✅@2023-07-14T08:50D90 AM4-2023F-13

$ (D^2+2h\omega_0D+{\omega_0}^2)G(t)=\frac{\delta(t)}{m}

を満たすと定義できる

理由

$ (D^2+2h\omega_0D+{\omega_0}^2)x(t)=\frac{\delta(t)}{m}

$ \implies {\cal F}(x)(\omega)={\cal F}(G)(\omega){\cal F}(\delta)(\omega)={\cal F}(G)(\omega)

$ \iff x=G

$ \therefore (D^2+2h\omega_0D+{\omega_0}^2)G(t)=\frac{\delta(t)}{m}

最初から振り返ってみる

$ {\cal F}(f')(\omega)=\int_\R f'(t)e^{-i\omega t}\mathrm dt

$ = \int_{t\in\R}e^{-i\omega t}\mathrm df(t)

$ =0-\int_{t\in\R}f(t)\mathrm de^{-i\omega t}

積の積分を使った

$ =i\omega\int_\R f(t)e^{-i\omega t}\mathrm d t

$ =i\omega{\cal F}(f)(\omega)

この変換を使えば、微分を代数方程式に変換できる

$ (D^2+2h\omega_0D+{\omega_0}^2)x(t)=\frac{f(t)}{m}

$ \implies {\cal F}((D^2+2h\omega_0D+{\omega_0}^2)x)(\omega)=\frac1m{\cal F}(f)(\omega)

$ \implies (-\omega^2+2ih\omega_0\omega+{\omega_0}^2){\cal F}(x)(\omega)=\frac1m{\cal F}(f)(\omega)

$ \iff \left((h\omega_0+i\omega)^2-i^2(1-h^2){\omega_0}^2\right){\cal F}(x)(\omega)=\frac1m{\cal F}(f)(\omega)

$ \iff \left(i\omega+h\omega_0+i\omega_0\sqrt{1-h^2}\right)\left(i\omega+h\omega_0-i\omega_0\sqrt{1-h^2}\right){\cal F}(x)(\omega)=\frac1m{\cal F}(f)(\omega)

$ \iff -\left(\omega-\left(ih\omega_0-\omega_0\sqrt{1-h^2}\right)\right)\left(\omega-\left(ih\omega_0+\omega_0\sqrt{1-h^2}\right)\right){\cal F}(x)(\omega)=\frac1m{\cal F}(f)(\omega)

$ \iff -(\omega-\omega_\alpha)(\omega-\omega_\beta){\cal F}(x)(\omega)=\frac1m{\cal F}(f)(\omega)

$ \iff {\cal F}(x)(\omega)=-\frac1m\frac1{\omega-\omega_\alpha}\frac1{\omega-\omega_\beta}{\cal F}(f)(\omega)=:{\cal F}(G)(\omega){\cal F}(f)(\omega)

$ \implies x(t)=\int_\R f(\tau)G(t-\tau)\mathrm d\tau

$ G(t)=\frac{-1}{2\pi m}\frac1{\omega_\alpha-\omega_\beta}\int_\R\left(\frac{1}{\omega-\omega_\alpha}-\frac{1}{\omega-\omega_\beta}\right)e^{i\omega t}\mathrm d\omega

$ = \frac{-1}{2\pi m}\frac{2\pi i}{\omega_\alpha-\omega_\beta}(e^{i\omega_\alpha t}-e^{i\omega_\beta t})\llbracket t>0\rrbracket

$ \int_\R\frac{1}{\omega-\omega_\alpha}e^{i\omega t}\mathrm d\omega= \int_\R\frac{1}{\omega-\omega_\alpha}e^{i\omega t}\mathrm d\omega\llbracket t>0\rrbracket+\int_\R\frac{1}{\omega-\omega_\alpha}\mathrm d\omega\llbracket t=0\rrbracket+\int_\R\frac{1}{\omega-\omega_\alpha}e^{i\omega t}\mathrm d\omega\llbracket t<0\rrbracket

ここで場合分けする

$ =\lim_{R\to\infty}\left(\oint_{M_+(R)\cup]-R,R[}\frac{1}{\omega-\omega_\alpha}e^{i\omega t}\mathrm d\omega-\int_{M_+(R)}\frac{1}{\omega-\omega_\alpha}e^{i\omega t}\mathrm d\omega\right)\llbracket t>0\rrbracket

$ +\lim_{R\to\infty}\ln\left|\frac{-R-\omega_\alpha}{R-\omega_\alpha}\right|\llbracket t=0\rrbracket

$ +\lim_{R\to\infty}\left(\oint_{M_-(R)\cup]-R,R[}\frac{1}{\omega-\omega_\alpha}e^{i\omega t}\mathrm d\omega-\int_{M_-(R)}\frac{1}{\omega-\omega_\alpha}e^{i\omega t}\mathrm d\omega\right)\llbracket t<0\rrbracket

直線=円周-円盤

$ =\lim_{R\to\infty}\left(2\pi ie^{i\omega_\alpha t}-0\right)\llbracket t>0\rrbracket

$ +0

$ +\lim_{R\to\infty}\left(0-0\right)\llbracket t<0\rrbracket

$ =2\pi ie^{i\omega_\alpha t}\llbracket t>0\rrbracket

$ \int_\R\frac{1}{\omega-\omega_\beta}e^{i\omega t}\mathrm d\omega=2\pi ie^{i\omega_\beta t}\llbracket t>0\rrbracket

計算過程は同じ

$ = \frac1m\frac{-i}{\omega_\alpha-\omega_\beta}(e^{i\omega_\alpha t}-e^{i\omega_\beta t})\llbracket t>0\rrbracket

$ = \frac1m\frac{-i}{2\sqrt{1-h^2}\omega_0}e^{-h\omega_0t}(e^{i\sqrt{1-h^2}\omega_0t}-e^{-i\sqrt{1-h^2}\omega_0t})\llbracket t>0\rrbracket

$ = \frac1m\frac1{\sqrt{1-h^2}\omega_0}e^{-h\omega_0t}\frac1{2i}(e^{i\sqrt{1-h^2}\omega_0t}-e^{-i\sqrt{1-h^2}\omega_0t})\llbracket t>0\rrbracket

$ = \frac1m\frac1{\sqrt{1-h^2}\omega_0}e^{-h\omega_0t}\Im e^{i\sqrt{1-h^2}\omega_0t}\llbracket t>0\rrbracket

$ \underline{= \frac1m\frac1{\sqrt{1-h^2}\omega_0}e^{-h\omega_0t}\sin\left(\sqrt{1-h^2}\omega_0t\right)\llbracket t>0\rrbracket\quad}_\blacksquare

2023-08-04 07:59:23 ここ書き途中だった

書いたらおさらいは終わりにしよう

外力の振動成分のうち、固有振動数部分を伝達関数が増幅する

練習問題

これ、Fourier変換で求まるのかな？

$ \because (D^2+2h\omega_0D+{\omega_0}^2)Ae^{i\omega_\alpha t}=(-{\omega_\alpha}^2+2ih\omega_0\omega_\alpha+{\omega_0}^2)Ae^{i\omega_\alpha t}=0

$ \begin{pmatrix}1&1\\i\omega_\alpha&i\omega_\beta\end{pmatrix}\cdot\begin{pmatrix}A\\B\end{pmatrix}=\begin{pmatrix}0\\\frac1m\end{pmatrix}

$ \iff\begin{pmatrix}A\\B\end{pmatrix}=\frac1m\frac{i}{\omega_\alpha-\omega_\beta}\begin{pmatrix}-1\\1\end{pmatrix}

以上より

$ x(t)=\frac1m\frac{i}{\omega_\alpha-\omega_\beta}\left(e^{i\omega_\beta t}-e^{i\omega_\alpha t}\right)

$ = \frac1m\frac1{2i}\frac{1}{\sqrt{1-h^2}\omega_0}e^{-h\omega_0t}2i\Im e^{i\sqrt{1-h^2}\omega_0t}

$ = \frac1m\frac{1}{\sqrt{1-h^2}\omega_0}e^{-h\omega_0t}\sin(\sqrt{1-h^2}\omega_0t)