What is inverse oflaplace 1/(2002s^2+4) ?

The answer to inverse oflaplace 1/(2002s^2+4) is 1/(2sqrt(2002))sin(sqrt(2/1001)t)

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inverse oflaplace 1/(2002s^2+4)

inverse laplace

\frac{1}{2002 s ^{2} + 4}

\frac{1}{2 2002} sin (\frac{2}{1001} t)

Solution steps

L^{- 1} {\frac{1}{2002 s ^{2} + 4}}

= L^{- 1} ⎩ ⎨ ⎧ \frac{1}{2 2002} \cdot \frac{\frac{2}{1001}}{s ^{2} + ( \frac{2}{1001} ) ^{2}} ⎭ ⎬ ⎫

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= \frac{1}{2 2002} L^{- 1} ⎩ ⎨ ⎧ \frac{\frac{2}{1001}}{s ^{2} + ( \frac{2}{1001} ) ^{2}} ⎭ ⎬ ⎫

Use Inverse Laplace Transform table:

L^{- 1} {\frac{a}{s ^{2} + a ^{2}}} = sin (a t)

L^{- 1} {\frac{\frac{2}{1001}}{s ^{2} + ( \frac{2}{1001} ) ^{2}}} = sin (\frac{2}{1001} t)

= \frac{1}{2 2002} sin (\frac{2}{1001} t)

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What is inverse oflaplace 1/(2002s^2+4) ?
The answer to inverse oflaplace 1/(2002s^2+4) is 1/(2sqrt(2002))sin(sqrt(2/1001)t)