H (t) - e^{- \frac{t}{2}} cos (\frac{t}{2}) - e^{- \frac{t}{2}} sin (\frac{t}{2})

Solution steps

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

Take the partial fraction of

\frac{1}{s ( 2 s ^{2} + 2 s + 1 )} : \frac{1}{s} + \frac{- 2 s - 2}{2 s ^{2} + 2 s + 1}

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

Expand

\frac{- 2 s - 2}{2 s ^{2} + 2 s + 1} : - \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}} - \frac{1}{2} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}

= L^{- 1} {\frac{1}{s} - \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}} - \frac{1}{2} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}}

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{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}} - \frac{1}{2} L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}}

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

L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}} : e^{- \frac{t}{2}} cos (\frac{t}{2})

L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}} : e^{- \frac{t}{2}} \cdot 2 sin (\frac{t}{2})

= H (t) - e^{- \frac{t}{2}} cos (\frac{t}{2}) - \frac{1}{2} e^{- \frac{t}{2}} \cdot 2 sin (\frac{t}{2})

Refine

H (t) - e^{- \frac{t}{2}} cos (\frac{t}{2}) - \frac{1}{2} e^{- \frac{t}{2}} 2 sin (\frac{t}{2}) : H (t) - e^{- \frac{t}{2}} cos (\frac{t}{2}) - e^{- \frac{t}{2}} sin (\frac{t}{2})

= H (t) - e^{- \frac{t}{2}} cos (\frac{t}{2}) - e^{- \frac{t}{2}} sin (\frac{t}{2})

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