Modulation 6 dB margin QPSK-16QAM-64QAM...

Hi,

QPSK has - 4 constellation points; 16 QAM -16 constellation points; 64 -
64 constellation points. 

16QAM vs QPSK -> Now with the 16 possible points in the constellation
diagram we have 16 possible symbols. For this 16 symbols we need 4 bits for
coding . Compared to the QPSK modulation now we have doubled the transfer
rate when using the same symbol clock.But compared to the QPSK, points are
closer to each other and so the allowed noise circle radius is decreased.
Noise can be interpreted as a vector which turns around the points of the
constellation diagram producing a circle with a noise amplitude dependend
radius. If we have normalized constellation (i.e. max distance is 1 - i.e.
distance between two outer edges of circle of outer constellation points),
then: 
required distance QAM16 -> 1/(SQRT(QAM  res.)) = 1/SQRT(16) = 0.25
required distance QPSK  -> 1/(SQRT(QPSK res.)) = 1/SQRT(4) = 0.5
So we get a ratio of QPSK/QAM distance required to get no overlap of the
points of: 20 * log(o.5/0.25) = 6dB. 
So to have the same S/N ratio with the same noise level the signal for
16QAM has to be 6 dB stronger than QPSK. The same case is between 64QAM and
16 QAM(6db difference), 256QAM and 64QAM(6db difference) ....
We see that with each bit we add to the symbol rate we need a 3 dB better
S/N of the received signal. So we see the tradeoff we have to do between
the increasing transfer rate and the required S/N ratio.
Of course another way to increase the transfer rate is a higher symbol
clock with the same symbol width.
(PS All this I found in AN97047.pdf
http://www.nxp.com/acrobat_download/applicationnotes/AN97047_1.pdf)
===================================================================

But, I saw also formula(in MATLAB help) Es/No = Eb/No + 10log10(k);  where
k is the number of information bits per symbol.
By this formula we get: 
QPSK:  Es/No = Eb/No + 10log10(2)=Eb/No + 3.0103;
16QAM: Es/No = Eb/No + 10log10(4)=Eb/No + 6.0206;
64QAM: Es/No = Eb/No + 10log10(6)=Eb/No + 7.7815;

For BER of 10e-5, we get for QPSK that required Eb/No is about 9.5 dB, for
16QAM is about 13dB, for 64QAM is about 17,5dB...(I get this by MATLAB
bertool). So we now have:
QPSK:  Es/No = Eb/No + 10log10(2)=9.5 + 3.0103  ~ 12.5 dB;
16QAM: Es/No = Eb/No + 10log10(4)=13 + 6.0206   ~ 20   dB;
64QAM: Es/No = Eb/No + 10log10(6)=17.5 + 7.7815 ~ 26.2 dB;

So as You see the difference between 12.5 and 20 is not 6dB, in case QPSK
vs 16QAM!

(PS: I think that this is a very important line: "So to have the same S/N
ratio with the same noise level the signal for 16QAM has to be 6 dB
stronger than QPSK." - in AN97047.pdf)

So what is correct? Is this the same thing I try to correlate or...
If so, why is that?

Thanks and best regards 

For QPSK, the *maximum* power and the
*average* power are the same.  For larger
QAM constellations (yes, QPSK is 4QAM),
these two quantities are different.  So the
comparison of maximum power and a
comparison of average power give different
answers.

Hope this helps

--Dilip Sarwate

>For QPSK, the *maximum* power and the
>*average* power are the same.  For larger
>QAM constellations (yes, QPSK is 4QAM),
>these two quantities are different.  So the
>comparison of maximum power and a
>comparison of average power give different
>answers.
>
>Hope this helps
>
>--Dilip Sarwate
>
>
Hi,
Thanks for replay but I didn't understand very well.
What is a *maximum* and what is *average* in formulas above?
i.e. which value is the *maximum* and which *average* in  formulas above?
Can You clear this up for me...

and also You said "For QPSK, the *maximum* power and the *average* power
are the same." . how You see this from formulas above can You tell me which
values, and also how for QAM constellations these two quantities are
different.? 

Best regards