Литература / UMTS-Report

2)'0$ (YDOXDWLRQ 5HSRUW

The Multiple Access Scheme Proposal for the UMTS Terrestrial Radio Air Interface (UTRA)

Part 3

OFDMA Concept – Frequently asked Questions (FAQ)

Summary:

This document describes details which were requested in many questions to the Beta Concept Group.

The covered areas are:

∙SFH/TDMA operation performance versus doppler frequency

∙SFH/TDMA operation performance versus hopping bandwidth

∙Fast Fourier Transform (FFT/IFFT) complexity as main element of the OFDMA system

∙Feasibility and importance of antenna diversity reception system in hand-portable Mobile Station

∙Detailed Handover procedures

∙Additional Information on Time and Frequency Synchronisation

∙Power Amplifier Requirements

∙Multiband Reception and Filter Requirements

∙Frequency Hopping Feasibility

Table of Contents

3. OFDMA receiver complexity

The main complexity of the signal processing elements for the OFDMA receiver is the FFT. (This is ignoring the processing needed for channel decoding. To calculate the number of operations needed for the FFT, the analysis presented by McDonnell and Wilkinson [1] is used.

The size of the FFT needed at the receiver depends on the service required (scalability). For the case of the low date rate service (speech), only a 32 point FFT is required. This is sufficient for one band slot with 24 carriers and DQPSK modulation. For the highest data rate service (2 Mbit/s) we shall assume a bandwidth of 1.6 MHz and 8-DPSK modulation. This service requires a 512 point FFT.

The total number of real multiplications for an FFT is given by [1]

2F log2 F

were F is the size of the FFT. At the receiver an FFT has to be performed at the same rate as the time slot duration (288.46 µs). For speech only every fourth time slot is used so we shall derive an average and peak multiplications per second figure.

For speech therefore,

Peak no. of real multiplications per second = 2 x 32 x 5 x (1.0 / 288.46 x 10-6 ) = 1.109 x

106

Average no. of real multiplications per second (1 FFT operation per frame) = 277.33 x 103

For one frame (4*288.46µs=1.154ms) 2 IFFT operations (diversity reception) and 1 FFT (TX burst construction) are required. This results in 3*0.27733MOPS = 0.832 MOPS.

For the highest data rate (2 Mbit/s) service every 7 out of 8 time slots are used.

Peak no. of real multiplications per second = 2 x 512 x 9 x (1.0 / 288.46 x 10-6) = 31.94 x

106

For one frame (8*288.46µs=2.307ms) 7*2 IFFT operations (7 used timeslots and 2 diversity reception) and 7 FFT (TX burst construction) are required.

This results in 3*(7/8)*31.94 MOPS = 83.9 MOPS.

This number can be reduced if only one Rx branch is used in the indoor environment (better C/I condition expected as compared to outdoor).

It is also important to note that the main processing element of the OFDMA receiver is a readily available FFT.

The following table summarizes the complexity of the FFT/IFFT processing. Please note the table gives ‘peak’ processing requirement which have to be divided by the acual used timeslots in the given TDMA structure.

Conclusions

It can be concluded that the complexity of the OFDMA depends upon the service required (almost linear complexity (MOPS) versus supported data rate). This offers benefits in terms of terminal cost and standby time for a given level of service. Even for the highest data rate service the complexity of the receiver is reasonable.

References

[1]J.T.E. McDonnell, T .A. Wilkinson, “Comparison of computation complexity of adaptive equaliser and OFDM for indoor wireless networks”, Proceedings IEEE Personal Indoor Mobile Radio Conference (PIMRC) 1996, pp. 1088-1091