OFDM Burst Frequency Synchronization by Single Carrier ... - ICE

We use a fading channel with rays and ray power according to table 1 represented at the samplingrate 1=T s . Any rayofthe time{variant channel impulse response h k =[h 0k :::h 31k ] is subject toJakes fading with bandwith f d =4kHz. Due to the band limitation in the transmit lter, the tapsof the channel impulse response h k are correlated with each other. The channel impulse responseis 32 taps long.Coded data information symbols b k are transmitted by the symbols a k by means of dierentialmodulation, which extends over the frequency domain a k+1 = b k a k , only a nNu;Nu=2 must beknown to the receiver a priori.Let s be the vector of one burst's samples' according to gure 1, f the frequency oset, k =0the correct frame start position, L B =2BS(M +1)+M T OSymT sthe burst and L = T OSym =T s theOFDM symbol length. Samples n k are taken from complex gaussian noise with variance such that(E S =N 0 ) (T sub =T OSym ) is the signal - to noise ratio of the subcarriers. The received signal is givenbys = [c 0 0 :::c S;1 0 :::c 0 0 :::c S;1 0 w o :::w L;1 c 0 0 :::c S;1 0 :::c 0 0 :::c S;1 0] (3)t k =TXj=;Tr k = exp(j2kfT s )s k;j f j (4)31Xj=03 Frequency Synchronizationt k;j h jk + n k (5)Weuseatwo{stage frequency synchronization algorithm. The rst stage provides a coarse frequencyestimate which is based on computing the average phase change during a single CAZAC block. Theestimate is computed under the assumption of the presence of correct frame synchronization. Asimple way to compute a frequency estimate in this case is to compute[9]d fT s (k) coa = 12 arg MXm=02(B;2)S;1Xl=0r k+kp r k;2S+k pk p = m (2BS + L)+4S + l (6)where only the self{interfering part of the CAZAC blocks is taken into consideration and thismotivates a periodic repetition of the CAZAC blocks. However, it turns out that the variance of3