I have a system using 16-QAM, with 2.4 GHz, working up to 2 km at different bit rates, e.g. 1Mbit/s. My task is to provide graph, which will show relation between BER and distance for different bit rates. However, based on that article: https://www.mathworks.com/help/comm/ug/bit-error-rate-ber.html#bq4230l and adding bit rates as errors in dB, I am able to provide the graph for BER vs. Eb/N0, I wonder- what is relation BER to distance?


In short, increasing the distance results in a decrease in $E_b$.

Consider two antennas in an otherwise empty universe (this is known as "free space"). The transmitter antenna radiates $P_t$ watts, and has gain $G_t$ in the direction of the receiver. The transmitted signal has wavelength $\lambda_0$. Then, the receiver antenna, with gain $G_r$ and at distance $d$ from the transmitter, receives a signal with power given by $$P_r = P_tG_rG_t\left(\frac{\lambda_0}{4\pi d}\right)^2.$$

You are assuming transmission at 1 Mb/s. Then, $E_b$ is given by $$E_b = \frac{P_r}{10^6}\frac{\text{joule}}{\text{bit}}.$$ You can see that, as $d$ increases, both $P_r$ and $E_b$ decrease. This causes an increase in the bit error rate.

In environments other than free space, the calculation of $P_r$ is different, since the presence of reflectors, obstacles, etcetera will influence how much power is received. A popular model for urban/suburban/rural environments is the Okomura-Hata model, which is valid for the range of frequencies, distances and antenna heights commonly used in cellular communications. There are many other models that apply in different scenarios, and with varying degrees of complexity and accuracy.

  • $\begingroup$ I'd like to ask the downvoter to explain their reasons. Perhaps I misinterpreted the question? $\endgroup$
    – MBaz
    Jan 6 at 17:04
  • $\begingroup$ I don't have Pt or Gr, Gt given. How should I cope with that? I mean, I don't know the distance or bit rate in theoretical system. How should I find the ratio with which Eb increases/ decreases? Eb_new = x* Eb, where x is ratio x=1/(d^2* Br) $\endgroup$
    – Magda
    Jan 11 at 18:19
  • $\begingroup$ Yes, that sounds reasonable. Your formula describes the fundamental issue here, which is that power decreases as the square of the distance. $\endgroup$
    – MBaz
    Jan 11 at 22:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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