laplacetransform f(t)=7sin(4(t+2))+3cos(4t-2)

Lösung

laplace transformation

f (t) = 7 sin (4 (t + 2)) + 3 cos (4 t - 2)

\frac{7 ( sin ( 8 ) s + 4 cos ( 8 ) )}{s ^{2} + 16} + \frac{3 ( cos ( 2 ) s + 4 sin ( 2 ) )}{s ^{2} + 16}

Schritte zur Lösung

L {7 sin (4 (t + 2)) + 3 cos (4 t - 2)}

Verwende die lineare Eigenschaft der Laplace-Transformation:

Für Funktionen

f (t), g (t)

und Konstanten

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= 7 L {sin (4 (t + 2))} + 3 L {cos (4 t - 2)}

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

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

= 7 \cdot \frac{sin ( 8 ) s + 4 cos ( 8 )}{s ^{2} + 16} + 3 \cdot \frac{cos ( 2 ) s + 4 sin ( 2 )}{s ^{2} + 16}

Fasse

7 \frac{sin ( 8 ) s + 4 cos ( 8 )}{s ^{2} + 16} + 3 \frac{cos ( 2 ) s + 4 sin ( 2 )}{s ^{2} + 16}

zusammen:

\frac{7 ( sin ( 8 ) s + 4 cos ( 8 ) )}{s ^{2} + 16} + \frac{3 ( cos ( 2 ) s + 4 sin ( 2 ) )}{s ^{2} + 16}

= \frac{7 ( sin ( 8 ) s + 4 cos ( 8 ) )}{s ^{2} + 16} + \frac{3 ( cos ( 2 ) s + 4 sin ( 2 ) )}{s ^{2} + 16}

\int \frac{2 x}{x ^{2} + 3} d x

\int sin^{- 4} (\frac{x}{5}) cos^{3} (\frac{x}{5}) d x

f (x) = (0.46) + (- 4.39 x) + (- 206 x^{2}) + (2.8 x^{3})

x \to - 1 lim (\frac{x ^{2}}{x ^{2} - 1})

\frac{d}{d x} (cos (20 x))