You might hear about the data rates in LTE is 300 Mbps or 100 Mbps. Whatever the assumption behind this is not practical but in lab conditions how it may possible is i am going to explore it in this article. Let's assume 20 MHz channel bandwidth, normal CP and 4x4 MIMO.
1. Calculate the number of resource elements (RE) in a subframe with 20 MHz channel bandwidth:
12 subcarriers x 7 OFDMA symbols x 100 resource blocks x 2 slots= 16800 REs per subframe. Each RE can carry a modulation symbol.
2. Assume 64 QAM modulation, no coding and one modulation symbol will carry 6 bits: The total bits in a subframe (1ms) over 20 MHz channel is 16800 modulation symbols x 6 bits / modulation symbol = 100800 bits. So the data rate is 100800 bits / 1 ms = 100.8 Mbps.
3. We know with 4x4 MIMO, the peak data rate goes up to 100.8 Mbps x 4 = 403 Mbps.
4. Estimate about 25% overhead: such as PDCCH, reference signal, sync signals, PBCH, and some coding. We get 403 Mbps x 0.75 = 302 Mbps.
This has been calculated by estimation above but we have defined table by 3gpp specifications 36.213 table 7.1.7.1-1 and table 7.1.7.2.1-1.
Table 7.1.7.1-1: - It shows tha maping between MCS (Modulation and Coding scheme index) and TBS (Transport Block Size index).
Table 7.1.7.2.1-1 - It shows the transport block size (TBS). It indicates the number of bits that can be transmitted in a subframe/TTI (Transmit Time Interval). For example, with 100 RBs and for TBS index of 26 the TBS is 75376.
Another detailed example: - You must be familiar with the technology i am giving the direct overview to understand the calculations.
let’s assume a 2×5 MHz LTE system. We first calculate the number of resource elements (RE) in a subframe (a subframe is of 1 msec):
12 Subcarriers x 7 OFDMA Symbols x 25 Resource Blocks x 2 slots = 4,200 RE's
*Then we calculate the data rate assuming 64 QAM with no coding (64QAM is the highest modulation for downlink LTE):*
6 bits per 64QAM symbol x 4,200 Re's / 1 msec = 25.2 Mbps
The MIMO data rate is then 2 x 25.2 = 50.4 Mbps.
We now have to subtract the overhead related to control signaling such as PDCCH and PBCH channels, reference & synchronization signals, and coding.
These estimated are as follows: -
- PDCCH channel can take 1 to 3 symbols out of 14 in a subframe. Assuming that on average it is 2.5 symbols, the amount of overhead due to PDCCH becomes 2.5/14 = 17.86 %.
- Downlink RS signal uses 4 symbols in every third subcarrier resulting in 16/336 = 4.76% overhead for 2×2 MIMO configuration.
- The other channels (PSS, SSS, PBCH, PCFICH, PHICH) added together amount to ~2.6% of overhead
The total approximate overhead for the 5 MHz channel is 17.86% + 4.76% + 2.6% = 25.22%.
- The peak data rate is then 0.75 x 50.4 Mbps = 37.8 Mbps.
Note: - The uplink would have lower throughput because the modulation scheme for most device classes is 16QAM in SISO mode only.
There is another technique to calculate the peak capacity which I include here as well for a 2×20 MHz LTE system with 4×4 MIMO configuration and 64QAM code rate 1:
Downlink data rate: -
1. Pilot overhead (4 Tx antennas) = 14.29%
2. Common channel overhead (adequate to serve 1 UE/subframe) = 10%
3. CP overhead = 6.66%
4. Guard band overhead = 10%
Downlink data rate = 4 x 6 bps/Hz x 20 MHz x (1-14.29%) x (1-10%) x (1-6.66%) x (1-10%) = 298 Mbps.
Uplink data rate: -
1 Tx antenna (no MIMO), 64 QAM code rate 1 (Note that typical UEs can support only 16QAM)
1. Pilot overhead = 14.3%
2. Random access overhead = 0.625%
3. CP overhead = 6.66%
4. Guard band overhead = 10%
Uplink data rate = 1 * 6 bps/Hz x 20 MHz x (1-14.29%) x (1-0.625%) x (1-6.66%) x (1-10%) = 82 Mbps.
