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Beliebt Rechnen >

inverselaplace (2s-3)/(2s^2-8s+16)

Lösung

inverse laplace transformation

\frac{2 s - 3}{2 s ^{2} - 8 s + 16}

Lösung

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

Schritte zur Lösung

L^{- 1} {\frac{2 s - 3}{2 s ^{2} - 8 s + 16}}

Schreibe

\frac{2 s - 3}{2 s ^{2} - 8 s + 16}

um:

\frac{s - 2}{( s - 2 ) ^{2} + 4} + \frac{1}{2} \cdot \frac{1}{( s - 2 ) ^{2} + 4}

= L^{- 1} {\frac{s - 2}{( s - 2 ) ^{2} + 4} + \frac{1}{2} \cdot \frac{1}{( s - 2 ) ^{2} + 4}}

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

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

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

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

Fasse

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

zusammen:

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

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

Beliebte Beispiele

f (x) = 4 x^{3} ln (x)

limit as x approaches 1 of tan(x/2)

x \to 1 lim (tan (\frac{x}{2}))

y^{''}-y=e^t(t+e^t)

y^{^{''}} - y = e^{t} (t + e^{t})

derivative 2t(3t^3-7)^4

d er i v a t i v e 2 t (3 t^{3} - 7)^{4}

limit as x approaches infinity of (ln(x))/(e^x)

x \to \infty lim (\frac{ln ( x )}{e ^{x}})