What is QAM in LTE

Exercise 4.4: For modulation with LTE

(1) Only that is correct Suggested solution 2:

  • The best reception conditions usually prevail in the area $ \ rm A $ close to the transmitter. The modulation method 64-QAM can be used here, which enables the highest throughput under ideal conditions, but also degrades the most when the SNR drops.
  • For the remote area $ \ rm C $, however, 64-QAM is not suitable. Here it is better to use the lowest level modulation method 4-QAM.
Throughput of the QAM variants

(2) The two curves labeled “$ 2 \ \ rm bit / symbol $” and “$ 4 \ \ rm bit / symbol $” are to be compared here.

  • The intersection is at $ 10 \ cdot {\ rm lg \ SNR_ {1}} \ hspace {0.15cm} \ underline {\ approx 15 \ \ rm dB} $ (red marking).
  • From this it follows directly: The 16 – QAM only leads to a higher throughput than the 4 – QAM for $ 10 \ cdot {\ rm lg \ SNR}> 15 \ \ rm dB $.

(3) The result $ 10 \ cdot {\ rm lg \ SNR_ {2}} \ hspace {0.15cm} \ underline {\ approx 22 \ \ rm dB} $ results from the intersection of the two curves
"$ 4 \ \ rm bit / symbol $" and "$ 6 \ \ rm bit / symbol $" (blue marking).

(4) The illustration shows that with 4 – QAM $ (2 \ \ rm bit / symbol) $ the throughput is (almost) zero.

  • In this comparison, the QPSK is identical to the 4 – QAM and is therefore also unsuitable.
  • It would be better here Binary phase shift keying $ \ rm (BPSK) $, which corresponds to the bottom curve “$ 1 \ \ rm bit / symbol $”
    ⇒   Suggested solution 1.