两同样大小的正态分布阻值电阻并联总电阻均值证明

命题

两个阻值\(\displaystyle R_{1} ,R_{2} \sim N( R ,\sigma )\)的电阻并联,求总电阻的期望。(注:过程极不严谨,充满主值积分)

证明

总电阻为: \[R=\frac{R_{1} R_{2}}{R_{1} +R_{2}}\] 则其期望为: \[\begin{gathered} \overline{R} =\int _{-\infty }^{+\infty }\int_{-\infty }^{+\infty }\frac{R_{1} R_{2}}{R_{1} +R_{2}}\frac{1}{\sqrt{2\pi } \sigma } e^{-\frac{( R_{1} -R)^{2}}{2\sigma ^{2}}}\frac{1}{\sqrt{2\pi } \sigma } e^{-\frac{( R_{2} -R)^{2}}{2\sigma ^{2}}}\mathrm{d} R_{1}\mathrm{d} R_{2}\\ =\frac{1}{2\pi \sigma ^{2}}\int _{-\infty }^{+\infty }\int_{-\infty }^{+\infty }\frac{R_{1} R_{2}}{R_{1} +R_{2}} e^{-\frac{( R_{1} -R)^{2}}{2\sigma ^{2}}} e^{-\frac{( R_{2} -R)^{2}}{2\sigma ^{2}}}\mathrm{d} R_{1}\mathrm{d} R_{2}\\ =\frac{1}{2\pi \sigma ^{2}}\int _{-\infty }^{+\infty }\int_{-\infty }^{+\infty }\frac{R_{1} R_{2}}{R_{1} +R_{2}} e^{-\frac{( R_{1} -R)^{2} +( R_{2} -R)^{2}}{2\sigma ^{2}}}\mathrm{d} R_{1}\mathrm{d} R_{2}\end{gathered}\]\(\displaystyle x=R_{1} +R_{2} ,y=R_{1} -R_{2}\) \[\begin{gathered} \frac{1}{4\pi \sigma ^{2}}\int _{-\infty }^{+\infty }\int_{-\infty }^{+\infty }\frac{x^{2} -y^{2}}{4x} e^{-\frac{y^{2}}{4\sigma ^{2}}} e^{-\frac{( x-2R)^{2}}{4\sigma ^{2}}}\mathrm{d} x\mathrm{d} y \notag\\ =\frac{1}{4\pi \sigma ^{2}}\int _{-\infty }^{+\infty } e^{-\frac{( x-2R)^{2}}{4\sigma ^{2}}}\left[\int _{-\infty }^{+\infty }\left(\frac{x}{4} -\frac{y^{2}}{4x}\right) e^{-\frac{y^{2}}{4\sigma ^{2}}}\mathrm{d} y\right]\mathrm{d} x\end{gathered}\] (1)中方括号中的式子可化为: \[\begin{gathered} \int_{-\infty }^{+\infty }\left(\frac{x}{4} -\frac{y^{2}}{4x}\right) e^{-\frac{y^{2}}{4\sigma ^{2}}}\mathrm{d} y=\frac{x}{4}\int _{-\infty }^{+\infty } e^{-\frac{y^{2}}{2\left(\sqrt{2} \sigma \right) \ ^{2}}}\mathrm{d} y-\int_{-\infty }^{+\infty }\frac{y^{2}}{4x} e^{-\frac{y^{2}}{4\sigma ^{2}}}\mathrm{d} y \notag\\ =\frac{x}{4} 2\sqrt{\pi } \sigma \frac{1}{\sqrt{2\pi }\sqrt{2} \sigma }\int _{-\infty }^{+\infty } e^{-\frac{y^{2}}{2\left(\sqrt{2} \sigma \right) \ ^{2}}}\mathrm{d} y-\int_{-\infty }^{+\infty }\frac{y^{2}}{4x} e^{-\frac{y^{2}}{4\sigma ^{2}}}\mathrm{d} y \notag\\ =\frac{\sqrt{\pi } \sigma x}{2} -\int _{-\infty }^{+\infty }\frac{y^{2}}{4x} e^{-\frac{y^{2}}{4\sigma ^{2}}}\mathrm{d} y\end{gathered}\] (2)式中的余项可化为: \[\begin{gathered} \int_{-\infty }^{+\infty }\frac{y^{2}}{4x} e^{-\frac{y^{2}}{4\sigma ^{2}}}\mathrm{d} y=2\int _{0}^{+\infty }\frac{y^{2}}{4x} e^{-\frac{y^{2}}{4\sigma ^{2}}}\mathrm{d} y=\int_{0}^{+\infty }\frac{\sigma ^{2} y}{x}\mathrm{d}\left( e^{-\frac{y^{2}}{4\sigma ^{2}}}\right)\\ =-\frac{\sigma ^{2} y}{x} e^{-\frac{y^{2}}{4\sigma ^{2}}}| _{0}^{+\infty } +\int_{-\infty }^{+\infty }\frac{\sigma ^{2}}{2x} e^{-\frac{y^{2}}{4\sigma ^{2}}}\mathrm{d} y\\ =2\sqrt{\pi } \sigma \frac{\sigma ^{2}}{2x}\\ =\frac{\sqrt{\pi } \sigma ^{3}}{x}\end{gathered}\] 故(2)式可化为: \[( 2) =\frac{\sqrt{\pi } \sigma x}{2} -\frac{\sqrt{\pi } \sigma ^{3}}{x} =\frac{\sqrt{\pi } \sigma }{2}\left( x-\frac{2\sigma ^{2}}{x}\right)\] 故(1)式可化为: \[\begin{gathered} =\frac{1}{4\pi \sigma ^{2}}\int _{-\infty }^{+\infty } e^{-\frac{( x-2R)^{2}}{4\sigma ^{2}}}\left[\frac{\sqrt{\pi } \sigma x}{2} -\frac{\sqrt{\pi } \sigma ^{3}}{x}\right]\mathrm{d} x \notag\\ =\frac{1}{4\sqrt{2\pi }\sqrt{2} \sigma }\int_{-\infty }^{+\infty }\left( x-\frac{2\sigma ^{2}}{x}\right) e^{-\frac{( x-2R)^{2}}{4\sigma ^{2}}}\mathrm{d} x \notag\\ =\frac{R}{2} -\frac{\sigma }{4\sqrt{\pi }}\int _{-\infty }^{+\infty }\frac{1}{x} e^{-\frac{( x-2R)^{2}}{4\sigma ^{2}}}\mathrm{d} x\end{gathered}\]\(\displaystyle u=\frac{x}{2\sigma } ,v=u-\frac{R}{\sigma } ,w=v^{2}\)(3)中的积分表达式可进一步化为: \[\begin{gathered} \int_{-\infty }^{+\infty }\frac{1}{x} e^{-\frac{( x-2R)^{2}}{4\sigma ^{2}}}\mathrm{d} x=\int _{-\infty }^{+\infty }\frac{1}{u} e^{-\left( u-\frac{R}{\sigma }\right)^{2}}\mathrm{d} u\\ =\int_{-\infty }^{+\infty }\frac{1}{v+\frac{R}{\sigma }} e^{-v^{2}}\mathrm{d} v\\ =\int _{-\infty }^{0}\frac{1}{v+\frac{R}{\sigma }} e^{-v^{2}}\mathrm{d} v+\int_{0}^{+\infty }\frac{1}{v+\frac{R}{\sigma }} e^{-v^{2}}\mathrm{d} v\\ =\int _{0}^{+\infty }\left(\frac{1}{v-\frac{R}{\sigma }} +\frac{1}{v+\frac{R}{\sigma }}\right) e^{-v^{2}}\mathrm{d} v\\ =\int_{0}^{+\infty }\frac{2v}{v^{2} -\frac{R^{2}}{\sigma ^{2}}} e^{-v^{2}}\mathrm{d} v\\ =\int _{0}^{+\infty }\frac{e^{-v^{2}}}{v^{2} -\frac{R^{2}}{\sigma ^{2}}}\mathrm{d} v^{2}\\ =\int_{0}^{+\infty }\frac{e^{-w}}{w-\frac{R^{2}}{\sigma ^{2}}}\mathrm{d} w\\ =-e^{-\frac{R^{2}}{\sigma ^{2}}} \mathrm{Ei}\left(\frac{R^{2}}{\sigma ^{2}}\right)\end{gathered}\] 其中,\(\mathrm{Ei}(x)=-\int_{x}^{+\infty}\frac{e^{-t}}{t}\mathrm{d}t\)是指数积分函数。

故所求总电阻为: \[\overline{R} =\frac{R}{2} +\frac{\sigma }{4\sqrt{\pi }} e^{-\frac{R^{2}}{\sigma ^{2}}} E_{i}\left(\frac{R^{2}}{\sigma ^{2}}\right)\]\(\displaystyle \frac{R^{2}}{\sigma ^{2}}\rightarrow \infty\)时,\(\displaystyle ( 4)\rightarrow 0\),与忽略阻值涨落相一致。

因为阻值涨落导致的阻值偏差比率\(\displaystyle \frac{\Delta \overline{R}}{\frac{R}{2}} =\frac{\sigma }{2\sqrt{\pi } R} e^{-\frac{R^{2}}{\sigma ^{2}}} E_{i}\left(\frac{R^{2}}{\sigma ^{2}}\right)\),记\(\displaystyle x=\frac{\sigma }{R}\),则有\(\displaystyle \frac{\Delta \overline{R}}{\frac{R}{2}} =g( x) =\frac{x}{2\sqrt{\pi }} e^{-\frac{1}{x^{2}}} E_{i}\left(\frac{1}{x^{2}}\right)\),如图

阻值涨落导致的阻值偏差比率

\(\displaystyle x\rightarrow 0\)时,\(\displaystyle g( x)\rightarrow \frac{x^{3}}{2\sqrt{\pi }}\)