By which value can the constellation of a 32QAM be normalized? for example 16QAM is normalized by squareroot of 10.

  • 2
    $\begingroup$ The question of normalization is meaningless if you don't tell what you want to normalize for. Average Power? What is the goal of the normalization? $\endgroup$
    – jan
    May 18 '14 at 13:05
  • $\begingroup$ the constellations points const_anti = [-1+1i -1-1i 1+1i 1-1i -1+3i -1-3i 1+3i 1-3i -3+3i -3-3i 3+3i 3-3i -3+1i -3-1i 3+1i 3-1i -5-1i -5-3i -5+1i -5+3i 5+1i 5+3i 5-1i 5-3i -1+5i -3+5i -1-5i -3-5i 1-5i 3-5i 1+5i 3+5i]; $\endgroup$ May 18 '14 at 13:15
  • $\begingroup$ That still does not answer the question of what you want to normalize for. Do you know why you want to normalize? What do you want to achieve by the normalization? That is the missing information here. $\endgroup$
    – jan
    May 18 '14 at 13:43
  • $\begingroup$ i want to get the points under 1 $\endgroup$ May 18 '14 at 14:28
  • $\begingroup$ As @jan says, it's not entirely clear what you're asking here. Please review Matt L.'s answer and see if that helps. $\endgroup$
    – Peter K.
    May 26 '14 at 15:51

If we assume that the symbols are $\{\pm a,\pm 3a, \dots\}+i\{\pm a,\pm 3a, \dots\}$ - for 16-QAM and for the 32-QAM cross-constellation, where in the latter case there are no symbols at $(\pm 5\pm i5)a$ - then the average energy per symbol is

$$E_s=10a^2\quad\textrm{(16-QAM)}\\ E_s=20a^2\quad\textrm{(32-QAM)}$$

where the last equality was shown in this answer. So if you choose $a=1/\sqrt{10}$ for 16-QAM, and $a=1/\sqrt{20}$ for 32-QAM cross-constellation, you normalize the average energy per symbol to unity. Note that the maximum energy per symbol in both cases is of course larger than the average energy:

$$E_{s,max}=18a^2=1.8E_s\quad\textrm{(16-QAM)}\\ E_{s,max}=34a^2=1.7E_s\quad\textrm{(32-QAM)}$$


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