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PhD Student in ECE @ MSU

LTE Telecommunication Primer - Part 3

2017-11-10

Fundamental of the LTE (long-term evolution) system, which is my note for a related video on Youtube. This is the third part. Heterogeneous network (small cell and relay node), coordinated multipoint, throughput calculation.

Portal to other posts in this series

  1. Part 1.
  2. Part 2.

Video 10 - Hetnet (Small Cell and Relay Node)

Video link : 10 - Hetnet (Small Cell & Relay Node) - Capacity & Coverage Enhancement

Hetnet

Hetnet is short for “heterogeneous network”. Because there are multiple types of cells distributed in the area.

Basic techniques for coverage improvement

Relay Node

Relay node connects to a macro node (eNB) and it transmits data to UE. eNB is called “donor”. Donor treats the relay node as a UE, however, the actual UE treats relay node as an eNB.

Difference between a repeater and a relay node:

Advantages of relay node

Video 11 - CoMP (Coordinate Multipoint)

Video link: 11 - CoMP (Coordinate Multipoint) - Capacity & Coverage Enhancement

Intercell interference is a big issue, where the user is at the edge of two adjacent cells.

In 2G, we used frequency re-use to avoid this issue. However, as the user becomes data hungry, frequency bandwidth is limited.

So how can we mitigate this issues?

CoMP

Multiple cells communicate and coordinate with each other to determine a way to provide user consistent and correct data.

The prime motivation of CoMP:

Ther are two types of CoMP: intrasite CoMP and intersite CoMP.

Intrasite CoMP

Multiple sectors in single eNB coordinates to improve the edge throughput.

The advantages of intrasite CoMP:

Intersite CoMP

This requires coordination between multiple cells. Since backhauls exist in such system, a lot of information needs to be exchanged.

2 ways by which multipoint coordinate:

In this method, multiple cells transmit their signals in the DL to UE to improve the received signal quality and strength. In UL transmission, UE utilizes antennae at different sites to coordinate between the different eNBs thus a virtual antenna array is formed. Signals received by the eNBs are then combined and processed to produce the final signal.

Joint processing places a high demand onto the backhaul network for both UL and DL.

They communicate with each other and adjust their sectors such that there is only one sector facing the user with a reliable and solid connection. Thus the interference between cells is mitigated. In the UL, UE only transmits PUSCH to a single cell.

This method significantly reduced the load in the backhaul network. Because only the scheduling data needs to be transferred between the different eNBs.

Challenges:

(A <- B means A is addressed by B.)

Video 12 - Throughput calculation

Video link: 12 - Throughput (Speed) Calculation in 4G

Shannon-hartley formula for channel capacity:

\[C=B\log_2(1+S/N)\]

where \(C\) is the throughput in bits/second. This formula gives the highest possible throughput of a noisy channel. It works on every specification such as GSM/2G or LTE/4G.

How this formula works?

Factor 1 - Modulation Scheme

Factor 2 - Number of RBs

Higher bandwidth allows more RB. Simple case.

Factor 3 - MIMO

By Nyquist theory, a channel of bandwidth \(B\) (MHz) can provide the maximum symbol rate \(2B\) symbol/second. However, we usually consider \(B\) symbol/second in practice.

Calculation Example

Assume in LTE system, the bandwidth is 20 MHz, with 64QAM modulation, 2x2 MIMO, the throughput is

\[ 20 \times 6 \times 2 = 240 Mbps\]

where:

This is the rough estimation of throughput. However, eNB can provide over one of its sectors.

From LTE perspective, we need to first find the number of resource elements for different bandwidth.

Each RB has 12 sub-carrier x 6 or 7 symbols = 84 resource elements. Each RE has 1 symbol.

For the 20 MHz bandwidth, we have 100 RBs, thus 8400 REs. This is what we have in a single slot,

Since 1 second has 2000 slots (each 0.5 ms), we have the number of REs

\[ 84 \times 100 \times 2000 = 16800000 \]

Still, assume 64QAM and 2x2 MIMO, the throughput is

\[ 6\times 2\times 16800000 = 201.6 Mbps \]

We can see this is actually smaller than the rough calculation above. However, this is only an ideal case in the physical layer, assuming no physical layer overhead, and every RE is being used for data transmission, which is impossible.

From IP layer to physical layer, PDCP Header, RLC Header, MAC Header etc. are added to the package. Thus, IP layer throughput is reduced compared to Physical layer.

Based on the results of 201.6 Mbps,

which leads to 60% data occupation. Therefore, 120 Mbps is the real data rate.

However, this is the maximum rate that a cell can provide. If we have \(N\) number of users, each user only has \(\frac{120}{N}\) Mbps.

The uplink throughput is calculated likewise. For accurate calculation, refer 3GPP - 36.213.



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