inverselaplace 6/s+821/(s^2+16)

Lösung

inverse laplace transformation

\frac{6}{s} + 8 \cdot 2 \cdot \frac{1}{s ^{2} + 16}

6 H (t) + 4 sin (4 t)

Schritte zur Lösung

L^{- 1} {\frac{6}{s} + 8 \cdot 2 \cdot \frac{1}{s ^{2} + 16}}

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{6}{s}} + 8 \cdot 2 L^{- 1} {\frac{1}{s ^{2} + 16}}

L^{- 1} {\frac{6}{s}} : 6 H (t)

L^{- 1} {\frac{1}{s ^{2} + 16}} : \frac{1}{4} sin (4 t)

= 6 H (t) + 8 \cdot 2 \cdot \frac{1}{4} sin (4 t)

Fasse

6 H (t) + 82 \frac{1}{4} sin (4 t)

zusammen:

6 H (t) + 4 sin (4 t)

= 6 H (t) + 4 sin (4 t)

t an g e n t f (x) = (3 x - 5) (x + 4), (2, 6)

\int_{3}^{\infty} \frac{x}{x ^{2} + 1} d x

x \to - 3 lim (\frac{5 x ^{2}}{x + 3})

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

\int 6 x \cdot csc^{2} (8 x^{2}) d x