I've searched for the theoretical equations of bit error rate over Rayleigh flat fading channel basically for binary phase shift key and 16-Quadrature modulation but i didn't find them, could any one give me their equations or a reference of a paper for their equations?

  • $\begingroup$ What research have you done? Typing in "Rayleigh BER BPSK" into google instantly yielded dsplog.com/2008/08/10/ber-bpsk-rayleigh-channel $\endgroup$ Commented Dec 2, 2018 at 20:35
  • $\begingroup$ You can find it here M. K. Simon and M. S. Alouini, Digital Communication over Fading Channels, 2nd edition, Wiley and Sons, Inc,pp.255-256, 2005 $\endgroup$
    – Am Ki
    Commented May 31, 2019 at 2:09

2 Answers 2


OK, let me explain that in easier way, As you know, in a communication system, to transmit information from transmitter to receiver, we use the binary information system. which means, for example, this message will be transmitted 1000101010.

But in receiver, this message will not be received correctly because of noise of the channel, so some bits will be changed. So, the BER is the average rate of bit error for a particular communication system. for instant, our message example was received as 1000101011, with an error of last bit was changed from 0 to 1. So the average of BER is 1/10 = 10%. which means one bit our of 10 was received wrongly.

That's the basic meaning of BER, let's now explain it regarding BPSK first, in BPSK information symbol 0 is modulated as the amplitude level √P (square root of signal power) , and The information symbol 1 is modulated as an amplitude level -√P which represents 2 voltage levels. So as you see you will have two phases 0 to 180.

We are setting the average power to $p$. Therefore, the average power of this modulation format, so since we are transmitting the amplitude $p^{1/2}$ and $-p^{1/2}$ through a channel, the received symbol will be,

$y = hx + n ............(1)$

where y is the received symbol, h is the channel, x is the transmitted symbol and n is the noise considered as Additive white Gaussian noise channel with the following probability density function that is:

enter image description here

So in the receiver side, we just detect those two levels of voltage such that if the received symbol $p >= 0 $ so the bit is 0, and if it's $p<0$, the bit is 1. and so on.

Do you what is $x$ in the above equation (1)? it's nothing by either $p^{1/2}$ which represent the bit 1 or $-p^{1/2}$ which represent the bit 0.

So based on that detected symbol $y$, we are going to decide if the received symbol is 0 or 1. and then calculate the potability of error and BER according to simple steps in this link http://www.dsplog.com/2008/08/10/ber-bpsk-rayleigh-channel/

So you asked what's about QPSK or 16-Quadrature modulation and so on. When understanding the above explanation, you will see that alone difference will be just in transmitted signal power, it won't be $p^{1/2}$ and -$p^{1/2}$ anymore. it means the phase will not be 180, it will be 90 in case of QPSK, representing four possible values of detected power of signal $p$ and similar to 1616-Quadrature. we usually say M-ary signal constellation, M represent that order of QPSK, M=4, 8PSK which M = 8, and so on.

hope that will help, Good luck

  • $\begingroup$ And how does one set thresholds in QAM when the signal strength fades? $\endgroup$ Commented May 31, 2019 at 2:38

For BPSK it's

$$P_e = \frac{1}{2}\left[1 - \sqrt{\frac{\gamma}{\gamma+1}}\right]$$

where $\gamma$ is the average SNR.

For 16-QAM, I am not sure if there is a closed-form solution. Check this book. The authors use the moment generation function to find the ABER for all modulation schemes.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.