If the resistance values of a single resistor have the same normal distribution, then what about the mean and variance of two independent resistors in parallel?
I have done this:
$$
R_x,R_y\sim N(R,\sigma)
$$
$$
R_p=\frac{R_xR_y}{R_x+R_y}
$$
$$
\bar{R_p}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{xy}{x+y}\frac{1}{2\pi\sigma^2}e^{-\frac{(x-R)^2+(y-R)^2}{2\sigma^2}}\mathrm{d}x\mathrm{d}y
$$
If I get $u=x+y,v=x-y$:
$$
\bar{R_p}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{4\pi\sigma^2}\frac{u^2-v^2}{4u}e^{-\frac{v^2}{4\sigma^2}}\mathrm{d}ve^{-\frac{(u-2R)^2}{4\sigma^2}}\mathrm{d}u
$$
$$
=\frac{1}{8\sqrt{\pi}\sigma}\int_{-\infty}^{\infty}(u-\frac{2\sigma^2}{u})e^{-\frac{(u-2R)^2}{4\sigma^2}}\mathrm{d}u
$$
$$
=\frac{R}{2}-\frac{\sigma}{4\sqrt{\pi}}\int_{-\infty}^{\infty}\frac{e^{-\frac{(u-2R)^2}{4\sigma^2}}}{u}\mathrm{d}u
$$
Of course, the simple result of two independent resistors(R) in parallel is $\frac{R}{2}$, but what is the rest integral? Is it converge?
And I use residue theorem (if I don't make mistake), the integral become:($f(z)=\frac{e^{-\frac{(z-2R)^2}{4\sigma^2}}}{z}$)
$$
\int_{-\infty}^{\infty}\frac{e^{-\frac{(u-2R)^2}{4\sigma^2}}}{u}\mathrm{d}u=\pi i Res(f(0))=\frac{e^{-\frac{R^2}{\sigma^2}}}{2}
$$ Is there anyone can check it?
emm, because $lim_{a\rightarrow \infty} e^{-(ia)^2}\rightarrow \infty$ ,so the residue theorem can't be used…
Best Answer
OK, I have solved this problem and I guess there is nobody who wants to answer this problem...
make$w=\frac{u}{2\sigma}$ $$ \int_{-\infty}^{\infty}\frac{e^{-\frac{(u-2R)^2}{4\sigma^2}}}{u}\mathrm{d}u=\int_{-\infty}^{\infty}\frac{e^{-(w-\frac{R}{\sigma})^2}}{w}\mathrm{d}w $$ $$ =\int_{-\infty}^{\infty}\frac{e^{-w^2}}{w+\frac{R}{\sigma}}\mathrm{d}w=\int_{0}^{\infty}(\frac{1}{w+\frac{R}{\sigma}}-\frac{1}{w-\frac{R}{\sigma}})e^{-w^2}\mathrm{d}w=\int_{0}^{\infty}\frac{-2\frac{R}{\sigma}}{w^2-\frac{R^2}{\sigma^2}}e^{-w^2}\mathrm{d}w $$ make$p=w^2$ $$ =-2\frac{R}{\sigma}\int_{0}^{\infty}\frac{e^{-p}}{2(p-\frac{R^2}{\sigma^2})\sqrt{p}}\mathrm{d}p=\pi e^{-\frac{R^2}{\sigma^2}}Erfi(\frac{R}{\sigma}) $$ So, the mean value of two independent resistors in parallel with normal distribution is$\frac{R}{2}-\frac{\sigma\sqrt{\pi}}{4}e^{-\frac{R^2}{\sigma^2}}Erfi(\frac{R}{\sigma})$
And the variance is the same method, but it's really divergent.