状態空間表現，バネマスダンパー系，状態ベクトルの変換
状態空間表現
$$\left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t}\boldsymbol{x}&=&\boldsymbol{A}\boldsymbol{x}&+&\boldsymbol{B}\boldsymbol{u} \;\ldots\;\boldsymbol{A}:状態行列,\;\boldsymbol{B}:入力行列,\boldsymbol{x}:状態ベクトル,\;\boldsymbol{u}:入力ベクトル \\\boldsymbol{y}&=&\boldsymbol{C}\boldsymbol{x}&+&\boldsymbol{D}\boldsymbol{u} \;\ldots\;\boldsymbol{C}:出力行列,\;\boldsymbol{D}:直達行列,\;\boldsymbol{y}:出力ベクトル \end{eqnarray} \right.$$バネマスダンパー系(入力なし)
$$\left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t} \begin{bmatrix} x\\ v\\ \end{bmatrix} &=& \begin{bmatrix} 0&1\\ -\frac{k}{m}&-\frac{c}{m}\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix}\;\ldots\;k:バネ,\;m:マス,\;c:ダンパー,\;x:位置,\;v=\frac{\mathrm{d}x}{\mathrm{d}t}:速度 \\ \\&=& \begin{bmatrix} 0&1\\ -\omega_0^2&-2\gamma\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix} \;\ldots\;\gamma=\frac{c}{2m},\;\omega_0=\sqrt{\frac{k}{m}} \\ \begin{bmatrix} y \end{bmatrix} &=& \begin{bmatrix} 1&0\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix} \end{eqnarray} \right.$$状態ベクトルの変換
$$ \begin{eqnarray} \boldsymbol{x}^{\prime}=\href{https://shikitenkai.blogspot.com/2021/04/blog-post_4.html}{\boldsymbol{T}^{-1}\boldsymbol{x}} &=& \frac{1}{-2\xi} \begin{bmatrix} -\gamma-\xi & -1 \\\gamma-\xi& 1 \end{bmatrix} \begin{bmatrix} x\\v \end{bmatrix} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/blog-post_56.html}{\boldsymbol{T}^{-1}=\frac{1}{-2\xi} \begin{bmatrix} -\gamma-\xi & -1 \\\gamma-\xi& 1 \end{bmatrix}} \\&&\;\ldots\;\xi=\sqrt{\gamma^2-\omega_0^2} \\&=& \frac{1}{-2\xi} \begin{bmatrix} \left(-\gamma-\xi\right)\cdot x - v \\\left(\gamma-\xi\right)\cdot x + v \end{bmatrix} \\&=& \begin{bmatrix} \frac{-\gamma x-\xi x - v}{-2\xi} \\\frac{\gamma x-\xi x + v}{-2\xi} \end{bmatrix} \\&=& \begin{bmatrix} \frac{\gamma x+\xi x + v}{2\xi} \\\frac{-\gamma x+\xi x - v}{2\xi} \end{bmatrix} \\&=& \begin{bmatrix} \frac{x}{2} + \frac{v+\gamma x}{2\xi} \\\frac{x}{2} - \frac{v+\gamma x}{2\xi} \end{bmatrix} \end{eqnarray}$$ $$\begin{eqnarray} \boldsymbol{x}^{\prime}\left(0\right)&=& \begin{bmatrix} \frac{x_0}{2} + \frac{v_0+\gamma x_0}{2\xi} \\\frac{x_0}{2} - \frac{v_0+\gamma x_0}{2\xi} \end{bmatrix}\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/c1-c2.html}{要素が微分方程式として解いた際の初期値C_1,C_2に等しい} \end{eqnarray}$$\(\boldsymbol{T}\)で戻して確認
$$\begin{eqnarray}
\boldsymbol{T}\boldsymbol{x}^{\prime}
&=&
\begin{bmatrix}
1 & 1
\\-\gamma+\xi & -\gamma-\xi
\end{bmatrix}
\begin{bmatrix}
\frac{x}{2} + \frac{v+\gamma x}{2\xi}
\\\frac{x}{2} - \frac{v+\gamma x}{2\xi}
\end{bmatrix}
\\&=&\begin{bmatrix}
1\cdot\left(\frac{x}{2} + \frac{v+\gamma x}{2\xi}\right)+1\cdot\left(\frac{x}{2} - \frac{v+\gamma x}{2\xi}\right)
\\\left(-\gamma+\xi\right)\cdot\left(\frac{x}{2} + \frac{v+\gamma x}{2\xi}\right)+\left(-\gamma-\xi\right)\cdot\left(\frac{x}{2} - \frac{v+\gamma x}{2\xi}\right)
\end{bmatrix}
\\&=&\begin{bmatrix}
\frac{x}{2} \color{red}{+ \frac{v+\gamma x}{2\xi}}\color{black}{+}\frac{x}{2}\color{red}{- \frac{v+\gamma x}{2\xi}}
\\\frac{\left(-\gamma+\xi\right)x}{2}\frac{\xi}{\xi} + \frac{\left(-\gamma+\xi\right)\left(v+\gamma x\right)}{2\xi}
+\frac{\left(-\gamma-\xi\right)x}{2}\frac{\xi}{\xi} - \frac{\left(-\gamma-\xi\right)\left(v+\gamma x\right)}{2\xi}
\end{bmatrix}
\\&=&\begin{bmatrix}
\frac{x}{2}+\frac{x}{2}
\\\frac{\color{red}{-\gamma \xi x} \color{blue}{+\xi^2 x} \color{green}{-\gamma v} \color{black}{+\xi v} \color{magenta}{-\gamma^2 x}\color{red}{+\gamma\xi x}}{2\xi}
+ \frac{\color{red}{-\gamma \xi x} \color{blue}{-\xi^2 x} \color{green}{+\gamma v} \color{black}{+\xi v} \color{magenta}{+\gamma^2 x}\color{red}{+\gamma\xi x}}{2\xi}
\end{bmatrix}
\\&=&\begin{bmatrix}
x
\\\frac{\xi v}{2\xi}
+ \frac{\xi v}{2\xi}
\end{bmatrix}
=\begin{bmatrix}
x\\v
\end{bmatrix}
=\boldsymbol{x}
\end{eqnarray}$$
$$\begin{eqnarray} \boldsymbol{T}\boldsymbol{x}^{\prime}&=&\boldsymbol{T}\boldsymbol{T}^{-1}\boldsymbol{x} \\&=& \begin{bmatrix} 1 & 1 \\-\gamma+\xi & -\gamma-\xi \end{bmatrix} \frac{1}{-2\xi} \begin{bmatrix} -\gamma-\xi & -1 \\\gamma-\xi& 1 \end{bmatrix} \begin{bmatrix} x\\v \end{bmatrix} \\&=& \frac{1}{-2\xi} \begin{bmatrix} 1 & 1 \\-\gamma+\xi & -\gamma-\xi \end{bmatrix} \begin{bmatrix} -\gamma-\xi & -1 \\\gamma-\xi& 1 \end{bmatrix} \begin{bmatrix} x\\v \end{bmatrix} \\&=& \begin{bmatrix} 1 & 0 \\0& 1 \end{bmatrix} \begin{bmatrix} x\\v \end{bmatrix} =\begin{bmatrix} x\\v \end{bmatrix} =\boldsymbol{x} \end{eqnarray}$$
$$\begin{eqnarray} \boldsymbol{T}\boldsymbol{x}^{\prime}&=&\boldsymbol{T}\boldsymbol{T}^{-1}\boldsymbol{x} \\&=& \begin{bmatrix} 1 & 1 \\-\gamma+\xi & -\gamma-\xi \end{bmatrix} \frac{1}{-2\xi} \begin{bmatrix} -\gamma-\xi & -1 \\\gamma-\xi& 1 \end{bmatrix} \begin{bmatrix} x\\v \end{bmatrix} \\&=& \frac{1}{-2\xi} \begin{bmatrix} 1 & 1 \\-\gamma+\xi & -\gamma-\xi \end{bmatrix} \begin{bmatrix} -\gamma-\xi & -1 \\\gamma-\xi& 1 \end{bmatrix} \begin{bmatrix} x\\v \end{bmatrix} \\&=& \begin{bmatrix} 1 & 0 \\0& 1 \end{bmatrix} \begin{bmatrix} x\\v \end{bmatrix} =\begin{bmatrix} x\\v \end{bmatrix} =\boldsymbol{x} \end{eqnarray}$$
状態空間表現，バネマスダンパー系，状態行列の対角化
状態空間表現
$$\left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t}\boldsymbol{x}&=&\boldsymbol{A}\boldsymbol{x}&+&\boldsymbol{B}\boldsymbol{u} \;\ldots\;\boldsymbol{A}:状態行列,\;\boldsymbol{B}:入力行列,\boldsymbol{x}:状態ベクトル,\;\boldsymbol{u}:入力ベクトル \\\boldsymbol{y}&=&\boldsymbol{C}\boldsymbol{x}&+&\boldsymbol{D}\boldsymbol{u} \;\ldots\;\boldsymbol{C}:出力行列,\;\boldsymbol{D}:直達行列,\;\boldsymbol{y}:出力ベクトル \end{eqnarray} \right.$$バネマスダンパー系(入力なし)
$$\left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t} \begin{bmatrix} x\\ v\\ \end{bmatrix} &=& \begin{bmatrix} 0&1\\ -\frac{k}{m}&-\frac{c}{m}\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix}\;\ldots\;k:バネ,\;m:マス,\;c:ダンパー,\;x:位置,\;v=\frac{\mathrm{d}x}{\mathrm{d}t}:速度 \\ \\&=& \begin{bmatrix} 0&1\\ -\omega_0^2&-2\gamma\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix} \;\ldots\;\gamma=\frac{c}{2m},\;\omega_0=\sqrt{\frac{k}{m}} \\ \begin{bmatrix} y \end{bmatrix} &=& \begin{bmatrix} 1&0\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix} \end{eqnarray} \right.$$状態行列の対角化(基底の変換)とそれに伴う出力行列の変換
$$\begin{eqnarray} \boldsymbol{\bar{A}}=\href{https://shikitenkai.blogspot.com/2021/04/blog-post_4.html} {\boldsymbol{T}^{-1}\boldsymbol{A}\boldsymbol{T}} &=& \frac{1}{-2\xi} \begin{bmatrix} -\gamma-\xi & -1 \\\gamma-\xi& 1 \end{bmatrix} \begin{bmatrix} 0&1\\ -\omega_0^2&-2\gamma\\ \end{bmatrix} \begin{bmatrix} 1 & 1 \\-\gamma+\xi & -\gamma-\xi \end{bmatrix} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/blog-post_56.html}{\boldsymbol{T}=\begin{bmatrix} 1 & 1 \\-\gamma+\xi & -\gamma-\xi \end{bmatrix}} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/blog-post_56.html}{\boldsymbol{T}^{-1}=\frac{1}{-2\xi} \begin{bmatrix} -\gamma-\xi & -1 \\\gamma-\xi& 1 \end{bmatrix}} \\&&\;\ldots\;\xi=\sqrt{\gamma^2-\omega_0^2},\;\xi^2=\gamma^2-\omega_0^2,\;\omega_0^2=\gamma^2-\xi^2 \\ \\&=& \frac{1}{-2\xi} \begin{bmatrix} -\left(\gamma+\xi\right) & -1 \\\gamma-\xi& 1 \end{bmatrix} \begin{bmatrix} 0&1\\ -\left(\gamma^2-\xi^2\right)&-2\gamma\\ \end{bmatrix} \begin{bmatrix} 1 & 1 \\-\gamma+\xi & -\left(\gamma+\xi\right) \end{bmatrix} \\ \\&=& \frac{1}{-2\xi} \begin{bmatrix} -\left(\gamma+\xi\right) & -1 \\\gamma-\xi& 1 \end{bmatrix} \begin{bmatrix} 0\cdot1+1\cdot\left(-\gamma+\xi\right) & 0\cdot1+1\cdot-\left(\gamma+\xi\right)\\ -\left(\gamma^2-\xi^2\right)\cdot1+\left(-2\gamma\right)\cdot\left(-\gamma+\xi\right) & -\left(\gamma^2-\xi^2\right)\cdot1+\left(-2\gamma\right)\cdot-\left(\gamma+\xi\right)\\ \end{bmatrix} \\ \\&=& \frac{1}{-2\xi} \begin{bmatrix} -\left(\gamma+\xi\right) & -1 \\\gamma-\xi& 1 \end{bmatrix} \begin{bmatrix} -\gamma+\xi & -\left(\gamma+\xi\right)\\ -\gamma^2+\xi^2+2\gamma^2-2\gamma\xi & -\gamma^2+\xi^2+2\gamma^2+2\gamma\xi\\ \end{bmatrix} \\ \\&=& \frac{1}{-2\xi} \begin{bmatrix} -\left(\gamma+\xi\right) & -1 \\\gamma-\xi& 1 \end{bmatrix} \begin{bmatrix} -\gamma+\xi & -\left(\gamma+\xi\right)\\ \xi^2+\gamma^2-2\gamma\xi & \xi^2+\gamma^2+2\gamma\xi\\ \end{bmatrix} \\ \\&=& \frac{1}{-2\xi} \begin{bmatrix} -\left(\gamma+\xi\right) & -1 \\\gamma-\xi& 1 \end{bmatrix} \begin{bmatrix} -\gamma+\xi & -\left(\gamma+\xi\right)\\ \left(\gamma-\xi\right)^2 & \left(\gamma+\xi\right)^2\\ \end{bmatrix} \\ \\&=& \frac{1}{-2\xi} \begin{bmatrix} -\left(\gamma+\xi\right)\cdot\left(-\gamma+\xi\right)+(-1)\cdot\left(\gamma-\xi\right)^2 & -\left(\gamma+\xi\right)\cdot\left\{-\left(\gamma+\xi\right)\right\}+(-1)\cdot\left(\gamma+\xi\right)^2 \\\left(\gamma-\xi\right)\cdot\left(-\gamma+\xi\right)+(1)\cdot\left(\gamma-\xi\right)^2 & \left(\gamma-\xi\right)\cdot\left\{-\left(\gamma+\xi\right)\right\}+(1)\cdot\left(\gamma+\xi\right)^2 \end{bmatrix} \\ \\&=& \frac{1}{-2\xi} \begin{bmatrix} -\left(\xi^2-\gamma^2\right)-\left(\gamma-\xi\right)^2 & \left(\gamma+\xi\right)^2-\left(\gamma+\xi\right)^2 \\-\left(\gamma-\xi\right)^2+\left(\gamma-\xi\right)^2 & -\left(\gamma^2-\xi^2\right)+\left(\gamma+\xi\right)^2 \end{bmatrix} \\ \\&=& \frac{1}{-2\xi} \begin{bmatrix} -\xi^2+\gamma^2-\left(\gamma^2-2\gamma\xi+\xi^2\right) & 0 \\0 & -\gamma^2+\xi^2+\left(\gamma^2+2\gamma\xi+\xi^2\right) \end{bmatrix} \\ \\&=& \frac{1}{-2\xi} \begin{bmatrix} -\xi^2\color{red}{+\gamma^2-\gamma^2}\color{black}{+}2\gamma\xi-\xi^2 & 0 \\0 & \color{red}{-\gamma^2}\color{black}{+}\xi^2\color{red}{+\gamma^2}\color{black}{+}2\gamma\xi+\xi^2 \end{bmatrix} \\ \\&=& \frac{1}{-2\xi} \begin{bmatrix} 2\gamma\xi-2\xi^2 & 0 \\0 & 2\gamma\xi+2\xi^2 \end{bmatrix} = \begin{bmatrix} \frac{2\gamma\xi-2\xi^2}{-2\xi} & 0 \\0 & \frac{2\gamma\xi+2\xi^2}{-2\xi} \end{bmatrix} \\&=& \begin{bmatrix} -\gamma+\xi & 0 \\0 & -\gamma-\xi \end{bmatrix} \\ \\&=& \begin{bmatrix} \lambda_1& 0 \\0 & \lambda_2 \end{bmatrix} \\\boldsymbol{\bar{C}}=\href{https://shikitenkai.blogspot.com/2021/04/blog-post_4.html}{\boldsymbol{C}\boldsymbol{T}} &=& \begin{bmatrix} 1&0\\ \end{bmatrix} \begin{bmatrix} 1 & 1 \\-\gamma+\xi & -\gamma-\xi \end{bmatrix} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/blog-post_56.html}{\boldsymbol{T}=\begin{bmatrix} 1 & 1 \\-\gamma+\xi & -\gamma-\xi \end{bmatrix}} \\&=& \begin{bmatrix} 1 \cdot 1 + 0 \cdot \left(-\gamma+\xi\right) & 1 \cdot 1 + 0 \cdot \left(-\gamma-\xi\right) \end{bmatrix} \\&=& \begin{bmatrix} 1 & 1 \end{bmatrix} \end{eqnarray}$$状態空間表現，バネマスダンパー系，状態行列の基底の変換行列
状態空間表現
$$\left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t}\boldsymbol{x}&=&\boldsymbol{A}\boldsymbol{x}&+&\boldsymbol{B}\boldsymbol{u} \;\ldots\;\boldsymbol{A}:状態行列,\;\boldsymbol{B}:入力行列,\boldsymbol{x}:状態ベクトル,\;\boldsymbol{u}:入力ベクトル \\\boldsymbol{y}&=&\boldsymbol{C}\boldsymbol{x}&+&\boldsymbol{D}\boldsymbol{u} \;\ldots\;\boldsymbol{C}:出力行列,\;\boldsymbol{D}:直達行列,\;\boldsymbol{y}:出力ベクトル \end{eqnarray} \right.$$バネマスダンパー系(入力なし)
$$\left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t} \begin{bmatrix} x\\ v\\ \end{bmatrix} &=& \begin{bmatrix} 0&1\\ -\frac{k}{m}&-\frac{c}{m}\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix}\;\ldots\;k:バネ,\;m:マス,\;c:ダンパー,\;x:位置,\;v=\frac{\mathrm{d}x}{\mathrm{d}t}:速度 \\ \\&=& \begin{bmatrix} 0&1\\ -\omega_0^2&-2\gamma\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix} \;\ldots\;\gamma=\frac{c}{2m},\;\omega_0=\sqrt{\frac{k}{m}} \\ \begin{bmatrix} y \end{bmatrix} &=& \begin{bmatrix} 1&0\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix} \end{eqnarray} \right.$$基底の変換行列\(\begin{bmatrix} \boldsymbol{v}_1 &\boldsymbol{v}_2 \end{bmatrix}\)
$$\begin{eqnarray} \boldsymbol{T}&=& \href{ https://shikitenkai.blogspot.com/2021/04/blog-post_4.html } { \begin{bmatrix} \boldsymbol{v}_1 &\boldsymbol{v}_2 \end{bmatrix} } \\&=&\begin{bmatrix} \begin{bmatrix}v_{11} \\ v_{12}\end{bmatrix}& \begin{bmatrix}v_{21} \\ v_{22}\end{bmatrix} \end{bmatrix} \\&=&\begin{bmatrix} v_{11} & v_{21} \\v_{12} & v_{22} \end{bmatrix} \\&=&\begin{bmatrix} 1 & 1 \\-\gamma+\xi & -\gamma-\xi \end{bmatrix} \\&&\;\ldots\; \href{https://shikitenkai.blogspot.com/2021/04/blog-post_55.html}{\boldsymbol{v}_1 = \begin{bmatrix} v_{11} \\v_{12} \end{bmatrix} = t\begin{bmatrix} 1 \\-\gamma+\xi \end{bmatrix}} \\&&\;\ldots\; \href{https://shikitenkai.blogspot.com/2021/04/blog-post_55.html}{\boldsymbol{v}_2 = \begin{bmatrix} v_{21} \\v_{22} \end{bmatrix} = t\begin{bmatrix} 1 \\-\gamma-\xi \end{bmatrix}} \\&&\;\ldots\;\xi=\sqrt{\gamma^2-\omega_0^2}, \end{eqnarray}$$基底の変換行列の逆行列
$$\begin{eqnarray} \boldsymbol{T}^{-1}&=&\frac{1}{\begin{vmatrix}T\end{vmatrix}} \begin{bmatrix} v_{22} & -v_{21} \\-v_{12} & v_{11} \end{bmatrix} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/05/blog-post_96.html}{ \boldsymbol{A}= \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix},\; \boldsymbol{A}^{-1}= \frac{1}{ |A| } \begin{bmatrix} d & -b \\ -c & a \\ \end{bmatrix}} \\ \\&=&\frac{1}{1\cdot\left(-\gamma-\xi\right)-1\cdot\left(-\gamma+\xi\right)} \begin{bmatrix} -\gamma-\xi & -1 \\-\left(-\gamma+\xi\right) & 1 \end{bmatrix} \\&=&\frac{1}{-\gamma-\xi+\gamma-\xi} \begin{bmatrix} -\gamma-\xi & -1 \\\gamma-\xi& 1 \end{bmatrix} \\&=&\frac{1}{-2\xi} \begin{bmatrix} -\gamma-\xi & -1 \\\gamma-\xi& 1 \end{bmatrix} \end{eqnarray}$$状態空間表現，対角正準形式
状態空間表現
$$ \left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t}\boldsymbol{x}&=&\boldsymbol{A}\boldsymbol{x}&+&\boldsymbol{B}\boldsymbol{u} \\\boldsymbol{y}&=&\boldsymbol{C}\boldsymbol{x}&+&\boldsymbol{D}\boldsymbol{u} \end{eqnarray} \right. $$対角正準形式
状態ベクトル\(\boldsymbol{x}\)が正則行列\(\boldsymbol{T}\)と\(\boldsymbol{x}^{\prime}\)を用いて表せるとする． $$ \begin{eqnarray} \boldsymbol{x}&=&\boldsymbol{T}\boldsymbol{x^{\prime}} \end{eqnarray} $$ これを代入すると状態空間表現は以下のようになる． $$ \left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t}\boldsymbol{T}\boldsymbol{x^{\prime}}&=&\boldsymbol{A}\boldsymbol{T}\boldsymbol{x^{\prime}}&+&\boldsymbol{B}\boldsymbol{u} \\\boldsymbol{y}&=&\boldsymbol{C}\boldsymbol{T}\boldsymbol{x^{\prime}}&+&\boldsymbol{D}\boldsymbol{u} \end{eqnarray} \right. $$ この時，一つ目の式に対して\(\boldsymbol{T}\)の逆行列である\(\boldsymbol{T}^{-1}\)を左から掛けると以下のようになる． $$ \left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t}\boldsymbol{T}^{-1}\boldsymbol{T}\boldsymbol{x^{\prime}}&=&\boldsymbol{T}^{-1}\boldsymbol{A}\boldsymbol{T}&\boldsymbol{x^{\prime}}&+&\boldsymbol{T}^{-1}&\boldsymbol{B}\boldsymbol{u} \\\boldsymbol{y}&=&\boldsymbol{C}\boldsymbol{T}&\boldsymbol{x^{\prime}}&+&\boldsymbol{D}\boldsymbol{u} \end{eqnarray} \right. $$ \(\boldsymbol{T}^{-1}\boldsymbol{T}=\boldsymbol{I}\)なので以下のようになる． $$ \left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t}\boldsymbol{x^{\prime}}&=&\boldsymbol{T}^{-1}\boldsymbol{A}\boldsymbol{T}&\boldsymbol{x^{\prime}}&+&\boldsymbol{T}^{-1}&\boldsymbol{B}\boldsymbol{u} \\\boldsymbol{y}&=&\boldsymbol{C}\boldsymbol{T}&\boldsymbol{x^{\prime}}&+&\boldsymbol{D}\boldsymbol{u} \end{eqnarray} \right. $$ 各行列部を\( \boldsymbol{\bar{A}}, \boldsymbol{\bar{B}}, \boldsymbol{\bar{C}} \)で置き直すことで以下のようになる． $$ \left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t}\boldsymbol{x^{\prime}}&=&\boldsymbol{\bar{A}}&\boldsymbol{x^{\prime}}&+&\boldsymbol{\bar{B}}\boldsymbol{u} \;\ldots\;\boldsymbol{\bar{A}}=\boldsymbol{T}^{-1}\boldsymbol{A}\boldsymbol{T},\;\boldsymbol{\bar{B}}=\boldsymbol{T}^{-1}&\boldsymbol{B} \\\boldsymbol{y}&=&\boldsymbol{\bar{C}}&\boldsymbol{x^{\prime}}&+&\boldsymbol{D}\boldsymbol{u} \;\ldots\;\boldsymbol{\bar{C}}=\boldsymbol{C}\boldsymbol{T} \end{eqnarray} \right. $$ この時に\(\boldsymbol{T}\)として\(\boldsymbol{A}\)の固有ベクトル(基底)を並べた行列(変換行列)を用いれば\(\boldsymbol{T}^{-1}\boldsymbol{A}\boldsymbol{T}=\boldsymbol{\bar{A}}\)は対角化され,対角行列となり，このような状態空間表現を対角正準形と呼ぶ． $$ \begin{eqnarray} \boldsymbol{T}&=&\begin{bmatrix}\boldsymbol{v}_{1}&\boldsymbol{v}_{2}&\cdots&\boldsymbol{v}_{n}\end{bmatrix} \end{eqnarray} $$ 対角正準形では状態ベクトル\(\boldsymbol{x}\)は状態ベクトル\(\boldsymbol{x}^{\prime}\)へと変換されている． $$ \begin{eqnarray} \boldsymbol{x}&=&\boldsymbol{T}\boldsymbol{x^{\prime}} \\\boldsymbol{T}^{-1}\boldsymbol{x}&=&\boldsymbol{T}^{-1}\boldsymbol{T}\boldsymbol{x^{\prime}} \\\boldsymbol{x^{\prime}}&=&\boldsymbol{T}^{-1}\boldsymbol{x} \end{eqnarray} $$状態空間表現，バネマスダンパー系(入力なし)
状態空間表現
$$\left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t}\boldsymbol{x}&=&\boldsymbol{A}\boldsymbol{x}&+&\boldsymbol{B}\boldsymbol{u} \;\ldots\;\boldsymbol{A}:状態行列,\;\boldsymbol{B}:入力行列,\boldsymbol{x}:状態ベクトル,\;\boldsymbol{u}:入力ベクトル \\\boldsymbol{y}&=&\boldsymbol{C}\boldsymbol{x}&+&\boldsymbol{D}\boldsymbol{u} \;\ldots\;\boldsymbol{C}:出力行列,\;\boldsymbol{D}:直達行列,\;\boldsymbol{y}:出力ベクトル \end{eqnarray} \right.$$ $$\left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t} \begin{bmatrix} x_1\\ x_2\\ \end{bmatrix} &=& \begin{bmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ \end{bmatrix} &+& \begin{bmatrix} b_{11}&b_{12}\\ b_{21}&b_{22}\\ \end{bmatrix} \begin{bmatrix} u_1\\ u_2\\ \end{bmatrix}\;\ldots\;状態ベクトルが2次の場合 \\ \begin{bmatrix} y_1\\ y_2\\ \end{bmatrix} &=& \begin{bmatrix} c_{11}&c_{12}\\ c_{21}&c_{22}\\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ \end{bmatrix} &+& \begin{bmatrix} d_{11}&d_{12}\\ d_{21}&d_{22}\\ \end{bmatrix} \begin{bmatrix} u_1\\ u_2\\ \end{bmatrix} \end{eqnarray} \right.$$
バネマスダンパー系
$$\begin{eqnarray} m\ddot{x}+c\dot{x}+kx&=&0\;\ldots\;k:バネ,\;m:マス,\;c:ダンパー,\;x:位置,\;v=\frac{\mathrm{d}x}{\mathrm{d}t}:速度 \\\frac{\mathrm{d^2}x}{\mathrm{d^2}t} +\frac{c}{m}\frac{\mathrm{d}x}{\mathrm{d}t} +\frac{k}{m}x &=&0 \end{eqnarray}$$ $$\begin{eqnarray} \frac{\mathrm{d^2}x}{\mathrm{d^2}t} &=&-\frac{c}{m}\frac{\mathrm{d}x}{\mathrm{d}t}-\frac{k}{m}x \\\frac{\mathrm{d}v}{\mathrm{d}t} &=&-\frac{c}{m}v-\frac{k}{m}x \\&&\;\ldots\;v=\frac{\mathrm{d}x}{\mathrm{d}t} \\&&\;\ldots\;\frac{\mathrm{d^2}x}{\mathrm{d^2}t}=\frac{\mathrm{d}}{\mathrm{d}t}\frac{\mathrm{d}}{\mathrm{d}t}x=\frac{\mathrm{d}}{\mathrm{d}t}v \end{eqnarray}$$ $$\left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t} \begin{bmatrix} x\\ v\\ \end{bmatrix} &=& \begin{bmatrix} 0&1\\ -\frac{k}{m}&-\frac{c}{m}\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix}\;\ldots\;k:バネ,\;m:マス,\;c:ダンパー \\ \begin{bmatrix} y\\ \end{bmatrix} &=& \begin{bmatrix} 1&0\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix} \end{eqnarray} \right.$$ 入力行列\(\boldsymbol{B}\)と直達行列\(\boldsymbol{D}\)はともに\(\boldsymbol{0}\)とする(入力なし)．状態空間表現，バネマスダンパー系，状態行列の固有値
状態空間表現
$$\left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t}\boldsymbol{x}&=&\boldsymbol{A}\boldsymbol{x}&+&\boldsymbol{B}\boldsymbol{u} \;\ldots\;\boldsymbol{A}:状態行列,\;\boldsymbol{B}:入力行列,\boldsymbol{x}:状態ベクトル,\;\boldsymbol{u}:入力ベクトル \\\boldsymbol{y}&=&\boldsymbol{C}\boldsymbol{x}&+&\boldsymbol{D}\boldsymbol{u} \;\ldots\;\boldsymbol{C}:出力行列,\;\boldsymbol{D}:直達行列,\;\boldsymbol{y}:出力ベクトル \end{eqnarray} \right.$$バネマスダンパー系(入力なし)
$$\left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t} \begin{bmatrix} x\\ v\\ \end{bmatrix} &=& \begin{bmatrix} 0&1\\ -\frac{k}{m}&-\frac{c}{m}\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix}\;\ldots\;k:バネ,\;m:マス,\;c:ダンパー,\;x:位置,\;v=\frac{\mathrm{d}x}{\mathrm{d}t}:速度 \\ \begin{bmatrix} y\\ \end{bmatrix} &=& \begin{bmatrix} 1&0\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix} \end{eqnarray} \right.$$状態行列の固有値
$$\begin{eqnarray} \left|\lambda\boldsymbol{I}-\boldsymbol{A}\right|&=&0 \\\begin{vmatrix} \lambda-0&1\\ -\frac{k}{m}&\lambda-\left(-\frac{c}{m}\right)\\ \end{vmatrix} &=&\lambda\left\{\lambda-\left(-\frac{c}{m}\right)\right\}-\left(-\frac{k}{m}\right) \\&=&\lambda\left(\lambda+\frac{c}{m}\right)+\frac{k}{m} \\&=&\lambda^2+\frac{c}{m}\lambda+\frac{k}{m} \\\lambda^2+\frac{c}{m}\lambda+\frac{k}{m}&=&0 \\\lambda_{1,2}&=&\frac{-\frac{c}{m}\pm\sqrt{\left(\frac{c}{m}\right)^2-4\frac{k}{m}}}{2} \\&=&-\frac{c}{2m}\pm\sqrt{\frac{c^2}{4m^2}-\frac{k}{m}} \\&=&-\gamma\pm\sqrt{\gamma^2-\omega_0^2}\;\ldots\;\gamma=\frac{c}{2m},\;\omega_0=\sqrt{\frac{k}{m}} \\&=&-\gamma\pm\frac{\omega_0}{\omega_0} \sqrt{\gamma^2-\omega_0^2} \\&=&-\gamma\pm\omega_0 \sqrt{\frac{1}{\omega_0^2}\left(\gamma^2-\omega_0^2\right)} \\&=&-\gamma\pm\omega_0 \sqrt{\frac{\gamma^2}{\omega_0^2}-\frac{\omega_0^2}{\omega_0^2}} \\&=&-\gamma\pm\omega_0 \sqrt{\frac{\gamma^2}{\omega_0^2}-1} \\&=&-\gamma\pm\omega_0 \sqrt{\left(\frac{\gamma}{\omega_0}\right)^2-1} \\&=&-\gamma\pm\omega_0 \sqrt{\zeta^2-1}\;\ldots\;\zeta=\frac{\gamma}{\omega_0} \\&=&-\gamma\pm\xi\;\ldots\;\xi=\omega_0 \sqrt{\zeta^2-1} \end{eqnarray}$$状態空間表現，バネマスダンパー系，状態行列の固有ベクトル
状態空間表現
$$\left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t}\boldsymbol{x}&=&\boldsymbol{A}\boldsymbol{x}&+&\boldsymbol{B}\boldsymbol{u} \;\ldots\;\boldsymbol{A}:状態行列,\;\boldsymbol{B}:入力行列,\boldsymbol{x}:状態ベクトル,\;\boldsymbol{u}:入力ベクトル \\\boldsymbol{y}&=&\boldsymbol{C}\boldsymbol{x}&+&\boldsymbol{D}\boldsymbol{u} \;\ldots\;\boldsymbol{C}:出力行列,\;\boldsymbol{D}:直達行列,\;\boldsymbol{y}:出力ベクトル \end{eqnarray} \right.$$バネマスダンパー系(入力なし)
$$\left\{ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t} \begin{bmatrix} x\\ v\\ \end{bmatrix} &=& \begin{bmatrix} 0&1\\ -\frac{k}{m}&-\frac{c}{m}\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix}\;\ldots\;k:バネ,\;m:マス,\;c:ダンパー,\;x:位置,\;v=\frac{\mathrm{d}x}{\mathrm{d}t}:速度 \\ \\&=& \begin{bmatrix} 0&1\\ -\omega_0^2&-2\gamma\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix} \;\ldots\;\gamma=\frac{c}{2m},\;\omega_0=\sqrt{\frac{k}{m}} \\ \begin{bmatrix} y\\ \end{bmatrix} &=& \begin{bmatrix} 1&0\\ \end{bmatrix} \begin{bmatrix} x\\ v\\ \end{bmatrix} \end{eqnarray} \right.$$固有値\(\lambda_{1,2}\)
$$\begin{eqnarray} \lambda_{1,2}&=&-\gamma\pm\sqrt{\gamma^2-\omega_0^2}&\;\ldots\;\gamma=\frac{c}{2m},\;\omega_0=\sqrt{\frac{k}{m}} \\&=&-\gamma\pm\omega_0\sqrt{\zeta^2-1}&\;\ldots\;\zeta=\frac{\gamma}{\omega_0} \\&=&-\gamma\pm\xi&\;\ldots\;\xi=\omega_0 \sqrt{\zeta^2-1}=\sqrt{\gamma^2-\omega_0^2}, \end{eqnarray}$$固有値\(\lambda_1\)に対応する固有ベクトル\(\boldsymbol{v}_1\)
$$\begin{eqnarray} \left(\lambda_1\boldsymbol{I}-\boldsymbol{A}\right)\boldsymbol{v}_1&=&0 \\\left( \begin{bmatrix} \lambda_1&0\\ 0&\lambda_1\\ \end{bmatrix} - \begin{bmatrix} 0&1\\ -\omega_0^2&-2\gamma\\ \end{bmatrix} \right)\boldsymbol{v}_1&=&0 \\\begin{bmatrix} \lambda_1&-1\\ \omega_0^2&\lambda_1+2\gamma\\ \end{bmatrix}\boldsymbol{v}_1 &=&0 \\\begin{bmatrix} \left(-\gamma+\xi\right)&-1\\ \omega_0^2&\left(-\gamma+\xi\right)+2\gamma\\ \end{bmatrix}\boldsymbol{v}_1 &=&0\;\ldots\;\lambda_1=-\gamma+\xi \\\begin{bmatrix} -\gamma+\xi&-1\\ \omega_0^2&\gamma+\xi\\ \end{bmatrix}\boldsymbol{v}_1 &=&0 \end{eqnarray}$$ $$\begin{eqnarray} \\\left\{\begin{matrix} \left(-\gamma+\xi\right) &v_{11}&-&&v_{12}&=&0 \\\omega_0^2 &v_{11}&+&\left(\gamma+\xi\right)&v_{12}&=&0 \end{matrix}\right. \end{eqnarray}$$ $$\begin{eqnarray} v_{11}&=&t \\\left(-\gamma+\xi\right) t-v_{12}&=&0 \\-v_{12}&=&-\left(-\gamma+\xi\right) t \\v_{12}&=&\left(-\gamma+\xi\right) t \end{eqnarray}$$ $$\begin{eqnarray} \boldsymbol{v}_1 = \begin{bmatrix} v_{11} \\v_{12} \end{bmatrix} = t\begin{bmatrix} 1 \\-\gamma+\xi \end{bmatrix} \end{eqnarray}$$2番目の式で確認
$$\begin{eqnarray}
\omega_0^2 v_{11}+\left(\gamma+\xi\right)v_{12}&=&0
\\\omega_0^2 t+\left(\gamma+\xi\right)v_{12}&=&0\;\ldots\;v_{11}=t
\\\left(\gamma+\xi\right)v_{12}&=&-\omega_0^2 t
\\v_{12}&=&\frac{-\omega_0^2}{\gamma+\xi}t
\\&=&\frac{-\left(\gamma^2-\xi^2\right)}{\gamma+\xi}t
\\&&\;\ldots\;\xi=\sqrt{\gamma^2-\omega_0^2},\;\xi^2=\gamma^2-\omega_0^2,\;\omega_0^2=\gamma^2-\xi^2
\\&=&\frac{-\left(\gamma+\xi\right)\left(\gamma-\xi\right)}{\gamma+\xi}t
\\&=&-\left(\gamma-\xi\right)t
\\&=&\left(-\gamma+\xi\right)t
\end{eqnarray}$$
固有値\(\lambda_2\)に対応する固有ベクトル\(\boldsymbol{v}_2\)
$$\begin{eqnarray} \left(\lambda_2\boldsymbol{I}-\boldsymbol{A}\right)\boldsymbol{v}_2&=&0 \\\left( \begin{bmatrix} \lambda_2&0\\ 0&\lambda_2\\ \end{bmatrix} - \begin{bmatrix} 0&1\\ -\omega_0^2&-2\gamma\\ \end{bmatrix} \right)\boldsymbol{v}_2&=&0 \\\begin{bmatrix} \lambda_2&-1\\ \omega_0^2&\lambda_2+2\gamma\\ \end{bmatrix}\boldsymbol{v}_2&=&0 \\\begin{bmatrix} -\gamma-\xi&-1\\ \omega_0^2&\left(-\gamma-\xi\right)+2\gamma\\ \end{bmatrix}\boldsymbol{v}_2&=&0\;\ldots\;\lambda_2=-\gamma-\xi \\\begin{bmatrix} -\gamma+\xi&-1\\ \omega_0^2&\gamma-\xi\\ \end{bmatrix}\boldsymbol{v}_2&=&0 \end{eqnarray}$$ $$\begin{eqnarray} \\\left\{\begin{matrix} \left(-\gamma-\xi\right)&v_{21}&-&&v_{22}&=0 \\\omega_0^2 &v_{21}&+&\left(\gamma-\xi\right)&v_{22}&=0 \end{matrix}\right. \end{eqnarray}$$ $$\begin{eqnarray} \\v_{21}&=&t \\\left(-\gamma-\xi\right) t-v_{22}&=&0 \\-v_{22}&=&-\left(-\gamma-\xi\right) t \\v_{22}&=&\left(-\gamma-\xi\right) t \end{eqnarray}$$ $$\begin{eqnarray} \\\boldsymbol{v}_2 = \begin{bmatrix} v_{21} \\v_{22} \end{bmatrix} = t\begin{bmatrix} 1 \\-\gamma-\xi \end{bmatrix} \end{eqnarray}$$2番目の式で確認
$$\begin{eqnarray}
\omega_0^2 v_{21}+\left(\gamma-\xi\right)v_{22}&=&0
\\\omega_0^2 t+\left(\gamma-\xi\right)v_{22}&=&0\;\ldots\;v_{21}=t
\\\left(\gamma-\xi\right)v_{22}&=&-\omega_0^2 t
\\v_{22}&=&\frac{-\omega_0^2}{\gamma-\xi}t
\\&=&\frac{-\left(\gamma^2-\xi^2\right)}{\gamma-\xi}t
\\&&\;\ldots\;\xi=\sqrt{\gamma^2-\omega_0^2},\;\xi^2=\gamma^2-\omega_0^2,\;\omega_0^2=\gamma^2-\xi^2
\\&=&\frac{-\left(\gamma+\xi\right)\left(\gamma-\xi\right)}{\gamma-\xi}t
\\&=&-\left(\gamma+\xi\right)t
\\&=&\left(-\gamma-\xi\right)t
\end{eqnarray}$$
e^xとマクローリン展開と双曲線凾数
\(e^{x}\)とマクローリン展開と双曲線凾数
\begin{eqnarray} e^{x} &=&1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\frac{1}{4!} x^4+\frac{1}{5!} x^5+\frac{1}{6!}x^6+\frac{1}{7!}x^7+\frac{1}{8!}x^8 +\cdots \\&=&\left(1+\frac{1}{2!}x^2+\frac{1}{4!} x^4+\frac{1}{6!}x^6+\frac{1}{8!}x^8+\cdots\right) +\left(x+\frac{1}{3!}x^3+\frac{1}{5!} x^5+\frac{1}{7!}x^7+\cdots\right) \\&=&\cosh{\left(x\right)}+\sinh{\left(x\right)}\;\ldots\;xが複素数でない=実数の場合 \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/coshx.html}{\cosh{\left(x\right)}=1+\frac{1}{2!}x^2+\frac{1}{4!} x^4+\frac{1}{6!}x^6+\frac{1}{8!}x^8+\cdots} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/sinhx.html}{\sinh{\left(x\right)}=x+\frac{1}{3!}x^3+\frac{1}{5!} x^5+\frac{1}{7!}x^7+\cdots} \\&=&\frac{e^{x}+e^{-x}}{2}+\frac{e^{x}-e^{-x}}{2}=\frac{e^{x}+e^{-x}+e^{x}-e^{-x}}{2}=\frac{2e^{x}}{2} \\&=&e^{x} \end{eqnarray}バネマスダンパー系，過減衰の式からの他の式の導出
バネマスダンパー系
運動方程式
$$\begin{eqnarray} m\frac{\mathrm{d^2}x}{\mathrm{d^2}t} &+&c\frac{\mathrm{d}x}{\mathrm{d}t} &+&kx &=&0 \;\ldots\;m:マス,\;c:ダンパー,\;バネ \\\frac{\mathrm{d^2}x}{\mathrm{d^2}t} &+&\frac{c}{m}\frac{\mathrm{d}x}{\mathrm{d}t} &+&\frac{k}{m}x &=&0 \\\frac{\mathrm{d^2}x}{\mathrm{d^2}t} &+&2\gamma\frac{\mathrm{d}x}{\mathrm{d}t} &+&\omega_0^2x &=&0 \;\ldots\;\gamma=\frac{c}{2m},\;\omega_0=\sqrt{\frac{k}{m}} \end{eqnarray}$$単振動，振動減衰，臨界減衰，過減衰
\(\gamma=\frac{c}{2m},\;\omega_0=\sqrt{\frac{k}{m}},\;x_0=x\left(0\right),\;v_0=\left.\frac{\mathrm{d}x(t)}{\mathrm{d}t}\right|_{t=0}\)| \(\gamma\) | \(\zeta=\frac{\gamma}{\omega_0}\) | \(\xi=\omega_0\sqrt{\zeta^2-1}\) | \(e^{-\gamma t}\left[x_0 \cosh\left(\xi t\right)+\frac{v_0 +\gamma x_0 }{\xi}\sinh{\left(\xi t \right)}\right]\) | \(x(t)\) | |
|---|---|---|---|---|---|
| 単振動 | \(0\) | \(0\) | \(\omega_0i\) | \(e^{-\color{red}{0}\color{black}{t}}\left[x_0 \cosh\left(\color{red}{\omega_0 i}\color{black}{t}\right)+\frac{v_0 +\color{red}{0}\color{black}{ x_0} }{\color{red}{\omega_0 i}} \sinh{\left(\color{red}{\omega_0 i}\color{black}{t}\right)}\right]\) | \(x_0 \cos\left(\omega_0 t\right)+\frac{v_0}{\omega_0 } \sin{\left(\omega_0 t \right)}\) |
| 振動減衰 | \(0\lt\gamma\lt\omega_0\) | \(0\lt\zeta\lt1\) | \(\omega i\) \(ただし0\lt\omega\lt1\) |
\(e^{-\color{red}{\gamma}\color{black}{t}}\left[x_0 \cosh\left(\color{red}{\omega i}\color{black}{t}\right)+\frac{v_0 +\color{red}{\gamma}\color{black}{x_0} }{\color{red}{\omega i}}\sinh{\left(\color{red}{\omega i}\color{black}{t}\right)}\right]\) | \(e^{-\gamma t}\left[x_0 \cos\left(\omega i t \right)+\frac{v_0 +\gamma x_0 }{\omega } \sin{\left(\omega t \right)}\right]\) |
| 臨界減衰 | \(\omega_0\) | \(1\) | \(0\) | \(e^{-\color{red}{\gamma}\color{black}{t}}\left[x_0 \color{red}{\left(1+\frac{1}{2!}\left(\xi t\right)^2+\cdots\right)} \color{black}{+}\frac{v_0 +\color{red}{\gamma}\color{black}{x_0} }{\xi} \color{red}{\left( \xi t + \frac{1}{3!}\left(\xi t\right)^3 + \cdots\right)} \right]\) | \(e^{-\gamma t}\left[x_0 +\left(v_0 +\gamma x_0\right) t\right]\) |
| 過減衰 | \(\omega_0\lt\gamma\) | \(1\lt\zeta\) | \(\omega\) \(ただし1\lt\omega\) |
\(e^{-\color{red}{\gamma}\color{black}{t}}\left[x_0 \cosh\left(\color{red}{\omega}\color{black}{t}\right)+\frac{v_0 +\color{red}{\gamma}\color{black}{x_0} }{\color{red}{\omega}}\sinh{\left(\color{red}{\omega}\color{black}{t}\right)}\right]\) | \(e^{-\gamma t}\left[x_0 \cosh\left(\omega t\right)+\frac{v_0 +\gamma x_0 }{\omega}\sinh{\left(\omega t \right)}\right]\) |
過減衰の式
$$\begin{eqnarray} x(t)&=&\href{https://shikitenkai.blogspot.com/2021/04/0_17.html}{e^{-\gamma t}\left[x_0 \cosh\left(\xi t\right)+\frac{v_0 +\gamma x_0 }{\xi}\sinh{\left(\xi t \right)}\right]}\;\ldots\;(\href{https://shikitenkai.blogspot.com/2021/04/0_17.html}{導出}) \end{eqnarray}$$\(\omega_0\lt\gamma\)
$$\begin{eqnarray} x(t)&=&\left.e^{-\gamma t}\left[x_0 \cosh\left(\xi t\right)+\frac{v_0 +\gamma x_0 }{\xi}\sinh{\left(\xi t \right)}\right]\right|_{1\lt\gamma,\;\xi=\omega} \\&&\;\ldots\;\omega_0\lt\gamma,\;1\lt\zeta,\;\xi=\omega_0\sqrt{\zeta^2-1}=\omega \\&&\;\ldots\;\omega=\omega_0\sqrt{\left|\zeta^2-1\right|} \\&=&e^{-\gamma t}\left[x_0 \cosh\left(\omega t\right)+\frac{v_0 +\gamma x_0 }{\omega}\sinh{\left(\omega t \right)}\right] \end{eqnarray}$$\(0\lt\gamma\lt\omega_0\)
$$\begin{eqnarray} \\x(t)&=&\left.e^{-\gamma t}\left[x_0 \cosh\left(\xi t\right)+\frac{v_0 +\gamma x_0 }{\xi}\sinh{\left(\xi t \right)}\right]\right|_{0\lt\gamma\lt1,\;\xi=\omega i} \\&&\;\ldots\;0\lt\gamma\lt\omega_0,\;0\lt\zeta\lt1,\;\xi=\omega_0\sqrt{\zeta^2-1}=\omega i \\&&\;\ldots\;\omega=\omega_0\sqrt{\left|\zeta^2-1\right|} \\&=&e^{-\gamma t}\left[x_0 \cosh\left(\omega i t\right)+\frac{v_0 +\gamma x_0 }{\omega i}\sinh{\left(\omega i t \right)}\right] \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/coshi-x-sinhi-x-cosh-sinh.html}{\cosh{\left(i x\right)}=\cos{\left(x\right)},\;\sinh{\left(i x\right)}=i\sin{\left(x\right)}} \\&=&e^{-\gamma t}\left[x_0 \cos\left(\omega t\right)+\frac{v_0 +\gamma x_0 }{\omega i} i \sin{\left(\omega t \right)}\right] \\&=&e^{-\gamma t}\left[x_0 \cos\left(\omega t\right)+\frac{v_0 +\gamma x_0 }{\omega } \sin{\left(\omega t \right)}\right] \end{eqnarray}$$\(\gamma=0\)
$$\begin{eqnarray} \\x(t)&=&\left.e^{-\gamma t}\left[x_0 \cosh\left(\xi t\right)+\frac{v_0 +\gamma x_0 }{\xi}\sinh{\left(\xi t \right)}\right]\right|_{\gamma=0,\;\xi=\omega_0 i} \\&&\;\ldots\;\gamma=0,\;\zeta=0,\;\xi=\omega_0\sqrt{0-1}=\omega_0 i \\&&\;\ldots\;\omega=\omega_0\sqrt{\left|\zeta^2-1\right|} \\&=&e^{-0 t}\left[x_0 \cosh\left(\omega_0 i t\right)+\frac{v_0 +0 x_0 }{\omega_0 i} \sinh{\left(\omega_0 i t \right)}\right] \\&=&x_0 \cos\left(\omega_0 t\right)+\frac{v_0 }{\omega_0 i} i \sin{\left(\omega_0 t \right)} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/coshi-x-sinhi-x-cosh-sinh.html}{\cosh{\left(i x\right)}=\cos{\left(x\right)},\;\sinh{\left(i x\right)}=i\sin{\left(x\right)}} \\&&\;\ldots\;e^0=1 \\&=&x_0 \cos\left(\omega_0 t\right)+\frac{v_0}{\omega_0 } \sin{\left(\omega_0 t \right)} \end{eqnarray}$$\(\gamma\rightarrow\omega_0\)
$$\begin{eqnarray} \\x(t)&=&\lim_{\gamma\rightarrow\omega_0,\;\xi \rightarrow 0}{e^{-\gamma t}\left[x_0 \cosh\left(\xi t\right)+\frac{v_0 +\gamma x_0 }{\xi}\sinh{\left(\xi t \right)}\right]} \\&&\;\ldots\;\gamma\rightarrow\omega_0,\;\zeta\rightarrow1,\;\xi\rightarrow0 \\&=&\lim_{\gamma\rightarrow\omega_0,\;\xi \rightarrow 0} {e^{-\gamma t}\left[ x_0 \left(1+\frac{1}{2!}\left(\xi t\right)^2+\cdots\right) +\frac{v_0 +\gamma x_0 }{\xi} \left( \xi t + \frac{1}{3!}\left(\xi t\right)^3 + \cdots\right) \right]} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/coshx.html}{\cosh\left(\xi t\right)=\frac{1}{0!}\left(\xi t\right)^0+\frac{1}{2!}\left(\xi t\right)^2+\cdots} ,\;\href{https://shikitenkai.blogspot.com/2021/04/sinhx.html}{\sinh{\left(\xi t \right)}=\frac{1}{1!}\left(\xi t\right)^1 + \frac{1}{3!}\left(\xi t\right)^3 + \cdots} \\&=&\lim_{\gamma\rightarrow\omega_0,\;\xi \rightarrow 0} {e^{-\gamma t}\left[ x_0 \left(1+\frac{1}{2!}\left(\xi t\right)^2+\cdots\right) +\left(v_0 +\gamma x_0\right) \left( t + \frac{1}{3!}\xi^2 t^3 + \cdots\right) \right]} \\&=&e^{-\omega_0 t}\left[ x_0 \cdot \left(1\right) +\left(v_0 +\omega_0 x_0\right) \cdot \left( t \right) \right] \\&=&e^{-\omega_0 t}\left[x_0 +\left(v_0 +\omega_0 x_0\right) t\right] \\&=&e^{-\gamma t}\left[x_0 +\left(v_0 +\gamma x_0\right) t\right] \end{eqnarray}$$sinh(x)のマクローリン展開
a点まわりのテイラー展開
\begin{eqnarray} f(x) &=& \sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{x!}(x-a)^k \;\ldots\;a点まわりのテイラー展開 \\&=& \frac{1}{0!}f^{(0)}(a)(x-a)^0+\frac{1}{1!}f^{(1)}(a)(x-a)^1+\frac{1}{2!}f^{(2)}(a)(x-a)^2+\dotsb \\&&\;\ldots\;f^{(n)}(x): f(x)のn階微分 \end{eqnarray}マクローリン展開(0点まわりのテイラー展開)
\begin{eqnarray} f(x) &=& \left.\sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{x!}(x-a)^k\right|_{a=0} \\&=& \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{x!}(x-0)^k \\&=& \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{x!}(x)^k \\&=& \frac{1}{0!}f^{(0)}(0)x^0+\frac{1}{1!}f^{(1)}(0)x+\frac{1}{2!}f^{(2)}(0)x^2+\cdots \end{eqnarray}\(f(x)=\sinh{\left(x\right)}\)のマクローリン展開(0点まわりのテイラー展開)
\begin{eqnarray} \sinh{\left(x\right)} &=& \frac{1}{0!}f^{(0)}(0)x^0&+\frac{1}{1!}f^{(1)}(0)x&+\frac{1}{2!}f^{(2)}(0)x^2+\cdots \\&=&\frac{1}{0!}\left\{\sinh{(0)}\right\}x^0&+\frac{1}{1!}\left\{\cosh{(0)}\right\}x &+\frac{1}{2!}\left\{\sinh{(0)}\right\}x^2 \\&&+\frac{1}{3!}\left\{\cosh{(0)}\right\}x^3&+\frac{1}{4!}\left\{\sinh{(0)}\right\}x^4&+\frac{1}{5!}\left\{\cosh{(0)}\right\}x^5 \\&&+\frac{1}{6!}\left\{\sinh{(0)}\right\}x^6&+\frac{1}{7!}\left\{\cosh{(0)}\right\}x^7&+\frac{1}{8!}\left\{\sinh{(0)}\right\}x^8 +\cdots \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/cosh-sinh.html}{\frac{\mathrm{d}}{\mathrm{d}x}\cosh{\left(x\right)}=\sinh{\left(x\right)},\;\frac{\mathrm{d}}{\mathrm{d}x}\sinh{\left(x\right)}=\cosh{\left(x\right)}} \\&=&\frac{1}{0!}\cdot 0\cdot x^0&+\frac{1}{1!}\cdot 1\cdot x &+\frac{1}{2!}\cdot 0\cdot x^2 \\&&+\frac{1}{3!}\cdot 1\cdot x^3&+\frac{1}{4!}\cdot 0\cdot x^4&+\frac{1}{5!}\cdot 1\cdot x^5 \\&&+\frac{1}{6!}\cdot 0\cdot x^6&+\frac{1}{7!}\cdot 1\cdot x^7&+\frac{1}{8!}\cdot 0\cdot x^8 +\cdots \;\ldots\;\cosh{(0)}=\frac{e^0+e^{-0}}{2}=\frac{1+1}{2}=1,\;\sinh{(0)}=\frac{e^0-e^{-0}}{2}=\frac{1-1}{2}=0 \\&=&\frac{1}{1!}x^1+\frac{1}{3!}x^3&+\frac{1}{5!} x^5+\frac{1}{7!}x^7 +\cdots \\&=&x+\frac{1}{3!}x^3+\frac{1}{5!} x^5&+\frac{1}{7!}x^7 +\cdots \end{eqnarray}cosh(x)のマクローリン展開
a点まわりのテイラー展開
\begin{eqnarray} f(x) &=& \sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{x!}(x-a)^k \;\ldots\;a点まわりのテイラー展開 \\&=& \frac{1}{0!}f^{(0)}(a)(x-a)^0+\frac{1}{1!}f^{(1)}(a)(x-a)^1+\frac{1}{2!}f^{(2)}(a)(x-a)^2+\dotsb \\&&\;\ldots\;f^{(n)}(x): f(x)のn階微分 \end{eqnarray}マクローリン展開(0点まわりのテイラー展開)
\begin{eqnarray} f(x) &=& \left.\sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{x!}(x-a)^k\right|_{a=0} \\&=& \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{x!}(x-0)^k \\&=& \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{x!}(x)^k \\&=& \frac{1}{0!}f^{(0)}(0)x^0+\frac{1}{1!}f^{(1)}(0)x+\frac{1}{2!}f^{(2)}(0)x^2+\cdots \end{eqnarray}\(f(x)=\cosh{\left(x\right)}\)のマクローリン展開(0点まわりのテイラー展開)
\begin{eqnarray} \cosh{\left(x\right)} &=& \frac{1}{0!}f^{(0)}(0)x^0&+\frac{1}{1!}f^{(1)}(0)x&+\frac{1}{2!}f^{(2)}(0)x^2+\cdots \\&=&\frac{1}{0!}\left\{\cosh{(0)}\right\}x^0&+\frac{1}{1!}\left\{\sinh{(0)}\right\}x &+\frac{1}{2!}\left\{\cosh{(0)}\right\}x^2 \\&&+\frac{1}{3!}\left\{\sinh{(0)}\right\}x^3&+\frac{1}{4!}\left\{\cosh{(0)}\right\}x^4&+\frac{1}{5!}\left\{\sinh{(0)}\right\}x^5 \\&&+\frac{1}{6!}\left\{\cosh{(0)}\right\}x^6&+\frac{1}{7!}\left\{\sinh{(0)}\right\}x^7&+\frac{1}{8!}\left\{\cosh{(0)}\right\}x^8 +\cdots \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/cosh-sinh.html}{\frac{\mathrm{d}}{\mathrm{d}x}\cosh{\left(x\right)}=\sinh{\left(x\right)},\;\frac{\mathrm{d}}{\mathrm{d}x}\sinh{\left(x\right)}=\cosh{\left(x\right)}} \\&=&\frac{1}{0!}\cdot 1\cdot x^0&+\frac{1}{1!}\cdot 0\cdot x &+\frac{1}{2!}\cdot 1\cdot x^2 \\&&+\frac{1}{3!}\cdot 0\cdot x^3&+\frac{1}{4!}\cdot 1\cdot x^4&+\frac{1}{5!}\cdot 0\cdot x^5 \\&&+\frac{1}{6!}\cdot 1\cdot x^6&+\frac{1}{7!}\cdot 0\cdot x^7&+\frac{1}{8!}\cdot 1\cdot x^8 +\cdots \;\ldots\;\cosh{(0)}=\frac{e^0+e^{-0}}{2}=\frac{1+1}{2}=1,\;\sinh{(0)}=\frac{e^0-e^{-0}}{2}=\frac{1-1}{2}=0 \\&=&\frac{1}{0!}x^0+\frac{1}{2!}x^2&+\frac{1}{4!} x^4+\frac{1}{6!}x^6&+\frac{1}{8!}x^8 +\cdots \\&=&1+\frac{1}{2!}x^2+\frac{1}{4!} x^4&+\frac{1}{6!}x^6+\frac{1}{8!}x^8 +\cdots \end{eqnarray}cosh, sinhの微分
\(\cosh{\left(x\right)}\)の微分
\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}x}\cosh{\left(x\right)} &=& \frac{\mathrm{d}}{\mathrm{d}x}\frac{e^x+e^{-x}}{2} \\&=& \frac{1}{2}\left(\frac{\mathrm{d}}{\mathrm{d}x}e^x+\frac{\mathrm{d}}{\mathrm{d}x}e^{-x}\right) \\&&\;\ldots\;\frac{\mathrm{d}}{\mathrm{d}x}Cf(x)=C\frac{\mathrm{d}}{\mathrm{d}x}f(x) \\&&\;\ldots\;\frac{\mathrm{d}}{\mathrm{d}x}\left(f(x)+g(x)\right)=\frac{\mathrm{d}}{\mathrm{d}x}f(x)+\frac{\mathrm{d}}{\mathrm{d}x}g(x) \\&=& \frac{1}{2}\left(\frac{\mathrm{d}}{\mathrm{d}x}e^x+\frac{\mathrm{d}}{\mathrm{d}x}e^{-x}\right) \\&=& \frac{1}{2}\left\{e^x+(-1)e^{-x}\right\} \\&=& \frac{1}{2}\left(e^x-e^{-x}\right) \\&=& \sinh{\left(x\right)} \end{eqnarray}\(\sinh{\left(x\right)}\)の微分
\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}x}\sinh{\left(x\right)} &=& \frac{\mathrm{d}}{\mathrm{d}x}\frac{e^x-e^{-x}}{2} \\&=& \frac{1}{2}\left(\frac{\mathrm{d}}{\mathrm{d}x}e^x-\frac{\mathrm{d}}{\mathrm{d}x}e^{-x}\right) \\&&\;\ldots\;\frac{\mathrm{d}}{\mathrm{d}x}Cf(x)=C\frac{\mathrm{d}}{\mathrm{d}x}f(x) \\&&\;\ldots\;\frac{\mathrm{d}}{\mathrm{d}x}\left(f(x)+g(x)\right)=\frac{\mathrm{d}}{\mathrm{d}x}f(x)+\frac{\mathrm{d}}{\mathrm{d}x}g(x) \\&=& \frac{1}{2}\left(\frac{\mathrm{d}}{\mathrm{d}x}e^x-\frac{\mathrm{d}}{\mathrm{d}x}e^{-x}\right) \\&=& \frac{1}{2}\left\{e^x-(-1)e^{-x}\right\} \\&=& \frac{1}{2}\left(e^x+e^{-x}\right) \\&=& \cosh{\left(x\right)} \end{eqnarray}cosh(i x), sinh(i x) (純虚数に対するcosh, sinh)
\(\cosh{\left(ix\right)}\)
$$\begin{eqnarray} \cosh{\left(x\right)}&=&\frac{e^x+e^{-x}}{2} \\\cosh{\left(ix\right)}&=&\frac{e^{ix}+e^{-ix}}{2} \\&=&\frac{\left\{\cos{\left(x\right)}+i\sin{\left(x\right)}\right\}+\left\{\cos{\left(-x\right)}+i\sin{\left(-x\right)}\right\}}{2} \\&&\;\ldots\;e^{ix}=\cos{\left(x\right)}+i\sin{\left(x\right)} \\&=&\frac{\left\{\cos{\left(x\right)}+i\sin{\left(x\right)}\right\}+\left\{\cos{\left(x\right)}-i\sin{\left(x\right)}\right\}}{2} \\&&\;\ldots\;\cos{\left(-x\right)}=\cos{\left(x\right)},\;\sin{\left(-x\right)}=-\sin{\left(-x\right)} \\&=&\frac{\cos{\left(x\right)}\color{red}{+i\sin{\left(x\right)}}\color{black}{+\cos{\left(x\right)}}\color{red}{-i\sin{\left(x\right)}}}{2} \\&=&\frac{\cos{\left(x\right)}+\cos{\left(x\right)}}{2} \\&=&\frac{2\cos{\left(x\right)}}{2} \\&=&\cos{\left(x\right)} \end{eqnarray}$$\(\sinh{\left(ix\right)}\)
$$\begin{eqnarray} \sinh{\left(x\right)}&=&\frac{e^x-e^{-x}}{2} \\\sinh{\left(ix\right)}&=&\frac{e^{ix}-e^{-ix}}{2} \\&=&\frac{\left\{\cos{\left(x\right)}+i\sin{\left(x\right)}\right\}-\left\{\cos{\left(-x\right)}+i\sin{\left(-x\right)}\right\}}{2} \\&&\;\ldots\;e^{ix}=\cos{\left(x\right)}+i\sin{\left(x\right)} \\&=&\frac{\left\{\cos{\left(x\right)}+i\sin{\left(x\right)}\right\}-\left\{\cos{\left(x\right)}-i\sin{\left(x\right)}\right\}}{2} \\&&\;\ldots\;\cos{\left(-x\right)}=\cos{\left(x\right)},\;\sin{\left(-x\right)}=-\sin{\left(-x\right)} \\&=&\frac{\color{red}{\cos{\left(x\right)}}\color{black}{+i\sin{\left(x\right)}}\color{red}{-\cos{\left(x\right)}}\color{black}{+i\sin{\left(x\right)}}}{2} \\&=&\frac{i\sin{\left(x\right)}+i\sin{\left(x\right)}}{2} \\&=&\frac{2i\sin{\left(x\right)}}{2} \\&=&i\sin{\left(x\right)} \end{eqnarray}$$\(cos{(i x)}, sin{(i x)}\) (純虚数に対する\(\cos, \sin\))
\(cos{(i x)}, sin{(i x)}\) (純虚数に対する\(\cos, \sin\))e^(ix)のマクローリン展開
\(e^{ix}\)のマクローリン展開(0点まわりのテイラー展開)
\begin{eqnarray} e^{ix} &=&1+(ix)+\frac{1}{2!}(ix)^2+\frac{1}{3!}(ix)^3+\frac{1}{4!} (ix)^4+\frac{1}{5!} (ix)^5+\frac{1}{6!}(ix)^6+\frac{1}{7!}(ix)^7+\frac{1}{8!}(ix)^8 +\cdots \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/ex.html}{e^x=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\frac{1}{4!} x^4+\frac{1}{5!} x^5+\frac{1}{6!}x^6+\frac{1}{7!}x^7+\frac{1}{8!}x^8} +\cdots \\&=&1+ix+\frac{1}{2!}i^2x^2+\frac{1}{3!}i^3x^3+\frac{1}{4!} i^4x^4+\frac{1}{5!} i^5x^5+\frac{1}{6!}i^6x^6+\frac{1}{7!}i^7x^7+\frac{1}{8!}i^8x^8 +\cdots \\&=&1+ix+\frac{1}{2!}(-1)x^2+\frac{1}{3!}(-i)x^3+\frac{1}{4!} x^4+\frac{1}{5!} ix^5+\frac{1}{6!}(-1)x^6+\frac{1}{7!}(-i)x^7+\frac{1}{8!}x^8 +\cdots \\&=&1+ix-\frac{1}{2!}x^2-i\frac{1}{3!}x^3+\frac{1}{4!} x^4+i\frac{1}{5!} x^5-\frac{1}{6!}x^6-i\frac{1}{7!}x^7+\frac{1}{8!}x^8 +\cdots \\&=&1-\frac{1}{2!}x^2+\frac{1}{4!} x^4-\frac{1}{6!}x^6+\frac{1}{8!}x^8+\cdots +i\left(x-\frac{1}{3!}x^3+\frac{1}{5!} x^5-\frac{1}{7!}x^7+\cdots\right) \\&=&\cos{\left(x\right)}+i\sin{\left(x\right)} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/cosx.html}{\cos{\left(x\right)}=1-\frac{1}{2!}x^2+\frac{1}{4!} x^4-\frac{1}{6!}x^6+\frac{1}{8!}x^8+\cdots} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/sinx.html}{\sin{\left(x\right)}=x-\frac{1}{3!}x^3+\frac{1}{5!} x^5-\frac{1}{7!}x^7+\cdots} \end{eqnarray}eを底とした指数凾数のマクローリン展開
a点まわりのテイラー展開
\begin{eqnarray} f(x) &=& \sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{x!}(x-a)^k \;\ldots\;a点まわりのテイラー展開 \\&=& \frac{1}{0!}f^{(0)}(a)(x-a)^0+\frac{1}{1!}f^{(1)}(a)(x-a)^1+\frac{1}{2!}f^{(2)}(a)(x-a)^2+\dotsb \\&&\;\ldots\;f^{(n)}(x): f(x)のn階微分 \end{eqnarray}マクローリン展開(0点まわりのテイラー展開)
\begin{eqnarray} f(x) &=& \left.\sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{x!}(x-a)^k\right|_{a=0} \\&=& \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{x!}(x-0)^k \\&=& \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{x!}(x)^k \\&=& \frac{1}{0!}f^{(0)}(0)x^0+\frac{1}{1!}f^{(1)}(0)x+\frac{1}{2!}f^{(2)}(0)x^2+\cdots \end{eqnarray}\(f(x)=e^{x}\)のマクローリン展開(0点まわりのテイラー展開)
\begin{eqnarray} e^{x} &=& \frac{1}{0!}f^{(0)}(0)x^0&+\frac{1}{1!}f^{(1)}(0)x&+\frac{1}{2!}f^{(2)}(0)x^2+\cdots \\&=&\frac{1}{0!}\left( e^0\right)x^0&+\frac{1}{1!}\left( e^0\right)x &+\frac{1}{2!}\left( e^0\right)x^2 \\&&+\frac{1}{3!}\left( e^0\right)x^3&+\frac{1}{4!}\left( e^0\right)x^4&+\frac{1}{5!}\left( e^0\right)x^5 \\&&+\frac{1}{6!}\left( e^0\right)x^6&+\frac{1}{7!}\left( e^0\right)x^7&+\frac{1}{8!}\left( e^0\right)x^8 +\cdots \\&=&\frac{1}{0!}\cdot 1\cdot x^0&+\frac{1}{1!}\cdot 1\cdot x &+\frac{1}{2!}\cdot 1\cdot x^2 \\&&+\frac{1}{3!}\cdot 1\cdot x^3&+\frac{1}{4!}\cdot 1\cdot x^4&+\frac{1}{5!}\cdot 1\cdot x^5 \\&&+\frac{1}{6!}\cdot 1\cdot x^6&+\frac{1}{7!}\cdot 1\cdot x^7&+\frac{1}{8!}\cdot 1\cdot x^8 +\cdots \\&=&\frac{1}{0!}x^0+\frac{1}{1!}x^1&+\frac{1}{2!}x^2+\frac{1}{3!}x^3&+\frac{1}{4!} x^4+\frac{1}{5!} x^5+\frac{1}{6!}x^6+\frac{1}{7!}x^7+\frac{1}{8!}x^8 +\cdots \\&=&1+x+\frac{1}{2!}x^2&+\frac{1}{3!}x^3+\frac{1}{4!} x^4&+\frac{1}{5!} x^5+\frac{1}{6!}x^6+\frac{1}{7!}x^7+\frac{1}{8!}x^8 +\cdots \end{eqnarray}sin(x)のマクローリン展開
a点まわりのテイラー展開
\begin{eqnarray} f(x) &=& \sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{x!}(x-a)^k \;\ldots\;a点まわりのテイラー展開 \\&=& \frac{1}{0!}f^{(0)}(a)(x-a)^0+\frac{1}{1!}f^{(1)}(a)(x-a)^1+\frac{1}{2!}f^{(2)}(a)(x-a)^2+\dotsb \\&&\;\ldots\;f^{(n)}(x): f(x)のn階微分 \end{eqnarray}マクローリン展開(0点まわりのテイラー展開)
\begin{eqnarray} f(x) &=& \left.\sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{x!}(x-a)^k\right|_{a=0} \\&=& \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{x!}(x-0)^k \\&=& \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{x!}(x)^k \\&=& \frac{1}{0!}f^{(0)}(0)x^0+\frac{1}{1!}f^{(1)}(0)x+\frac{1}{2!}f^{(2)}(0)x^2+\cdots \end{eqnarray}\(f(x)=\sin{\left(x\right)}\)のマクローリン展開(0点まわりのテイラー展開)
\begin{eqnarray} \sin{\left(x\right)}&=& \frac{1}{0!}f^{(0)}(0)x^0&+\frac{1}{1!}f^{(1)}(0)x&+\frac{1}{2!}f^{(2)}(0)x^2+\cdots \\&=&\frac{1}{0!}\left\{ \sin{(0)}\right\}x^0&+\frac{1}{1!}\left\{ \cos{(0)}\right\}x &+\frac{1}{2!}\left\{-\sin{(0)}\right\}x^2 \\&&+\frac{1}{3!}\left\{-\cos{(0)}\right\}x^3&+\frac{1}{4!}\left\{ \sin{(0)}\right\}x^4&+\frac{1}{5!}\left\{ \cos{(0)}\right\}x^5 \\&&+\frac{1}{6!}\left\{-\sin{(0)}\right\}x^6&+\frac{1}{7!}\left\{-\cos{(0)}\right\}x^7&+\frac{1}{8!}\left\{ \sin{(0)}\right\}x^8 +\cdots \\&=&\frac{1}{0!}\cdot 0\cdot x^0&+\frac{1}{1!}\cdot 1\cdot x &+\frac{1}{2!}\cdot 0\cdot x^2 \\&&+\frac{1}{3!}\cdot-1\cdot x^3&+\frac{1}{4!}\cdot 0\cdot x^4&+\frac{1}{5!}\cdot 1\cdot x^5 \\&&+\frac{1}{6!}\cdot 0\cdot x^6&+\frac{1}{7!}\cdot-1\cdot x^7&+\frac{1}{8!}\cdot 0\cdot x^8 +\cdots \\&=&\frac{1}{1!}x^1-\frac{1}{3!}x^3+\frac{1}{5!} x^5&-\frac{1}{7!}x^7 +\cdots \\&=&x-\frac{1}{3!}x^3+\frac{1}{5!} x^5&-\frac{1}{7!}x^7 +\cdots \end{eqnarray}cos(x)のマクローリン展開
a点まわりのテイラー展開
\begin{eqnarray} f(x) &=& \sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{x!}(x-a)^k \;\ldots\;a点まわりのテイラー展開 \\&=& \frac{1}{0!}f^{(0)}(a)(x-a)^0+\frac{1}{1!}f^{(1)}(a)(x-a)^1+\frac{1}{2!}f^{(2)}(a)(x-a)^2+\dotsb \\&&\;\ldots\;f^{(n)}(x): f(x)のn階微分 \end{eqnarray}マクローリン展開(0点まわりのテイラー展開)
\begin{eqnarray} f(x) &=& \left.\sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{x!}(x-a)^k\right|_{a=0} \\&=& \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{x!}(x-0)^k \\&=& \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{x!}(x)^k \\&=& \frac{1}{0!}f^{(0)}(0)x^0+\frac{1}{1!}f^{(1)}(0)x+\frac{1}{2!}f^{(2)}(0)x^2+\cdots \end{eqnarray}\(f(x)=\cos{\left(x\right)}\)のマクローリン展開(0点まわりのテイラー展開)
\begin{eqnarray} \cos{\left(x\right)} &=& \frac{1}{0!}f^{(0)}(0)x^0&+\frac{1}{1!}f^{(1)}(0)x&+\frac{1}{2!}f^{(2)}(0)x^2+\cdots \\&=&\frac{1}{0!}\left\{ \cos{(0)}\right\}x^0&+\frac{1}{1!}\left\{-\sin{(0)}\right\}x &+\frac{1}{2!}\left\{-\cos{(0)}\right\}x^2 \\&&+\frac{1}{3!}\left\{ \sin{(0)}\right\}x^3&+\frac{1}{4!}\left\{ \cos{(0)}\right\}x^4&+\frac{1}{5!}\left\{ \sin{(0)}\right\}x^5 \\&&+\frac{1}{6!}\left\{-\cos{(0)}\right\}x^6&+\frac{1}{7!}\left\{ \sin{(0)}\right\}x^7&+\frac{1}{8!}\left\{ \cos{(0)}\right\}x^8 +\cdots \\&=&\frac{1}{0!}\cdot 1\cdot x^0&+\frac{1}{1!}\cdot 0\cdot x &+\frac{1}{2!}\cdot-1\cdot x^2 \\&&+\frac{1}{3!}\cdot 0\cdot x^3&+\frac{1}{4!}\cdot 1\cdot x^4&+\frac{1}{5!}\cdot 0\cdot x^5 \\&&+\frac{1}{6!}\cdot-1\cdot x^6&+\frac{1}{7!}\cdot 0\cdot x^7&+\frac{1}{8!}\cdot 1\cdot x^8 +\cdots \\&=&\frac{1}{0!}x^0-\frac{1}{2!}x^2+\frac{1}{4!} x^4&-\frac{1}{6!}x^6+\frac{1}{8!}x^8 +\cdots \\&=&1-\frac{1}{2!}x^2+\frac{1}{4!} x^4&-\frac{1}{6!}x^6+\frac{1}{8!}x^8 +\cdots \end{eqnarray}二階微分のラプラス変換
ラプラス変換
$$\begin{eqnarray} \mathfrak{L}\left[ {f\left( t \right)} \right] &=&\int_0^\infty {f\left( t \right){e^{–st}}}\mathrm{d}t \end{eqnarray}$$二階微分のラプラス変換
$$\begin{eqnarray} f\left(t\right)&=&\frac{\mathrm{d^2}g\left(t\right)}{\mathrm{d^2}t} \\\mathfrak{L}\left[ {f\left( t \right)} \right] &=&\int_0^\infty {f\left( t \right){e^{–st}}}\mathrm{d}t \\&=&\int_0^\infty {\frac{\mathrm{d^2}g\left(t\right)}{\mathrm{d^2}t}{e^{–st}}}\mathrm{d}t \\&=&\left[\frac{\mathrm{d}g\left(t\right)}{\mathrm{d}t}{e^{–st}}\right]_0^\infty-\int_0^\infty {\frac{\mathrm{d}g\left(t\right)}{\mathrm{d}t}{\left(-s\right)e^{–st}}}\mathrm{d}t \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2020/02/blog-post_7.html}{\int_a^b{f'\left( t \right) g\left( t \right) }\mathrm{d}t=\left[f\left( t \right) g\left( t \right)\right]_a^b-\int_a^b{f\left( t \right) g'\left( t \right) }\mathrm{d}t\;\;(f':fの微分,\;g':gの部分)} \\&=&\left[\frac{\mathrm{d}g\left(\infty\right)}{\mathrm{d}t}{e^{–s\infty}} -\left\{\left.\frac{\mathrm{d}g\left(t\right)}{\mathrm{d}t}\right|_{t=0}\right\}{e^{–s0}}\right] -\left(-s\right)\int_0^\infty{\frac{\mathrm{d}g\left(t\right)}{\mathrm{d}t}{e^{–st}}}\mathrm{d}t \\&=&\left[0-g^{\prime}\left(0\right)\right] +s\int_0^\infty{\frac{\mathrm{d}g\left(t\right)}{\mathrm{d}t}{e^{–st}}}\mathrm{d}t \\&&\;\ldots\;\left.\frac{\mathrm{d}g\left(t\right)}{\mathrm{d}t}\right|_{t=0}=g^{\prime}\left(0\right) \\&=&s\int_0^\infty{\frac{\mathrm{d}g\left(t\right)}{\mathrm{d}t}{e^{–st}}}\mathrm{d}t -g^{\prime}\left(0\right) \\&=&s\left\{s\mathfrak{L}\left[g(t)\right]- g\left(0\right)\right\}-g^{\prime}\left(0\right) \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/fracmathrmdfmathrmdt.html}{\int_0^\infty{\frac{\mathrm{d}g\left(t\right)}{\mathrm{d}t}{e^{–st}}}\mathrm{d}t =\mathfrak{L}\left[\frac{\mathrm{d}g\left(t\right)}{\mathrm{d}t}\right] =s\mathfrak{L}\left[g(t)\right]- g\left(0\right)} \\&=&s^2\mathfrak{L}\left[ {g\left( t \right)} \right] - sg\left(0\right) -g^{\prime}\left(0\right) \end{eqnarray}$$バネマスダンパー系，γ,ω0による比較
バネマスダンパー系
\(\gamma, \omega_0\)による比較
$$\begin{eqnarray} \ddot{x}(t) &+&\frac{c}{m}\dot{x}(t) &+&\frac{k}{m}x(t) &=&0 \\ \frac{\mathrm{d^2}}{\mathrm{d^2}t}x(t) &+&\frac{c}{m}\frac{\mathrm{d}}{\mathrm{d}t}x(t) &+&\frac{k}{m}x(t) &=&0 \end{eqnarray}$$ $$\begin{eqnarray} \gamma&=&\frac{c}{2m} \\\omega_0^2&=&\frac{k}{m} \\\omega^2&=&\left|\gamma^2-\omega_0^2\right| \\&=&\left|\frac{\omega_0^2}{\omega_0^2}\left(\gamma^2-\omega_0^2\right)\right| \\&=&\omega_0^2\left|\left(\frac{\gamma}{\omega_0}\right)^2-1\right| \\&=&\omega_0^2\left|\zeta^2-1\right| \;\ldots\;\zeta=\frac{\gamma}{\omega_0} \\\omega&=&\sqrt{\omega_0^2\left|\zeta^2-1\right|} \\&=&\omega_0\sqrt{\left|\zeta^2-1\right|} \end{eqnarray}$$ $$\begin{eqnarray} x(t)&=& &\left.x_0\cos{\left(\omega_0 t\right)}\right. &\left.+\frac{v_0 }{\omega_0} \sin{\left(\omega_0 t\right)}\right. &\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/0.html}{\gamma=0} &\href{https://shikitenkai.blogspot.com/2021/04/0.html}{単振動}&\zeta=0 \\&=& e^{-\gamma t} &\left[x_0 \cos{\left(\omega t\right)}\right. &\left.+\frac{v_0 +\gamma x_0 }{\omega}\sin{\left(\omega t\right)}\right] &\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/00.html}{0\lt\gamma\lt\omega_0} &\href{https://shikitenkai.blogspot.com/2021/04/00.html}{減衰振動}&0\lt\zeta\lt1 \\&=&e^{-\gamma t} &\left[x_0\right. &\left.+\left(v_0+\gamma x_0\right)t\right] &\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/0_56.html}{\gamma=\omega_0} &\href{https://shikitenkai.blogspot.com/2021/04/0_56.html}{臨界減衰}&\zeta=1 \\&=&e^{-\gamma t} &\left[x_0 \cosh\left(\omega t\right)\right. &\left.+\frac{v_0 +\gamma x_0 }{\omega}\sinh{\left(\omega t \right)}\right] &\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/0_17.html}{\omega_0\lt\gamma} &\href{https://shikitenkai.blogspot.com/2021/04/0_17.html}{過減衰}&1\lt\zeta \end{eqnarray}$$
バネマスダンパー系，臨界減衰(γ=ω0)の場合
バネマスダンパー系
\(\gamma=\omega_0\)
$$\begin{eqnarray} \lambda_{1,2}&=&\left.-\gamma\pm\sqrt{\gamma^2-\omega_0^2}\right|_{\gamma=\omega_0} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/12.html}{\lambda_{1,2}=-\gamma\pm\sqrt{\gamma^2-\omega_0^2}} \\&=&-\omega_0\pm\sqrt{\omega_0^2-\omega_0^2} \\&=&-\omega_0 \end{eqnarray}$$ $$\begin{eqnarray} x(t)&=&C(t) e^{-\omega_0 t} \;\ldots\;定数変化法 \\\dot{x}&=&C'(t)e^{-\omega_0 t}-\omega_0 C(t) e^{-\omega_0 t}&\;\ldots\;\href{https://shikitenkai.blogspot.com/2020/02/blog-post.html}{(fg)'=f'g+fg'} \\\ddot{x}&=&\left\{ C''(t)e^{-\omega_0 t}- \omega_0 C'(t) e^{-\omega_0 t} \right\} +\left\{ -\omega_0 C(t)' e^{-\omega_0 t} + \omega_0^2 C(t) e^{-\omega_0 t} \right\}&\;\ldots\;\href{https://shikitenkai.blogspot.com/2020/02/blog-post.html}{(fg)'=f'g+fg'} \\&=&C''(t)e^{-\omega_0 t}-2\omega_0 C'(t) e^{-\omega_0 t}+\omega_0^2 C(t) e^{-\omega_0 t} \end{eqnarray}$$ $$\begin{eqnarray} \left.\ddot{x}+2\gamma\dot{x}+\omega_0^2 x\right|_{\gamma=\omega_0}&=&0 \\\ddot{x}+2\omega_0\dot{x}+\omega_0^2 x&=&0 \\\left[C''(t)e^{-\omega_0 t}-2\omega_0 C'(t) e^{-\omega_0 t}+\omega_0^2 C(t) e^{-\omega_0 t}\right] +2\omega_0\left[C'(t)e^{-\omega_0 t}-\omega_0 C(t) e^{-\omega_0 t}\right] +\omega_0^2 C(t) e^{-\omega_0 t} &=&0 \\C''(t)e^{-\omega_0 t} \color{red} {-2\omega_0 C'(t)e^{-\omega_0 t}} \color{blue}{+ \omega_0^2 C(t) e^{-\omega_0 t}} \color{red} {+2\omega_0 C'(t)e^{-\omega_0 t}} \color{blue}{-2\omega_0^2 C(t) e^{-\omega_0 t}} \color{blue}{+ \omega_0^2 C(t) e^{-\omega_0 t}} &=&0 \\C''(t)e^{-\omega_0 t} &=&0 \\C''(t)&=&0\;\ldots\;e^{-\omega_0 t}\gt 0 \end{eqnarray}$$ $$\begin{eqnarray} C''(t)&=&0 \\C'(t)&=&C_1 \\C(t)&=&C_1 t + C_2 \\x(t)&=&\left(C_1 t + C_2\right)e^{-\omega_0 t} \end{eqnarray}$$ $$\begin{eqnarray} x(0)&=&\left.\left(C_1 t + C_2\right)e^{-\omega_0 t}\right|_{t=0} \\&=&\left(C_1 0 + C_2\right)e^{-\omega_0 0} \\&=&C_2\cdot 1=C_2 \\C_2&=&x(0)=x_0 \\x'(0)&=&\left.\left\{C_1e^{-\omega_0 t}-\omega_0\left(C_1 t + C_2\right)e^{-\omega_0 t}\right\}\right|_{t=0,C_2=x_0} \\&=&C_1e^{-\omega_0 0}-\omega_0\left(C_1 0 + x_0\right)e^{-\omega_0 0} \\&=&C_1\cdot1-\omega_0\left(x_0\right)\cdot1 \\&=&C_1-\omega_0x_0 \\C_1&=&x'(0)+\omega_0x_0=v_0+\omega_0x_0 \end{eqnarray}$$ $$\begin{eqnarray} x(t)&=&C(t)e^{-\omega_0 t} \\&=&\left(C_1 t+C_2\right)e^{-\omega_0 t} \\&=&\left\{\left(v_0+\omega_0x_0\right)t+x_0\right\}e^{-\omega_0 t} \\&=&\left\{\left(v_0+\gamma x_0\right)t+x_0\right\}e^{-\gamma t}\;\ldots\;\omega_0=\gamma \\&=&e^{-\gamma t}\left\{x_0+\left(v_0+\gamma x_0\right)t\right\} \\&=&\color{red}{x_0e^{-\gamma t} \left(1+\gamma t\right)} \color{blue}{+v_0e^{-\gamma t}\left( t\right)}\color{black}{} \;\ldots\;\color{red}{初期位置x_0による臨界減衰}\color{black}{},\;\color{blue}{初期速度v_0による臨界減衰} \end{eqnarray}$$バネマスダンパー系，過減衰(ω0<γ)の場合
バネマスダンパー系
\(\omega_0\lt\gamma\)
$$\begin{eqnarray} \lambda_{1,2}&=&-\gamma\pm\sqrt{\gamma^2-\omega_0^2} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/12.html}{\lambda_{1,2}=-\gamma\pm\sqrt{\gamma^2-\omega_0^2}} \\&=&-\gamma\pm\sqrt{\omega^2} \;\ldots\;\omega^2=\left|\gamma^2-\omega_0^2\right|,\;\gamma^2-\omega_0^2=\omega^2 \\&=&-\gamma\pm\omega \end{eqnarray}$$ $$\begin{eqnarray} C_{1,2} &=&\left.\frac{x_0 }{2}\pm\frac{v_0 +\gamma x_0 }{2\sqrt{\gamma^2-\omega_0^2}}\right|_{\gamma\gt\omega_0} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/c1-c2.html}{ C_{1,2} =\frac{x_0 }{2}\pm\frac{v_0 +\gamma x_0 }{2\sqrt{\gamma^2-\omega_0^2}} } \\&=&\frac{x_0 }{2}\pm\frac{v_0 +\gamma x_0 }{2\sqrt{\omega^2}} \\&=&\frac{x_0 }{2}\pm\frac{v_0 +\gamma x_0 }{2\omega}\;\ldots\;ただし\omega\ne0 \end{eqnarray}$$ $$\begin{eqnarray} x(t)&=&C_1 e^{\lambda_1 t}+C_2 e^{\lambda_2 t} \\&=&C_1 e^{\left(-\gamma+\omega\right)t}+C_2 e^{\left(-\gamma-\omega\right)t} \\&=&C_1 e^{-\gamma t}e^{\omega t}+C_2 e^{-\gamma t}e^{-\omega t} \\&=&e^{-\gamma t}\left\{C_1 e^{\omega t}+C_2 e^{-\omega t}\right\} \\&=&e^{-\gamma t}\left[ \left\{\frac{x_0 }{2}+\frac{v_0 +\gamma x_0 }{2\omega}\right\}e^{\omega t} +\left\{\frac{x_0 }{2}-\frac{v_0 +\gamma x_0 }{2\omega}\right\}e^{-\omega t} \right] \\&=&e^{-\gamma t}\left\{ \frac{x_0 }{2}e^{\omega t}+\frac{v_0 +\gamma x_0 }{2\omega}e^{\omega t} +\frac{x_0 }{2}e^{-\omega t}-\frac{v_0 +\gamma x_0 }{2\omega}e^{-\omega t} \right\} \\&=&e^{-\gamma t}\left\{ \frac{x_0 }{2}\left(e^{\omega t}+e^{-\omega t}\right)+\frac{v_0 +\gamma x_0 }{2\omega}\left(e^{\omega t}-e^{-\omega t}\right) \right\} \\&=&e^{-\gamma t}\left\{ x_0 \frac{e^{\omega t}+e^{-\omega t}}{2}+\frac{v_0 +\gamma x_0 }{\omega}\frac{e^{\omega t}-e^{-\omega t}}{2} \right\} \\&=&e^{-\gamma t}\left\{ x_0 \cosh\left(\omega t\right)+\frac{v_0 +\gamma x_0 }{\omega}\sinh{\left(\omega t \right)} \right\} \\&&\;\ldots\;\cosh{\left(\omega t\right)}=\frac{e^{\omega t}+e^{-\omega t}}{2}, \sinh{\left(\omega t \right)}=\frac{e^{\omega t}-e^{-\omega t}}{2} \\&=&\color{red}{x_0 e^{-\gamma t}\left\{ \cosh\left(\omega t\right)+\frac{\gamma }{\omega}\sinh{\left(\omega t \right)} \right\}} \color{blue}{+v_0 e^{-\gamma t}\left\{ \frac{1}{\omega}\sinh{\left(\omega t \right)} \right\}} \;\ldots\;\color{red}{初期位置x_0による過減衰}\color{black}{},\;\color{blue}{初期速度v_0による過減衰} \\&=& x_0 e^{-\gamma t}\left\{ \cosh\left(\omega t\right)+\frac{\zeta }{\sqrt{\zeta^2-1}}\sinh{\left(\omega t \right)} \right\} +v_0 e^{-\gamma t}\left\{ \frac{1}{\omega}\sinh{\left(\omega t \right)} \right\} \\&&\;\ldots\;\frac{\gamma}{\omega} =\frac{\gamma}{\sqrt{\left|\gamma^2-\omega_0^2\right|}} =\frac{\gamma}{\sqrt{\frac{\omega_0^2}{\omega_0^2}\left|\gamma^2-\omega_0^2\right|}} =\frac{\gamma}{\sqrt{\omega_0^2\left|\frac{\gamma^2}{\omega_0^2}-\frac{\omega_0^2}{\omega_0^2}\right|}} =\frac{\gamma}{\omega_0\sqrt{\left|\zeta^2-1\right|}} =\frac{\zeta}{\sqrt{\left|\zeta^2-1\right|}} \\&&\;\ldots\;\omega=\sqrt{\left|\gamma^2-\omega_0^2\right|},\;\zeta=\frac{\gamma}{\omega_0} \end{eqnarray}$$バネマスダンパー系，減衰振動(0<γ<ω0)の場合
バネマスダンパー系
\(0\lt\gamma\lt\omega_0\)
$$\begin{eqnarray} \lambda_{1,2}&=&\left.-\gamma\pm\sqrt{\gamma^2-\omega_0^2}\right|_{0\lt\gamma\lt\omega_0} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/12.html}{\lambda_{1,2}=-\gamma\pm\sqrt{\gamma^2-\omega_0^2}} \\&=&-\gamma\pm\sqrt{\gamma^2-\omega_0^2} \\&=&-\gamma\pm\sqrt{-\omega^2} \;\ldots\;\omega^2=\left|\gamma^2-\omega_0^2\right|,\;\gamma^2-\omega_0^2=-\omega^2 \\&=&-\gamma\pm\omega\sqrt{-1} \\&=&-\gamma\pm\omega i \end{eqnarray}$$ $$\begin{eqnarray} C_{1,2} &=&\left.\frac{x_0 }{2}\pm\frac{v_0 +\gamma x_0 }{2\sqrt{\gamma^2-\omega_0^2}}\right|_{0\lt\gamma\lt\omega_0} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/c1-c2.html}{ C_{1,2} =\frac{x_0 }{2}\pm\frac{v_0 +\gamma x_0 }{2\sqrt{\gamma^2-\omega_0^2}} } \\&=&\frac{x_0 }{2}\pm\frac{v_0 +\gamma x_0 }{2\omega\sqrt{-1}} \\&=&\frac{x_0 }{2}\pm\frac{v_0 +\gamma x_0 }{2\omega i}\frac{i}{i} \\&=&\frac{x_0 }{2}\pm\frac{v_0 +\gamma x_0 }{2\omega\cdot(-1)}i \\&=&\frac{x_0 }{2}\mp\frac{v_0 +\gamma x_0 }{2\omega}i\;\ldots\;ただし\omega\ne0 \end{eqnarray}$$ $$\begin{eqnarray} x(t)&=&C_1 e^{\lambda_1 t}+C_2 e^{\lambda_2 t} \\&=&C_1 e^{\left(-\gamma+\omega i\right) t}+C_2 e^{\left(-\gamma-\omega i\right) t} \\&=&C_1 e^{-\gamma t}e^{\omega i t}+C_2 e^{-\gamma t}e^{-\omega i t} \\&=&e^{-\gamma t}\left\{C_1 e^{\omega i t}+C_2 e^{-\omega i t}\right\} \\\left\{C_1 e^{\omega i t}+C_2 e^{-\omega i t}\right\}&=& \left\{\frac{x_0 }{2}-\frac{v_0 +\gamma x_0 }{2\omega}i\right\}e^{\omega i t} +\left\{\frac{x_0 }{2}+\frac{v_0 +\gamma x_0 }{2\omega}i\right\}e^{-\omega i t} \\&=&\left\{\frac{x_0 }{2}-\frac{v_0 +\gamma x_0 }{2\omega}i\right\} \left\{ \cos{\left(\omega t\right)}+i\sin{\left(\omega t\right)}\right\} \\&&+\left\{\frac{x_0 }{2}+\frac{v_0 +\gamma x_0 }{2\omega}i\right\} \left\{ \cos{\left(-\omega t\right)}+i\sin{\left(-\omega t\right)}\right\} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/eix.html}{e^{ix}= \cos{\left(x\right)}+i \sin{\left(x\right)}} \\&=&\left\{\frac{x_0 }{2}-\frac{v_0 +\gamma x_0 }{2\omega}i\right\} \left\{ \cos{\left(\omega t\right)}+i\sin{\left(\omega t\right)}\right\} \\&&+\left\{\frac{x_0 }{2}+\frac{v_0 +\gamma x_0 }{2\omega}i\right\} \left\{ \cos{\left(\omega t\right)}-i\sin{\left(\omega t\right)}\right\} \\&&\;\ldots\;\cos{\left(-\theta\right)}=\cos{\left(\theta\right)},\;\sin{\left(-\theta\right)}=-\sin{\left(\theta\right)} \\&=&\left\{\frac{x_0 }{2}-\frac{v_0 +\gamma x_0 }{2\omega}i\right\}\cos{\left(\omega t\right)} +i\left\{\frac{x_0 }{2}-\frac{v_0 +\gamma x_0 }{2\omega}i\right\}\sin{\left(\omega t\right)} \\&&+\left\{\frac{x_0 }{2}+\frac{v_0 +\gamma x_0 }{2\omega}i\right\}\cos{\left(\omega t\right)} -i\left\{\frac{x_0 }{2}+\frac{v_0 +\gamma x_0 }{2\omega}i\right\}\sin{\left(\omega t\right)} \\&=&\left\{\frac{x_0 }{2}-\frac{v_0 +\gamma x_0 }{2\omega}i\right\}\cos{\left(\omega t\right)} +\left\{\frac{x_0 }{2}i+\frac{v_0 +\gamma x_0 }{2\omega}\right\}\sin{\left(\omega t\right)} \\&&+\left\{\frac{x_0 }{2}+\frac{v_0 +\gamma x_0 }{2\omega}i\right\}\cos{\left(\omega t\right)} -\left\{\frac{x_0 }{2}i-\frac{v_0 +\gamma x_0 }{2\omega}\right\}\sin{\left(\omega t\right)} \\&&\;\ldots\;i\cdot i=-1 \\&=&\left\{\frac{x_0 }{2}\color{red}{-\frac{v_0 +\gamma x_0 }{2\omega}i}\color{black}{+\frac{x_0 }{2}}\color{red}{+\frac{v_0 +\gamma x_0 }{2\omega}i}\right\}\cos{\left(\omega t\right)} \\&&+\left\{\color{blue}{\frac{x_0 }{2}i}\color{black}{+\frac{v_0 +\gamma x_0 }{2\omega}}\color{blue}{-\frac{x_0 }{2}i}\color{black}{+\frac{v_0 +\gamma x_0 }{2\omega}}\right\}\sin{\left(\omega t\right)} \\&=&\left\{\frac{x_0 }{2}+\frac{x_0 }{2}\right\}\cos{\left(\omega t\right)}+\left\{\frac{v_0 +\gamma x_0 }{2\omega}+\frac{v_0 +\gamma x_0 }{2\omega}\right\}\sin{\left(\omega t\right)} \\&=&x_0 \cos{\left(\omega t\right)}+\frac{v_0 +\gamma x_0 }{\omega}\sin{\left(\omega t\right)} \end{eqnarray}$$ $$\begin{eqnarray} x(t)&=&e^{-\gamma t}\left\{C_1 e^{\omega i t}+C_2 e^{-\omega i t}\right\} \\&=&e^{-\gamma t}\left\{ x_0 \cos{\left(\omega t\right)}+\frac{v_0 +\gamma x_0 }{\omega}\sin{\left(\omega t\right)} \right\} \\&=& \color{red}{ x_0 e^{-\gamma t}\left\{ \cos{\left(\omega t\right)}+\frac{\gamma}{\omega}\sin{\left(\omega t\right)} \right\} } \color{blue}{ +v_0e^{-\gamma t}\left\{ \frac{1}{\omega}\sin{\left(\omega t\right)} \right\} }\color{black}{} \;\ldots\;\color{red}{初期位置による減衰振動}\color{black}{},\;\color{blue}{初期速度による減衰振動} \\&=& x_0 e^{-\gamma t}\left\{ \cos{\left(\omega t\right)}+\frac{\zeta}{\sqrt{\zeta^2-1}}\sin{\left(\omega t\right)} \right\} +v_0e^{-\gamma t}\left\{ \frac{1}{\omega}\sin{\left(\omega t\right)} \right\} \\&&\;\ldots\;\frac{\gamma}{\omega} =\frac{\gamma}{\sqrt{\left|\gamma^2-\omega_0^2\right|}} =\frac{\gamma}{\sqrt{\frac{\omega_0^2}{\omega_0^2}\left|\gamma^2-\omega_0^2\right|}} =\frac{\gamma}{\sqrt{\omega_0^2\left|\frac{\gamma^2}{\omega_0^2}-\frac{\omega_0^2}{\omega_0^2}\right|}} =\frac{\gamma}{\omega_0\sqrt{\left|\zeta^2-1\right|}} =\frac{\zeta}{\sqrt{\left|\zeta^2-1\right|}} \\&&\;\ldots\;\omega=\sqrt{\left|\gamma^2-\omega_0^2\right|},\;\zeta=\frac{\gamma}{\omega_0} \end{eqnarray}$$バネマスダンパー系，単振動(γ=0 / バネマス系)の場合
バネマスダンパー系
\(\gamma=0\)
$$\begin{eqnarray} \lambda_{1,2}&=&\left.-\gamma\pm\sqrt{\gamma^2-\omega_0^2}\right|_{\gamma=0} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/12.html}{\lambda_{1,2}=-\gamma\pm\sqrt{\gamma^2-\omega_0^2}} \\&=&-0\pm\sqrt{0^2-\omega_0^2} \\&=&\pm\sqrt{-\omega_0^2} \\&=&\pm\omega_0\sqrt{-1} \\&=&\pm\omega_0i \end{eqnarray}$$ $$\begin{eqnarray} C_{1,2} &=&\left.\frac{x_0 }{2}\pm\frac{v_0 +\gamma x_0 }{2\sqrt{\gamma^2-\omega_0^2}}\right|_{\gamma=0} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/c1-c2.html}{ C_{1,2} =\frac{x_0 }{2}\pm\frac{v_0 +\gamma x_0 }{2\sqrt{\gamma^2-\omega_0^2}} } \\&=&\frac{x_0 }{2}\pm\frac{v_0 +0 x_0 }{2\sqrt{0^2-\omega_0^2}} \\&=&\frac{x_0 }{2}\pm\frac{v_0 }{2\omega_0\sqrt{-1}} \\&=&\frac{x_0 }{2}\pm\frac{v_0 }{2\omega_0i}\frac{i}{i} \\&=&\frac{x_0 }{2}\pm\frac{v_0 }{2\omega_0\cdot(-1)}i \\&=&\frac{x_0 }{2}\mp\frac{v_0 }{2\omega_0}i\;\ldots\;ただし\omega_0\ne0 \end{eqnarray}$$ $$\begin{eqnarray} x(t)&=&C_1 e^{\lambda_1 t}+C_2 e^{\lambda_2 t} \\&=&\left\{\frac{x_0 }{2}-\frac{v_0 }{2\omega_0}i\right\}e^{\omega_0i t} +\left\{\frac{x_0 }{2}+\frac{v_0 }{2\omega_0}i\right\}e^{-\omega_0i t} \\&=&\left\{\frac{x_0 }{2}-\frac{v_0 }{2\omega_0}i\right\} \left\{ \cos{\left(\omega_0 t\right)}+i\sin{\left(\omega_0 t\right)}\right\} \\&&+\left\{\frac{x_0 }{2}+\frac{v_0 }{2\omega_0}i\right\} \left\{ \cos{\left(-\omega_0 t\right)}+i\sin{\left(-\omega_0 t\right)}\right\} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/eix.html}{e^{ix}= \cos{\left(x\right)}+i \sin{\left(x\right)}} \\&=&\left\{\frac{x_0 }{2}-\frac{v_0 }{2\omega_0}i\right\} \left\{ \cos{\left(\omega_0 t\right)}+i\sin{\left(\omega_0 t\right)}\right\} \\&&+\left\{\frac{x_0 }{2}+\frac{v_0 }{2\omega_0}i\right\} \left\{ \cos{\left(\omega_0 t\right)}-i\sin{\left(\omega_0 t\right)}\right\} \\&&\;\ldots\;\cos{\left(-\theta\right)}=\cos{\left(\theta\right)},\;\sin{\left(-\theta\right)}=-\sin{\left(\theta\right)} \\&=&\left\{\frac{x_0 }{2}-\frac{v_0 }{2\omega_0}i\right\}\cos{\left(\omega_0 t\right)} +i \left\{\frac{x_0 }{2}-\frac{v_0 }{2\omega_0}i\right\} \sin{\left(\omega_0 t\right)} \\&&+\left\{\frac{x_0 }{2}+\frac{v_0 }{2\omega_0}i\right\} \cos{\left(\omega_0 t\right)} -i \left\{\frac{x_0 }{2}+\frac{v_0 }{2\omega_0}i\right\}\sin{\left(\omega_0 t\right)} \\&=&\left\{\frac{x_0 }{2}-\frac{v_0 }{2\omega_0}i\right\}\cos{\left(\omega_0 t\right)} +\left\{\frac{x_0 }{2}i+\frac{v_0 }{2\omega_0}\right\} \sin{\left(\omega_0 t\right)} \\&&+\left\{\frac{x_0 }{2}+\frac{v_0 }{2\omega_0}i\right\} \cos{\left(\omega_0 t\right)} -\left\{\frac{x_0 }{2}i-\frac{v_0 }{2\omega_0}\right\} \sin{\left(\omega_0 t\right)} \\&&\;\ldots\;i\cdot i=-1 \\&=&\left\{\frac{x_0 }{2}\color{red}{ -\frac{v_0 }{2\omega_0}i }\color{black}{ +\frac{x_0 }{2} }\color{red}{ +\frac{v_0 }{2\omega_0}i}\right\} \cos{\left(\omega_0 t\right)} \\&&+\left\{\color{blue}{\frac{x_0 }{2}i}\color{black}{+\frac{v_0 }{2\omega_0}}\color{blue}{-\frac{x_0 }{2}i}\color{black}{+\frac{v_0 }{2\omega_0}}\right\} \sin{\left(\omega_0 t\right)} \\&=&\left\{\frac{x_0 }{2}+\frac{x_0 }{2}\right\}\cos{\left(\omega t\right)}+\left\{\frac{v_0 }{2\omega_0}+\frac{v_0 }{2\omega_0}\right\}\sin{\left(\omega t\right)} \\&=&x_0 \cos{\left(\omega_0 t\right)}+\frac{v_0 }{\omega_0} \sin{\left(\omega_0 t\right)} \\&=&\color{red}{x_0 \cos{\left(\omega_0 t\right)}}\color{blue}{+v_0 \frac{1}{\omega_0} \sin{\left(\omega_0 t\right)}}\color{black}{} \;\ldots\;\color{red}{初期位置による単振動}\color{black}{},\;\color{blue}{初期速度による単振動} \end{eqnarray}$$バネマスダンパー系，C1, C2 を求める
バネマスダンパー系
\(C_1 ,C_2 を求める\)
$$\begin{eqnarray} \href{https://shikitenkai.blogspot.com/2021/04/blog-post_17.html}{ \frac{sx_0 +v_0 +\frac{c}{m}x_0 }{s^2+s\frac{c}{m}+\frac{k}{m}}=\frac{(C_1 +C_2 )s-(C_1 \lambda_2+C_2 \lambda_1)}{\left(s-\lambda_1\right)\left(s-\lambda_2\right)} } \end{eqnarray}$$ $$\begin{eqnarray} \\sx_0 +v_0 +\frac{c}{m}x_0&=&(C_1 +C_2 )s-(C_1 \lambda_2+C_2 \lambda_1)\;\cdots\;分子について考える \end{eqnarray}$$ $$\left\{ \begin{eqnarray} C_1 +C_2 &=&x_0 \\-(C_1 \lambda_2+C_2 \lambda_1)&=&v_0 +\frac{c}{m}x_0 \\&=&v_0 +2\gamma x_0 \end{eqnarray} \right.\;\ldots\;sの係数を比較し連立させる$$ $$\begin{eqnarray} C_1 &=&x_0 -C_2 \\-(C_1 \lambda_2+C_2 \lambda_1)&=&v_0 +2\gamma x_0 \\C_1 \lambda_2+C_2 \lambda_1&=&-\left(v_0 +2\gamma x_0 \right) \\\left(x_0 -C_2 \right)\lambda_2+C_2 \lambda_1&=&-v_0 -2\gamma x_0 \\x_0 \lambda_2-C_2 \lambda_2+C_2 \lambda_1&=&-v_0 -2\gamma x_0 \\C_2 \left(\lambda_1-\lambda_2\right)&=&-v_0 -2\gamma x_0 -\lambda_2 x_0 \\C_2 \left(\lambda_1-\lambda_2\right)&=&-v_0 -x_0 \left(2\gamma +\lambda_2 \right) \\ \\C_2 &=&\frac{-v_0 -x_0 \left(2\gamma + \lambda_2 \right)}{\lambda_1-\lambda_2} \\&=&-\frac{v_0 +x_0 \left\{2\gamma + \left(-\gamma-\sqrt{\gamma^2-\omega_0^2}\right)\right\}} {\left(-\gamma+\sqrt{\gamma^2-\omega_0^2}\right)-\left(-\gamma-\sqrt{\gamma^2-\omega_0^2}\right)} \\&=&-\frac{v_0 +x_0 \left(\gamma-\sqrt{\gamma^2-\omega_0^2}\right)} {2\sqrt{\gamma^2-\omega_0^2}} \\&=&-\frac{v_0 +\gamma x_0 -x_0 \sqrt{\gamma^2-\omega_0^2}} {2\sqrt{\gamma^2-\omega_0^2}} \\&=&-\frac{v_0 +\gamma x_0 }{2\sqrt{\gamma^2-\omega_0^2}} + \frac{x_0 \sqrt{\gamma^2-\omega_0^2}}{2\sqrt{\gamma^2-\omega_0^2}} \\&=&-\frac{v_0 +\gamma x_0 }{2\sqrt{\gamma^2-\omega_0^2}} + \frac{x_0 }{2} \\&=&\frac{x_0 }{2}-\frac{v_0 +\gamma x_0 }{2\sqrt{\gamma^2-\omega_0^2}} \\ \\C_1 &=&x_0 -C_2 \\&=&x_0 -\left\{\frac{x_0 }{2}-\frac{v_0 +\gamma x_0 }{2\sqrt{\gamma^2-\omega_0^2}} \right\} \\&=&\frac{x_0 }{2}+\frac{v_0 +\gamma x_0 }{2\sqrt{\gamma^2-\omega_0^2}} \end{eqnarray}$$ $$\begin{eqnarray} C_{1,2}&=&\frac{x_0 }{2}\pm\frac{v_0 +\gamma x_0 }{2\sqrt{\gamma^2-\omega_0^2}}\;\ldots\;ただし\gamma\ne\omega_0 \end{eqnarray}$$バネマスダンパー系，λ1,λ2を求める
バネマスダンパー系
\(\lambda_{1,2}を求める\)
$$\begin{eqnarray} \href{https://shikitenkai.blogspot.com/2021/04/blog-post_17.html}{ \frac{sx_0 +v_0 +\frac{c}{m}x_0 }{s^2+s\frac{c}{m}+\frac{k}{m}}=\frac{(C_1 +C_2 )s-(C_1 \lambda_2+C_2 \lambda_1)}{\left(s-\lambda_1\right)\left(s-\lambda_2\right)} } \end{eqnarray}$$ $$\begin{eqnarray} \\s^2+s\frac{c}{m}+\frac{k}{m}&=&\left(s-\lambda_1\right)\left(s-\lambda_2\right)\;\cdots\;分母について考える \\\lambda_{1,2}&=&\frac{-\frac{c}{m}\pm\sqrt{\left(\frac{c}{m}\right)^2-4\cdot1\cdot\frac{k}{m}}}{2\cdot1} \\&=&-\frac{c}{2m}\pm\frac{1}{2}\sqrt{\left(\frac{c}{m}\right)^2-4\frac{k}{m}} \\&=&-\frac{c}{2m}\pm\sqrt{\frac{1}{4}\left(\left(\frac{c}{m}\right)^2-4\frac{k}{m}\right)} \\&=&-\frac{c}{2m}\pm\sqrt{\left(\frac{c}{2m}\right)^2-\frac{k}{m}} \\&=&-\gamma\pm\sqrt{\gamma^2-\omega_0^2} \\&&\;\ldots\;\gamma=\frac{c}{2m},\;\omega_0^2=\frac{k}{m} \end{eqnarray}$$バネマスダンパー系，運動方程式，ラプラス変換，逆ラプラス変換，入力なし
バネマスダンパー系
運動方程式
$$\begin{eqnarray} \ddot{x} &+&\frac{c}{m}\dot{x} &+&\frac{k}{m}x &=&0 \\ \frac{\mathrm{d^2}x}{\mathrm{d^2}t} &+&\frac{c}{m}\frac{\mathrm{d}x}{\mathrm{d}t} &+&\frac{k}{m}x &=&0 \end{eqnarray}$$ラプラス変換と計算
$$\begin{eqnarray} \mathfrak{L}\left[ \frac{\mathrm{d^2}x}{\mathrm{d^2}t} \right.&+&\left.\frac{c}{m}\frac{\mathrm{d}x}{\mathrm{d}t} \right.&+&\left.\frac{k}{m}x \right]&=0 \\\mathfrak{L}\left[ \frac{\mathrm{d^2}x}{\mathrm{d^2}t} \right] &+&\mathfrak{L}\left[ \frac{c}{m}\frac{\mathrm{d}x}{\mathrm{d}t} \right] &+&\mathfrak{L}\left[ \frac{k}{m}x\right] &=0 \;\ldots\; \href{https://shikitenkai.blogspot.com/2021/04/gtb-ht.html}{ \mathfrak{L}\left[ a g(t)+b h(t) \right]=a\mathfrak{L}\left[ {g\left( t \right)} \right] + b\mathfrak{L}\left[ {h\left( t \right)} \right] } \\ s^2X-sx_0 -v_0 &+&\frac{c}{m}\left(sX-x_0 \right) &+&\frac{k}{m}X &=0 \\ \\&&&&&\;\ldots\;\mathfrak{L}\left[x\right]=X \\&&&&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/fracmathrmdfmathrmdt.html}{\mathfrak{L}\left[ \frac{\mathrm{d}x}{\mathrm{d}t}\right] =s^2X-x_0,\;x_0=x(0)} \\&&&&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/blog-post_62.html}{\mathfrak{L}\left[ \frac{\mathrm{d^2}x}{\mathrm{d^2}t}\right] =s^2X-sx_0 -v_0,\;v_0=x'(0)} \\ s^2X-sx_0 -v_0 &+&s\frac{c}{m}X-\frac{c}{m}x_0 &+&\frac{k}{m}X &=0 \end{eqnarray}$$ $$\begin{eqnarray} s^2X+s\frac{c}{m}X+\frac{k}{m}X &=& sx_0 +v_0 +\frac{c}{m}x_0 \\ \left(s^2+s\frac{c}{m}+\frac{k}{m}\right)X &=& sx_0 +v_0 +\frac{c}{m}x_0 \\ X&=&\frac{sx_0 +v_0 +\frac{c}{m}x_0 }{s^2+s\frac{c}{m}+\frac{k}{m}} \\&=&\frac{sx_0 +v_0 +\frac{c}{m}x_0 }{\left(s-\lambda_1\right)\left(s-\lambda_2\right)} \;\ldots\;\lambda_{1,2}=s^2+s\frac{c}{m}+\frac{k}{m}の解 \\&=&\frac{C_1 }{s-\lambda_1}+\frac{C_2 }{s-\lambda_2} \\&=&\frac{C_1 \left(s-\lambda_2\right)+C_2 \left(s-\lambda_1\right)}{\left(s-\lambda_1\right)\left(s-\lambda_2\right)} \\&=&\frac{(C_1 +C_2 )s-(C_1 \lambda_2+C_2 \lambda_1)}{\left(s-\lambda_1\right)\left(s-\lambda_2\right)} \end{eqnarray}$$逆ラプラス変換
$$\begin{eqnarray} \mathfrak{L}^{-1}\left[X\right]&=&\mathfrak{L}^{-1}\left[C_1 \frac{1}{s-\lambda_1}+C_2 \frac{1}{s-\lambda_2}\right] \\&=&C_1 \mathfrak{L}^{-1}\left[\frac{1}{s-\lambda_1}\right]+C_2 \mathfrak{L}^{-1}\left[\frac{1}{s-\lambda_2}\right] \\x(t)&=&C_1 e^{\lambda_1 t}+C_2 e^{\lambda_2 t} \\&&\;\ldots\;\mathfrak{L}^{-1}\left[X\right]=x(t) \\&&\;\ldots\;\mathfrak{L}^{-1}\left[ \frac{1}{s+a} \right]=e^{-at} \end{eqnarray}$$eを底とした指数凾数(at乗)のラプラス変換
ラプラス変換
$$\begin{eqnarray} \mathfrak{L}\left[ {f\left( t \right)} \right] &=&\int_0^\infty {f\left( t \right){e^{–st}}}\mathrm{d}t \end{eqnarray}$$\( e^{-at} \)のラプラス変換
$$\begin{eqnarray} f\left( t \right)&=&e^{at} \\\mathfrak{L}\left[ {f\left( t \right)} \right] &=& \int_0^\infty {f\left( t \right){e^{ –st}}}\mathrm{d}t \\&=& \int_0^\infty {e^{at}e^{ –st}}\mathrm{d}t \\&=& \int_0^\infty {e^{-(s-a)t}}\mathrm{d}t \\&&\;\cdots\;u=-(s-a)t,\;t:0\rightarrow \infty,\;u:0\rightarrow -\infty \\&&\;\cdots\;\frac{\mathrm{d}u}{\mathrm{d}t}=-(s-a),\;\mathrm{d}t=\frac{-1}{s-a}\mathrm{d}u \\&=&\int_0^{-\infty} {e^{u}} \frac{-1}{s-a} \mathrm{d}u \\&=&\frac{-1}{s-a}\int_0^{-\infty} {e^{u}} \mathrm{d}u \\&=&\frac{-1}{s-a}\left[ {e^{u}} \right]_0^{-\infty} \\&=&\frac{-1}{s-a}\left[ {e^{-\infty}} - {e^{0}}\right] \\&=&\frac{-1}{s-a}\left[ 0 - 1\right] \\&=&\frac{-1}{s-a}\left(-1\right) \\&=&\frac{1}{s-a} \end{eqnarray}$$sin凾数のラプラス変換
ラプラス変換
$$\begin{eqnarray} \mathfrak{L}\left[ {f\left( t \right)} \right] &=&\int_0^\infty {f\left( t \right){e^{–st}}}\mathrm{d}t \end{eqnarray}$$\( \sin{\left(\omega t\right)} \)のラプラス変換
$$\begin{eqnarray} f\left( t \right)&=&\sin{\left(\omega t\right)} \\\mathfrak{L}\left[ {f\left( t \right)} \right] &=& \int_0^\infty {\sin{\left(\omega t\right)}{e^{ –st}}}\mathrm{d}t \\&=& \left[ \sin{\left(\omega t\right)} \cdot {\frac{–1}{s}e^{-st}} \right]_0^{\infty} -\int_0^\infty {\omega \cos{\left(\omega t\right)} \cdot {\frac{–1}{s}e^{–st}}}\mathrm{d}t \\&&\;\cdots\;\int_a^b{f'\left( t \right) g\left( t \right) }\mathrm{d}t=\left[f\left( t \right) g\left( t \right)\right]_a^b-\int_a^b{f\left( t \right) g'\left( t \right) }\mathrm{d}t\;\;(f':fの微分,\;g':gの部分) \\&&\;\cdots\;\int e^{at} \mathrm{d}t =\frac{1}{a}e^{at}+C, \frac{\mathrm{d}}{\mathrm{d}t}\sin{\left(\omega t\right)}=\omega \cos{\left(\omega t\right)} \\&=& \left[ \sin{\left(\omega \infty\right)} \cdot {\frac{–1}{s}e^{-s\infty}} - \sin{\left(\omega 0\right)} \cdot {\frac{–1}{s}e^{-s0}}\right] -\left(\frac{-\omega}{s}\right)\int_0^\infty {\cos{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t \\&=& \left[ 0 - 0\right] +\frac{\omega}{s}\int_0^\infty {\cos{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t \\&&\;\cdots\;e^{-\infty}=0,\;\sin{\left( 0\right)}=0 \\&=& \frac{\omega}{s}\int_0^\infty {\cos{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t \\&=& \frac{\omega}{s}\left[ \left[ \cos{\left(\omega t\right)} \cdot {\frac{–1}{s}e^{-st}} \right]_0^{\infty} -\int_0^\infty {-\omega \sin{\left(\omega t\right)} \cdot {\frac{–1}{s}e^{–st}}}\mathrm{d}t \right] \\&&\;\cdots\;\int_a^b{f'\left( t \right) g\left( t \right) }\mathrm{d}t=\left[f\left( t \right) g\left( t \right)\right]_a^b-\int_a^b{f\left( t \right) g'\left( t \right) }\mathrm{d}t\;\;(f':fの微分,\;g':gの微分) \\&&\;\cdots\;\int e^{at} \mathrm{d}t =\frac{1}{a}e^{at}+C, \frac{\mathrm{d}}{\mathrm{d}t}\cos{\left(\omega t\right)}=-\omega \sin{\left(\omega t\right)} \\&=& \frac{\omega}{s}\left[ \left[ \cos{\left(\omega \infty\right)} \cdot {\frac{–1}{s}e^{-s\infty}} - \cos{\left(\omega 0\right)} \cdot {\frac{–1}{s}e^{-s0}}\right] -\frac{\omega}{s}\int_0^\infty {\sin{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t \right] \\&=& \frac{\omega}{s}\left[ \left[0 - 1 \cdot \frac{–1}{s}\cdot 1\right] -\frac{\omega}{s}\int_0^\infty {\sin{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t \right] \\&&\;\cdots\;e^{-\infty}=0,\; \cos{\left( 0\right)}=1,\; e^{0}=1 \\&=& \frac{\omega}{s}\left[ \frac{1}{s} -\frac{\omega}{s}\int_0^\infty {\sin{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t \right] \\&=& \frac{\omega}{s^2}-\frac{\omega^2}{s^2}\int_0^\infty {\sin{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t \\&=& \frac{\omega}{s^2}-\frac{\omega^2}{s^2}\mathfrak{L}\left[ {f\left( t \right)} \right] \\\mathfrak{L}\left[ {f\left( t \right)} \right]+\frac{\omega^2}{s^2}\mathfrak{L}\left[ {f\left( t \right)} \right] &=& \frac{\omega}{s^2} \\\mathfrak{L}\left[ {f\left( t \right)} \right]\left(1+\frac{\omega^2}{s^2}\right) &=& \frac{\omega}{s^2} \\\mathfrak{L}\left[ {f\left( t \right)} \right]&=& \frac{\omega}{s^2}\frac{1}{1+\frac{\omega^2}{s^2}} \\&=&\frac{\omega}{s^2\left(1+\frac{\omega^2}{s^2}\right)} \\&=&\frac{\omega}{s^2+\omega^2} \end{eqnarray}$$冪凾数のラプラス変換
ラプラス変換
$$\begin{eqnarray} \mathfrak{L}\left[ {f\left( t \right)} \right] &=&\int_0^\infty {f\left( t \right){e^{–st}}}\mathrm{d}t \end{eqnarray}$$\( t^n \)のラプラス変換
$$\begin{eqnarray} f\left( t \right)&=&t^n \\\mathfrak{L}\left[ {f\left( t \right)} \right] &=& \int_0^\infty {t^n{e^{ –st}}}\mathrm{d}t \\&=& \left[ t^n \cdot {\frac{–1}{s}e^{-st}} \right]_0^{\infty} -\int_0^\infty {nt^{n-1} \cdot {\frac{–1}{s}e^{–st}}}\mathrm{d}t \\&&\;\cdots\;\int_a^b{f'\left( t \right) g\left( t \right) }\mathrm{d}t=\left[f\left( t \right) g\left( t \right)\right]_a^b-\int_a^b{f\left( t \right) g'\left( t \right) }\mathrm{d}t\;\;(f':fの微分,\;g':gの部分) \\&&\;\cdots\;\int e^{at} \mathrm{d}t =\frac{1}{a}e^{at}+C \\&&\;\cdots\; \frac{\mathrm{d}}{\mathrm{d}t}t^n=nt^{n-1} \\&=& \left[ \infty^n \cdot {\frac{–1}{s}e^{-s\infty}} - 0^n \cdot {\frac{–1}{s}e^{-s0}} \right] -\left(\frac{–n}{s}\right)\int_0^\infty {t^{n-1} {e^{–st}}}\mathrm{d}t \\&=& \left[ 0 - 0\right] +\frac{n}{s}\int_0^\infty {t^{n-1} {e^{–st}}}\mathrm{d}t \\&&\;\cdots\;e^{-\infty}=0,\;\sin{\left( 0\right)}=0 \\&=& \frac{n}{s}\int_0^\infty {t^{n-1} {e^{–st}}}\mathrm{d}t \\&=& \frac{n}{s}\frac{n-1}{s}\int_0^\infty {t^{n-2} {e^{–st}}}\mathrm{d}t \\&&\vdots \\&=& \frac{n}{s}\frac{n-1}{s}\cdots\frac{2}{s}\int_0^\infty {te^{–st}}\mathrm{d}t \\&=& \frac{n}{s}\frac{n-1}{s}\cdots\frac{2}{s}\frac{1}{s}\int_0^\infty {e^{–st}}\mathrm{d}t \\&=& \frac{n}{s}\frac{n-1}{s}\cdots\frac{2}{s}\frac{1}{s}\left[ \frac{–1}{s}e^{-st} \right]_0^{\infty} \\&=& \frac{n!}{s^n}\frac{–1}{s}\left[ e^{-s\infty} - e^{-s0} \right] \\&&\;\cdots\;n(n-1)\cdots2\;1=n! \\&=& \frac{n!}{s^n}\frac{–1}{s}\left[ 0 - 1\right] \\&&\;\cdots\;e^{-\infty}=0,\;e^{0}=1 \\&=& \frac{n!}{s^n}\frac{–1}{s}\left(-1\right) \\&=& \frac{n!}{s^{n+1}} \end{eqnarray}$$積分のラプラス変換
ラプラス変換
$$\begin{eqnarray} \mathfrak{L}\left[ {f\left( t \right)} \right] &=&\int_0^\infty {f\left( t \right){e^{–st}}}\mathrm{d}t \end{eqnarray}$$\( \int_0^t g(u) \mathrm{d}u \)のラプラス変換
$$\begin{eqnarray} f\left( t \right)&=&\int_0^t g(u) \mathrm{d}u \\\mathfrak{L}\left[ {f\left( t \right)} \right] &=& \int_0^\infty {\left\{\int_0^t g(u) \mathrm{d}u\right\}{e^{ –st}}}\mathrm{d}t \\&=& \left[ \left\{\int_0^t g(u) \mathrm{d}u\right\} {\frac{–1}{s}e^{-st}} \right]_0^{\infty} -\int_0^\infty {g(t) \cdot {\frac{–1}{s}e^{–st}}}\mathrm{d}t \\&&\;\cdots\;\int_a^b{f'\left( t \right) g\left( t \right) }\mathrm{d}t=\left[f\left( t \right) g\left( t \right)\right]_a^b-\int_a^b{f\left( t \right) g'\left( t \right) }\mathrm{d}t\;\;(f':fの微分,\;g':gの部分) \\&&\;\cdots\;\int e^{at} \mathrm{d}t =\frac{1}{a}e^{at}+C \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2021/04/blog-post.html}{\frac{\mathrm{d}}{\mathrm{d}t}\int_0^t g\left(u\right)\mathrm{d}t =\frac{\mathrm{d}}{\mathrm{d}t}\left[G\left(t\right)-G\left(0\right)\right] =\frac{\mathrm{d}}{\mathrm{d}t}G\left(t\right)-\frac{\mathrm{d}}{\mathrm{d}t}G\left(0\right) =g\left(t\right)-0 =g\left(t\right)} \\&=& \left[ \left\{\int_0^\infty g(u) \mathrm{d}u\right\} {\frac{–1}{s}e^{-s\infty}} -\left\{\int_0^0 g(u)\mathrm{d}u\right\} {\frac{–1}{s}e^{-s0}} \right] +\frac{1}{s}\int_0^\infty {g(t) {e^{–st}}}\mathrm{d}t \\&&\;\cdots\;e^{-\infty}=0,\;\int_t^t f(x)\mathrm{d}x=\left[F(t)-F(t)\right]=0 \\&=& \left[0-0 \right] +\frac{1}{s}\int_0^\infty {g(t) {e^{–st}}}\mathrm{d}t \\&=& \frac{1}{s} \mathfrak{L}\left[ {g\left( t \right)} \right] \end{eqnarray}$$微分のラプラス変換
ラプラス変換
$$\begin{eqnarray} \mathfrak{L}\left[ {f\left( t \right)} \right] &=&\int_0^\infty {f\left( t \right){e^{–st}}}\mathrm{d}t \end{eqnarray}$$\( \frac{\mathrm{d}f}{\mathrm{d}t} \)のラプラス変換
$$\begin{eqnarray} f\left( t \right)&=&\frac{\mathrm{d}g\left( t \right)}{\mathrm{d}t} \\\mathfrak{L}\left[ {g\left( t \right)} \right] &=& \int_0^\infty {\frac{\mathrm{d}g\left( t \right)}{\mathrm{d}t} {e^{ –st}}}\mathrm{d}t \\&=& \left[ g\left( t \right) {e^{-st}} \right]_0^{\infty} -\int_0^\infty {\frac{\mathrm{d}g\left( t\right)}{\mathrm{d}t} {\left(–s\right)e^{ –st}}}\mathrm{d}t \\&&\;\cdots\;\int_a^b{f'\left( t \right) g\left( t \right) }\mathrm{d}t=\left[f\left( t \right) g\left( t \right)\right]_a^b-\int_a^b{f\left( t \right) g'\left( t \right) }\mathrm{d}t\;\;(f':fの微分,\;g':gの部分) \\&&\;\cdots\;\frac{\mathrm{d}}{\mathrm{d}t}e^{at}=ae^{at} \\&=& \left[ g\left( \infty \right) {e^{-s\infty}} - g\left( 0 \right) {e^{-s0}} \right] -\left(–s\right)\int_0^\infty {\frac{\mathrm{d}g\left( t\right)}{\mathrm{d}t} {e^{ –st}}}\mathrm{d}t \\&=& \left[ 0 - g\left( 0 \right) \cdot {1} \right] -\left(–s\right)\int_0^\infty {\frac{\mathrm{d}g\left( t\right)}{\mathrm{d}t} {e^{ –st}}}\mathrm{d}t \\&&\;\cdots\;e^{-\infty}=0,\;e^{0}=1 \\&=&-g\left( 0 \right) +s\int_0^\infty {\frac{\mathrm{d}g\left( t\right)}{\mathrm{d}t} {e^{ –st}}}\mathrm{d}t \\&=&-g\left( 0 \right)+s\mathfrak{L}\left[ {g\left( t \right)} \right] \\&=&s\mathfrak{L}\left[ {g\left( t \right)} \right]-g\left( 0 \right) \end{eqnarray}$$線形結合のラプラス変換
ラプラス変換
$$\begin{eqnarray} \mathfrak{L}\left[ {f\left( t \right)} \right] &=&\int_0^\infty {f\left( t \right){e^{–st}}}\mathrm{d}t \end{eqnarray}$$\( a g(t)+b h(t) \)のラプラス変換
$$\begin{eqnarray} f\left( t \right)&=&a g(t)+b h(t) \;\cdots\;a, b:tによらない(定数) \\\mathfrak{L}\left[ {f\left( t \right)} \right] &=& \int_0^\infty {\left\{a g(t)+b h(t)\right\}{e^{ –st}}}\mathrm{d}t \\&=& \int_0^\infty { \left\{ a g(t) e^{ –st} +b h(t)e^{ –st} \right\} }\mathrm{d}t \\&=& \int_0^\infty a g(t) e^{–st} \mathrm{d}t + \int_0^\infty b h(t) e^{–st} \mathrm{d}t \\&=& a\int_0^\infty g(t) e^{–st} \mathrm{d}t + b\int_0^\infty h(t) e^{–st} \mathrm{d}t \\&=& a\mathfrak{L}\left[ {g\left( t \right)} \right] + b\mathfrak{L}\left[ {h\left( t \right)} \right] \end{eqnarray}$$定積分の微分
定積分の微分
$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t}\int_0^t g\left(x\right)\mathrm{d}x &=&\frac{\mathrm{d}}{\mathrm{d}t}\left[G\left(x\right)\right]_0^t \\&=&\frac{\mathrm{d}}{\mathrm{d}t}\left[G\left(t\right)-G\left(0\right)\right] \\&=&\frac{\mathrm{d}}{\mathrm{d}t}G\left(t\right)-\frac{\mathrm{d}}{\mathrm{d}t}G\left(0\right) \\&=&g\left(t\right)-0\;\cdots\;0を代入したG\left(0\right)は定数であり，定数の微分は0 \\&=&g\left(t\right) \end{eqnarray}$$
登録:
投稿 (Atom)