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Derivation of BPSK BER in Rayleigh channel

Posted By Krishna Sankar On January 22, 2009 @ 2:56 am In Channel,Modulation | 49 Comments

This is a guest post by Jose Antonio Urigüen who is an Electrical and Electronic Engineer currently studying an MSc in Communications and Signal Processing at Imperial College in London. This guest post has been created due to his own curiosity when reviewing some concepts of BER for BPSK in Rayleigh channnel [1] published in the dsplog.com

From the post on BER for BPSK in Rayleigh channnel [1], it was shown that, in the presence of channel $h$, the effective bit energy to noise ratio is $\frac{|h|^2E_b}{N_0}$.

The bit error probability for a given value of $h$is,

$P_{b|h}=\frac{1}{2}erfc\left({\sqrt{\frac{|h|^2E_b}{N_0}}}\right)=\frac{1}{2}erfc\left(\sqrt{\gamma}\right)$, where

$\gamma = \frac{|h|^2E_b}{N_0}$ .

The resulting BER in a communications sytem in the presence of a channel $h$, for any random values of $|h|^2$, must be calculated evaluating the conditional probability density function $P_{b|h}$ over the probability density function of $\gamma$.

$P_{b}=\int_0^\infty\frac{1}{2}erfc\left(\sqrt{\gamma}\right)p\left(\gamma\right)d\gamma$ where,

the probability density function of $\gamma$ is

$p\left(\gamma\right) = \frac{1}{\bar{\gamma}}e^{\frac{-\gamma}{\bar{\gamma}}},\ \gamma \ge 0$ and $\bar{\gamma} = \frac{E_b}{N_0}$.

First, we are going to derive the result for the definite integral using a different notation, and then we will apply the result to the concrete expression obtained for the BER.

Using the definitions for the ‘erfc’ and ‘erf’ functions, we can directly compute the first derivative:

$erfc(x) = \frac{2}{\sqrt{\pi}}\int_x^{\infty}e^{-t^2}dt$

$erf(x) = \frac{2}{\sqrt{\pi}}\int_0^{x}e^{-t^2}dt$

$\frac{d}{dx}erf(x) = -\frac{2}{\sqrt{\pi}}e^{-x^2}$

And also the definite integral, simply calculating it by parts:

$\begin{eqnarray}\int erfc(x) dx & = & x erfc(x) + \int x \frac{2}{\sqrt{\pi}}e^{-x^2}dx \\
& = & x erfc(x) -\frac{1}{\sqrt{\pi}}e^{-x^2}\end{eqnarray}$

In the following steps we will use the next also straightforward results:

$\frac{d}{dx} erfc (\sqrt{x}) = -\frac{1}{\sqrt{\pi}}e^{-x}x^{-1/2}$

$\frac{d}{dx} erf (\sqrt{x}) = \frac{1}{\sqrt{\pi}}e^{-x}x^{-1/2}$

In total, we want to derive the following integral, which can be solved by parts:
$\int erfc(\sqrt{x})e^{-\frac{x}{a}}dx = -a erfc({\sqrt{x}})e^{-\frac{x}{a}} - \int (-a)e^{-\frac{x}{a}} (-)\frac{1}{\sqrt{\pi}}e^{-x}x^{-\frac{1}{2}}dx$

Lets find the last term applying the previous result for the ‘erf’ function, and doing a change of variable:

$\begin{eqnarray}\int e^{-\frac{x}{a}}e^{-x}x^{-\frac{1}{2}}dx & = & \int e ^{-x\left(\frac{a+1}{a}\right)}x^{-\frac{1}{2}}dx\\ & = & \int e^{-u} \left(\frac{a}{a+1}\right)^{-\frac{1}{2}}u^{-\frac{1}{2}}\left(\frac{a}{a+1}\right)du\\ & = & \left(\frac{a}{a+1}\right)^{\frac{1}{2}} \int e^{-u}u^{-\frac{1}{2}}du\\ & = & \sqrt{\pi}\sqrt{\frac{a}{a+1}} erf(\sqrt{u})\\&=&\sqrt{\pi}\sqrt{\frac{a}{a+1}} erf\left(\sqrt{\frac{a+1}{a}}\sqrt{x}\right)\end{eqnarray}$.

Finally, we conclude the expected result:
$\begin{eqnarray}\int erfc (\sqrt{x}) e^{-\frac{x}{a}}dx & = &-a erfc(\sqrt{x})e^{-\frac{x}{a}}-\frac{a}{\sqrt{\pi}}\sqrt{\pi}\sqrt{\frac{a}{a+1}}erf\left(\sqrt{\frac{a+1}{a}}\sqrt{x}\right)\\ & = & -a erfc(\sqrt{x})e^{-\frac{x}{a}}-{a}\sqrt{\frac{a}{a+1}}erf\left(\sqrt{\frac{a+1}{a}}\sqrt{x}\right) \end{eqnarray}$

With the above demonstration, we can easily derive the BER for a Rayleigh channel using BPSK modulation:

$\begin{eqnarray}P_{b} & = & \frac{1}{2\bar{\gamma}}\int_0^{\infty} erfc (\sqrt{\gamma}) e^{-\frac{\gamma}{\bar{\gamma}}}d\gamma \\ & = & \frac{1}{2\bar{\gamma}}\left[\bar{\gamma}erfc(\sqrt{\gamma})e^{-\frac{\gamma}{\bar{\gamma}}} + \bar{\gamma}\sqrt{\frac{\bar{\gamma}}{\bar{\gamma}+1}}erf\left(\sqrt{\frac{\bar{\gamma}}{\bar{\gamma}+1}}\sqrt{{\gamma}}\right) \right]_\infty^0\\ & = & \frac{1}{2} \lef(1-\sqrt{\frac{\bar{\gamma}}{\bar{\gamma}+1}}\right)\\ & = & \frac{1}{2} \lef(1-\sqrt{\frac{(E_b/N_0)}{(E_b/N_0)+1}}\right)\end{eqnarray}$.

This forms the proof for BER for BPSK modulation in Rayleigh channel.