間違いしかありません.コメントにてご指摘いただければ幸いです(気が付いた点を特に断りなく頻繁に書き直していますのでご注意ください).

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,cos凾数,その2

バネマスダンパー系

運動方程式

$$\begin{eqnarray} m\ddot{x} &+&c\dot{x} &+&kx &=&F\cos{\left(\omega_f t\right)} \\ \frac{\mathrm{d^2}x}{\mathrm{d^2}t} &+&\frac{c}{m}\frac{\mathrm{d}x}{\mathrm{d}t} &+&\frac{k}{m}x &=&\frac{F}{m}\cos{\left(\omega_f t\right)} \\ \frac{\mathrm{d^2}x}{\mathrm{d^2}t} &+&2\gamma\frac{\mathrm{d}x}{\mathrm{d}t} &+&\omega_0^2x &=&\frac{F}{m}\cos{\left(\omega_f t\right)} \;\cdots\;\gamma=\frac{c}{2m},\;\omega_0^2=\frac{k}{m} \end{eqnarray}$$

ラプラス変換

$$\begin{eqnarray} \mathfrak{L}\left[ \frac{\mathrm{d^2}x}{\mathrm{d^2}t} \right.&+&\left.2\gamma\frac{\mathrm{d}x}{\mathrm{d}t} \right.&+&\left.\omega_0^2x \right]&=\mathfrak{L}\left[\frac{F}{m}\cos{\left(\omega_f t\right)}\right] \\\mathfrak{L}\left[ \frac{\mathrm{d^2}x}{\mathrm{d^2}t} \right] &+&\mathfrak{L}\left[ 2\gamma\frac{\mathrm{d}x}{\mathrm{d}t} \right] &+&\mathfrak{L}\left[ \omega_0^2 x\right] &=\mathfrak{L}\left[\frac{F}{m}\cos{\left(\omega_f t\right)}\right] \\\mathfrak{L}\left[ \frac{\mathrm{d^2}x}{\mathrm{d^2}t} \right] &+&2\gamma\mathfrak{L}\left[ \frac{\mathrm{d}x}{\mathrm{d}t} \right] &+&\omega_0^2\mathfrak{L}\left[ x\right] &=\mathfrak{L}\left[\frac{F}{m}\cos{\left(\omega_f t\right)}\right] \\ s^2X-sx_0 -v_0 &+&2\gamma\left(sX-x_0 \right) &+&\omega_0^2X &=\frac{F}{m}\frac{s}{s^2+\omega_f^2} \\ \\&&&&&\;\ldots\;\mathfrak{L}\left[x\right]=X \\&&&&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/fracmathrmdfmathrmdt.html}{\mathfrak{L}\left[ \frac{\mathrm{d}x}{\mathrm{d}t}\right] =s^2X-x_0,\;x_0=x(0)} \\&&&&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/blog-post_62.html}{\mathfrak{L}\left[ \frac{\mathrm{d^2}x}{\mathrm{d^2}t}\right] =s^2X-sx_0 -v_0,\;v_0=x'(0)} \\&&&&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/05/cos.html}{\mathfrak{L}\left[\cos{\left(\omega_f t\right)}\right]=\frac{s}{s^2+\omega_f^2}} \end{eqnarray}$$

Xについて解く

$$\begin{eqnarray} s^2X+2\gamma Xs+\omega_0^2X &=& sx_0 +v_0 +2\gamma x_0 +\frac{F}{m}\frac{s}{s^2+\omega_f^2} \\ \left(s^2+2\gamma s+\omega_0^2\right)X &=& sx_0 +v_0 +2\gamma x_0 +\frac{F}{m}\frac{s}{s^2+\omega_f^2} \\ X&=&\frac{sx_0 +v_0 +2\gamma x_0 }{s^2+2\gamma s+\omega_0^2} +\frac{F}{m}\frac{1}{s^2+2\gamma s+\omega_0^2}\frac{s}{s^2+\omega_f^2} \\&=&\frac{sx_0 +v_0 +2\gamma x_0 }{\left(s-\lambda_1\right)\left(s-\lambda_2\right)} +\frac{F}{m}\frac{1}{\left(s-\lambda_1\right)\left(s-\lambda_2\right)}\frac{s}{s^2+\omega_f^2} \\&&\;\ldots\;\left(s-\lambda_1\right)\left(s-\lambda_2\right)=s^2+2\gamma s+\omega_0^2 \\&&\;\ldots\;\lambda_{1,2}=\href{https://shikitenkai.blogspot.com/2020/11/blog-post.html}{\frac{-2\gamma\pm\sqrt{\left(2\gamma\right)^2-4\cdot1\cdot\omega_0^2}}{2\cdot1}}=-\gamma\pm\sqrt{\gamma^2-\omega_0^2}=-\gamma\pm\xi \end{eqnarray}$$

部分分数分解 準備

$$\begin{eqnarray} X&=&\frac{sx_0 +v_0 +2\gamma x_0 }{\left(s-\lambda_1\right)\left(s-\lambda_2\right)} +\frac{F}{m}\frac{1}{\left(s-\lambda_1\right)\left(s-\lambda_2\right)}\frac{s}{s^2+\omega_f^2} \\&=&\frac{sx_0 +v_0 +2\gamma x_0 }{\left(s-\lambda_1\right)\left(s-\lambda_2\right)} +\frac{F}{m}\frac{1}{s^2-\left(\lambda_1+\lambda_2\right)s+\lambda_1\lambda_2}\frac{s}{s^2+\omega_f^2} \\&=&\frac{sx_0 +v_0 +2\gamma x_0 }{\left(s-\lambda_1\right)\left(s-\lambda_2\right)} +\frac{F}{m}\frac{s}{s^4-\left(\lambda_1+\lambda_2\right)s^3+\lambda_1\lambda_2s^2+\omega_f ^2s^2-\omega_f ^2\left(\lambda_1+\lambda_2\right)s+\omega_f ^2\lambda_1\lambda_2} \\&=&\frac{sx_0 +v_0 +2\gamma x_0 }{\left(s-\lambda_1\right)\left(s-\lambda_2\right)} +\frac{F}{m}\frac{s}{s^4-\left(\lambda_1+\lambda_2\right)s^3+\left(\lambda_1\lambda_2+\omega_f ^2\right)s^2-\omega_f ^2\left(\lambda_1+\lambda_2\right)s+\omega_f ^2\lambda_1\lambda_2} \\&=&\frac{sx_0 +v_0 +2\gamma x_0 }{\left(s-\lambda_1\right)\left(s-\lambda_2\right)} +\frac{F}{m}\frac{s}{\left\{\left(s-a\right)^2+b^2\right\}\left\{\left(s-c\right)^2+d^2\right\}} \\&=&\frac{sx_0 +v_0 +2\gamma x_0 }{\left(s-\lambda_1\right)\left(s-\lambda_2\right)} +\frac{F}{m}\left\{ C_A\frac{s-a+b}{\left(s-a\right)^2+b^2} + C_B \frac{s-c+d}{\left(s-c\right)^2+d^2} \right\} \\&=&\frac{C_1}{s-\lambda_1}+\frac{C_2}{s-\lambda_2} +\frac{F}{m}\left\{ C_3 \frac{s-a}{\left(s-a\right)^2+b^2} + C_4 \frac{b} {\left(s-a\right)^2+b^2} + C_5 \frac{s-c}{\left(s-c\right)^2+d^2} + C_6 \frac{d} {\left(s-c\right)^2+d^2} \right\} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/e-at.html}{\mathfrak{L}\left[ e^{a t} \right]=\frac{1}{s-a}} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/05/eatcos.html}{\mathfrak{L}\left[ e^{a t}\cos{\left( b t \right) } \right]=\frac{s-a}{(s-a)^2+b^2}} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/05/eatsin.html}{\mathfrak{L}\left[ e^{a t}\sin{\left( b t \right) } \right]=\frac{b}{(s-a)^2+b^2}} \\&=&\frac{C_1\left(s-\lambda_2\right)+C_2\left(s-\lambda_1\right)}{\left(s-\lambda_2\right)\left(s-\lambda_1\right)} +\frac{F}{m}\left\{ \frac{C_3\left(s-a\right)+C_4b}{\left(s-a\right)^2+b^2} + \frac{C_5\left(s-c\right)+C_6d}{\left(s-c\right)^2+d^2} \right\} \\&=&\frac{\left(C_1+C_2\right)s-\left(C_1\lambda_2+C_2\lambda_1\right)}{\left(s-\lambda_2\right)\left(s-\lambda_1\right)} +\frac{F}{m}\frac{ \left\{C_3\left(s-a\right)+C_4b\right\}\left\{\left(s-c\right)^2+d^2\right\} +\left\{C_5\left(s-c\right)+C_6d\right\}\left\{\left(s-a\right)^2+b^2\right\} }{\left\{\left(s-a\right)^2+b^2\right\}\left\{\left(s-c\right)^2+d^2\right\}} \end{eqnarray}$$

部分分数分解 第1項分子の係数比較

$$\begin{eqnarray} sx_0 +v_0 +2\gamma x_0&=&\left(C_1 +C_2 \right)s-\left(C_1 \lambda_2+C_2 \lambda_1\right) \end{eqnarray}$$ $$\left\{\begin{eqnarray} x_0&=&C_1 +C_2 \\v_0 +2\gamma x_0&=&-\left(C_1 \lambda_2+C_2 \lambda_1\right) \end{eqnarray}\right.$$

解く

$$\begin{eqnarray} \begin{bmatrix} x_0\\v_0 +2\gamma x_0 \end{bmatrix} &=& \begin{bmatrix} 1&1 \\-\lambda_2&-\lambda_1 \end{bmatrix} \begin{bmatrix} C_1\\C_2 \end{bmatrix} \\&=&\boldsymbol{A} \begin{bmatrix} C_1\\C_2 \end{bmatrix} \\\boldsymbol{A}^{-1} &=&\href{https://shikitenkai.blogspot.com/2021/05/blog-post_96.html}{\frac{1}{\begin{vmatrix}\boldsymbol{A}\end{vmatrix}}\tilde{\boldsymbol{A}}} \\&=&\frac{1}{\begin{vmatrix}\boldsymbol{A}\end{vmatrix}} \href{https://shikitenkai.blogspot.com/2021/05/blog-post_3.html}{ \begin{bmatrix} (-1)^{1+1}\left|\boldsymbol{M}_{11}\right| &(-1)^{1+2}\left|\boldsymbol{M}_{21}\right| \\ (-1)^{2+1}\left|\boldsymbol{M}_{12}\right| &(-1)^{2+2}\left|\boldsymbol{M}_{22}\right| \end{bmatrix}} \\&=& \frac{1}{\left(1\cdot-\lambda_1\right)-\left(1\cdot-\lambda_2\right)} \begin{bmatrix} (-1)^2\cdot-\lambda_1&(-1)^3\cdot1 \\(-1)^3\cdot-\lambda_2&(-1)^2\cdot1 \end{bmatrix} \\&=& \frac{1}{\lambda_2-\lambda_1} \begin{bmatrix} -\lambda_1&-1 \\\lambda_2&1 \end{bmatrix} \\&=& \frac{1}{(-\gamma-\xi)-(-\gamma+\xi)} \begin{bmatrix} -(-\gamma+\xi)&-1 \\(-\gamma-\xi)&1 \end{bmatrix} \;\ldots\;\lambda_{1,2}=-\gamma\pm\xi \\&=& \frac{1}{-2\xi} \begin{bmatrix} \gamma-\xi&-1 \\-\gamma-\xi&1 \end{bmatrix} \\\begin{bmatrix} C_1\\C_2 \end{bmatrix} &=&\boldsymbol{A}^{-1} \begin{bmatrix} x_0\\v_0 +2\gamma x_0 \end{bmatrix} \\&=&\frac{1}{-2\xi} \begin{bmatrix} \gamma-\xi&-1 \\-\gamma-\xi&1 \end{bmatrix} \begin{bmatrix} x_0\\v_0 +2\gamma x_0 \end{bmatrix} \\&=&\frac{1}{-2\xi} \begin{bmatrix} (\gamma-\xi) \cdot x_0 + -1 \cdot (v_0 +2\gamma x_0) \\(-\gamma-\xi) \cdot x_0 + 1 \cdot (v_0 +2\gamma x_0) \end{bmatrix} \\&=&\frac{1}{-2\xi} \begin{bmatrix} \gamma x_0-\xi x_0 - v_0 - 2\gamma x_0 \\-\gamma x_0-\xi x_0 + v_0 +2\gamma x_0 \end{bmatrix} \\&=&\frac{1}{-2\xi} \begin{bmatrix} -\xi x_0 - v_0 - \gamma x_0 \\-\xi x_0 + v_0 + \gamma x_0 \end{bmatrix} \\&=& \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} \end{eqnarray}$$

\(C_1, C_2\)

$$\begin{eqnarray} \begin{bmatrix} C_1\\C_2 \end{bmatrix}&=& \begin{bmatrix} \frac{x_0}{2} + \frac{v_0 + \gamma x_0}{2\xi} \\\frac{x_0}{2} - \frac{v_0 + \gamma x_0}{2\xi} \end{bmatrix} \end{eqnarray}$$

部分分数分解 第2項分母の係数比較

$$\begin{eqnarray} \\&&s^4-\left(\lambda_1+\lambda_2\right)s^3 +\left(\lambda_1\lambda_2+\omega_f ^2\right)s^2 -\omega_f ^2\left(\lambda_1+\lambda_2\right)s +\omega_f ^2\lambda_1\lambda_2 \\&=&\left\{\left(s-a\right)^2+b^2\right\}\left\{\left(s-c\right)^2+d^2\right\} \\&=&\left(s^2-2as+a^2+b^2\right)\left(s^2-2cs+c^2+d^2\right) \\&=&s^4-2cs^3+(c^2+d^2)s^2-2as^3+4ac s^2-2as(c^2+d^2)+(a^2+b^2)s^2-2(a^2+b^2)cs+(a^2+b^2)(c^2+d^2) \\&=&s^4-2(a+c)s^3+(a^2+b^2+c^2+d^2+4ac)s^2-2(ac^2+ad^2+a^2c+b^2c)s+a^2c^2+a^2d^2+b^2c^2+b^2d^2 \end{eqnarray}$$ $$ \left\{\begin{eqnarray} -2(a+c)&=&-\left(\lambda_1+\lambda_2\right) \\a^2+b^2+c^2+d^2+4ac&=&\lambda_1\lambda_2+\omega_f ^2 \\-2(ac^2+ad^2+a^2c+b^2c)&=&-\omega_f ^2\left(\lambda_1+\lambda_2\right) \\a^2c^2+a^2d^2+b^2c^2+b^2d^2&=&\omega_f ^2\lambda_1\lambda_2 \end{eqnarray} \right.$$

解く

$$\begin{eqnarray} a&=&0を仮定 \\c&=&\frac{\lambda_1+\lambda_2}{2}-a&\;\ldots\;一つ目の式より \\&=&\frac{\lambda_1+\lambda_2}{2}-0&\;\ldots\;a=0を代入 \\&=&\frac{\lambda_1+\lambda_2}{2} \\\omega_f^2\frac{\lambda_1+\lambda_2}{2}&=&ac^2+ad^2+a^2c+b^2c&\;\ldots\;三つ目の式より \\&=&0\cdot c^2+0\cdot d^2+0^2\cdot c+b^2c&\;\ldots\;a=0を代入 \\&=&b^2c \\&=&b^2\frac{\lambda_1+\lambda_2}{2}&\;\ldots\;c=\frac{\lambda_1+\lambda_2}{2}を代入 \\\omega_f^2&=&b^2&\;\ldots\;両辺\frac{\lambda_1+\lambda_2}{2}で割る \\b&=&\omega_f&\;\ldots\;角周波数なので正を利用 \\a^2c^2+a^2d^2+b^2c^2+b^2d^2&=&\omega_f^2\lambda_1\lambda_2&\;\ldots\;四つ目の式 \\0^2c^2+0^2d^2+b^2c^2+b^2d^2&=&&\;\ldots\;a=0を代入 \\b^2\left(c^2+d^2\right)&=&& \\\omega_f^2\left(c^2+d^2\right)&=&&\;\ldots\;b=\omega_fを代入 \\c^2+d^2&=&\lambda_1\lambda_2\;\ldots\;両辺\omega_f^2で割る \\d^2&=&\lambda_1\lambda_2-c^2 \\&=&\lambda_1\lambda_2-\left(\frac{\lambda_1+\lambda_2}{2}\right)^2&\;\ldots\;c=\frac{\lambda_1+\lambda_2}{2}を代入 \\&=&\frac{1}{4}\left\{4\lambda_1\lambda_2-\left(\lambda_1+\lambda_2\right)^2\right\} \\&=&\frac{1}{4}\left(4\lambda_1\lambda_2-\lambda_1^2-2\lambda_1\lambda_2-\lambda_2^2\right) \\&=&-\frac{1}{4}\left(\lambda_1^2-2\lambda_1\lambda_2+\lambda_2^2\right) \\&=&-\frac{\left(\lambda_1-\lambda_2\right)^2}{4} \\&=&-\left(\frac{\left(\lambda_1-\lambda_2\right)}{2}\right)^2 \\d&=&\sqrt{-\left(\frac{\left(\lambda_1-\lambda_2\right)}{2}\right)^2}&\;\ldots\;角周波数なので正を利用 \\&=& \sqrt{-1}\frac{\lambda_1-\lambda_2}{2} \\&=& i\frac{\lambda_1-\lambda_2}{2} \end{eqnarray}$$
$$ \left\{ \begin{eqnarray} a&=&0 \\b&=&\omega_f \\c&=&\frac{\lambda_1+\lambda_2}{2}=\frac{(-\gamma+\xi)+(-\gamma-\xi)}{2}=\frac{-2\gamma}{2}=-\gamma \\d&=&i\frac{\lambda_1-\lambda_2}{2}=i\frac{(-\gamma+\xi)-(-\gamma-\xi)}{2}=i\frac{2\xi}{2}=i\xi \end{eqnarray} \right.$$

部分分数分解 第2項分子の係数比較

$$\begin{eqnarray} s &=&\left\{C_3\left(s-a\right)+C_4b\right\}\left\{\left(s-c\right)^2+d^2\right\} +\left\{C_5\left(s-c\right)+C_6d\right\}\left\{\left(s-a\right)^2+b^2\right\} \\&=&\left\{C_3s-C_3a+C_4b\right\}\left\{s^2-2cs+c^2+d^2\right\} +\left\{C_5s-C_5c+C_6d\right\}\left\{s^2-2as+a^2+b^2\right\} \\&=& C_3s^3 -2C_3cs^2 +C_3(c^2+d^2)s -C_3as^2 +2C_3acs -C_3a(c^2+d^2) +C_4bs^2 -2C_4bcs +C_4b(c^2+d^2) \\&& +C_5s^3 -2C_5as^2 +C_5(a^2+b^2)s -C_5cs^2 +2C_5acs -C_5c(a^2+b^2) +C_6ds^2 -2C_6ads +C_6d(a^2+b^2) \\&=&C_3s^3 +C_5s^3 \\&&-2C_3cs^2-C_3as^2+C_4bs^2 -2C_5as^2-C_5cs^2+C_6ds^2 \\&&+C_3(c^2+d^2)s+2C_3acs-2C_4bcs +C_5(a^2+b^2)s+2C_5acs-2C_6ads \\&&-C_3a(c^2+d^2)+C_4b(c^2+d^2) -C_5c(a^2+b^2)+C_6d(a^2+b^2) \\&=&(C_3+C_5)s^3 \\&&+(-2C_3c-C_3a+C_4b-2C_5a-C_5c+C_6d)s^2 \\&&+(C_3(c^2+d^2)+2C_3ac-2C_4bc+C_5(a^2+b^2)+2C_5ac-2C_6ad)s \\&&+\left\{-C_3a(c^2+d^2)+C_4b(c^2+d^2)-C_5c(a^2+b^2)+C_6d(a^2+b^2)\right\} \\&=&(C_3+C_5)s^3 \\&&+(C_3(-2c-a)+C_4b+C_5(-2a-c)+C_6d)s^2 \\&&+(C_3((c^2+d^2)+2ac)+C_4(-2bc)+C_5((a^2+b^2)+2ac)+C_6(-2ad))s \\&&+\left\{C_3(-a(c^2+d^2))+C_4b(c^2+d^2)+C_5(-c(a^2+b^2))+C_6d(a^2+b^2)\right\} \end{eqnarray}$$ $$\left\{ \begin{eqnarray} 0&=&C_3+C_5 \\0&=&C_3(-2c-a)+C_4b+C_5(-2a-c)+C_6d \\1&=&C_3(c^2+d^2+2ac)+C_4(-2bc)+C_5((a^2+b^2)+2ac)+C_6(-2ad) \\0 &=&C_3(-a(c^2+d^2))+C_4b(c^2+d^2)+C_5(-c(a^2+b^2))+C_6d(a^2+b^2) \end{eqnarray} \right.$$

解く

$$\begin{eqnarray} \begin{bmatrix} 0\\0\\1\\0 \end{bmatrix} &=& \begin{bmatrix} 1&0&1&0 \\-2c-a&b&-2a-c&d \\c^2+d^2+2ac&-2bc&(a^2+b^2)+2ac&-2ad \\-a(c^2+d^2)&b(c^2+d^2)&-c(a^2+b^2)&d(a^2+b^2) \end{bmatrix} \begin{bmatrix} C_3\\C_4\\C_5\\C_6 \end{bmatrix} \\&=& \begin{bmatrix} 1&0&1&0 \\-2c-0&b&-2\cdot0-c&d \\c^2+d^2+2\cdot 0\cdot c&-2bc&(0^2+b^2)+2\cdot 0\cdot c&-2\cdot 0\cdot d \\-0(c^2+d^2)&b(c^2+d^2)&-c(0^2+b^2)&d(0^2+b^2) \end{bmatrix} \begin{bmatrix} C_3\\C_4\\C_5\\C_6 \end{bmatrix} \\&=& \begin{bmatrix} 1&0&1&0 \\-2c&b &-c&d \\c^2+d^2&-2bc&b^2&0 \\0&b(c^2+d^2)&-b^2c&b^2d \end{bmatrix} \begin{bmatrix} C_3\\C_4\\C_5\\C_6 \end{bmatrix} \\&=& \begin{bmatrix} 1&0&1&0 \\-2(-\gamma)&(\omega_f) &-(-\gamma)&(i\xi) \\(-\gamma)^2+(i\xi)^2&-2(\omega_f)(-\gamma)&(\omega_f)^2&0 \\0&(\omega_f)((-\gamma)^2+(i\xi)^2)&-(\omega_f)^2(-\gamma)&(\omega_f)^2(i\xi) \end{bmatrix} \begin{bmatrix} C_3\\C_4\\C_5\\C_6 \end{bmatrix} \\&=& \begin{bmatrix} 1&0&1&0 \\2\gamma&\omega_f&\gamma&i\xi \\\gamma^2-\xi^2&2\gamma\omega_f&\omega_f^2&0 \\0&\omega_f(\gamma^2-\xi^2)&\gamma\omega_f^2&i\xi\omega_f^2 \end{bmatrix} \begin{bmatrix} C_3\\C_4\\C_5\\C_6 \end{bmatrix} \\&=& \begin{bmatrix} 1&0&1&0 \\2\gamma&\omega_f&\gamma&i\xi \\\omega_0^2&2\gamma\omega_f&\omega_f^2&0 \\0&\omega_0^2\omega_f&\gamma\omega_f^2&i\xi\omega_f^2 \end{bmatrix} \begin{bmatrix} C_3\\C_4\\C_5\\C_6 \end{bmatrix}\;\ldots\;\omega_0^2=\gamma^2-\xi^2\;\;(\xi^2=\gamma^2-\omega_0^2より) \\&=&\boldsymbol{A}\begin{bmatrix} C_3\\C_4\\C_5\\C_6 \end{bmatrix} \end{eqnarray}$$ $$\begin{eqnarray} \left|\boldsymbol{M}_{31}\right| &=&\begin{vmatrix} 0&1&0 \\\omega_f&\gamma&i\xi \\\omega_0^2\omega_f&\gamma\omega_f^2&i\xi\omega_f^2 \end{vmatrix} \\&=&-\begin{vmatrix} \omega_f&i\xi \\\omega_0^2\omega_f&i\xi\omega_f^2 \end{vmatrix} \\&=&-\left\{ \omega_f \cdot i\xi\omega_f^2 -i\xi \cdot \omega_0^2\omega_f \right\} \\&=&-\left\{ i\xi\omega_f^3 - i\xi\omega_0^2\omega_f \right\} \\&=&-i\xi\omega_f\left(\omega_f^2 - \omega_0^2\right) \\&=&i\xi\omega_f\left(\omega_0^2 - \omega_f^2\right) \\ \\\left|\boldsymbol{M}_{32}\right| &=&\begin{vmatrix} 1&1&0 \\2\gamma&\gamma&i\xi \\0&\gamma\omega_f^2&i\xi\omega_f^2 \end{vmatrix} \\&=&\begin{vmatrix} \gamma&i\xi \\\gamma\omega_f^2&i\xi\omega_f^2 \end{vmatrix} -\begin{vmatrix} 2\gamma&i\xi \\0&i\xi\omega_f^2 \end{vmatrix} \\&=& ( \gamma \cdot i\xi\omega_f^2 - i\xi \cdot \gamma\omega_f^2 ) -( 2\gamma \cdot i\xi\omega_f^2 -i\xi\cdot 0 ) \\&=&i\gamma\xi\omega_f^2 - i\gamma\xi\omega_f^2 - i2\gamma\xi\omega_f^2 \\&=&-i2\gamma\xi\omega_f^2 \\ \\\left|\boldsymbol{M}_{33}\right| &=&\begin{vmatrix} 1&0&0 \\2\gamma&\omega_f&i\xi \\0&\omega_0^2\omega_f&i\xi\omega_f^2 \end{vmatrix} \\&=&\begin{vmatrix} \omega_f&i\xi \\\omega_0^2\omega_f&i\xi\omega_f^2 \end{vmatrix} \\&=&\omega_f \cdot i\xi\omega_f^2 - i\xi \cdot \omega_0^2\omega_f \\&=& i\xi\omega_f^3- i\xi\omega_0^2\omega_f \\&=& i\xi\omega_f(\omega_f^2-\omega_0^2) \\&=& -i\xi\omega_f(\omega_0^2-\omega_f^2) \\ \\\left|\boldsymbol{M}_{34}\right| &=&\begin{vmatrix} 1&0&1 \\2\gamma&\omega_f&\gamma \\0&\omega_0^2\omega_f&\gamma\omega_f^2 \end{vmatrix} \\&=&\begin{vmatrix} \omega_f&\gamma \\\omega_0^2\omega_f&\gamma\omega_f^2 \end{vmatrix} +\begin{vmatrix} 2\gamma&\omega_f \\0&\omega_0^2\omega_f \end{vmatrix} \\&=& \omega_f \cdot \gamma\omega_f^2 - \gamma \cdot \omega_0^2\omega_f +2\gamma \cdot \omega_0^2\omega_f -\omega_f \cdot 0 \\&=&\gamma\omega_f^3-\gamma\omega_0^2\omega_f+2\gamma\omega_0^2\omega_f \\&=&\gamma\omega_f(\omega_0^2 + \omega_f^2) \\ \\ \\ \\&&(-1)^{3+1}a_{31}\left|\boldsymbol{M}_{31}\right|+(-1)^{3+2}a_{32}\left|\boldsymbol{M}_{32}\right|+(-1)^{3+3}a_{33}\left|\boldsymbol{M}_{33}\right|+(-1)^{3+4}a_{34}\left|\boldsymbol{M}_{34}\right| \\&=& \omega_0^2 \cdot ( i\xi\omega_f\left(\omega_0^2 - \omega_f^2\right) ) \\&& -2\gamma\omega_f \cdot ( -i2\gamma\xi\omega_f^2 ) \\&& +\omega_f^2 \cdot ( -i\xi\omega_f\left(\omega_0^2 - \omega_f^2\right) ) \\&& -0 \cdot ( \gamma\omega_f(\omega_0^2 + \omega_f^2) ) \\&=&i\xi\omega_0^2\omega_f\left(\omega_0^2 - \omega_f^2\right) +i4\gamma^2\xi\omega_f^3 -i\xi\omega_f^3\left(\omega_0^2 - \omega_f^2\right) -0 \\&=&i\xi\omega_f\left(\omega_0^2 - \omega_f^2\right)\left(\omega_0^2-\omega_f^2\right) +i4\gamma^2\xi\omega_f^3 \\&=&i\xi\omega_f\left\{\left(\omega_0^2 - \omega_f^2\right)^2+(2\gamma\omega_f)^2\right\} \end{eqnarray}$$ $$\begin{eqnarray} (-1)^{1+3}\frac{\left|\boldsymbol{M}_{31}\right|}{\left|A\right|} &=&\frac{i\xi\omega_f\left(\omega_0^2 - \omega_f^2\right) } {i\xi\omega_f\left\{\left(\omega_0^2 - \omega_f^2\right)^2+(2\gamma\omega_f)^2\right\}} \\&=&\frac{\omega_0^2 - \omega_f^2} {\left(\omega_0^2 - \omega_f^2\right)^2+(2\gamma\omega_f)^2} \\(-1)^{2+3}\frac{\left|\boldsymbol{M}_{32}\right|}{\left|A\right|} &=&-1\cdot\frac{-i2\gamma\xi\omega_f^2} {i\xi\omega_f\left\{\left(\omega_0^2 - \omega_f^2\right)^2+(2\gamma\omega_f)^2\right\}} \\&=&\frac{2\gamma\omega_f } {\left(\omega_0^2 - \omega_f^2\right)^2+(2\gamma\omega_f)^2} \\(-1)^{3+3}\frac{\left|\boldsymbol{M}_{33}\right|}{\left|A\right|} &=&\frac{-i\xi\omega_f\left(\omega_0^2 - \omega_f^2\right)} {i\xi\omega_f\left\{\left(\omega_0^2 - \omega_f^2\right)^2+(2\gamma\omega_f)^2\right\}} \\&=&\frac{-\left(\omega_0^2 - \omega_f^2\right)} {\left(\omega_0^2+\omega_f^2\right)^2-\left(2\gamma\omega_f\right)^2} \\(-1)^{4+3}\frac{\left|\boldsymbol{M}_{34}\right|}{\left|A\right|} &=&-1\cdot\frac{\gamma\omega_f(\omega_0^2 + \omega_f^2) } {i\xi\omega_f\left\{\left(\omega_0^2 - \omega_f^2\right)^2+(2\gamma\omega_f)^2\right\}} \\&=&\frac{-\gamma(\omega_0^2 + \omega_f^2)} {i\xi\left\{\left(\omega_0^2 - \omega_f^2\right)^2-(2\gamma\omega_f)^2\right\}} \end{eqnarray}$$ $$\begin{eqnarray} \begin{bmatrix} C_3\\C_4\\C_5\\C_6 \end{bmatrix} &=&\boldsymbol{A}^{-1} \begin{bmatrix} 0\\0\\1\\0 \end{bmatrix} \\&=&\begin{bmatrix} (-1)^{1+1}\frac{\left|\boldsymbol{M}_{11}\right|}{\left|A\right|} &(-1)^{1+2}\frac{\left|\boldsymbol{M}_{21}\right|}{\left|A\right|} &(-1)^{1+3}\frac{\left|\boldsymbol{M}_{31}\right|}{\left|A\right|} &(-1)^{1+4}\frac{\left|\boldsymbol{M}_{41}\right|}{\left|A\right|} \\(-1)^{2+1}\frac{\left|\boldsymbol{M}_{12}\right|}{\left|A\right|} &(-1)^{2+2}\frac{\left|\boldsymbol{M}_{22}\right|}{\left|A\right|} &(-1)^{2+3}\frac{\left|\boldsymbol{M}_{32}\right|}{\left|A\right|} &(-1)^{2+4}\frac{\left|\boldsymbol{M}_{42}\right|}{\left|A\right|} \\(-1)^{3+1}\frac{\left|\boldsymbol{M}_{13}\right|}{\left|A\right|} &(-1)^{3+2}\frac{\left|\boldsymbol{M}_{23}\right|}{\left|A\right|} &(-1)^{3+3}\frac{\left|\boldsymbol{M}_{33}\right|}{\left|A\right|} &(-1)^{3+4}\frac{\left|\boldsymbol{M}_{43}\right|}{\left|A\right|} \\(-1)^{4+1}\frac{\left|\boldsymbol{M}_{14}\right|}{\left|A\right|} &(-1)^{4+2}\frac{\left|\boldsymbol{M}_{24}\right|}{\left|A\right|} &(-1)^{4+3}\frac{\left|\boldsymbol{M}_{34}\right|}{\left|A\right|} &(-1)^{4+4}\frac{\left|\boldsymbol{M}_{44}\right|}{\left|A\right|} \end{bmatrix} \begin{bmatrix} 0\\0\\1\\0 \end{bmatrix} \\&=& \begin{bmatrix} (-1)^{1+3}\frac{\left|\boldsymbol{M}_{31}\right|}{\left|A\right|}\cdot 1 \\(-1)^{2+3}\frac{\left|\boldsymbol{M}_{32}\right|}{\left|A\right|}\cdot 1 \\(-1)^{3+3}\frac{\left|\boldsymbol{M}_{33}\right|}{\left|A\right|}\cdot 1 \\(-1)^{4+3}\frac{\left|\boldsymbol{M}_{34}\right|}{\left|A\right|}\cdot 1 \end{bmatrix} = \begin{bmatrix} \frac{\omega_0^2 - \omega_f^2}{\left(\omega_0^2 - \omega_f^2\right)^2+(2\gamma\omega_f)^2} \\\frac{2\gamma\omega_f }{\left(\omega_0^2 - \omega_f^2\right)^2+(2\gamma\omega_f)^2} \\\frac{-\left(\omega_0^2 - \omega_f^2\right)}{\left(\omega_0^2 - \omega_f^2\right)^2-\left(2\gamma\omega_f\right)^2} \\\frac{-\gamma(\omega_f^2 + \omega_0^2)}{i\xi\left\{\left(\omega_0^2 - \omega_f^2\right)^2-(2\gamma\omega_f)^2\right\}} \end{bmatrix} \\&=& \frac{1}{\left(\omega_0^2-\omega_f^2\right)^2-\left(2\gamma\omega_f\right)^2} \begin{bmatrix} \omega_0^2 - \omega_f^2\\2\gamma\omega_f\\-\left(\omega_0^2 - \omega_f^2\right)\\-\frac{\gamma}{i\xi}(\omega_0^2 + \omega_f^2) \end{bmatrix} \end{eqnarray}$$

\(C_3,C_4,C_5,C_6\)

$$\begin{eqnarray} \begin{bmatrix} C_3\\C_4\\C_5\\C_6 \end{bmatrix} &=& \frac{1}{\left(\omega_0^2-\omega_f^2\right)^2-\left(2\gamma\omega_f\right)^2} \begin{bmatrix} \omega_0^2 - \omega_f^2\\2\gamma\omega_f\\-\left(\omega_0^2 - \omega_f^2\right)\\-\frac{\gamma}{i\xi}(\omega_0^2 + \omega_f^2) \end{bmatrix} \end{eqnarray}$$

求まった係数を用いて表す

$$\begin{eqnarray} X&=&\frac{C_1}{s-\lambda_1}+\frac{C_2}{s-\lambda_2} +\frac{F}{m}\left\{ C_3 \frac{s-a}{\left(s-a\right)^2+b^2} + C_4 \frac{b} {\left(s-a\right)^2+b^2} + C_5 \frac{s-c}{\left(s-c\right)^2+d^2} + C_6 \frac{d} {\left(s-c\right)^2+d^2} \right\} \\&=& \frac{\frac{x_0}{2}+\frac{v_0+\gamma x_0}{2\xi }}{s-\lambda_1}+\frac{\frac{x_0}{2}-\frac{v_0+\gamma x_0}{2\xi }}{s-\lambda_2} \\&&+\frac{F}{m} \frac{1}{\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \left\{ \left(\omega_0^2 - \omega_f^2\right) \frac{s-0} {\left(s-0\right)^2+\omega_f^2} + 2\gamma\omega_f \frac{\omega_f} {\left(s-0\right)^2+\omega_f^2} - \left(\omega_0^2 - \omega_f^2\right) \frac{s-(-\gamma)} {\left(s-(-\gamma)\right)^2+(i\xi)^2} - \frac{\gamma}{i\xi}(\omega_0^2 + \omega_f^2) \frac{(i\xi)} {\left(s-(-\gamma)\right)^2+(i\xi)^2} \right\} \\&=& \frac{\frac{x_0}{2}+\frac{v_0+\gamma x_0}{2\xi }}{s-\lambda_1}+\frac{\frac{x_0}{2}-\frac{v_0+\gamma x_0}{2\xi }}{s-\lambda_2} \\&&+\frac{F}{m} \frac{1}{\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \left\{ \left(\omega_0^2 - \omega_f^2\right) \frac{s-0} {\left(s-0\right)^2+\omega_f^2} + 2\gamma\omega_f \frac{\omega_f} {\left(s-0\right)^2+\omega_f^2} \right\} \\&&+\frac{F}{m} \frac{1}{\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \left\{ - \left(\omega_0^2 - \omega_f^2\right) \frac{s-(-\gamma)} {\left(s-(-\gamma)\right)^2+(i\xi)^2} - \frac{\gamma}{i\xi}(\omega_0^2 + \omega_f^2) \frac{(i\xi)} {\left(s-(-\gamma)\right)^2+(i\xi)^2} \right\} \end{eqnarray}$$

第1,2項の逆ラプラス変換

$$\begin{eqnarray} &&\mathfrak{L}^{-1}\left[ \frac{\frac{x_0}{2}+\frac{v_0+\gamma x_0}{2\xi }}{s-\lambda_1}+\frac{\frac{x_0}{2}-\frac{v_0+\gamma x_0}{2\xi }}{s-\lambda_2} \right] \\&=& \left(\frac{x_0}{2}+\frac{v_0+\gamma x_0}{2\xi }\right)\mathfrak{L}^{-1}\left[\frac{1}{s-\lambda_1}\right] +\left(\frac{x_0}{2}-\frac{v_0+\gamma x_0}{2\xi }\right)\mathfrak{L}^{-1}\left[\frac{1}{s-\lambda_2}\right] \\&=& \left(\frac{x_0}{2}+\frac{v_0+\gamma x_0}{2\xi }\right)e^{\lambda_1 t} +\left(\frac{x_0}{2}-\frac{v_0+\gamma x_0}{2\xi }\right)e^{\lambda_2 t} \\&=& \left(\frac{x_0}{2}+\frac{v_0+\gamma x_0}{2\xi }\right)e^{\left(-\gamma+\xi\right)t} +\left(\frac{x_0}{2}-\frac{v_0+\gamma x_0}{2\xi }\right)e^{\left(-\gamma-\xi\right)t} \;\ldots\;\lambda_{1,2}=-\gamma\pm\xi \\&=& \left(\frac{x_0}{2}+\frac{v_0+\gamma x_0}{2\xi }\right) e^{-\gamma t}e^{\xi t} +\left(\frac{x_0}{2}-\frac{v_0+\gamma x_0}{2\xi }\right) e^{-\gamma t}e^{-\xi t} \;\ldots\;e^{A+B}=e^Ae^B \\&=&e^{-\gamma t}\left\{ \frac{x_0}{2} e^{\xi t} +\frac{v_0+\gamma x_0}{2\xi } e^{\xi t} +\frac{x_0}{2} e^{-\xi t} -\frac{v_0+\gamma x_0}{2\xi } e^{-\xi t} \right\} \\&=&e^{-\gamma t}\left\{ \frac{x_0}{2} \left(e^{\xi t}+e^{-\xi t}\right) +\frac{v_0+\gamma x_0}{2\xi } \left(e^{\xi t}-e^{-\xi t}\right) \right\} \\&=&e^{-\gamma t}\left\{ \frac{x_0}{2} \left(e^{i\omega t}+e^{-i\omega t}\right) +\frac{v_0+\gamma x_0}{2i\omega } \left(e^{i\omega t}-e^{-i\omega t}\right) \right\} \;\ldots\;\xi=\sqrt{\gamma^2-\omega_0^2}=\sqrt{-1}\sqrt{\left|\gamma^2-\omega^2\right|}=i\omega\;(\gamma \lt \omega_0の場合) \\&=&e^{-\gamma t}\left\{ \frac{x_0}{2} 2\cos{\left(\omega t\right)} +\frac{v_0+\gamma x_0}{2i\omega } 2i\sin{\left(\omega t\right)} \right\} \;\ldots\;\cos{\left(\theta\right)}=\frac{e^{i\theta}+e^{-i\theta}}{2},\;\sin{\left(\theta\right)}=\frac{e^{i\theta}+e^{-i\theta}}{2i} \\&=&e^{-\gamma t}\left\{ x_0 \cos{\left(\omega t\right)} +\frac{v_0+\gamma x_0}{\omega } \sin{\left(\omega t\right)} \right\} \\&=& x_0 e^{-\gamma t} \left\{ \cos{\left(\omega t\right)} + \frac{\gamma}{\omega } \sin{\left(\omega t\right)} \right\} +v_0 e^{-\gamma t} \frac{1}{\omega } \sin{\left(\omega t\right)} \end{eqnarray}$$

第3項の逆ラプラス変換

$$\begin{eqnarray} &&\mathfrak{L}^{-1}\left[ \frac{F}{m} \frac{1}{\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \left\{ \left(\omega_0^2 - \omega_f^2\right) \frac{s-0} {\left(s-0\right)^2+\omega_f^2} + 2\gamma\omega_f \frac{\omega_f} {\left(s-0\right)^2+\omega_f^2} \right\} \right] \\&=& \frac{F}{m} \frac{1}{\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \left\{ \left(\omega_0^2 - \omega_f^2\right) e^{0 t}\cos{\left(\omega_f t\right)} + 2\gamma\omega_f e^{0 t}\sin{\left(\omega_f t\right)} \right\} \\&=& \frac{F}{m} \frac{1}{\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \left\{ \left(\omega_0^2 - \omega_f^2\right) \cos{\left(\omega_f t\right)} + 2\gamma\omega_f \sin{\left(\omega_f t\right)} \right\} \end{eqnarray}$$

第4項の逆ラプラス変換

$$\begin{eqnarray} &&\mathfrak{L}^{-1}\left[ \frac{F}{m} \frac{1}{\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \left\{ - \left(\omega_0^2 - \omega_f^2\right) \frac{s-(-\gamma)} {\left(s-(-\gamma)\right)^2+(i\xi)^2} - \frac{\gamma}{i\xi}(\omega_0^2 + \omega_f^2) \frac{(i\xi)} {\left(s-(-\gamma)\right)^2+(i\xi)^2} \right\} \right] \\&=& \frac{F}{m}\frac{- \left(\omega_0^2 - \omega_f^2\right)} {\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \mathfrak{L}^{-1}\left[\frac{s-(-\gamma)} {\left(s-(-\gamma)\right)^2+(i\xi)^2} \right] +\frac{F}{m}\frac{- \frac{\gamma}{i\xi}(\omega_0^2 + \omega_f^2)} {\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \mathfrak{L}^{-1}\left[\frac{(i\xi)} {\left(s-(-\gamma)\right)^2+(i\xi)^2} \right] \\&=& \frac{F}{m}\frac{- \left(\omega_0^2 - \omega_f^2\right)} {\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} e^{-\gamma t}\cos{\left(i\xi t\right)} +\frac{F}{m}\frac{- \frac{\gamma}{i\xi}(\omega_0^2 + \omega_f^2)} {\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} e^{-\gamma t}\sin{\left(i\xi t\right)} \\&=& \frac{F}{m} \frac{1}{\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \left\{ - \left(\omega_0^2 - \omega_f^2\right) e^{-\gamma t}\cos{\left(i\xi t\right)} - \frac{\gamma}{i\xi}(\omega_0^2 + \omega_f^2) e^{-\gamma t}\sin{\left(i\xi t\right)} \right\} \\&=& \frac{F}{m} \frac{-e^{-\gamma t}}{\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \left\{ \left(\omega_0^2 - \omega_f^2\right) \cos{\left(i\xi t\right)} +\frac{\gamma}{i\xi}(\omega_0^2 + \omega_f^2) \sin{\left(i\xi t\right)} \right\} \\&=& \frac{F}{m} \frac{-e^{-\gamma t}}{\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \left\{ \left(\omega_0^2 - \omega_f^2\right) \cos{\left(i\cdot i\omega t\right)} +\frac{\gamma}{i\cdot i\omega}(\omega_0^2 + \omega_f^2) \sin{\left(i\cdot i\omega t\right)} \right\}\;\ldots\;\xi=i\omega \\&=& \frac{F}{m} \frac{-e^{-\gamma t}}{\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \left\{ \left(\omega_0^2 - \omega_f^2\right) \cos{\left(-\omega t\right)} +(-1)\frac{\gamma}{\omega}(\omega_0^2 + \omega_f^2) \sin{\left(-\omega t\right)} \right\}\;\ldots\;i\cdot i=-1 \\&=& \frac{F}{m} \frac{-e^{-\gamma t}}{\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \left\{ \left(\omega_0^2 - \omega_f^2\right) \cos{\left(\omega t\right)} +(-1)\frac{\gamma}{\omega}(\omega_0^2 + \omega_f^2) (-1) \sin{\left(\omega t\right)} \right\}\;\ldots\;\cos{\left(-\theta\right)}=\cos{\left(\theta\right)},\;\sin{\left(-\theta\right)}=-\sin{\left(\theta\right)} \\&=& \frac{F}{m} \frac{-e^{-\gamma t}}{\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \left\{ \left(\omega_0^2 - \omega_f^2\right) \cos{\left(\omega t\right)} +\frac{\gamma}{\omega}(\omega_0^2 + \omega_f^2) \sin{\left(\omega t\right)} \right\}\;\ldots\;-1\cdot -1=1 \end{eqnarray}$$

全体の逆ラプラス変換(1,2,4,3項の順に入れ替え)

$$\begin{eqnarray} \mathfrak{L}^{-1}\left[X\right]=x\left(t\right) &=& x_0 e^{-\gamma t} \left\{ \cos{\left(\omega t\right)} + \frac{\gamma}{\omega } \sin{\left(\omega t\right)} \right\} \\&&+v_0 e^{-\gamma t} \frac{1}{\omega } \sin{\left(\omega t\right)} \\&&+\frac{F}{m}\frac{-e^{-\gamma t}}{\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \left\{ \left(\omega_0^2 - \omega_f^2\right) \cos{\left(\omega t\right)} +\frac{\gamma}{\omega}(\omega_0^2 + \omega_f^2) \sin{\left(\omega t\right)} \right\} \\&&+\frac{F}{m}\frac{1}{\left(\omega_0^2-\omega_f^2\right)^2+\left(2\gamma\omega_f\right)^2} \left\{ \left(\omega_0^2 - \omega_f^2\right) \cos{\left(\omega_f t\right)} + 2\gamma\omega_f \sin{\left(\omega_f t\right)} \right\} \end{eqnarray}$$

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