inverselaplace (1+e^{-pis})/(s^2+1)

Lösung

inverse laplace transformation

\frac{1 + e ^{- π s}}{s ^{2} + 1}

sin (t) + H (t - π) sin (t - π)

Schritte zur Lösung

L^{- 1} {\frac{1 + e ^{- π s}}{s ^{2} + 1}}

Schreibe um

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

Wende die lineare Eigenschaftder inversen Laplace-Transformation an:

Für Funktionen

f (s), g (s)

und Konstanten

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 ^{2} + 1}} + L^{- 1} {\frac{e ^{- π s}}{s ^{2} + 1}}

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

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

= sin (t) + H (t - π) sin (t - π)

t an g e n t f (x) = 2 sec (x), at x = \frac{π}{3}

in v erse l a pl a ce e^{3 s}

\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d x

x \to \infty lim (\frac{π}{x})

\frac{\partial}{\partial x} (4 cos^{3} (k x - wt) - 3 cos (k x - wt))