inverselaplace (1/7)/(s^2+19980/7)

Lösung

inverse laplace transformation

\frac{\frac{1}{7}}{s ^{2} + \frac{19980}{7}}

\frac{1}{6 3885} sin (\frac{6 555 t}{7})

Schritte zur Lösung

L^{- 1} {\frac{\frac{1}{7}}{s ^{2} + \frac{19980}{7}}}

= L^{- 1} ⎩ ⎨ ⎧ \frac{1}{6 3885} \cdot \frac{\frac{6 555}{7}}{s ^{2} + ( \frac{6 555}{7} ) ^{2}} ⎭ ⎬ ⎫

Wende die konstante Multiplikationseigenschaft der inversen Laplace-Transformation an:

Für Funktionen

f (t)

und Konstante

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

= \frac{1}{6 3885} L^{- 1} ⎩ ⎨ ⎧ \frac{\frac{6 555}{7}}{s ^{2} + ( \frac{6 555}{7} ) ^{2}} ⎭ ⎬ ⎫

Verwende inverse Laplace-Transformationstabelle:

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

L^{- 1} {\frac{\frac{6 555}{7}}{s ^{2} + ( \frac{6 555}{7} ) ^{2}}} = sin (\frac{6 555}{7} t)

= \frac{1}{6 3885} sin (\frac{6 555}{7} t)

= \frac{1}{6 3885} sin (\frac{6 555 t}{7})

\int t^{- \frac{1}{2}} (t^{4} + 2 t^{2} - 1) d t

\int sin (s) d s

l a pl a ce t r an s f or m - 4 t^{2} + 16 t + 9

\frac{\partial}{\partial x} (4 x (2 + y)^{- 1})

t an g e n t f (x) = \frac{x}{1 + x ^{2}}, at x = 2