inverselaplace 1/(s(s^2+6pis+pi^2))

פתרון

inverse laplace

\frac{1}{s ( s ^{2} + 6 π s + π ^{2} )}

\frac{1}{π ^{2}} H (t) - \frac{1}{π ^{2}} e^{- 3 π t} cosh (22 π t) - \frac{3}{2 2 π ^{2}} e^{- 3 π t} sinh (22 π t)

צעדי פתרון

L^{- 1} {\frac{1}{s ( s ^{2} + 6 π s + π ^{2} )}}

\frac{1}{s ( s ^{2} + 6 π s + π ^{2} )}

קח את השבר החלקי של:

\frac{1}{π ^{2} s} + \frac{- s - 6 π}{π ^{2} ( s ^{2} + 6 π s + π ^{2} )}

= L^{- 1} {\frac{1}{π ^{2} s} + \frac{- s - 6 π}{π ^{2} ( s ^{2} + 6 π s + π ^{2} )}}

\frac{- s - 6 π}{π ^{2} ( s ^{2} + 6 π s + π ^{2} )}

הרחב את:

- \frac{1}{π ^{2}} \cdot \frac{s + 3 π}{( s + 3 π ) ^{2} - 8 π ^{2}} - \frac{3}{π} \cdot \frac{1}{( s + 3 π ) ^{2} - 8 π ^{2}}

= L^{- 1} {\frac{1}{π ^{2} s} - \frac{1}{π ^{2}} \cdot \frac{s + 3 π}{( s + 3 π ) ^{2} - 8 π ^{2}} - \frac{3}{π} \cdot \frac{1}{( s + 3 π ) ^{2} - 8 π ^{2}}}

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}{π ^{2} s}} - \frac{1}{π ^{2}} L^{- 1} {\frac{s + 3 π}{( s + 3 π ) ^{2} - 8 π ^{2}}} - \frac{3}{π} L^{- 1} {\frac{1}{( s + 3 π ) ^{2} - 8 π ^{2}}}

L^{- 1} {\frac{1}{π ^{2} s}} : \frac{1}{π ^{2}} H (t)

L^{- 1} {\frac{s + 3 π}{( s + 3 π ) ^{2} - 8 π ^{2}}} : e^{- 3 π t} cosh (22 π t)

L^{- 1} {\frac{1}{( s + 3 π ) ^{2} - 8 π ^{2}}} : e^{- 3 π t} \frac{1}{2 2 π} sinh (22 π t)

= \frac{1}{π ^{2}} H (t) - \frac{1}{π ^{2}} e^{- 3 π t} cosh (22 π t) - \frac{3}{π} e^{- 3 π t} \frac{1}{2 2 π} sinh (22 π t)

\frac{1}{π ^{2}} H (t) - \frac{1}{π ^{2}} e^{- 3 π t} cosh (22 π t) - \frac{3}{π} e^{- 3 π t} \frac{1}{2 2 π} sinh (22 π t)

פשט את:

\frac{1}{π ^{2}} H (t) - \frac{1}{π ^{2}} e^{- 3 π t} cosh (22 π t) - \frac{3}{2 2 π ^{2}} e^{- 3 π t} sinh (22 π t)

= \frac{1}{π ^{2}} H (t) - \frac{1}{π ^{2}} e^{- 3 π t} cosh (22 π t) - \frac{3}{2 2 π ^{2}} e^{- 3 π t} sinh (22 π t)

\int \frac{3 tan ( x ) - 4 cos ^{2} ( x )}{cos ( x )} d x

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