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Max bit rate and guaranteed bit rate in LTE

+2 votes

Is there any way to calculate the max bit rate and guaranteed bit rate in LTE, any pointer.

posted Jul 29, 2013 by Sheetal Chauhan

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2 Answers

+2 votes

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.

More detail please check at

Guaranteed bit rate is not calculated it is the operator obligation.

answer Jul 29, 2013 by Nora Jones
0 votes

In order to get a raw peak data rate at the PHY layer e.g. in the DL you do the following:
1) assume we have 20MHz band, so the number of PRBs in the frequency domain is: PRBno = 100
2) assume we have 1 OFDM symbol for control region (for PHICH, PCFICH and PDCCH) in each subframe, so number of OFDM symbols per subframe for user plane data (PDSCH) is: NoOFDMSymbols = 13 (for normal CP)
3) assume we have SISO case (one antenna), so the number of Cell RS for the PDSCH per 2PRBs is: NoRS = 6
4) the number of subcarriers per PRB is: NoSubcarriers = 12
5) the number of RE (resource elements) available for carrying PDSCH per 2PRBs is: NoREs = NoOFDMSymbols * NoSubcarriers – NoRS = 13 * 12 – 6 = 150
6) the number of REs for subframe is: NoREPDSCH = NoREs * PRBno = 150 * 100 = 15000
7) for peak datarate we use 64QAM, which gives the number of bits per RE: bitsRE = 6
8) the number of bits for the whole subframe is: NoBitsPDSCH = NoREPDSCH * bitsRE = 15000 * 6 = 90 000
9) the number of subframes in one sec is: NoSFs = 1000 [SFs/Sec]
10) the max throughput then (raw, ie. without FEC) is: RawThrpt = NoBitsPDSCH [bits/SF] * NoSFs [SFs/Sec] = 90 000 * 1000 = 90 000 000 bits/sec = 90 Mbits/s
11) if you add then the typical FEC rate for good channel conditions of: FECrate = 5/6
12) you end up at: PHYThrpt = RawThrpt * FECrate = 90 Mbits/s * 5/6 = 75Mbit/s

So that’s your thrpt for DL at PHY. Then additionally you need to get out of it some % for RRC signalling and system info, and overhead of MAC/RLC/PDCP and TCP/IP stack.

For UL, the situation is similar, but you need to take into account the typical UL frame PHY overhead (DRS, SRS, PUCCH, PRACH)


answer Aug 29, 2013 by Salil Agrawal
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