inverselaplace 1/(19s^2+5s+19990)

解

逆ラプラス

\frac{1}{19 s ^{2} + 5 s + 19990}

\frac{2 e ^{- \frac{5 t}{38}} sin ( \frac{1519215 t}{38} )}{1519215}

解答ステップ

L^{- 1} {\frac{1}{19 s ^{2} + 5 s + 19990}}

拡張

\frac{1}{19 s ^{2} + 5 s + 19990} : \frac{1}{19} \cdot \frac{1}{( s + \frac{5}{38} ) ^{2} + \frac{1519215}{1444}}

= L^{- 1} {\frac{1}{19} \cdot \frac{1}{( s + \frac{5}{38} ) ^{2} + \frac{1519215}{1444}}}

逆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^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= \frac{1}{19} L^{- 1} {\frac{1}{( s + \frac{5}{38} ) ^{2} + \frac{1519215}{1444}}}

L^{- 1} {\frac{1}{( s + \frac{5}{38} ) ^{2} + \frac{1519215}{1444}}} : e^{- \frac{5 t}{38}} \frac{38}{1519215} sin (\frac{1519215 t}{38})

= \frac{1}{19} e^{- \frac{5 t}{38}} \frac{38}{1519215} sin (\frac{1519215 t}{38})

改良

\frac{1}{19} e^{- \frac{5 t}{38}} \frac{38}{1519215} sin (\frac{1519215 t}{38}) : \frac{2 e ^{- \frac{5 t}{38}} sin ( \frac{1519215 t}{38} )}{1519215}

= \frac{2 e ^{- \frac{5 t}{38}} sin ( \frac{1519215 t}{38} )}{1519215}