Article #360

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posted by sakurai on February 10, 2021 #360

引き続き、前稿の続きの計算をします。本稿では次の(360.1)及び(360.2)を求めます。 $$ \frac{1}{T_\text{lifetime}}\int_0^{T_\text{lifetime}}F_\text{SM}(t)dt\tag{360.1} $$ 及び $$ \frac{1}{T_\text{lifetime}}\int_0^{T_\text{lifetime}}F_\text{SM}(u)dt,\ \ s.t.\ \ u:=t\bmod\tau\tag{360.2} $$ まず、(360.1)式に、$F_\text{SM}(t)=1-e^{-\lambda_\text{SM}t}$を代入し、 $$ \require{cancel} \begin{eqnarray} (360.1)&=&\frac{1}{T_\text{lifetime}}\int_0^{T_\text{lifetime}}(1-e^{-\lambda_\text{SM}t})dt=\frac{1}{T_\text{lifetime}}\left[t+\frac{e^{-\lambda_\text{SM}t}}{\lambda_\text{SM}}\right]^{T_\text{lifetime}}_0\\ &=&\frac{1}{T_\text{lifetime}}\left(T_\text{lifetime}-\bcancel{0}+\frac{1}{\lambda_\text{SM}}(e^{-\lambda_\text{SM}T_\text{lifetime}}-1)\right)\\ \end{eqnarray}\tag{360.3} $$ ここで$\lambda t\ll 1$の条件で$e^{-\lambda t}$のMaclaurin展開は $$e^{-\lambda t}=1-\lambda t + \frac{1}{2}\lambda^2 t^2-O((\lambda t)^3)$$であるから、$O((\lambda t)^3)\approx 0$と近似し、これを(360.3)に代入すると(360.3)は、 $$ \begin{eqnarray} (360.3)&\approx&\frac{1}{\bcancel{T_\text{lifetime}}}\left(\bcancel{T_\text{lifetime}}+\frac{1}{\bcancel{\lambda_\text{SM}}}(-\bcancel{\lambda_\text{SM}}\bcancel{T_\text{lifetime}}+\frac{1}{2}\lambda_\text{SM}^\bcancel{2}T_\text{lifetime}^\bcancel{2})\right)\\ &=&\bcancel{1}-\bcancel{1}+\frac{1}{2}\lambda_\text{SM}T_\text{lifetime}=\frac{1}{2}\lambda_\text{SM}T_\text{lifetime} \end{eqnarray}\tag{360.4} $$ 以上から次のように(360.1)の値が求められました。

$$ \frac{1}{T_\text{lifetime}}\int_0^{T_\text{lifetime}}F_\text{SM}(t)dt\approx\frac{1}{2}\lambda_\text{SM}T_\text{lifetime}\tag{360.5} $$ 次に(360.2)は、 $$ \begin{eqnarray} (360.2)&=&\frac{1}{T_\text{lifetime}}\sum_{i=0}^{n-1}\int_{i\tau}^{(i+1)\tau}(1-e^{-\lambda_\text{SM}u})du =\frac{n}{T_\text{lifetime}}\int_0^\tau\left(1- e^{-\lambda_\text{SM}u}\right)du\\ &=&\frac{\bcancel{n}1}{\bcancel{T_\text{lifetime}}\tau}\left[u\bcancel{-}+\frac{e^{-\lambda_\text{SM}u}}{\bcancel{-}\lambda_\text{SM}}\right]^{\tau}_0 =\frac{1}{\tau}\left(\tau-\bcancel{0}+\frac{1}{\lambda_\text{SM}}\left(e^{-\lambda_\text{SM}\tau}-1\right)\right) \end{eqnarray}\tag{360.6} $$ ここで同様に、$$e^{-\lambda t}\approx1-\lambda t + \frac{1}{2}\lambda^2 t^2$$を用いて、 $$ \begin{eqnarray} (360.8)&=&\frac{1}{\bcancel{\tau}}\left(\bcancel{\tau}+\frac{1}{\bcancel{\lambda_\text{SM}}}\left(-\bcancel{\lambda_\text{SM}}\bcancel{\tau}+\frac{1}{2}\lambda_\text{SM}^\bcancel{2}\tau^\bcancel{2}\right)\right) &=&\bcancel{1}-\bcancel{1}+\frac{1}{2}\lambda_\text{SM}\tau\\ &=&\frac{1}{2}\lambda_\text{SM}\tau \end{eqnarray}\tag{360.7} $$ 以上から次のように(360.2)の値が求められました。 $$ \frac{1}{T_\text{lifetime}}\int_0^{T_\text{lifetime}}F_\text{SM}(u)dt\approx\frac{1}{2}\lambda_\text{SM}\tau,\ \ s.t.\ \ u:=t\bmod\tau\tag{360.8} $$


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