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Beliebt Rechnen >

inverselaplace s/((s-2)^2+16)

Lösung

inverse laplace transformation

\frac{s}{( s - 2 ) ^{2} + 16}

Lösung

e^{2 t} cos (4 t) + \frac{1}{2} e^{2 t} sin (4 t)

Schritte zur Lösung

L^{- 1} {\frac{s}{( s - 2 ) ^{2} + 16}}

Schreibe

\frac{s}{( s - 2 ) ^{2} + 16}

um:

\frac{s - 2}{( s - 2 ) ^{2} + 16} + 2 \cdot \frac{1}{( s - 2 ) ^{2} + 16}

= L^{- 1} {\frac{s - 2}{( s - 2 ) ^{2} + 16} + 2 \cdot \frac{1}{( s - 2 ) ^{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{s - 2}{( s - 2 ) ^{2} + 16}} + 2 L^{- 1} {\frac{1}{( s - 2 ) ^{2} + 16}}

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

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

= e^{2 t} cos (4 t) + 2 e^{2 t} \frac{1}{4} sin (4 t)

Fasse

e^{2 t} cos (4 t) + 2 e^{2 t} \frac{1}{4} sin (4 t)

zusammen:

e^{2 t} cos (4 t) + \frac{1}{2} e^{2 t} sin (4 t)

= e^{2 t} cos (4 t) + \frac{1}{2} e^{2 t} sin (4 t)

Beliebte Beispiele

integral from 1/3 to 5 of 16xln(3x)

\int_{\frac{1}{3}}^{5} 16 x ln (3 x) d x

tangent f(x)=x^2+4x-5,\at x=2

t an g e n t f (x) = x^{2} + 4 x - 5, at x = 2

integral of 1/(25-81x^2)

\int \frac{1}{25 - 81 x ^{2}} d x

integral of ((x^3+x))/(2x-1)

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

integral from 0 to pi/8 of (sin(2x))

\int_{0}^{\frac{π}{8}} (sin (2 x)) d x