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Timeline for The error probability of 16QAM

Current License: CC BY-SA 3.0

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Aug 23, 2018 at 11:52 vote accept Shine Sun
Dec 13, 2017 at 7:42 comment added Matt L. @ShineSun: Yes, that's what I mean.
Dec 13, 2017 at 7:13 comment added Shine Sun you mean $\frac{M}{M}$=$1$?
Dec 13, 2017 at 6:52 comment added Matt L. @ShineSun: Yes, there is. It's a sum over $M$ equal elements that's divided by $M$. What do you get if you sum up $M$ equal numbers and divide the result by $M$?
Dec 13, 2017 at 0:08 comment added Shine Sun don't understand,but there is also a $\frac{1}{M} \sum ^M_{i=1}$ in (5)
Dec 12, 2017 at 14:48 history edited Matt L. CC BY-SA 3.0
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Dec 12, 2017 at 14:45 comment added Matt L. @ShineSun: If this answer helped you, you can accept it.
Dec 12, 2017 at 13:49 comment added Matt L. @ShineSun: Just plug (6) into (5) and note that the sum $\sum_{j\neq i}$ is over $M-1$ equal numbers.
Dec 12, 2017 at 13:15 comment added Shine Sun ok! now i think my problem is (M-1),because the difference of union bound and mine is the value before Q function.How does (M-1) come from?
Dec 12, 2017 at 12:45 comment added Matt L. @ShineSun: The noise variance is indeed $N_0/2$, and the argument of the $Q$ function must be half the distance between the symbols divided by the square root of the noise variance: $\frac{d/2}{\sqrt{N_0/2}}=\sqrt{\frac{d^2}{2N_0}}$
Dec 12, 2017 at 11:45 comment added Shine Sun Is Eq. (7) a definition?if not,it must have a proof,because i don't understand why the enominator in Q function is $2N_0$,not $\frac{N_0}{2}$
Dec 12, 2017 at 10:11 comment added Shine Sun so union bound is one of method of calculating the error probability,and better than mine,but do you have the proof about $P_e \le (M-1)Q(\sqrt{\frac{d^2_{min}}{2N_0}})$ ?
Dec 12, 2017 at 9:17 history answered Matt L. CC BY-SA 3.0