1
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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

$\endgroup$
10
  • $\begingroup$ Still wrong answer $\endgroup$ Mar 4, 2017 at 16:26
  • $\begingroup$ Please tell us what the right answer is and maybe we can see what the difference is. $\endgroup$ Mar 4, 2017 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 $\endgroup$ Mar 4, 2017 at 16:53
  • $\begingroup$ I used your formula and got 6.77 for M; did you use 1000 for SNR? $\endgroup$ Mar 4, 2017 at 17:05
  • 1
    $\begingroup$ Yes that makes sense; note that M itself does not need to be a power of two, just an integer. k is a power of two and is the number of points in the constellation (in your case 2^6 = 64 (64-QAM). Glad it all worked out! $\endgroup$ Mar 4, 2017 at 23:32

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.