How to calculate peak data rate in LTE?

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.

Ans: 39.7 Mbps.