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Beliebt Rechnen >

inverselaplace e^{-3s}*5/(4s^2+16)

Lösung

inverse laplace transformation

e^{- 3 s} \cdot \frac{5}{4 s ^{2} + 16}

Lösung

H (t - 3) \frac{5}{8} sin (2 (t - 3))

Schritte zur Lösung

L^{- 1} {e^{- 3 s} \frac{5}{4 s ^{2} + 16}}

Wende inverse Transformationsregel an: if

L^{- 1} {F (s)} = f (t)

dann

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Wo

H (t)

Heaviside (Schritt) Funktion ist

= H (t - 3) L^{- 1} {\frac{5}{4 s ^{2} + 16}} (t - 3)

L^{- 1} {\frac{5}{4 s ^{2} + 16}} : \frac{5}{8} sin (2 t)

= H (t - 3) \frac{5}{8} sin (2 (t - 3))

Beliebte Beispiele

tangent 4x-x^2(1.3)

t an g e n t 4 x - x^{2} (1.3)

derivative of ln(e^x+e^{-x})

\frac{d}{d x} (ln (e^{x} + e^{- x}))

limit as x approaches 4+of 2/((x-4)^4)

x \to 4 + lim (\frac{2}{( x - 4 ) ^{4}})

16y^{''''}=256y

16 y^{^{''''}} = 256 y

integral of ln(1+1/(x^2))

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