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Considerations for STS for MIMO-OFDM

Considerations for STS for MIMO-OFDM. Seigo Nakao, snakao@gf.hm.rd.sanyo.co.jp Yoshiharu Doi, doi@gf.hm.rd.sanyo.co.jp. SANYO Electric Co., Ltd. Japan. Presented by Jon W. Rosdahl. Background & motivation Proposal for new STS STS example for 2TX mode STS example for 3TX mode Conclusion.

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Considerations for STS for MIMO-OFDM

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  1. Considerations for STSfor MIMO-OFDM Seigo Nakao, snakao@gf.hm.rd.sanyo.co.jp Yoshiharu Doi, doi@gf.hm.rd.sanyo.co.jp SANYO Electric Co., Ltd. Japan Presented by Jon W. Rosdahl

  2. Background & motivation Proposal for new STS STS example for 2TX mode STS example for 3TX mode Conclusion Contents

  3. TX1 STS DATA1 TX2 STS DATA2 TX3 STS DATA2 Background: Issues for STS The receive signal power of STS should be the same as that of DATA for ideal AGC control.

  4. Is only one legacy (.11a) STS enough for MIMO-OFDM? Assertion: Legacy STS (from .11a) is not preferable for MIMO applications (with more than one Tx antenna). Assuming legacy STS is not sufficient, what STSs are preferable? Legacy STS Questions

  5. Postulate: • Each MIMO TX antenna should have it’s own STS • The cross-correlation of 1 STS cycle for any pair of STSs should be 0 for • Easy synchronization, • Good frequency offset estimation, and • Optimum AGC implementation.

  6. Signal Calculation Assumptions h11 S1(t) h21 X1(t) = h11S1(t)+h21S2(t)+n1(t) h12 h22 S2(t) X2(t) = h12S1(t)+h22S2(t)+n2(t) S|Si(t)|2 = SSi(t)S*i(t) = 1, (i=1,2), SS*1(t)S2(t) = Xc, SS*i(t)nj(t) = 0 (i=1,2 j=1,2), nj(t) << 1 i.e. |nj(t)|2 ~= 0 (j=1,2) S means the sum of 1 STS cycle. (i.e. from t=0 to t=15)

  7. Receiving Signal Power of STS Calculating the receiving signal power of 1 STS cycle (16 FFT points). S|X1(t)|2 =SX1(t)X*1(t) =S{h11S1(t)+h21S2(t)+n1(t)}{h*11S*1(t)+h*21S*2(t)+n*1(t)} =h11h*11SS1(t)S*1(t) + h21h*21SS2(t)S*2(t) + h11h*21SS1(t)S*2(t) +h*11h21SS*1(t)S2(t) + h11SS1(t)n*1(t) + h21SS2(t)n*1(t) + h*11SS*1(t)n1(t) + h*21SS*2(t)n1(t) + Sn1(t)n*1(t) Here, SS*1(t)S2(t) = Xc, SS*i(t)nj(t) = 0 and |nj(t)|2 ~= 0, so S|X1(t)|2 =|h11|2+|h21|2+h11h*21X*c+h*11h21Xc =|h11|2+|h21|2+2Re[h11h*21X*c]

  8. Therefore New STS is necessary. Receiving Signal Power of STS If STS of TX1 is the same as that of TX2 (i.e. S1(t)=S2(t)), Xc = 1. The receiving signal power of 1 STS cycle is, |h11|2+|h21|2+2Re[h11h*21]. (/ 16 FFT points) When h11 = –h21, AGC can’t work correctly because the receiving signal power is 0. However, the receiving power of DATA sequence is |h11|2+|h21|2. (/ 16 FFT points) because Xc = 0 in the DATA sequence.

  9. Receiving Signal Power of STS In general, if the |Xc| is not small, the receiving power difference between STS and DATA (i.e. 2Re[h11h*21Xc]) becomes large depending on propagation situations. Therefore the optimum AGC is difficult. Xc should be 0 in 1 STS cycle. (i.e. the cross correlation between transmitted STSs should be small)

  10. Cross-correlation of X1 and S1 When finding synchronization symbol timing or estimating frequency offset of signals from Ant1, the calculation for cross-correlation of X1(t) and S1(t) is necessary. ( i.e. SX1(t)S*1(t) / sqrt{S|X1(t)|2}sqrt{S|S1(t)|2}) SX1(t)S*1(t) / sqrt{S|X1(t)|2}sqrt{S|S1(t)|2} =S{(h11S1(t)+h21S2(t)+n1(t)}S*1(t) / sqrt{S|X1(t)|2} =S{(h11S1(t)S*1(t)+h21S2(t)S*1(t)+S*1(t)n1(t)} / sqrt{S|X1(t)|2} ={h11SS1(t)S*1(t)+h21SS*1(t)S2(t)+SS*1(t)n1(t)}/sqrt{S|X1(t)|2} =(h11 + h21Xc) / sqrt{|h11|2+|h21|2+2Re{h11h*21Xc}} In case h11 = –h21Xc, the cross-correlation of X1(t) and S1(t) becomes 0. Thus, if Xc is small (preferably Xc = 0), then there is no case when the (h11 + h21Xc) term becomes 0.

  11. Single STS and single LTS (legacy) Multiple Simultaneous STS and LTS Multiple Simultaneous STS and staggered LTS Some Preamble & Frame Format combinations:

  12. Type 1 (single STS) TX1 STS LTS Signal DATA1 TX2 LTS Signal DATA2 2TX, Single STS example

  13. Type 2 (overlapped STS and LTS) TX1 STS LTS Signal DATA1 TX2 STS LTS Signal DATA2 Type 3 (overlapped STS, staggered LTS) TX1 STS LTS Signal DATA1 TX2 STS LTS Signal DATA2 2TX, Overlapped STS examples

  14. TX1 STS1 DATA1 TX2 STS2 DATA2 TX3 STS3 DATA2 Proposal for New STSs • Each TX antenna should have an unique STS. • The cross-correlations of one STS cycle (i.e. 16 FFT points) for STS pair should be 0. These STSs can be applied for all types of frame format. (e.g. overlapped LTSs, non-overlapped LTSs, overlapped Signals, non-overlapped Signals)

  15. How to find new STSs Generally, the cross-correlation of u(t) and v(t) is defined as Where, U(f) and V(f) are the Fourier transform of u(t) and v(t), respectively. • Therefore, the cross-correlation of following STSs is always 0. • STS1 which consists of 6 subcarriers of legacy STS. • (e.g. –24, –4, 8, 12, 16, 20) • STS2 which consists of 6 different subcarriers. • (e.g. –20, –16, –12, –8, 4, 24) • Moreover, both STS1 and STS2 have the same cycle length as the • Legacy .11a STS (16 FFT points).

  16. 2 TX and 3 TX STS Examples

  17. Example of New STS for 2TX STS1–26, 26 = sqrt(13/3){0, 0, –1–j, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, –1–j, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, –1–j, 0, 0, 0, 1+j, 0, 0, 0, –1–j, 0, 0, 0, 1+j, 0, 0, 0, 0, 0, 0} STS2–26, 26 = sqrt(13/3){0, 0, 0, 0, 0, 0, 1+j, 0, 0, 0, 1+j, 0, 0, 0, 1+j, 0, 0, 0, 1+j, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, –1–j, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, –1–j, 0, 0} Cross-correlation between STS1 and STS2 is 0. Cross-correlation between legacy STS and both STS1 and STS2 is non-zero but small enough to help identify legacy and 11n devices.

  18. Time domain of STS1 for 2TX Amplitude FFT point number

  19. Time domain of STS2 for 2TX Amplitude FFT point number

  20. Example of New STS for 3TX STS1–26, 26 = sqrt(13/2){0, 0, –1–j, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, –1–j, 0, 0, 0, 1+j, 0, 0, 0, 0, 0, 0, 0, –1–j, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0} STS2–26, 26 = sqrt(13/2){0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1+j, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, –1–j, 0, 0, 0, 1+j, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1+j, 0, 0} STS3–26, 26 = sqrt(13/2){0, 0, 0, 0, 0, 0, –1–j, 0, 0, 0, –1–j, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1+j, 0, 0, 0, 1+j, 0, 0, 0, 0, 0, 0}

  21. Time domain of STS1 for 3TX Amplitude FFT point number

  22. Time domain of STS2 for 3TX Amplitude FFT point number

  23. Time domain of STS3 for 3TX Amplitude FFT point number

  24. New STSs are necessary for MIMO-OFDM. Each TX antenna should have an unique STS. The cross-correlation of 1 STS cycle for any pair of STSs should be 0. Appropriate STSs are possible Preliminary examples of new STSs were shown. STS usage should be included in Comparison Criteria considerations Conclusion

  25. Work is in progress to find new STSs which meet this criteria and also provide good auto-correlation performance. Additional investigation

  26. Questions?

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