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Beliebt Rechnen >

inverselaplace (2e^{-2s})/(s(s+2)(s+3)^2)

Lösung

inverse laplace transformation

\frac{2 e ^{- 2 s}}{s ( s + 2 ) ( s + 3 ) ^{2}}

Lösung

H (t - 2) (\frac{1}{9} - e^{- 2 (t - 2)} + \frac{8}{9} e^{- 3 (t - 2)} + \frac{2 e ^{- 3 (t - 2)} ( t - 2 )}{3})

Schritte zur Lösung

L^{- 1} {\frac{2 e ^{- 2 s}}{s ( s + 2 ) ( s + 3 ) ^{2}}}

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 - 2) L^{- 1} {\frac{2}{s ( s + 2 ) ( s + 3 ) ^{2}}} (t - 2)

L^{- 1} {\frac{2}{s ( s + 2 ) ( s + 3 ) ^{2}}} : \frac{1}{9} H (t) - e^{- 2 t} + \frac{8}{9} e^{- 3 t} + \frac{2 e ^{- 3 t} t}{3}

= H (t - 2) (\frac{1}{9} H (t - 2) - e^{- 2 (t - 2)} + \frac{8}{9} e^{- 3 (t - 2)} + \frac{2 e ^{- 3 (t - 2)} ( t - 2 )}{3})

H (t - 2) = {01 t < 2 t \geq 2

= {0 \frac{1}{9} - e^{- 2 (t - 2)} + \frac{8}{9} e^{- 3 (t - 2)} + \frac{2 e ^{- 3 (t - 2)} ( t - 2 )}{3} t < 2 t \geq 2

= H (t - 2) (\frac{1}{9} - e^{- 2 (t - 2)} + \frac{8}{9} e^{- 3 (t - 2)} + \frac{2 e ^{- 3 (t - 2)} ( t - 2 )}{3})

Beliebte Beispiele

limit as x approaches a of 2f(x)-3g(x)

x \to a lim (2 f (x) - 3 g (x))

derivative of (sqrt(x^2+a^2)^{e^{x^2}})

\frac{d}{d x} ((x^{2} + a^{2})^{e^{x^{2}}})

d/(dy)((-y)/(x^2+y^2))

\frac{d}{d y} (\frac{- y}{x ^{2} + y ^{2}})

derivative of log_{2}(3x+1)

\frac{d}{d x} (lo g_{2} (3 x + 1))

x y^{^{'}} + y = x