inverselaplace 1/(4s^2+1)

Lösung

inverse laplace transformation

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

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

Schritte zur Lösung

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

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

Verwende inverse Laplace-Transformationstabelle:

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

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

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

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

\int cos (\frac{1}{2}) x d x

\frac{d}{d x} (ln (2 x^{4} - x^{3} + 3 x^{2} - 3 x))

im pl i c i t \frac{d y}{d x}, x^{3} y^{5} + 3 x = 8 y^{3} + 1

\int_{0}^{4} 1 + e^{- 2 y} d y

\int \frac{28 s + 28}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s