Multicarrier transmission DMT/OFDM

Transcription

1 W. Henkel, International University Bremen 1 Multicarrier transmission DMT/OFDM DMT: Discrete Multitone (wireline, baseband) OFDM: Orthogonal Frequency Division Multiplex (wireless, with carrier, passband) Idea: Sub-division of the frequency band in N narrow sub-bands PSD frequency multiplex Consequences: Symbol duration is N-times longer All reflection components (wireless) received within a single symbol = equalization inside a symbol Problems with narrow-band interference = Requires coding across carriers (frequencies) carrier no.

2 W. Henkel, International University Bremen 2 OFDM: u(t) = 2 R { } N 1 F i e j2π T t i e jω ct i=0, u k = 2 R { } N 1 F i e j 2π N ik e jω ckt /N i=0 DMT: u(t) = 2 N 1 i=0 F ie j2π t T i = 2 N/2 1 i=1 F i e j2π t T i + F i e j2π t T i u k = 2 N 1 i=0 F ie j 2π N ik = 2 N/2 1 i=1 F i e j 2π for F 0 = F N/2 = 0; F i = F N i, i = 1,...,N/2 1 N ik + F i e j 2π N ik m 1 bit Daten seriell parallel Wandler 1 m 2 bit 2 m N bit Modulator + Signal N

3 W. Henkel, International University Bremen 3 The Discrete Fourier Transform (DFT) f i = 1 N 1 N k=0 F k = 1 N 1 N i=0 F k e j 2π N ik = 1 N F ( x = (e j 2π N ) i) i = 0,1,...,N 1 f k e j 2π N ik = 1 ( f x = (e j 2π N ) k) N k = 0,1,...,N 1 IDFT DFT

4 W. Henkel, International University Bremen 4 Some properties: Original domain DFT domain c j = a j b j C(x) = 1 N A(x) B(x) c(x) = 1 N a(x) b(x) c j = 1 N N 1 l=0 a l b j l C k = 1 N N 1 l=0 A l B k l A k B k c j mod N = a j l mod N C k = A k ( ) e j 2π k l N c(x) = a(x) x l mod (x N 1) ( ) c j = a j e j 2π j l N C k mod N = A k l mod N, C(x) = A(x) x l mod (x N 1)

5 W. Henkel, International University Bremen 5 information bitstream serial m 1 m 2 modulation c 1 c 2 IDFT parallel transmitted signal parallel m N c N serial In case of DMT: 1 DFT domain M-QAM conjugates time domain

6 W. Henkel, International University Bremen 6 h(k) a) k x(k) y(k) * h(k) = k k b) x p (k) y p (k) * h(k) = k k c) x z (k) y z (k) * h(k) = k k

7 W. Henkel, International University Bremen 7 Receiver structure of DMT (OFDM) AGC 1AGC2 AGC 3AGC4 Time-domain equalizer (transversal) S P DFT AGC N/2-1 Decoder

8 W. Henkel, International University Bremen 8 Some considerations regarding the power density spectrum without Guard interval: with Guard interval: 1 PSD 1 PSD 1.E-1 1.E-1 1.E-2 1.E-2 1.E carrier no. Spectral influence of the D/A conversion (sample & hold): 1 1.E-1 1.E-2 PSD 1.E carrier no. 512 corresponds to the sampling frequency of ADSL 1.E carrier no.

9 W. Henkel, International University Bremen 9 Criteria for the parameter choices N as big as possible and G as short as possible, in order to minimize the rate loss G/N G roughly as long as the impulse response (delay spread) - otherwise, e.g., time-domain equalization (TEQ) Latency (N) will be limited. N limited by coherence time. Complexity increases with N log N (FFT) Error correcting codes almost always required!

10 W. Henkel, International University Bremen 10 Hughes-Hartogs algorithm Bit allocation 1. Determine the noise powers at all carrier positions 2. Determine the required transmit powers S i,r of carrier i to transmit a data rate of R b/symbol. R = 0,...,R max 3. Determine the incremental transmit powers S i,r = S i,r S i,r 1 WHILE (R < R tot ) { Matrix S i,r, i = 1..N carr, R = 0..R max 4. Search for the carrier i with the smallest incremental power S i,r 5. Increment the bit content of carrier i 6. Delete S i,r from the matrix of power increments } The algorithm delivers bit and power allocation.

11 W. Henkel, International University Bremen 11 Water filling for parallel AWGN channels Theorem: Let us consider n parallel time-discrete, memoryless Gaussian channels with statistically independent noise signals with variances N i, i = 1,2,...,n. Let the total power of the input signals be limited to E, i.e., The channel capacity will be C = n E{Xi 2 } E i=1 n i=1 1 2 log(1 + S i/n i ) and will be achieved, when all input signals are independent and zero-mean Gaussian distributed random variables with variances S i, i = 1,2,...,n, fulfilling the following conditions

12 W. Henkel, International University Bremen 12 choosing B such that n i=1 S i = E S i + N i = B, if N i < B S i = 0, if N i B, B S 1 N 1 S 2 S 3 S 5 N 4 S 6 S 7 N 6 N 3 N 5 N 2 N 7

13 W. Henkel, International University Bremen 13 Proof: I(X; Y) I(X; Y) n i=1 n i=1 I(X i ;Y i ) 1 2 log(1 + S i/n i ) Maximization under side-conditions by applying the Lagrange Multiplier Side condition: n i=1 S i = E Lagrange multiplier: S j [ n i=1 1 2 log(1 + S i/n i ) + λ = S j n i=1 ( E )] n S i = 0, i = 0,1,2,...,n i=1 1 2 log(1 + S i/n i ) = λ 1 2(S j + N j ) = λ = S j + N j = 1 2λ =: B

14 W. Henkel, International University Bremen 14 Algorithm by Chow et al. 1. Determine the SNRs P i /N i of all carrier positions i at constant transmit power density P i = P T /N carr 2. Initialize the margin γ with γ = γ 0 (e.g., 0.1), the iteration counter n iter to zero, and the number of actually used carriers N used to N carr. 3. Computation of the bit allocation and rounding ( R i = log P ) i N i γ ˆR i = R i + 0,5 R i = R i ˆR i Let N 0 be the number of carriers with ˆR i = 0. The number of actually used carriers becomes N used = N carr N 0 4. R Σ = i ˆR i. IF (R Σ = 0) THEN BAD CHANNEL, EXIT

15 W. Henkel, International University Bremen Update of γ: 6. Increment of iteration counter γ := γ 2 R Σ R T N used n iter := n iter + 1; 7. IF (R Σ R T ) (n iter < n max ) Reset of N used to N carr and GOTO 3 8. WHILE (R Σ > R T ) { Search for carrier i with minimum R i (only such i with ˆR i > 0) ˆR i := ˆR i 1 } R i := R i + 1 IF ( ˆR i = 0)N used := N used 1

16 W. Henkel, International University Bremen WHILE (R Σ < R T ) { Search for carrier i with maximum R i (only such i with ˆR i < R max ) ˆR i := ˆR i + 1 } R i := R i 1 IF ( ˆR i = 1)N used := N used Calculate all carrier powers P i of all active carriers according to the desired bit-error probability P b. For quadratic QAM-constellations with R i 4, this means P i = 2 N i (2 R i 1) 3 [ erfc 1 ( Pb R i 2 R i 2 ( 2 R i 1) )] Compute P Σ = i: ˆR i >0 P i and modify the power allocation according to P i := P T P Σ P i.

17 W. Henkel, International University Bremen 17 Explanation of Step 5 ( ) R/N used = log SNR γ. 2 R Σ/N used = 1 + SNR γ Σ, 2 R T /N used = 1 + SNR γ T. 2 (R Σ R T )/N used = 1 + SNR γ Σ 1 + SNR γ T γ T γ Σ,

18 W. Henkel, International University Bremen 18 Algorithm by Yaniv George and Ofer Amrani (ISIT 2004) Essential idea: keep the power allocation constant and let the bit-error rate vary along carriers. the average BER is minimized. This is a very relevant algorithm in practice, since spectral masks are defined in standards. No tolerance band of a few db would be needed any more! One can minimize the BER for a given throughput and power, maximize the throughput for a given power and target BER.

19 W. Henkel, International University Bremen 19 The alg. requires the BERs P b (i, j) for different (quanized) SNR values (index j = 1,...,ν) and bit numbers (index i = 1,...,m), i.e., a table in the form m\snrs 1 2 ν 1 P b (1,1) P b (1,2) P b (1,ν) 2 P b (2,1) P b (2,2) P b (2,ν) 3 P b (3,1) P b (3,2) P b (3,ν). m P b (m,1) P b (m,2) P b (m,ν)

20 W. Henkel, International University Bremen 20 For minimizing the average bit-error rate, we use a table of incremental bit-error rates weighted by the actual number of bits b i We obtain a table in the form P(i, j) = b i P b (i, j) b i 1 P b (i 1, j) m\snrs 1 2 ν 1 P(1, 1) P(1, 2) P(1, ν) 2 P(2, 1) P(2, 2) P(2, ν) 3 P(3, 1) P(3, 2) P(3, ν). m P(m, 1) P(m, 2) P(m, ν)

21 W. Henkel, International University Bremen 21 We accumulate smallest-value entries until the desired target rate (target number of bits) is reached. m\snrs 1 2 ν 1 P(1, 1) P(1, 2) P(1, ν) 2 P(2, 1) P(2, 2) P(2, ν) 3 P(3, 1) P(3, 2) P(3, ν). m P(m, 1) P(m, 2) P(m, ν)

22 W. Henkel, International University Bremen 22 Maximum throughput is obtained when accumulating until the acceptable maximum average bit-error rate is reached m\snrs 1 2 ν i init i 1 i 2 i ν b P b init b 1 P b1 b 2 P b2 b ν P bν 1 P(i 1,1) P(i 2,2) P(i ν,ν) 2 P(i 1 + 1,1) P(i 2 + 1,2) P(i ν + 1,ν) 3 P(i 1 + 2,1) P(i 2 + 2,2) P(i ν + 2,ν). (i is a counter, b the corresponding number of bits. Usually, they may be identical. However, keeping them separate is more general and allows for other increments.)