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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?

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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

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  • $\begingroup$ 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

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