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 😦 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


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 kth 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)


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

\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.


~ by overblaze on November 13, 2008.

One Response to “Probability of Error for BPSK over Rayleigh channel with Maximal Ratio Combiner”

  1. I am sorry. I don’t know

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: