Exercise 4.4: Modulation in LTE

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Throughput comparison for LTE

With LTE, the scheduler selects the appropriate modulation method depending on the nature of the environment and the distance of the user from the base station. In this task, we consider various QAM procedures, namely

• 4–QAM with  $b\text{ = 2 bit/Symbol}$,
• 16– QAM with  $b\text{ = 4 bit/Symbol}$,
• 64– QAM with  $b\text{ = 6 bit/Symbol}$.

Formally, these procedures can be described as  "$b^{2}$-QAM" . The signal space constellations for 16-QAM and 64-QAM are shown on the right. The yellow dots indicate the 4-QAMs.

The lower diagram from  [MG08]  shows for different  $b$–values the throughput depending on the signal-to-noise ratio   ⇒   $10 \cdot {\rm lg \ SNR}$. You can see that for a very good channel $($also very large  $\rm SNR)$  the throughput is approximately proportional to $b$ .

Notes:

• This task belongs to the chapter  Bitübertragungsschicht bei LTE.
• Reference is made in particular to the pages  Modulation bei LTE  as well as  Digitale Modulationsverfahren  in the book „Modulation Methods”.
• The areas  $\rm A$,  $\rm B$  and  $\rm C$  shown in the above sketch are to be assigned to the modulation methods 4-QAM, 16-QAM and 64-QAM in the subtask '(1)’.
• The literature reference [MG08] refers to:  
  Myung, H.; Goodman, D.: Single Carrier FDMA - A New Air Interface for Long Term Evolution. West Sussex: John Wiley & Sons, 2008.

Questionnaire

1

Which assignments apply in the areas  $\rm A$,  $\rm B$  and  $\rm C$?

	The modulation method for area  $\rm A$  is 4-QAM.
	The modulation method for area  $\rm B$  is 16-QAM.
	The modulation method for area  $\rm C$  is 64-QAM.

2

From what signal-to-noise ratio  $(\rm SNR_{1})$  is 16-QAM better than 4-QAM?

$10 \cdot \rm lg \ SNR_{1} \ = \ $ $\ \rm dB$

3

From what signal-to-noise ratio $(\rm SNR_{2})$ is 64-QAM better than 16-QAM?

$10 \cdot \rm lg \ SNR_{2} \ = \ $ $\ \rm dB$

4

Which modulation method would be most suitable for  $10 \cdot {\rm lg \ SNR} = 5 \ \rm dB$ ?

	BPSK (Binary Phase Shift Keying),
	QPSK (Quadrature Phase Shift Keying),
	4–QAM.

Sample solution

Solution

(1)  Correct is only the solution 2:

• The area close to the transmitter  $\rm A$  usually has the best reception conditions. Here the 64-QAM modulation method can be used, which allows the highest throughput under ideal conditions, but also degrades the most when the SNR decreases.
• 64-QAM, on the other hand, is not suitable for the remote area  $\rm C$ . Here it is better to use the lowest level modulation method 4-QAM.

Throughput of the QAM versions

(2)  Compare here the two curves labeled "$2 \ \ \rm bit/Symbol$" and "$4 \ \ \rm bit/Symbol$".

• The point of intersection is $10 \cdot {\rm lg \ SNR_{1}}\hspace{0.15cm}\underline{ \approx 15 \ \rm dB}$.
• It follows immediately: The 16-QAM only results in a higher throughput than the 4-QAM if $10 \cdot {\rm lg \ SNR} > 15 \ \ \rm dB$.

(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 „$6 \ \rm bit/Symbol$”.

(4)  From the diagram you can see that with 4-QAM ($2 \ \rm bit/symbol$) the throughput is (almost) $0$.

• In this comparison, the QPSK is identical to the 4-QAM and therefore also unsuitable.

• Better would be Binary Phase Shift Keying (BPSK), which corresponds to the lowest curve "$1\ \rm bit/Symbol$"   ⇒   solution 1.