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Beliebt Rechnen >

inverselaplace (10s)/(s^2+16)

Lösung

inverse laplace transformation

\frac{10 s}{s ^{2} + 16}

Lösung

10 cos (4 t)

Schritte zur Lösung

L^{- 1} {\frac{10 s}{s ^{2} + 16}}

Wende inverse Transformationsregel an: if

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

dann

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

Für

\frac{10 s}{s ^{2} + 16} : F (s) = \frac{10}{s ^{2} + 16}

= \frac{d}{d t} (L^{- 1} {\frac{10}{s ^{2} + 16}}) + L^{- 1} {\frac{10}{s ^{2} + 16}} (0)

L^{- 1} {\frac{10}{s ^{2} + 16}} : \frac{5}{2} sin (4 t)

\frac{d}{d t} (\frac{5}{2} sin (4 t)) = 10 cos (4 t)

Evaluiere

\frac{5}{2} sin (4 t)

am Punkt

0 = 0

= 10 cos (4 t)

Beliebte Beispiele

integral of (sec^2(x)+8)

integral of 2xsqrt(x^2+9)

derivative of (2x^2-4x^2)

inverselaplace 3/(ss^2+8ss+18s)

d/(d{x)}(4{x}-3{x}^2-{y}^2-({z}-1)^2)