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Just had a theoretical question. I'm reading a book on LTE, and I came across an introductory section that talks about using higher modulation with channel coding. I understand that usually using modulation schemes like 16 QAM and 64 QAM will require more Eb/No at the receiver end as compared to QPSK. But then they talk about the higher modulations being more efficient when combined with channel coding, that is, they require a lower Eb/No at the receiver as compared to QPSK. The example the books talks about is as follows:

"As an example, if a bandwidth utilization of close to 2 information bits per modulation symbol is required, QPSK modulation would allow for very limited channel coding (channel-coding rate close to 1). On the other hand, the use of 16QAM modulation would allow for a channel-coding rate of the order of one-half."

I'm confused about the second part of the statement. How do we get a channel-coding rate of one-half? Can anyone shed some light on this or perhaps point me to something I can read to get a better understanding?

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    \$\begingroup\$ First guess without looking into it: those quadrature amplitude modulation techniques might require higher signal/noise ratio to resolve the different levels, but since they cram more bits in a symbol, you can insert a more advanced encoder before the modulator to lower the required signal/noise ratio for a fixed error rate. \$\endgroup\$
    – Mister Mystère
    Commented Jun 28, 2015 at 0:14
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    \$\begingroup\$ @MisterMystère I agree, but I'm confused as to how it works. :) They mention in the book that for instance in the 16QAM case and a requirement of 2 bits per modulation symbol, channel coding reduces the data rate, and because of that the bandwidth utilization also decreases to the level of an uncoded QPSK. \$\endgroup\$
    – Nick
    Commented Jun 28, 2015 at 0:34
  • \$\begingroup\$ Perhaps 16QAM and variants only become interesting if the signal/noise ratio is sufficiently improved - otherwise heavy encoding is required which may become as big as the actual data. QPSK is a rather legacy modulation now, the sharp improvement of the equipment certainly enabled QAM to develop (by improving the S/N). I think it depends largely on how big your packets are, but I find it hard to believe that the data doubles in size after encoding (I don't have years of experience on that particular topic though). \$\endgroup\$
    – Mister Mystère
    Commented Jun 28, 2015 at 20:22

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The first part of the statement says 2 information bits per modulation symbol. QPSK transmits two bits per symbol, so you don't have any opportunity to add extra error correction bits (channel coding) while keeping to the 2 bits per symbol limit. Using 16 QAM, you have 4 bits per symbol, so you have spare bits to add error correcting bits (2 bits per symbol for data, and an extra two bits for error correction coding). The gain you get by adding these error correcting bits can be more than the loss in sensitivity that you get by increasing the modulation order.

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