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Re: [Discuss-gnuradio] benchmark_tx and rx : too many FALSE packet


From: Marcus Müller
Subject: Re: [Discuss-gnuradio] benchmark_tx and rx : too many FALSE packet
Date: Wed, 16 Mar 2016 15:10:24 +0100
User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:38.0) Gecko/20100101 Thunderbird/38.6.0

Hi SangHyuk

On 16.03.2016 02:21, SangHyuk Kim wrote:
Dear Nathan,

I know QAM is the fastest modulation because it can carry more bits per symbol. This is a reason why I use QAM modulation.

As bits per symbol level is higher, bit error rates will be increased. I expected it returns feasible error rate. However, error rates were too high. So I wondered "is this normal case or what is it I'm missing".
Well, how do you figure it is "too high"?

I mean, the math is all there: A packet of length N symbols for which we assume that the error probability is identical and independent fails when at least one of these flips (assuming CRC is collision free, which we can easily do).
The packet error probability is then the complementary to the packet success probility, which is the product of the individual symbol error probabilities. These being IID (as assumed above), this packet success probability is instantly clear:
$P_\text{packet
      good} = (P_\text{symbol good})^N$.


Just a quick example: If your channel's SNR and your modulation / pulse shaping choices lead to a symbol error probability of $10^{-4}$ (which is pretty good), and your packets are 1500B long, and you're using 16QAM, i.e. 4b/Symbol, then your packet has $N=3000$ symbols:

$P_\text{packet
      good} = (P_\text{symbol good})^N = (1-10^{-4})^{3000}\approx
      74\%$

But what symbol error probabilities are realistic?

Let's assume we're doing 16-QAM. One can, based on a noise model (which typically has Gaussian noise amplitude) derive the error probabilities for all 16 QAM points (using the Q function) that another point is detected instead, average that, and come to a formula that gives one symbol error probabilty $P_\text{symbol
      wrong} = 1-P_\text{symbol good}$. Not going to do that here. Many digital communication textbooks do that, and if you're used to dealing with probabilities, you can do that on your own. Usually a shortcut is taken and only the probabilities of switching to the nearest neighboring constellation points are considered. That approximation holds well for QAM.

The approximate solution given in such calculation is typically (you should really do some literature research on your own) something like

$P_\text{symbol
      wrong} \approx 3\, Q\left(\frac{2}{\sqrt
      5}\sqrt{\frac{E_b}{N_0}}\right)$

with the Q-Function $Q(x)$ being the probality that a normal Gaussian random variable takes a value larger than $x$,
$\frac{E_b}{N_0}$ being the bit energy to noise power density ratio; that should be defined as
$\frac{E_b}{N_0}
      = \text{SNR} \cdot \frac{B_\text{signal}}{r_\text{bits}}$, i.e. the SNR on the channel times the spectrum-per-bit-rate.

You do the math; is the packet error rate really so much higher than what you'd expect under these formulas? If it really is by more than a factor of 2, this is worth looking into any further. If it is not: well, you won't be much better with an SNR like that.

Best regards,
Marcus

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