From the table below (taken from the specs) you can see that using 10MHz results in having 50 PRBs. A PRB consists of 12 subcarriers and in each subcarrier we can transmit 7 symbols per slot (0.5ms). Using 64QAM means that each symbol represents 6 bits. So each PRB can deliver 12 x 7 x 6 x 2 x 1000 = 1008000 bits per second Since we have 50 PRBs in 10MHz we can transmit 50.4Mbps Since we are using 2x2 MIMO we can transmit two streams so 100.8Mbps However we have to use coding to protect the data, so at 3/4 that means 0.75 x 100.8Mbps = 75.6Mbps The above assumes no control channels etc, which of course is not possible so the "real" theoretical max throughput will be a bit lower.
In all documents I saw that theoretical rate for these conditions is 64.8 Mbps. So the difference between 75.6 Mbps and 64.8 Mbps is control channels etc . Next question - How to calculate "real" theoretical max throughput?
Each PRB looks like the picture below. As we said the PRB consists of 12 subcarriers and the slot (0.5ms) carries 7 symbols. This is represented as a two-dimensiona two-dimensionall matrix. Each subcarrier/symbol combination is referred to as a RB (Resource Element). In total we have 12 x 7 = 84 RE per PRB. Control channels will use 12 RE per PRB (6 for the PDCCH and 6 for the Reference Signals) So now you are left with 72 RE per PRB. 72 x 50 (total PRB per 10MHz) x 2 (to give value in ms) = 7200 RE per ms Each RE can carry 6 bits (64 QAM) so 43200 bits per ms or 43.2Mbps Since we are using MIMO 43.2 x 2 = 86.4Mbps But with coding 86.4 x 0.75 = 64.8MBps!
In addition, per 10ms frame additional Physical Channel Overheads are also present, i.e. the PSS, SSS and PBCH. This would bring the r ate slightly down 64.5516Mbps (assuming a 3/4 rate coding). Added: Note that this assumes a Normal Cyclic prefix. If the extended CP is used the rate for the same 3/4 mimo 2x2 would be 53.7516Mbps.
You may hear it many times that the peak data rate of LTE is about 300Mbps? How is the number calculated? What are the assumptions behind? Let's estimate it in a simple way. Assume 20 MHz channel bandwidth, normal CP, 4x4 MIMO.
First, 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. Second, assume 64 QAM modulation and no coding, 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. Third, with 4x4 MIMO, the peak data rate goes up to 100.8 Mbps x 4 = 403 Mbps. Fourth, estimate about 25% overhead such as PDCCH, reference signal, sync signals, PBCH, and some coding. We get 403 Mbps x 0.75 = 302 Mbps.
Ok, it is done through estimation. Is there a way to calculate it more accurately? If this is what you look for, you need to check the 3GPP specs 36.213, table 7.1.7.1-1 and table 7.1.7.2.1-1. Table 7.1.7.1-1 shows the mapping between MCS (Modulation and Coding Scheme) index and TBS (Transport Block Size) index. Let's pick the highest MCS index 28 (64 QAM with the least coding), which is mapping to TBS index of 26. Table 7.1.7.2.1-1 shows the transport block size. It indicates the number of bits that can be transmitted in a subframe/TTI (Transmit Time Interval). For example, with 100 RBs and TBS index of 26, the TBS is 75376. Assume 4x4 MIMO, the peak data rate will be 75376 x 4 = 301.5 Mbps. Table 7.1.7.1-1: Modulation and TBS index table for PDSCH (3GPP TX 36.213)
Table 7.1.7.2.1-1: Transport block size table (3GPP TS 36.213)
[Q: Want to try a small exercise? Here you are: what is the peak data rate if MCS 20 is used? Assume the channel bandwidth is 10 MHz and 2x2 MIMO is configured.] We also know that there are different device capabilities, which is defined in 3GPP TS 36.306, Table 4.1-1 and table 4.1-2. For example, with a cat 2 device, the supported peak data rate is about 50 Mbps in the DL and about 25 Mbps in the UL. All UE categories should support all channel bandwidths (1.4/3/5/10/15/20 MHz) and all duplex modes (FDD/TDD/H-FDD) in LTE. Cat 1~4 devices can support up to 2x2 MIMO in the DL. Only cat 5 device can support 4x4 MIMO in the DL and 64QAM in the UL.
Now, if a cat 3 device is used in a 10 MHz channel with 2x2 MIMO configuration, can we get the peak data rate of 100 Mbps in the DL? Let's calculate. We know that, in the network side, the peak data rate is 300 Mbps for 20MHz channel with 4x4 MIMO, so the peak data rate is 75 Mbps for 10 MHz with 2x2 MIMO. Therefore, in case of cat 3 device in 10 MHz channel with 2x2 MIMO, the expected peak data rate over the air interface is: min (device capability, network capability) = min (cat 3, 10MHz with 2x2 MIMO) = (100, 75) = 75 Mbps. One more thing before you claim you've mastered the peak data rate calculation, QoS profile can also impose constraint on the actual peak data rate a user expects.
