laplacetransform t^2cos(2t+30)

解

ラプラス変換

t^{2} cos (2 t + 3 0^{\circ})

\frac{3 s ^{3} - 6 s ^{2} - 12 3 s + 8}{( s ^{2} + 4 ) ^{3}}

解答ステップ

L {t^{2} cos (2 t + 3 0^{\circ})}

ラプラス変換表を使用する:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

t^{2} cos (2 t + 3 0^{\circ})

向け

: f (t) = cos (2 t + 3 0^{\circ}), k = 2

= (- 1)^{2} \frac{d ^{2}}{d s ^{2}} (L {cos (2 t + 3 0^{\circ})})

L {cos (2 t + 3 0^{\circ})} : \frac{3 s - 2}{2 ( s ^{2} + 4 )}

\frac{d ^{2}}{d s ^{2}} (\frac{3 s - 2}{2 ( s ^{2} + 4 )}) = \frac{3 s ^{3} - 6 s ^{2} - 12 3 s + 8}{( s ^{2} + 4 ) ^{3}}

= (- 1)^{2} \frac{3 s ^{3} - 6 s ^{2} - 12 3 s + 8}{( s ^{2} + 4 ) ^{3}}

= \frac{3 s ^{3} - 6 s ^{2} - 12 3 s + 8}{( s ^{2} + 4 ) ^{3}}