QUESTION: To deliver 100Mbit/s to the end user, how much bandwidth is needed in LTE spectrum? And which spectrum band should be used?
ANSWER:
Roger Jacques: This question represents one of the difficulties of wireless technologies. Considering peak data rate, 10 MHz bandwidth could be enough. However, this is a peak data rate that need specific conditions to be delivered, perfect path between user and site, short distance, as well as dedicating all the capacity to this user. Getting 100 Mbs average in real life, which means a cell with average conditions with (limited) mobility and when other users are also getting some capacity is quite difficult with LTE. This is something which is shared with the wideband standards, the average possibility is smaller that the peak rate.
The other part of the question is about the frequency range, again, this is simply an element of the conditions, peak data rate can be reached whatever the frequency band. However, lower frequencies are easier for improving the range. LTE 2.6 GHz is characterized by short ranges, 3GPP is often considering scenario with 500 or 1300 m inter site distance.
Fernando Martín: Trying to put all your more than important inputs together and making deeper calculations with real values, I think I am able to give a more visual case to clarify the original question. More specifically, I will try to demonstrate that 10MHz BW is not enough to achieve the 100Mbps desired.
I will base the calculations in the following scenario:
FDD access
- BW configuration, setting up the Resource Block number to 50
- MIMO
- Normal CP
- CFI=2
- Very goof RF conditions. CQI=15 => 64QAM & 948/1024 coding
* As for "100Mbps to the end user" I understand real data that the user is consuming (video streaming, voice), so I will focus on the PDSCH Resource Elements to calculate it.
* I will calculate the number of resource elements assigned to PDSCH by frame, since the pattern repeats each frame:
* x 12subcarriers x 7 symbols x 2 slots x 10 subframes= 84000 Resource Elements per Frame
If
you deduce the resource blocks per frame assigned to:
PCFICH, PHICH in each subframe
PDCCH in each subframe
RS1, RS2 in each slot
PBCH in second slot in the first subframe
P-SCH & S-SCH: in first slot in the 1st and 6th subframes.
The global idea of the mapping could be seen here
The number of resource elements assigned to PDSCH is around 60000 RE/10 ms.
With an average of:
1. MRE/s for PDSCH
* Applying the CQI value, due to nice RF path to the UE, and the MIMO condition, we have:
CQI=15 => 64QAM => 6 bits per RE
CQI=15 => 948/1024 bit coding rate.
So:
* x 6 bits/RE x (948/1024)= 33.3Mbps
Having 2x2 MIMO configuration:
* x 2 = 66.6Mbps
* Conclusion:
* would be the maximum DL real throughput (PDSCH) that the user could get with this configuration.
So either the number of antennas change into 4x4 (most difficult and expensive to implement) giving us 133.2Mbps,
or we select higher bandwidth like 20MHz which will increase the number of RE and the tput.
I hope this clarified that higher BW than 10MHz must be used for achieving 100Mbps at the moment since the 4x4 antenna solutions is still not really feasible nowadays.
Notice that very good RF conditions are needed to achieve those values.
If the UE is in the edge of the cell, the Tput values will decrease since the modulation scheme will change.
Another important fact that Jawad pointed out is that only 1 UE should be in the cell at the time to achieve such results.
Source: Linkedin Discussion from 'LTE (Long Term Evolution) - 3GPP' group