laplacetransform t*sin(5t+0.2pi)

解

ラプラス変換

t \cdot sin (5 t + 0.2 π)

- \frac{- 0.58778 \dots s ^{2} - 8.09016 \dots s + 14.69463 \dots}{( s ^{2} + 25 ) ^{2}}

解答ステップ

L {t sin (5 t + 0.2 π)}

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 sin (5 t + 0.2 π)

向け

: f (t) = sin (5 t + 0.2 π), k = 1

= (- 1)^{1} \frac{d}{d s} (L {sin (5 t + 0.2 π)})

L {sin (5 t + 0.2 π)} : \frac{sin ( 0.62831 \dots ) s + 5 cos ( 0.62831 \dots )}{s ^{2} + 25}

\frac{d}{d s} (\frac{sin ( 0.62831 \dots ) s + 5 cos ( 0.62831 \dots )}{s ^{2} + 25}) = \frac{- 0.58778 \dots s ^{2} - 8.09016 \dots s + 14.69463 \dots}{( s ^{2} + 25 ) ^{2}}

= (- 1)^{1} \frac{- 0.58778 \dots s ^{2} - 8.09016 \dots s + 14.69463 \dots}{( s ^{2} + 25 ) ^{2}}

= - \frac{- 0.58778 \dots s ^{2} - 8.09016 \dots s + 14.69463 \dots}{( s ^{2} + 25 ) ^{2}}