Coherent Turbo Coded MIMO OFDM

Transcription

1 Coherent Turbo Coded MIMO OFDM K Vasudevan Associate Professor Telematics Lab Department of EE Indian Institute of Technology Kanpur vasu@iitk.ac.in ICWMC Barcelona Spain, 13th 17th Nov 2016 CCISP Dubai, 18th 20th Nov 2016 November 13, / 50

2 An Open Question What is the operating SNR per bit of the present day mobile phones? No answer in the open literature Surprisingly, the SNR per bit has not been used as a performance measure in the context of wireless communications 2 / 50

3 A coherent receiver requires the minimum signal-to-noise ratio (SNR) per bit to achieve a given bit-error-rate (BER) This translates to a longer battery life in the mobile Sync and channel estimation required training (preamble) needs to be transmitted along with the data Data is organized into frames, QPSK modulation A rate-1/2, 4-state turbo code is used to improve the BER A frequency selective Rayleigh fading channel having a uniform power delay profile is assumed Channel is static over one frame, varies independently from frame-to-frame Channel taps C N (0, 2σ f 2 ) Channel span Lh 3 / 50

4 The other impairments Carrier frequency offset (CFO) for k th frame, ωk [ 0.04, 0.04] radians Additive white Gaussian noise (AWGN), C N (0, 2σ w) 2 Orthogonal frequency division multiplexing (OFDM) converts a frequency selective (multipath) channel into a frequency flat channel, thereby eliminating intersymbol interference (ISI) Two transmit and two receive antennas are considered (2 2 MIMO OFDM system) Channel is independent across different transmit and receive antennas 4 / 50

5 Discrete-time algorithms have been developed for carrier and timing synchronization and channel estimation The minimum SNR per bit for error-free transmission over fading channels has been derived and shown to be identical to that of the AWGN channel, that is, 1.6 db Simulations results for a 2 2 turbo coded MIMO OFDM system indicate that a BER of 10 5, is obtained at an SNR per bit of just 5.5 db, which is a 2.5 db improvement over the earlier work 1 The best so far in the open literature 1 K. Vasudevan, Coherent detection of turbo-coded ofdm signals transmitted through frequency selective rayleigh fading channels with receiver diversity and increased throughput, Wireless Personal Communications, Springer, vol. 82, no. 3, 2015, pp [Online]. Available: 5 / 50

6 KISS (Keep it Simple Stupid) Simple operations used Modulation, demodulation matched filtering FFT, IFFT Turbo decoding No matrix inversions in real-time Only matrix multiplications are used 6 / 50

7 s1,n,nt s4,n,nt sk,2,n,nt sk,3,n,nt (Lp) (Lcs) (Lcp) (Ld) Cyclic suffix Cyclic B Data Postamble B prefix s5,n,nt sk,n,nt (L) Ld Buffer (B) Data (Ld2) Postamble (Lo) Buffer (B) Sk,3,0,nt sk,3,0,nt Data Interleaver π( ) Ld point IFFT sk,3,l d 1,nt Sk,3,L d 1,nt 7 / 50

8 Frequency domain S1,0,nt QPSK symbols IFFT S1,Lp 1,nt Sk,3,0,nt Data QPSK symbols IFFT Sk,3,L d 1,nt rk,n,nr ỹk,n,nr e jω kn wk,n,nr (AWGN) Time domain s1,0,nt s1,lp 1,nt sk,3,0,nt sk,3,l d 1,nt Parallel to serial and add cyclic prefix and suffix Channel nt hk,n,nr,nt sk,n,nt 8 / 50

9 S1,i,nt nt = 1 nt = 2 Lp = 8 i (subcarrier) Note that S1,i,ntS 1,i,mt = (2NtLp/Ld)δK(nt mt) s1,n,nt Lp s 1, n,mt = for 0 i Lp 1 0 for nt mt, 0 n Lp 1 (2Lp/Ld)δK(n) for nt = mt (1) 9 / 50

10 Signal for the k th frame and receive antenna nr (for 0 n L+Lh 2): rk,n,nr = Nt nt=1 ( sk,n,nt hk,n,nr,nt ) e jω kn + w k,n,nr = ỹk,n,nr ejω kn + w k,n,nr (2) where denotes convolution hk,n,nr,nt denotes the channel wk,n,nr denotes AWGN 10 / 50

11 Choose that value of m and νk which maximizes ( ) rk,m,nr e jν km s 1,Lp 1 m,nt (3) Outcome: ˆmk(nr, nt) denotes the time instant and ˆνk(nr, nt) denotes the coarse estimate of the frequency offset, at which the maximum in (3) is obtained Note that Lp 1 ˆmk(nr, nt) Lp +Lh 2 (4) If ˆmk(nr, nt) lies outside the range in (4), the frame is considered as erased (not detected) 11 / 50

12 nt = 1, nr = 1 SNR per bit 0 db, Lp = nt = 1, nr = nt = 2, nr = 1 Time, frequency nt = 2, nr = / 50 Correlation magnitude

13 nt = 1, nr = 1 SNR per bit 0 db, Lp = nt = 1, nr = nt = 2, nr = 1 Time, frequency nt = 2, nr = / 50 Correlation magnitude

14 nt = 1, nr = 1 SNR per bit 0 db, Lp = 4096 nt = 1, nr = nt = 2, nr = 1 Time, frequency nt = 2, nr = / 50 Correlation magnitude

15 Table 1: Probability of frame erasure. Frame configuration Probability of erasure Lp = 512, Lo = Lp = 1024, Lo = / 50

16 Assume frequency offset has been perfectly canceled Let m0,k = ˆmk(1, 1) Lp +1 0 m0,k Lh 1 (5) Define m1,k = m0,k +Lh 1 (6) henceforth denoted by m1 The steady-state, preamble part of the received signal for the k th frame and receive antenna nr can be written as: rk,m1,nr = Nt s5,nt hk,nr,nt + w k,m1,nr (7) nt=1 16 / 50

17 where rk,m1,nr = wk,m1,nr = hk,nr,nt = s5, nt = [ rk,m1,nr... rk,m1+lp 1,nr ] T [Lp 1] vector [ wk,m1,nr... wk,m1+lp 1,nr ] T [Lp 1] vector [ hk,0,nr,nt... hk,lhr 1,nr,nt ] T [Lhr 1] vector s5,lhr 1,nt... s5,0,nt..... s5,lp+lhr 2,nt... s5,lp 1,nt [Lp Lhr] matrix (8) 17 / 50

18 The channel span assumed by the receiver is Lhr = 2Lh 1 (9) Note that { s H 5,mt s 5,nt = for 0Lhr Lhr nt mt (10) for (2Lp/Ld)ILhr nt = mt The channel estimate is ĥk,nr,mt = ( s H 5,mt s 5,mt) 1 s H 5,mt r k,m1,nr (11) 18 / 50

19 Note that ( s H 5,mt s 5,mt) 1 s H 5,mt (12) can be precomputed and stored The channel estimation error is ũ = ( s H 5,mt s 5,mt) 1 s H 5,mt w k,m1,nr (13) It can be shown that [ E ũũ H] = σ2 wld Lp ILhr = 2σ 2 u ILhr (14) 19 / 50

20 nt = 1, nr = 1 SNR per bit 0 db, Lp = 512 nt = 1, nr = 2 Actual 6 Estimated Actual Estimated nt = 2, nr = 1 0 Subcarrier nt = 2, nr = 2 Actual 6 Estimated Actual Estimated / 50 Channel magnitude response

21 nt = 1, nr = 1 SNR per bit 0 db, Lp = Actual Estimated 5 nt = 1, nr = 2 Actual Estimated nt = 2, nr = 1 Actual Estimated 0 Subcarrier nt = 2, nr = 2 Actual Estimated / 50 Channel magnitude response

22 nt = 1, nr = 1 SNR per bit 0 db, Lp = Actual Estimated 4 5 Actual Estimated nt = 1, nr = nt = 2, nr = 1 Actual Estimated 0 Subcarrier nt = 2, nr = Actual Estimated / 50 Channel magnitude response

23 Choose that value of time instant m and frequency offset νk, f which maximizes: ( rk,m,nr e j(ˆω k+νk,f)m ) ỹ 1,k,L2 1 m,nr,nt (15) where L2 = Lhr +Lp 1 ŷ1,k,m,nr,nt = s1,m,nt ĥk,m,nr,nt (16) where ĥk,m,nr,nt is obtained from (11). The average value of the fine frequency offset estimate is ˆωk,f = N r nr=1 N t nt=1 ˆν k,f(nr, nt) NrNt (17) 23 / 50

24 The fine frequency offset estimate in (17) is still inadequate for turbo decoding and data detection when Ld Lp The residual frequency offset is equal to: ωk ˆωk ˆωk,f (18) This residual frequency offset is estimated by interpolating the FFT output and performing postamble matched filtering at the receiver Let the interpolation factor be I 24 / 50

25 Let m2,k = m1,k +Lp +Lcs (19) where m1,k is defined in (6) Define the FFT input in the time domain as: rk,m2,nr = [ rk,m2,nr... rk,m2+ld 1,nr ] T (20) which is the data part of the received signal in (2) for the k th frame and receive antenna nr, assumed to have the residual frequency offset given by (18) Compute the ILd-point FFT of (20) Do postamble matched filtering 25 / 50

26 HiSi HiSi HiSi (a) i (b) i (c) i 26 / 50

27 Assume that the peak of the postamble matched filter output occurs at m3, k(nr) Ideally, in the absence of noise and frequency offset m3,k(nr) = ILd 1 (21) In the presence of the frequency offset, the peak occurs to the left or right of ILd 1 The average superfine estimate of the residual frequency offset is given by: ˆωk, sf = 2π/(ILdNr) Nr nr=1 [m3,k(nr) ILd +1] (22) 27 / 50

28 Table 2: RMS frequency offset estimation error (radians). Frame configuration Coarse Fine Superfine Lp = 512 Lo = Lp = 1024 Lo = / 50

29 Table 3: Maximum frequency offset estimation error (radians). Frame configuration Coarse Fine Superfine Lp = 512 Lo = Lp = 1024 Lo = / 50

30 The noise variance is estimated as follows, for the purpose of turbo decoding: ˆσ 2 w = 1 2LpNr Nr nr=1 rk,m1,nr Nt nt=1 rk,m1,nr Nt nt=1 s5,ntĥk,nr,nt s5,ntĥk,nr,nt H (23) 30 / 50

31 Assume Nr = Nt = 2 A 4-state turbo code is used The generating matrix for each of the constituent encoders is given by: G(D) = [ 1 1+D 2 1+D +D 2 ] (24) The output of each constituent encoder is mapped to QPSK The two QPSK symbols are transmitted from 2 antennas 31 / 50

32 Assume rk,m2,nr in (20) contains no frequency offset The output of the Ld-point FFT of rk,m2,nr is given by: Rk,i,nr = Nt Hk,i,nr,nt S + k,3,i,nt Wk,i,nr (25) nt=1 for 0 i Ld 1, where Hk,i,nr,nt is the L d-point FFT of hk,n,nr,nt Wk,i,nr is the L d-point FFT of wk,n,nr It can be shown that [ 1 2 E Wk,i,nr [ 1 2 E Hk,i,nr,nt 2 ] = Ldσ 2 w 2 ] = Lhσ 2 f (26) 32 / 50

33 For the transition from state m to state n, at decoder 1, for the k th frame, at time i, we define (for 0 i Ld2 1): γ1,k,i,m,n = exp ( Z1,k,i,m,n/ ( 2Ldˆσ 2 w )) (27) where Z1,k,i,m,n is given by min all Sm,n,2 2 nr=1 Rk,i,nr 2 nt=1ĥk,i,nr,nt S m,n,nt 2 (28) where Sm, n, nt denotes the QPSK symbol corresponding to the transition from state m to state n in the trellis, at transmit antenna nt 33 / 50

34 Forward recursion for decoder 1: α i+1,n = αi,mγ1,k,i,m,np (Sb,i,m,n) m Cn α0,n = 1 for 0 n S 1 = α / ( S 1 ) αi+1,n i+1,n α i+1,n n=0 (29) where P(Sb,i,m,n) = { F2,i+ if Sb,i,m,n = +1 F2,i if Sb,i,m,n = 1 (30) denotes the a priori probability of the systematic bit obtained from decoder 2 34 / 50

35 The backward recursion at decoder 1: β i,n = βi+1,mγ1,k,i,n,mp (Sb,i,n,m) m Dn βld2,n = 1 for 0 n S 1 = β / ( S 1 ) βi,n i,n n=0 β i,n (31) ρ + (n) denotes the state that is reached from state n when the input symbol is +1. ρ (n) denotes the state that can be reached from state n when the input symbol is / 50

36 Then (for 0 i Ld2 1) G1, i+ = G1, i = S 1 n=0 S 1 n=0 αi,nγ 1,k,i,n,ρ + (n) β i+1,ρ + (n) αi,nγ 1,k,i,n,ρ (n) β i+1,ρ (n) (32) The extrinsic information is F1,i+ = G1,i+/(G1,i+ +G1,i ) F1,i = G1,i /(G1,i+ +G1,i ) (33) F2, i+ and F2, i in (30) is defined similarly 36 / 50

37 Similarly for decoder 2 ( γ2,k,i,m,n = exp Z2,k,i,m,n/ ( )) 2Ldˆσ w 2 where Z2,k,i,m,n is given by min all Sm,n,1 2 nr=1 Rk,i,nr 2 nt=1ĥk,i,nr,nt S m,n,nt 2 Robust turbo decoding is used The exponents in (27) and (34) are normalized in the range [ 30, 0] (34) (35) 37 / 50

38 Table 4: Frame parameters. Parameter Value (QPSK symbols) Lp Ld B 4 Lo Ld2 Lh Lcp = Lcs 512, , , / 50

39 1.0e e e e e e e e-07 Lp=512, Lo=256, Pr Id Lp=1024, Lo=512, Pr SNR per bit (db) Figure 1: BER simulation results. 39 / 50 BER

40 The throughput is defined as T = Ld2 Ld +Lp +Lcp +Lcs. (36) Table 5: Throughput. Lp Lo Ld2 T % % 40 / 50

41 Consider the signal rn = xn + wn for 0 n < N (37) xn is the message wn denotes samples of zero-mean noise, not necessarily Gaussian All the terms in (37) are complex-valued or two-dimensional The term dimension refers to a communication link between the transmitter and the receiver carrying only real-valued signals xn and wn are ergodic random processes 41 / 50

42 The 2-D signal power: 1 N N 1 n=0 xn 2 = P av (38) The 1-D noise power: 1 2N N 1 n=0 wn 2 = σ 2 w = 1 2N N 1 n=0 rn xn 2 (39) 42 / 50

43 The 2-D received signal power is: 1 N N 1 n=0 rn 2 = 1 N = 1 N N 1 n=0 N 1 n=0 xn + wn 2 xn 2 + wn 2 = P av +2σ 2 w [ = E xn + wn 2] (40) Independence between xn and wn wn has zero-mean 43 / 50

44 Note that (39) is the expression for a 2N-dimensional noise hypersphere with radius σ w 2N Similarly, (40) is the expression for a 2N-dimensional received signal hypersphere with radius N(P av +2σ 2 w ) How many noise hyperspheres (messages) can fit into the received signal hypersphere Ratio of the volumes of the two hyperspheres Volume (radius) 2N Therefore, the number of possible messages is M = ( N (P av +2σ 2 w )) N (2Nσ 2 w ) N = ( P av +2σ 2 w 2σ 2 w ) N (41) over N samples (transmissions) 44 / 50

45 Number of bits per transmission C = 1 N log 2(M) ( = log 2 1+ P av 2σ 2 w ) (42) over two dimensions When xn = Nt nt=1 Hk,n,nr,nt S k,3,n,nt wn = Wk,n,nr (43) as in (25), the channel capacity remains the same as in (42) 45 / 50

46 Proposition 1 Clearly, the channel capacity is additive over the number of dimensions. In other words, channel capacity over D dimensions, is equal to the sum of the capacities over each dimension, provided the information is independent across dimensions Proposition 2 Conversely, if C bits per transmission are sent over 2Nr dimensions, (Nr complex dimensions), it seems reasonable to assume that each complex dimension receives C/Nr bits per transmission 46 / 50

47 The average SNR of (26) as: Rk,i,nr in (25) can be computed from SNRav = 2L hσ f 2 P avnt 2Ldσ w 2 = P av 2σ 2 (44) w for κnt/nr bits, where [ Pav = E Sk,3,i,nt 2] (45) κ bits sent from each transmit antenna 47 / 50

48 The average SNR per bit is SNRav,b = 2L hσ f 2 P avnt 2Ldσ w 2 Nr κnt = L hσ f 2 P avnr Ldσ wκ 2 = P av 2σ 2 w Nr κnt. (46) For each receive antenna we have C = κnt/nr bits per transmission (47) over two dimensions 48 / 50

49 Substituting (46) and (47) in (42) we get C = log 2 (1+C SNRav,b) SNRav,b = 2C 1 C. (48) Clearly as C 0, SNRav,b ln(2) 49 / 50

50 The concepts can be extended to massive MIMO systems The peak-to-average power ratio (PAPR) of the transmitted signal needs to be addressed 50 / 50