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Texによる数式表現31~ラプラス変換例

 前回のラプラス変換の公式に引き続き、今回はラプラス変換の例についてのTeXによる数式表現について紹介する。

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ラプラス変換の例

以下\delta(t)ディラックデルタ関数, u(t)はヘビサイド関数, \Gamma(z)はガンマ関数, \gammaオイラー定数

また、tは時間, sは複素数,α,β,τ,ωは実数, nは整数とする。

単位インパルス関数:

[tex: \displaystyle \mathcal{L} \\[ \delta(t) \\] = 1 ]

 \displaystyle \mathcal{L} [ \delta(t) ] = 1

単位ステップ関数:

[tex: \displaystyle \mathcal{L} \\[ u(t) \\] = \frac{1}{s} ]

 \displaystyle \mathcal{L} [ u(t) ] = \frac{1}{s}

ランプ関数

[tex: \displaystyle \mathcal{L} \\[ tu(t) \\] = \frac{1}{s^{2}} ]

 \displaystyle \mathcal{L} [ tu(t) ] = \frac{1}{s^{2}}

n乗関数

[tex: \displaystyle \mathcal{L} \\[ t^{n}u(t) \\] = n! \frac{1}{s^{n}} ]

 \displaystyle \mathcal{L} [ t^{n}u(t) ] = n! \frac{1}{s^{n}}

q(複素数)乗関数

[tex: \displaystyle \mathcal{L} \\[ t^{q}u(t) \\] = \Gamma (q+1) \frac{1}{s^{n}} ]

 \displaystyle \mathcal{L} [ t^{q}u(t) ] = \Gamma (q+1) \frac{1}{s^{n}}

n乗根

[tex: \displaystyle \mathcal{L} \\[ t^{1/n}u(t) \\] = \Gamma (1+\frac{1}{n}) \frac{1}{s^{1+1/n}} ]

 \displaystyle \mathcal{L} [ t^{1/n}u(t) ] = \Gamma (1+\frac{1}{n}) \frac{1}{s^{1+1/n}}

指数減衰関数

[tex: \displaystyle \mathcal{L} \\[ e^{-\alpha t}u(t) \\] = \frac{1}{s+\alpha} ]

 \displaystyle \mathcal{L} [ e^{-\alpha t}u(t) ] = \frac{1}{s+\alpha}

n乗指数減衰

[tex: \displaystyle \mathcal{L} \\[ t^{n} e^{-\alpha t}u(t) \\] =n! \frac{1}{(s+\alpha)^{n+1}} ]

 \displaystyle \mathcal{L} [ t^{n} e^{-\alpha t}u(t) ] =n! \frac{1}{(s+\alpha)^{n+1}}

理想遅延:

[tex: \displaystyle \mathcal{L} \\[ \delta(t-\tau) \\] = e^{-\tau s} ]

 \displaystyle \mathcal{L} [ \delta(t-\tau) ] = e^{-\tau s}

遅延単位ステップ関数

[tex: \displaystyle \mathcal{L} \\[ u(t-\tau) \\] = \frac{1}{s} e^{-\tau s} ]

 \displaystyle \mathcal{L} [ u(t-\tau) ] = \frac{1}{s} e^{-\tau s}

指数関数的接近

[tex: \displaystyle \mathcal{L} \\[ (1-e^{\alpha t} )u(t) \\] = \frac{\alpha}{s(s+\alpha)}  ]

 \displaystyle \mathcal{L} [ (1-e^{\alpha t} )u(t) ] = \frac{\alpha}{s(s+\alpha)}

sin関数

[tex: \displaystyle \mathcal{L} \\[ (\sin (\omega t) u(t) \\] = \frac{\omega}{s^{2}+\omega^{2}}  ]

 \displaystyle \mathcal{L} [ (\sin (\omega t) u(t) ] = \frac{\omega}{s^{2}+\omega^{2}}

cos関数

[tex: \displaystyle \mathcal{L} \\[ (\cos (\omega t) u(t) \\] = \frac{s}{s^{2}+\omega^{2}}  ]

 \displaystyle \mathcal{L} [ (\cos (\omega t) u(t) ] = \frac{s}{s^{2}+\omega^{2}}

sinh関数

[tex: \displaystyle \mathcal{L} \\[ (\sinh (\alpha t) u(t) \\] = \frac{\alpha}{s^{2}-\alpha^{2}}  ]

 \displaystyle \mathcal{L} [ (\sinh (\alpha t) u(t) ] = \frac{\alpha}{s^{2}-\alpha^{2}}

cosh関数

[tex: \displaystyle \mathcal{L} \\[ (\cosh (\alpha t) u(t) \\] = \frac{s}{s^{2}-\alpha^{2}}  ]

 \displaystyle \mathcal{L} [ (\cosh (\alpha t) u(t) ] = \frac{s}{s^{2}-\alpha^{2}}

sin減衰

[tex: \displaystyle \mathcal{L} \\[ (e^{-\alpha t} \sin (\omega t) u(t) \\] = \frac{\omega}{(s+\alpha)^{2}+\omega^{2}}  ]

 \displaystyle \mathcal{L} [ (e^{-\alpha t} \sin (\omega t) u(t) ] = \frac{\omega}{(s+\alpha)^{2}+\omega^{2}}

cos減衰

[tex: \displaystyle \mathcal{L} \\[ (e^{-\alpha t} \cos (\omega t) u(t) \\] = \frac{s+\alpha}{(s+\alpha)^{2}+\omega^{2}}  ]

 \displaystyle \mathcal{L} [ (e^{-\alpha t} \cos (\omega t) u(t) ] = \frac{s+\alpha}{(s+\alpha)^{2}+\omega^{2}}

対数関数

[tex: \displaystyle \mathcal{L} \\[ \ln ( \frac{t}{t\_{0}} ) u(t)  \\] = \frac{1}{s} \left( \ln (t\_{0}s ) +\gamma \right)  ]

 \displaystyle \mathcal{L} [ \ln ( \frac{t}{t_{0}} ) u(t)  ] = \frac{1}{s} \left( \ln (t_{0}s ) +\gamma \right)

第1種ベッセル関数

[tex:\displaystyle \mathcal{L} \\[ J\_{n}(\omega t ) \\] = \frac{\omega^{n} ( s+ \sqrt{s^{2}+\omega^{2} })^{-n} }{\sqrt{s^{2}+\omega^{2}}} ] 

\displaystyle \mathcal{L} [ J_{n}(\omega t ) ] = \frac{\omega^{n} ( s+ \sqrt{s^{2}+\omega^{2} })^{-n} }{\sqrt{s^{2}-\omega^{2}}}

第1種変形ベッセル関数

[tex:\displaystyle \mathcal{L} \\[ I\_{n}(\omega t ) \\] = \frac{\omega^{n} ( s+ \sqrt{s^{2}+\omega^{2} })^{-n} }{\sqrt{s^{2}+\omega^{2}}} ] 

\displaystyle \mathcal{L} [ I_{n}(\omega t ) ] = \frac{\omega^{n} ( s+ \sqrt{s^{2}+\omega^{2} })^{-n} }{\sqrt{s^{2}+\omega^{2}}}

第2種ベッセル関数

[tex:\displaystyle \mathcal{L} \\[ Y\_{0}(\alpha t ) \\] = - \frac{\sinh^{-1} (s/\alpha )}{\pi\sqrt{s^{2}+\omega^{2}}} ] 

\displaystyle \mathcal{L} [ Y_{0}(\alpha t ) ] = - \frac{\sinh^{-1} (s/\alpha )}{\pi\sqrt{s^{2}+\omega^{2}}}

誤差関数

[tex:\displaystyle \mathcal{L} \\[ {\rm{erf}}(t)u(t) \\] = - \frac{e^{-s^{2}/4} {\rm{erf}} (s/2)}{s} ] 

\displaystyle \mathcal{L} [ {\rm{erf}}(t)u(t) ] = - \frac{e^{-s^{2}/4} {\rm{erf}} (s/2)}{s}