Consider a QAM system designed to transmit over a bandwidth of $3 \ \mathrm{kHz}$. The channel's power constraint imposes a maximum $\mathrm{SNR}$ of $30 \ \mathrm{dB}$. The system can tolerate a probability of error of $10^{−6}$. I want to determine the maximum throughput of the system in bits per second.

Let's call the throughput $T$. If $M$ is the constellation size and $"$ is the bandwidth, then I know that $T=MW$. I've calculated the value of $M$ using theformula $$M = \log_2 \left( 1 - \frac{1.5\mathrm{SNR}}{\ln(p_e)}\right)$$ Then I approximated $M$ to the nearest power of $2$, which is $8$. Therefore, $T = 24000 \ \mathrm{\frac{bits}{sec}}$ but it is a wrong answer. Can anyone tell me what I have done wrong?


When $\log_2(M)$ is even, the relationship betweeen BER and QAM if an AWGN channel is given as below (Assuming Gray coding so that a symbol error in most cases is one bit error):

\begin{align} k&=\log_2(M)\\ y &= 10^{\frac{SNR}{10}}\\ P_e&=\frac{4}{k}\frac{\sqrt{M}-1}{2\sqrt{M}}\textrm{erfc}\left(\frac{\sqrt{3k\frac{y}{M-1}}}{\sqrt{2}}\right) \end{align}

Using the above formula to solve for $y$ given your SNR and constellation $M$ should achieve the result you are looking for.

This is an excellent explanation for the derivation of the formula, specific to QAM.

And here is a nice summary by the same author for BER vs SNR of many different modulation schemes

  • $\begingroup$ Still wrong answer $\endgroup$ – Mohamed Maherz Mar 4 '17 at 16:26
  • $\begingroup$ Please tell us what the right answer is and maybe we can see what the difference is. $\endgroup$ – Dan Boschen Mar 4 '17 at 16:36
  • $\begingroup$ I don't know the right answer... It is an online quiz where I input my solutions and it shows correct or wrong only $\endgroup$ – Mohamed Maherz Mar 4 '17 at 16:53
  • $\begingroup$ I used your formula and got 6.77 for M; did you use 1000 for SNR? $\endgroup$ – Dan Boschen Mar 4 '17 at 17:05
  • $\begingroup$ I forgot to convert SNR and used it in db .. you are right according to the formula M = 6.77 and approximating it to 8 so throughput will equal 24,000 bits/sec but it isn't also correct ... it is very frustrating and I've been searching all day but I can't solve it $\endgroup$ – Mohamed Maherz Mar 4 '17 at 17:28

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.