laplacetransform-3(sin(4(t-(5pi)/4)))

解

ラプラス変換

- 3 (sin (4 (t - \frac{5 π}{4})))

\frac{12}{s ^{2} + 16}

解答ステップ

L {- 3 (sin (4 (t - \frac{5 π}{4})))}

ラプラス変換の定数乗法を使用する:

関数 f(t) と定数

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= - 3 L {sin (4 (t - \frac{5 π}{4}))}

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

= - 3 (- \frac{4}{s ^{2} + 16})

改良

- 3 (- \frac{4}{s ^{2} + 16}) : \frac{12}{s ^{2} + 16}

= \frac{12}{s ^{2} + 16}