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I saw a code computing the error probability of QPSK, and I found that in this code the bit error rate is computed as symbol error rate divided by $2$.

I think that $2$ is because the QPSK symbol consists of two bits. But why can the bit error rate be equal to the symbol error rate divided by the number of bits per symbol?

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In general, the bit error probability depends on the symbol error probability and the coder, i.e., the mapping between bits and symbols. The formula you refer to

$$P[\textrm{bit error}]\approx\frac{1}{M}P[\textrm{symbol error}]\tag{1}$$

with $M$ being the number of bits per symbol ($M=2$ for QPSK) is an approximation for a certain type of coder under the assumption that the SNR is relatively high.

In $(1)$ it is assumed that the system implements a Gray code, where nearest neighbor symbols only differ by one bit. So if the SNR is high, most symbol errors occur due to confusing a symbol with one of its nearest neighbors, which causes only one out of $M$ bits to be wrong.

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  • $\begingroup$ so the formula (1) is right only if the SNR is high,so what is the value of SNR can be said the SNR is high?can i say when SNR = 10dB is high? $\endgroup$
    – shineele
    Commented May 5, 2019 at 11:45
  • $\begingroup$ @shineele: It's just an approximation, but it becomes more accurate for increasing SNR. An SNR of 10dB is indeed relatively high. $\endgroup$
    – Matt L.
    Commented May 5, 2019 at 12:09
  • $\begingroup$ I disagree with the blanket assertion that if there are $M$ bits per symbol, then the bit error probability is $1/M$ times the symbol error probability. The symbol error probability is dominantly the probability that the transmitted signal is mistakenly demodulated into one to its nearest neighbors in the signal constellation, and in QAM signal constellations, most constellation points have 4 nearest neighbors, not $M$ nearest neighbors. For QPSK (which is 4-QAM), all points have 2 nearest neighbors and 1 distant neighbor, and so the claimed result holds serendipitously. $\endgroup$
    – Dilip Sarwate
    Commented May 6, 2019 at 13:10
  • $\begingroup$ @DilipSarwate: It's of course an approximation, as shown by the $\approx$ sign in the formula. But for high SNR and Gray coding it holds pretty well for practical constellations I know of. Is there any realistic constellation for which the formula doesn't work? $\endgroup$
    – Matt L.
    Commented May 6, 2019 at 15:55
  • $\begingroup$ Any number can be claimed to be an approximation to any other number, and so your formula "works". My point is that for QAM in general, $\frac 14 P_{\text{e, symbol}}$ is a better approximation to $P_{\text{e, bit}}$ than $\frac 1M P_{\text{e, symbol}}$ but for QPSK, that $\frac 14$ is better replaced by $\frac 12$ for reasons stated in my previous comment (and my own answer too). That $M$ equals 2 for QPSK is a serendipitous coincidence, and not a reliable guide to the general case for QAM as well as other constellations such as MPSK for $M>4$ etc, or orthogonal signaling, and so on. $\endgroup$
    – Dilip Sarwate
    Commented May 6, 2019 at 17:48
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The exact symbol error rate in QPSK is $2p-p^2$ where $p$ is the bit error rate. When $p$ is small (which happens when the SNR is large), the symbol error rate is very slightly smaller than twice the bit error rate. If you write $p$ in scientific notation such as 0.314E-3 a.k.a. $0.314\times 10^{-3}$ correct to three significant digits, the symbol error rate is “exactly” twice the bit error rate because the difference between $2p-p^2$ and $2p$ is lost in the rounding off to three digits.

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Each symbol contains K bits, for different modulation schemes you can refer to this relation: K= log2(M) K number of bits/ symbol for BPSK: M=2, K=1 for QPSK: M=4, K=2 for 8PSK: M=8, K=3 and so on....

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