How to calculate peak data rate in LTE?

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How to calculate peak data rate in LTE?

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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 and table  Table 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 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 Modulation and TBS index table for PDSCH (3GPP TX 36.213)

Table 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.

  • Good article. But in LTE, even with 4 Tx configuration, we can have maximum of 2 codewords right? The data rate to one UE will then be 75376 x 2. Is my understanding correct? Thanks

  • If you have a 10Mhz BW channel (i.e 50 RB) and SIMO in the uplink / the UL MIMO is disabled ; assuming CAT 3 UE, then the max UL throughput based on TBS charts is 15.264 Mbps (Itbs= 15 based on Imcs=16) ?


  • Hi Hongyan Lei,

    Excellent work!

    Can you explain how a cat3 device in 20MHz network get peak rate 100Mpbs?

    From your theory above, the peak rate should bo 120Mbps (75Mbps*2*80/100=120, 80 RBs)

  • Good. Thanks

  • Lei,

    Thank you for the clarification.

    Could you please shed some light on LTE Advanced rates, with 3*20MHZ vednors are able to acheive around 1Gbps, is this due to UE limitations?

    60Mhz  = 300RB's

    128QAM = 7symbols

    12 = Subcarriers

    14 = slots

    8 = 8x8 MIMO

    the theory for 60Mhz based on 128QAM 8x8 Mimo = (300*14*12*7*8)*1000 = 2.82Gbps

    If we were to consider a 30% overhead = around 2Gbps.

    I can only assume the limitation is due to the UE.

    Thank you,

    Asad M Abdirahman

  • Hi Asad,

    With LTE advanced, there are 3 more categories of UE added, cat8 UE supports 8 x 8 MIMO.

    with 60 MHz = 300 RBs

    64 QAM = 6bits/RE (pls note no 128 QAM supported with LTE)

    Now if we take 8 x 8 MIMO, eNB can transmit 2 codeword on 4 antenna each. From table "Table One-layer to four-layer TBS translation table" in 36.213, the transport block size i.e, the data sent in one codeword will be 299856 bits/100 RB/codeword

    so eNB can transmit 299856 * 2 * 3 = 1799136 bits/ms = ~1.8Gbps

    From theory perspective = (300 * 14 * 12 * 6 * 8 ) * 1000 = ~2.4Gbps

    with control signalling overhead of 30%, the thruput will be ~1.7Gbps, which is close to practical based on the explanantion above (even more because the controlling overload on other 2 secondary 20 MHz channel is less).

    There is no other UE limitation that it should be cat8 UE....


  • How can you say that All UE categories should support all channel bandwidths (1.4/3/5/10/15/20 MHz) and all duplex modes (FDD/TDD/H-FDD) ?

    Eg: Ipad 3 & Many of the smart phones available now don't support TDD

  • how about 8 antennas?  

    also what is the right way to read Tables to 5 One-layer to X-layer TBS translation table?

  • Here you can also find similar calculation

  • Given the FEC 1/2  actually  we get 4,752 instead of 6 bits, and the result will be  lower

  • Hello Sir,

    I am new to LTE, and I have one question. When we say the peak data rate is 300Mbps Down link, is this the theoretical upper limit a specific user would get or is it the aggregate data rate of all users with in the cell? because, in the calculation you have used up all the  Resource Blocks (100 RBS)?