Probability of Error for BPSK over Rayleigh channel with Maximal Ratio Combiner

Aside from deriving BER for BPSK over rayleigh channel, I also got another assignment to derive the equation of probability of error (BER) for BPSK over rayleigh channel with maximal ratio combiner (mrc). After searching here and there, I found the solution in Digital Communications by Proakis page 825. But just like the previous assignment, Proakis didn’t derive the equation in detail. Even Mr. Google couldn’t help me this time ðŸ˜¦ So I tried to solve it myself, but I’m stuck. So I posted here my unfinished work, maybe you can help me figuring it out.

For a certain value of time, the bit error rate probability is

$P_b(\gamma_b)=Q(\sqrt{2\gamma_b})$

where $\gamma_b=\frac{E_b}{N_o}\sum_{k=1}^L {h_k}^2 = \sum_{k=1}^L \gamma_k$ . $\gamma_k$ is the instantaneous SNR on the $k$th channel.

When $k=1$, $\gamma_b\equiv\gamma_1$ and

$P(\gamma_1)=\frac{1}{\bar \gamma_c} e^{-\frac{\gamma_1}{\gamma_c}}$

where $\bar \gamma_c=\frac{E_b}{N_0}E({h_k}^2)$ (average SNR per channel). So the characteristic function is

$\psi_{\gamma_1}(jv) = E(e^{jv\gamma_1})$
$=\int_{-\infty}^{\infty} e^{jv\gamma_1}\frac{1}{\bar \gamma_c}e^{-\frac{\gamma_1}{\bar \gamma_c}}d\gamma_1$
$=\frac{1}{\bar \gamma_c}\int_{-\infty}^{\infty}e^{jv\gamma_1}\frac{1}{\bar \gamma_c}e^{-\frac{\gamma_1}{\bar \gamma_c}}d\gamma_1$
$=\left. -\frac{1}{\bar \gamma_c}\frac{\bar \gamma_c}{1-jv\bar \gamma_c}e^{-\gamma_1\left(\frac{1}{\bar \gamma_c}-jv \right) }\right|_{-\infty}^{\infty}$
$=\frac{1}{1-jv\bar \gamma_c}$

Since the fading on K channels is mutually statistically independent
$\psi_{\gamma_b}(jv)=\frac{1}{(1-jv\bar \gamma_c)^K}$

Using inverse fourier transform,

$p(\gamma_b) = \frac{1}{2\pi}\int_{-\infty}^\infty \psi_\gamma(jv)e^{-jv\gamma_b}$

$=\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{(1-jv\bar \gamma_c)}e^{-jv\gamma_b}$

$=\frac{1}{{\bar \gamma_c}^K \Gamma(K)}{\gamma_b}^{K-1}e^{-\frac{\gamma_b}{\bar \gamma_c}}$

$=\frac{1}{(K-1)!{\bar \gamma_c}^K}{\gamma_b}^{K-1}e^{-\frac{\gamma_b}{\bar \gamma_c}}$

So, the Bit error rate (error probability)

$P_2=\int_{-\infty}^\infty P_2(\gamma_b)p(\gamma_b)d\gamma_b$

$=\int_{-\infty}^\infty Q(\sqrt{2\gamma_b})p(\gamma_b)d\gamma_b$

To solve equation above, use integration by parts.
By definition, $\int f(x)g(x)dx = F(x)g(x)-\int F(x)g'(x)$
which is another way of rewriting the usual $\int udv=uv-\int vdu$

Let $f(\gamma_b)=p(\gamma_b)$.
And by definition

$\int x^ne^{ax}dx=\frac{e^{ax}}{a^{n+1}}\left[ (ax)^n-n(ax)^{n-1}+n(n-1)(ax)^{n-2}-\dots+(-1)^nn!\right]$

So we have

$F(\gamma_b) = \int f(\gamma_b)d\gamma_b = \int \frac{1}{(K-1)!{\bar \gamma_c}^K}{\gamma_b}^{K-1}e^{-\frac{\gamma_b}{\bar \gamma_c}}$

$=\frac{1}{(K-1)!{\bar \gamma_c}^K} \int {\gamma_b}^{K-1}e^{-\frac{\gamma_b}{\bar \gamma_c}}$

$=\frac{e^{-\frac{\gamma_b}{\bar \gamma_c}}}{(-1)^K(K-1)!}\left[ \left(-\frac{\gamma_b}{\bar \gamma_c} \right)^{K-1}-(K-1)\left(-\frac{\gamma_b}{\bar \gamma_c} \right)^{K-2}+\right.$

$\left.(K-1)(K-2)\left(-\frac{\gamma_b}{\bar \gamma_c} \right)^{K-3}-\dots+(-1)^{K-1}(K-1)!\right]$

Let $g(\gamma_b)=Q(\sqrt{2\gamma_b})$ also

$\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)dt=f(v(x))v'(x)-f(u(x))u'(x)$

So

$g'(\gamma_b)=\frac{d}{d\gamma_b}Q(\sqrt{2\gamma_b})=\left( -\frac{1}{\sqrt{2\pi}}e^{-\gamma_b}\right)\left( \frac{1}{\sqrt{2}}\gamma_b^{-\frac{1}{2}}\right)$

Probability of error becomes

$P_2=\int_0^\infty Q(\sqrt{2\gamma_b})p(\gamma_b)d\gamma_b=\left. F(\gamma_b)g(\gamma_b)\right|_0^\infty-\int_0^\infty F(\gamma_b)g'(\gamma_b)$

$=\frac{1}{2}-\frac{1}{(-1)^K(K-1)!}\frac{1}{2\sqrt{\pi}}\int_0^\infty \gamma_b^{-\frac{1}{2}}e^{-\gamma_b\left( 1+\frac{1}{\bar \gamma_c} \right)}\left[ \left(-\frac{\gamma_b}{\bar \gamma_c} \right)^{K-1}-(K-1)\left(-\frac{\gamma_b}{\bar \gamma_c} \right)^{K-2}\right.$

$\left.+(K-1)(K-2)\left(-\frac{\gamma_b}{\bar \gamma_c} \right)^{K-3}-\dots+(-1)^{K-1}(K-1)!\right]d\gamma_b$

And somehow the equation becomes

$P_2=\left[ \frac{1}{2}(1-\mu)\right]^K\sum_{k=0}^K \binom{K-1+k}{k}\left[ \frac{1}{2}(1+\mu)\right]^k$

where
$\mu=\sqrt{\frac{\bar \gamma_c}{1+\bar \gamma_c}}$

So that’s it. Anybody can help me on the last part of the derivation? You can download the pdf version of that equation here.