# 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.