Internal note 9: Blind carrier frequency offset estimation for ... - CiteSeerX

Internal note 9: Blind carrier frequency offset estimation for OFDM/OQAM over multipath channel∗ Gang Lin, Lars Lundheim, Nils Holte Department of Electronics and Telecommunications Norwegian University of Science and Technology (NTNU) 7491 Trondheim, Norway E-mail:{lingang; lundheim; holte}@iet.ntnu.no March 8, 2006

1

Introduction

It is well known that multicarrier systems are much more sensitive to carrier frequency offset (CFO) then single carrier systems. The effect caused by CFO for OFDM/QAM system was analyzed in [1]. In [1], the author indicated that CFO should be less than 2% of the band width of subchannel to guarantee the signal to interference ratio be higher than 30-dB. A critically sampled OFDM/OQAM system is also not robust to CFO [2], even when optimal pulses are used as shaping filters [3]. Bolcskei presented a blind CFO and STO estimation algorithm for OFDM system in [4], which is a natural extension of the estimator for single carrier QAM transmitting system designed by Gini [5]. For OFDM/OQAM system, Bolcskei’s estimator can estimate only CFO and has relatively large mse thus can be used only for coarse CFO estimation. A fine CFO estimator for OFDM/OQAM is presented by Ciblat and Serpedin in [6]. This estimator is an extension of the estimator introduced by Ciblat and Vandendorpe [7]. Both Bolcskei and Ciblat estimators are based the second order statistics of the received sequence before demodulator. For Bolcskei estimator, cyclic correlation functions are used to estimate CFO, while channel impulse response needs to be known over multipath fading channel. Ciblat estimator utilizes the cyclic conjugate correlation functions. While for non-weighting OFDM/OQAM system over AWGN channel, it can be strictly proved that the received sequence before demodulator is circular, which leads the conjugate cyclic correlation function to be zero. Thus subchannel weighting is also needed for Ciblat estimator, which is as same as Bolcskei estimator. For OFDM/OQAM systems, time-frequency well-localized shaping pulses can be used [8][9]. Thus large enough subchannel number, the equivalent channel can be approximated as flat-fading. It is obvious that flat-fading channel is easier to deal than frequency selective channel since flat-fading channel is just a special case of frequency selective channel. This motives us to estimate CFO based on the sequence out of receiver filters. In addition, since the sampling rate of the signal out of receiver filter is N/2 times lower than that of immediately received signal, lower implementation complexity can be achieved. In this note, we just use cyclic conjugate correlation functions to estimate CFO (the method based on correlation functions has been presented separately). It is (unfortunately) shown that our method also doesn’t work for non-weighting OFDM/OQAM systems over AWGN channel. While for practical OFDM systems, null-subchannels are always in presence, which can be used to estimate CFO blindly. Simulation shows that Ciblat estimator doesn’t work fine for OFDM/OQAM systems with sparsely distributed nullsubchannels, then we also consider interleave weighted OFDM/OQAM systems to make a comparison with Ciblat estimator. ∗ This

work is supported by BEATS. http://www.iet.ntnu.no/projects/beats

1

2 2.1

System model Overview and definitions

The discrete model of OFDM/OQAM is shown in Figure 1. This model has N subchannels, and subN −1 channels are weighted by factors {wk }k=0 . The weighting factor wk is real-valued to maintain the orthogonality.

Figure 1: The time discrete model of base band OFDM/OQAM with frequency offset I Each subchannel transmits one QAM symbol ak [n] = aR k [n] + j ak [n] per T seconds. The OQAM symbols are formed by shifting the imaginary part of QAM symbols by T /2. By summing up all the subchannels, the modulator generates a T /N sampled output sequence

x[l] =

N −1 X m=0

wm

¶ ∞ µ X 2π π N I aR [n] g[l − nN ] + j a [n] g[l − nN − ] ej( N l+ 2 )m m m 2 n=−∞

that is also the input to the channel. 2

(1)

Here we consider stationary channel. If the number of subchannels N is large enough, the equivalent channel of the each subchannel can be approximated as flat-fading. The fading factors are denoted by N −1 {µk }k=0 . The channel model also includes an independent zero-mean white Gaussian additive noise ν[l] with correlation function cν [τ ] = σν2 δ[τ ]. The carrier frequency offset fe is normalized with respect to 1/T . Then we can write the received sequence out of channel as ¶ N −1 ∞ µ X X 2π π 2π N I r[l] = ej N fe l wm µm aR [n] g[l − nN ] + j a [n] g[l − nN − ] ej( N l+ 2 )m + ν[l]. (2) m m 2 n=−∞ m=0 2π

π

In the subchannel k of receiver, the received sequence is first down-converted by modulator e−j( N l+ 2 )k , then filtered by the receiver filter f [l] and N/2 times down-sampled to generate a T /2 spaced sequence ¯ π 2π ¯ bk [s] = r[l] e−j( N l+ 2 )k ∗ f [l]¯ N l=s

=

∞ X

2

∞ X ¡ R N ¢ am [n] g[l − l0 − nN ] + j aIm [n] g[l − l0 − nN − ] 2 n=−∞ m=0 l0 =−∞ ¯ 0 π 2π π 2π ¯ × ej( N (l−l )+ 2 )(m−k) f [l0 ] + ν[l] e−j( N l+ 2 )k ∗ f [l]¯ N

ej

0 2π N fe (l−l )

N −1 X

wm µm

l=s

2π

= ej N f e l

N −1 X

wm µm

∞ X

¡

2

0 I 0 aR m [n] g[l − l − nN ] + j am [n] g[l − l − nN −

n=−∞ l0 =−∞

m=0

× e

∞ X

0 π j( 2π N (l−l )+ 2 )(m−k)

¯ 0 2π 2π π ¯ f [l0 ] e−j N fe l + ν[l] e−j( N l+ 2 )k ∗ f [l]¯

l=s N 2

N ¢ ] 2

.

(3)

Note that although the sequence immediately before decimator (or immediately out of receiver filter) contains more information than the N/2 down-sampled sequence bk [s], it is not possible in practice since for FFT based OFDM/OQAM systems [10][11][12][13], such decimator is merged into the FFT modular and the only output signal is bk [s]. We define (o)

2π

π

2π

pm,k [l] = g[l] ej( N l+ 2 )(m−k) ∗ f [l] e−j N fe l ,

(4)

and (o)

2π

π

νk [l] = ν[l] e−j( N l+ 2 )k ∗ f [l].

(5)

(o)

We can see that pm,k [l] is actually the equivalent overall impulse response from subchannel m in the (o)

transmitter side to the subchannel k in the receiver side over ideal transmitting channel, and νk [l] is the filtered noise. If g[l] and f [l] are bandlimited to [−1/T, 1/T ] (as usually assumed for OFDM/OQAM (o) systems) and |fe | < 1, pm,k [l] ≡ 0 for |m − k| > 2. Then we can simplify (3) as ¯ ¶ N −1 ∞ µ ¯ X X N ¯ (o) (o) (o) j 2π R (m−k) I f l e bk [s] = e N am [n] pm,k [l − nN ] + j (−1) am [n] pm,k [l − nN − ] + νk [l]¯ wm µm ¯ 2 n=−∞ m=0

l=s N 2

¶ N −1 ∞ µ X X N N N (o) N R (m−k) I jπfe s am [n] pm,k [s − nN ] + j (−1) am [n] pm,k [s − nN − ] + νk [s ] =e wm µm 2 2 2 2 n=−∞ m=0 = ejπfe s

N −1 X

wm µm

m=0 (o)

∞ ³ X

´ (m−k) I aR am [n] pm,k [s − 2n − 1] + νk [s], m [n] pm,k [s − 2n] + j (−1)

(6)

n=−∞ (o)

(o)

(o)

where pm,k [s] = pm,k [s N2 ] and νk [s] = νk [s N2 ] are N/2 times down-sampled versions of pm,k [l] and νk [l] respectively. We assume that the data symbols are statistical independent between different subchannels, different symbols, and the real and imaginary parts are i.i.d., i.e. ½ 2 £ R ¤ £ I ¤ σa /2, if m = k and n1 = n2 R I E am [n1 ] ak [n2 ] = E am [n1 ] ak [n2 ] = 0, otherwise, 3

and

£ ¤ I E aR m [n1 ] ak [n2 ] = 0, ∀ m, k, n1 , n2 ,

where the operator E[·] represents statistical expectation and σa2 is the average power of the QAM symbol.

2.2

The conjugate correlation function

It has been reported that very precise estimated CFO can be attained based on the cyclic conjugate correlation function of the received sequence r[l] [6]. Now we will check the conjugate correlation function of bk [s], which is the output sequence of receiver filter in subchannel k. The conjugated correlation function is defined as c˜k [s, τ ] = E [bk [s + τ ] bk [s]]. Based on the expression of bk [s] shown in (6) and the assumptions above, we have c˜k [s, τ ] =

N −1 ∞ σa2 jπfe (2s+τ ) X 2 2 X e wm µm (pm,k [s + τ − 2n] pm,k [s − 2n] − pm,k [s + τ − 2n − 1] pm,k [s − 2n − 1]) 2 n=−∞ m=0

+ E [νk [s + τ ] νk [s]] .

(7)

By further assuming that the real and imaginary parts of ν[l] are i.i.d., then based on (5), we find after a few intermediate steps that E [νk [s + τ ] νk [s]] = 0. It’s also possible to simplify the summation in (7). For s = 2q, i.e. s is even, the summation can be written as ∞ X

=

n=−∞ ∞ X

(pm,k [2q + τ − 2n] pm,k [2q − 2n] − pm,k [2q + τ − 2n − 1] pm,k [2q − 2n − 1]) (pm,k [2n + τ ] pm,k [2n] − pm,k [2n + τ + 1] pm,k [2n + 1]) .

n=−∞

Similarly, for s = 2q + 1, we have ∞ X

(pm,k [2q + 1 + τ − 2n] pm,k [2q + 1 − 2n] − pm,k [2q + τ − 2n] pm,k [2q − 2n])

n=−∞ ∞ X

=−

(pm,k [2n + τ ] pm,k [2n] − pm,k [2n + τ + 1] pm,k [2n + 1]) .

n=−∞

Thus we have that c˜k [s, τ ] = c˜k [s + 2q, τ ] and c˜k [s + 2q + 1, τ ] = −˜ ck [s + 2q, τ ]. Then by defining Am,k (τ, fe ) =

∞ X

(pm,k [2n + τ ] pm,k [2n] − pm,k [2n + τ + 1] pm,k [2n + 1]) ,

(8)

n=−∞

we can simplify (7) as c˜k [s, τ ] =

N −1 σa2 j2πfe (s+τ /2) X 2 2 e wm µm (−1)s Am,k (τ, fe ) 2 m=0

=

N −1 σa2 jπfe τ j2π(fe +1/2)s X 2 2 e e wm µm Am,k (τ, fe ) 2 m=0

= rk (τ, fe ) ej2π(fe +1/2)s , where rk (τ, fe ) =

N −1 σa2 jπfe τ X 2 2 e wm µm Am,k (τ, fe ). 2 m=0

(9)

(10)

We can see clearly that c˜k [s, τ ] is cyclic stationary with a period (fe + 1/2)−1 . Thus for |rk (τ, fe )| 6= 0, the spectrum (with respect to s) of c˜k [s, τ ] has a sharp peak at fe + 1/2, which can be used to estimate 4

fe . By assuming that the transmitter f [l] and receiver g[l] are identical real-valued and symmetric pulses, and band-limited to [−1/T, 1/T ] (as usually assumed in OFDM/OQAM systems), we find (unfortunately) PN −1 that (see appendix A) that m=0 Am,k (τ, fe ) = 0 for any value of fe . Then going back to (9), in the case of non-weighting system and ideal transmitting channel, i.e. wm = 1, µm = 1, we have that c˜[s, τ ] ≡ 0. This implies that the information of fe will be totally lost.

3

Estimation algorithm

In the previous section, we have proved that for the case of wk = 1 and µk = 1, the conjugate correlation function c˜k [s, τ ] will be zero. One method to remain frequency offset information is subchannel weighting, i.e., distributing individual subchannel different power. Here we consider two kinds of weighting methods: Method 1: Set wk = 0 for L selected subchannels k1 , k2 , · · · , kL , which can be referred as null-subchannel, while the other factors chosen as 1. If the null-subchannels are sparsely distributed and the kth subchannel is one null subchannel, it can be easily verified that c˜b,k−1 [s, τ ], c˜k [s, τ ] and c˜b,k+1 [s, τ ] contain the information of fe . This implies that we can estimate fe from subchannels k − 1, k and k + 1. While the useful signal power of subchannels k − 1 and k + 1 is much lower than k, here we use only the output sequence of subchannel k to estimate CFO. PN −1 Another practical problem is to choose τ . Recall that m=0 Am,k (τ, fe ) = 0. If subchannel k is ¯ σ2 ¯ null-subchannel, we have |˜ ck [s, τ ]| ' 2a ¯µ2k Ak,k (τ, fe )¯ (they are actually equal for flat-fading channel) immediately based on (9). If the shaping pulses are known, we can calculate Ak,k (τ, fe ) directly based on (28). Here we show an example of g[l] is the square root raised cosine pulse with a roll-off factor α. The 3-D curves of |Ak,k (τ, fe )| are shown in Figure 2. We can see that smaller roll off factor α leads to wider spread over τ . For the case of α = 1.0, |Ak,k (τ, fe )| is close to zero if |τ | > 2, thus we choose τ up to 2. Note that negative τ is just a shifted version of positive τ and thus contains no extra information. For the case of α = 0.2, we can choose τ up to 15.

0.15

0.6

Amplitude

Amplitude

0.5 0.1

0.05

0.4 0.3 0.2 0.1

0

0 5 20

0.5 τ

0

0.5 0 τ

0 fe

−20

0 fe −5

−0.5

(a) α = 0.2

−0.5

(b) α = 1.0

Figure 2: The amplitude of Ak,k (τ, fe ) M −1

In practice only one set of finite-length data records {bk [s]}s=0 is available, since c˜k [s, τ ] is a periodic with respect to s, it can be estimated by only one sample. We may denote this one sample estimation of c˜k [s, τ ] by yk [s, τ ] = bk [s + τ ] bk [s].

(11)

£ ¤T Then by defining yk [s] = yk [s, 0], yk [s, 1], · · · , yk [s, Lτ ] , we can write the estimation algorithm as ° °2 L ° M −1 ° X 1 def ° 1 X −j2πf s ° ˆ fe,M = arg max JM (f ) − , where JM (f ) = ykl [s] e ° ° . °M ° 2 f ∈(0,1) s=0 l=1

5

(12)

We can see that JM (f ) contains the contribution from different τ s and different null-subchannels. Method 2: w = {w1 , w2 , w1 , w2 , ...}, where w1 6= w2 , and they should be real-valued to guarantee the orthogonality between subchannels. It is shown in appendix A that for |fe | < 0.5 and |m − k| > 1, Am,k (τ, fe ) ≡ 0. ¯ σ2 ¯ Thus we have |˜ ck [s, τ ]| ' 2a ¯µ2k (w1 − w2 ) Ak,k (τ, fe )¯ (they are actually equal for flat-fading channel). Thus CFO can be estimated over all subchannels, and we can write the estimation algorithm as fˆe,M

° M −1 °2 N ° ° X 1 def ° 1 X −j2πf s ° = arg max JM (f ) − , where JM (f ) = yk [s] e ° ° . °M ° 2 f ∈(0,1) s=0

(13)

k=1

Weighting method 1 is more practical than method 2 since null-subchannel is always in presence for OFDM systems. Here we introduce weighting method 2 is just to make a comparison with Ciblat estimator. Thus in the next section, we will only give theoretical analysis for weighting method 1.

4

Asymptotic analysis

In previous section, we have suggested two weighting methods and the corresponding estimation algorithms (12) and (13). In this section, we will present the asymptotic analysis for weighting method 1, i.e. inserting of null-subchannels. Recall that yk [s, τ ] defined in (11) is just the estimation of c˜k [s, τ ] by one sample. It is obvious that E [yk [s, τ ]] = c˜k [s, τ ]. Thus we can write yk [s, τ ] as yk [s, τ ] = c˜k [s, τ ] + ek [s, τ ]. In fact ek [s, τ ] is just the estimation error and caused by two independent parts: one part comes from channel noise, another part comes from input symbol. It will be verified later that ek [s, τ ] is actually stationary (with respect to time instant s). £ ¤T In practice, the duration of g[l] should be finite. We define ek [s] = ek [s, 0] ek [s, 1] · · · ek [s, Lτ ] , then ek [s] satisfies mixing condition: ∀ L ∈ N, ∀ s1 ∈ Z, ∀ νi ∈ {0, 1} , ∞ ° ³ ´° X ° ° (ν ) (ν ) °cumL ek 1 [s1 ], · · · , ek L [sL ] ° = ML < ∞,

(14)

s2 ,··· ,sL =−∞ (0)

(1)

where ek [si ] = ek [si ], ek [si ] = e∗k [si ]. We may explain more about mixing condition (14). If´ek [s1 ] and ek [s2 ] become statistically indepen³ (ν1 ) (ν ) dent as |s1 − s2 | > P , then cumL ek [s1 ], · · · , ek L [sL ] = 0, for max{|si − sj | : i, j ∈ {1, · · · , L}} > P (ref. [14]: Theorem 2.3.1 (iii), p.19). This means a finite memory system can always satisfy mixing condition. Then based on this mixing condition and using a reasoning similar to Ciblat et al. [15], we find that (see appendix ) that fˆe,M is an unbiased estimation for fe , and the variance is asymptotically given by n o PL e l=1 Re Ψkl (fe ) − Ψkl (fe ) 3 γ= , (15) ³P ´2 2 π2 M 3 L Φ (f ) k e l l=1 where Ψk (fe ) = rH k (fe ) Pk rk (fe ) e ∗ e Ψk (fe ) = rH k (fe ) Pk rk (fe ) Φk (fe ) = rH k (fe ) rk (fe ). 6

(16)

£ ¤T e k are matrices Here rk (fe ) is a vector given by rk (fe ) = rk (0, fe ), rk (1, fe ), · · · , rk (Lτ , fe ) and Pk , P with entries given by £ ¤ Pk τ1 ,τ2 = Sek (fe + 1/2, τ1 , τ2 ) £ ¤ ek = Seek (fe + 1/2, τ1 , τ2 ) (17) P τ1 ,τ2 respectively, where Sek (fe + 1/2, τ1 , τ2 ) and Seek (fe + 1/2, τ1 , τ2 ) are given by (70) and (74) in appendix B respectively.

5

Simulation results

First we present some preliminary simulation conditions which are used through out all simulations: (1) 16OQAM modulation with power σa2 = 10; (2) g[l] and p[l] are square root raised cosine pulses with a roll off factor α = 1.0; (3) Each result is obtained by averaging over 1000 Monte Carlo trials; def

(4) SNR = σa2 /σν2 (5) The number of subchannels: N = 16. h¡ ¢4 i We further assume that the input QAM symbols are uniformly distributed, then E aR = k [n] 0.41 σa4 . To find the maximum point of JM (f ), 64 times over sampling FFT is used to find the coarse peak, then simplex method is used to find the precise maximum point. In Figure 3, examples of JM (f ) for M = 256, f e = 0.2, SNR = 0 and 30 dB are shown. We can see that for high SNR, there is a sharp peak around fe + 0.5, as shown in 3(a). While for low SNR shown in 3(b), the desired peak around fe + 0.5 is lower than peak(s) caused by noise, then false detection occurs. False detection can be avoided by appropriate initialization. 10

30

9 25 8 7 20 6 5

15

4 10 3 2 5 1 0 0

0.1

0.2

0.3

0.4

0.5 f (1/T)

0.6

0.7

0.8

0.9

0 0

1

0.1

0.2

(a) SNR=30dB

0.3

0.4

0.5 f (1/T)

0.6

0.7

0.8

0.9

1

(b) SNR=0dB

Figure 3: The curves of JM (f ) Another practical problem is to choose Lτ . Here we just show the theoretical mse versus Lτ of estimator (12) for a special case of M = 256 and SNR = 30 dB. The results are shown in Figure 4. We can see that for α = 1.0, the mse becomes saturated after Lτ = 2. For smaller roll-off factor α, we can gain more from increasing Lτ . This is accordant to the results shown in Figure 2. Since the roll-off factor α = 1.0, we will set Lτ = 2 for both estimator (12) and (13) in the following simulations. Simulation 1: false detection ratio versus SNR In this simulation, we set fe = 0.2, L = 1, Lτ = 3 and the false detection ratio is gotten over 30000 Monte Carlo trials. The simulated results are shown in Figure 5. We can see that the false detection ratio falls quickly with increasing SNR after a specific threshold, which can be used to explain the threshold effects of mse. 7

−8

10

mse

α = 1.0, fe = 0 α = 1.0, fe = 0.2 α = 1.0, fe = 0.4 α = 0.5, fe = 0 α = 0.5, fe = 0.2 α = 0.5, fe = 0.4 α = 0.2, fe = 0 α = 0.2, fe = 0.2 α = 0.2, fe = 0.4 −9

10

−10

10

0

1

2

3

4

5 Lτ

6

7

8

9

10

Figure 4: Theoretical mse versus Lτ for estimator (12)

0

10

M=128 M=256 M=512 −1

false detection ratio

10

−2

10

−3

10

−4

10

0

1

2

3

4

5 SNR (dB)

6

7

8

9

10

Figure 5: False detection ratio versus SNR for different M

Simulation 2: performance of estimator (12) versus SNR Here we choose M = 256, L = 1. The bias and mse versus SNR for different fe are shown in Figure 6. We can see that after a specific threshold, the theoretical curves matches well with the simulated results. The threshold effect is caused by the false peak detection of JM (f ) for high lever noise. Such false peak detection can be avoided by appropriate coarse CFO estimation. We also note that for fe = 0.2 and 0.4, the simulated values deviate from theoretical prediction for high SNR. It will be shown soon that such a deviation disappears asymptotically with increasing M . Simulation 3: performance of estimator (12) versus fe We still keep M = 256, L = 1. The bias and mse versus fe for different SNR are shown in Figure 7. We can see that in the acquisition range ( |fe | < 0.5), the estimator performance is almost independent to fe . It also shows that the estimator can work only after a specific SNR, which can be also observed in simulation 1. For high SNR, there exists gap between theoretical and simulated results. Simulation 4: performance of estimator (12) versus M In this simulation, we set L = 1. The bias and mse versus M for different SNR are shown in Figure 8. We can see that there exists obvious gap between theoretical and simulated results, while this gap decreases with increasing M . Simulation 5: performance of estimator (12) versus L From (49), we can see that the theoretical mse should be inversely proportional to the number of nullsubchannels if the null-subchannels are sparsely distributed. We simulate one case fe = 0.2, M = 256. In our simulation, subchannels 2,6,10,14 are null-subchannels for L = 4. The simulation results are shown 8

fe = 0: theoretical fe = 0.2: theoretical fe = 0.4: theoretical fe = 0: simulated fe = 0.2: simulated fe = 0.4: simulated

−2

10

−4

10

−6

mse

10

−8

10

−10

10

−12

10

0

10

20

30 SNR (dB)

40

50

60

Figure 6: Mse versus SNR for different fe

SNR = 0 dB: theoretical SNR = 20 dB: theoretical SNR = 40 dB: theoretical SNR = 60 dB: theoretical SNR = 0 dB: simulated SNR = 20 dB: simulated SNR = 40 dB: simulated SNR = 60 dB: simulated

−2

10

−4

10

−6

mse

10

−8

10

−10

10

−12

10

−0.5

−0.4

−0.3

−0.2

−0.1

0 fe (1/T)

0.1

0.2

0.3

0.4

0.5

Figure 7: Mse versus fe for different SNR

in Figure 9. We can see that more null-subchannels will lead to lower MSE and lower SNR threshold. It is interesting that although there exists obvious gap between theoretical and simulated results for high SNR, the relationship of simulated mse versus the number of null-subchannels is as same as theoretical prediction. Simulation 6: Comparison of estimators (19) and (20) In this simulation, we set (19) and (20). For estimator (20), we √ M = 256,√fe = 0.2 for both estimator √ simulate two cases: w1 = 2/2, w2 = 6/2; w1 = 1/2, w2 = 7/2. The total power is kept to the same. For estimator (19), we simulate the case of L = 1. The results are shown in Figure 10. We can see that estimator (20) is less sensitive to SNR if the SNR has exceeded the threshold. While the floor mse of estimator (19) is much lower than estimator (20). Simulation 7: performance of estimator (12) over multipath channel In this simulation, we set M = 256, fe = 0.2. First we simulate the estimator performance over time-invariant multipath channel. Here we consider a five paths channel, which has the impulse response h[l] =

4 X

λd δ[l − d],

(18)

d=0

where λd is the complex-valued amplitude. First we consider a specific for the purpose of simulation, we £ ¤ £ stationary multipath channel. Just ¤ set λ0 λ1 λ2 λ3 λ4 = 1 0.5 −0.125 −0.25 −0.125 . Its frequency response is shown in Figure 11. Here we just use the frequency response at f = k to approximate the flat-fading attenuation 9

SNR = 0dB: theoretical SNR = 20dB: theoretical SNR = 40dB: theoretical SNR = 60dB: theoretical SNR = 0dB: simulated SNR = 20dB: simulated SNR = 40dB: simulated SNR = 60dB: simulated

−5

−5

mse

10

mse

10

SNR = 0dB: theoretical SNR = 20dB: theoretical SNR = 40dB: theoretical SNR = 60dB: theoretical SNR = 0dB: simulated SNR = 20dB: simulated SNR = 40dB: simulated SNR = 60dB: simulated

−10

−10

10

10

−15

10

−15

500

1000

1500

2000

2500

3000

3500

10

4000

500

1000

1500

2000

M

2500

3000

3500

4000

M

(a) fe = 0.2

(b) fe = 0.4

Figure 8: Mse versus M for different SNR and fe

L = 1: theoretical L = 4: theoretical L = 1: simulated L = 4: simulated

−2

10

−4

10

−6

mse

10

−8

10

−10

10

−12

10

0

10

20

30 SNR (dB)

40

50

60

Figure 9: Mse versus SNR for different L P4 factor µk of subchannel k, i.e. µk = d=0 λd e−j2πkd/16 . Subchannel 4 is set as the only null-subchannel. The simulation results is shown in Figure 12. We can see that the simulated curve matches well with theoretical prediction. This verifies that the flat-fading approximation of each subchannel is reasonable. Comparing the mse for AWGN and multipath channels, we find they are close. For low SNR section, the mse for multipath channel is even better than AWGN. This can be explained by that the amplitude of frequency response of multipath channel around subchannel 4 is largely higher than 1 (as shown in Figure 11, which implies that the equivalent SNR is higher than AWGN channel. Anyway, we can conclude that estimator (12) is robust to multipath effects. Now we change the coefficients λd over each trial. Just for the purpose of simulation, we set the covariance of λd to be 1/5 to keep the average power of received signal as same as AWGN channel. The coefficients λd is independently randomly selected over each trial. The simulation results of estimator (12) and (13) are shown in Figure 13(a) and 13(b) respectively. We can see that the unknown multipath channel will degrade estimator performance which lies in two aspects: (1) higher SNR threshold; (2) slightly larger mse in high SNR section. For estimator (12), the degradation is less obvious for larger L. For large L, the probability of that all null-subchannels for CFO estimation are in low power is much smaller, then the performance should be less sensitive to frequency-selective fading. Then we simulate estimator (12) over multipath Rayleigh-fading channel. Now the amplitude λd in (18) will be relative to time instant l. In our simulation, λd [l] are identical distributed for all paths and modelled as a lowpass autoregressive precess with fifth order multiple poles at ρ, which can be expressed as λd [l] = 5ρλd [l − 1] − 10ρ2 λd [l − 2] + 10ρ3 λd [l − 3] − 5ρ4 λd [l − 4] + ρ5 λd [l − 5] + ns [l], 10

(19)

Estimator (19): L = 1 Estimator (20): w1=√2/2, w2=√6/2

−2

10

Estimator (20): w1=1/2, w2=√7/2

−4

mse

10

−6

10

−8

10

−10

10

0

10

20

30 SNR (dB)

40

50

60

Figure 10: Mse versus SNR for different methods

2 1.8 1.6 1.4

|H(f)|

1.2 1 0.8 0.6 0.4 0.2 0 0

2

4

6

8 f (1/T)

10

12

14

16

Figure 11: The frequency response of transmitting channel

where ns [l] is the complex-valued zero-mean Gaussian white noise with variance σn2 . 1 The frequency response of such a filter is H(f ) = (1−ρe−j2πf and the 3-dB bandwidth Bλ T is )5 √ √ √ 5 5 5 2 related to ρ by the equation ( 2 − 1)ρ − 2[ 2 − cos(2πBλ T )]ρ + 2 − 1 = 0. Since σλ2 d = cλd [0] = R 0.5 R 0.5 P4 σn2 −0.5 |H(f )|2 df , we set σn2 = 0.2/ −0.5 |H(f )|2 df to let s=0 σλ2 d = 1. We simulate two cases: Bλ T = 0.001 (slow fading) and Bλ T = 0.01 (fast fading). The simulation results are shown in Figure 14. We can see that estimator (12) is not robust to multipath Rayleigh-fading channel. However, for slow Rayleigh-fading channel, estimator (12) can still get reliable CFO estimation. Simulation 8: Comparison with Bolcskei estimator Here we compare estimator (12) to Bolcskei estimator [4]. The formula (15) in [4] is used for estimation of carrier frequency offset θe and timing offset ne . Note that for OFDM/OQAM system, M = N , which was clearly indicated in (3) of [4]. Replacing M by N , we copy (15) in [4] as ¶ · ¤ 1 j2πθe τ −j 2π kne k £ 2 (g,g) 2 N e e Cr [k, τ ] = ΓN [τ ] A τ, σc,R + (−1)k σc,I + cρ [τ ] δ[k], (20) N N where ΓN [τ ] =

N −1 X

2π

|wk |2 ej N kτ

k=0

· ¶ ∞ X 2π k g[l] g[l − τ ]e−j N kl , A(g,g) τ, = N l=−∞

11

(21)

AWGN: theoretical Multipath: theoretical AWGN: simulated Multipath: simulated

−2

10

−4

mse

10

−6

10

−8

10

−10

10

0

10

20

30 SNR (dB)

40

50

60

Figure 12: Mse versus SNR over multipath channel

−1

10 AWGN: L = 1 AWGN: L = 4 Multipath: L = 1 Multipath: L = 4

−2

10

AWGN: w =√2/2, w =√6/2 1

−2

10

2

AWGN: w =1/2, w =√7/2 1

2

Multipath: w1=√2/2, w2=√6/2

−3

10

Multipath: w =1/2, w =√7/2 1

−4

10

2

−4

mse

mse

10 −6

10

−5

10

−6

10 −8

10

−7

10

−8

10

−10

10

−9

0

10

20

30 SNR (dB)

40

(a) Estimator (12)

50

60

10

0

10

20

30 SNR (dB)

40

50

60

(b) Estimator (13)

Figure 13: The performance of estimator (12) and (13) over AWGN and time-invariant multipath channel

2 2 and σc,R and σc,I are respectively the average power of the real and imaginary parts of sent QAM symbols. The frequency offset θe in [4] is normalized with respect to N/T , while our expression fe is with respect to 1/T , thus fe = N θe . 2 2 Now we assume σc,R = σc,I and the transmitter pulse g(t) is band-limited to [−1/T, 1/T ]. By using £ k¢ Parseval’s relation, we can rewrite A(g,g) τ, N shown in (21) as · ¶ ∞ X 2π k A(g,g) τ, g[l] g[l − τ ] e−j N kl = N l=−∞ Z 0.5 k (22) = G(f ) G(f + ) e−j2πf τ df, N −0.5 P∞ where G(f ) = l=∞ g[l] e−j2πf l . Since g[l] is the discrete form of £g(t) with a sampling interval T /N , ¢ k G(f ) should be band-limited to [−1/N, 1/N ]. This leads to A(g,g) τ, N is nonzero only if k = 0, ±1. 2 2 2 2 While for k = ±1, σc,R + (−1)k σc,I = 0 because of the assumption of σc,R = σc,I . Thus we conclude that Cr [k, τ ] can be used for estimation only for k = 0. In this case, only frequency offset can be recovered and the accuracy will be affected by the term cρ [τ ] caused by noise. But the weighting factor wk is still important to keep ΓN [τ ] nonzero for τ ∈ [0, N − 1]. At last, the estimator can be expressed as ( ) Lτ X ˆr [0, τ ] C 1 1 arg . (23) θˆe = 2π τ =1 τ ΓN [τ ] £ ¢ Note that since A(g,g) τ, Nk is real-valued and positive, we don’t need to consider it’s effect. 12

−2

10

−4

mse

10

AWGN: L = 1 AWGN: L = 4 Multipath channel (BλT=0.001): L = 1

−6

10

Multipath channel (B T=0.001): L = 4 λ

Multipath channel (BλT=0.01): L = 1 Multipath channel (B T=0.01): L = 4

−8

λ

10

−10

10

0

10

20

30 SNR (dB)

40

50

60

Figure 14: Mse versus SNR of estimator (12) over AWGN and multipath Rayleigh-fading channel

In simulation, g[l] is square root raised cosine pulse with a roll off factor 1 and Lτ = 15. The weighting factors are chosen as: wk = 1 except w9 = 0. One can verify that ΓN [τ ] 6= 0 for all τ ∈ [0, N − 1]. The comparison of performance versus SNR is shown in Figure 15. The simulation conditions are the same: M = 256, fe = 0.2. From the simulation results, we can see that estimator (12) is far more accurate than Bolcskei estimator except the low SNR section of SNR< 2dB.

−2

10

−4

10

mse

Estimator (19) with L=1 Bolcskei estimator −6

10

−8

10

−10

10

0

10

20

30 SNR (dB)

40

50

60

Figure 15: The comparison of performance versus SNR between estimator (19) and Bolcskei estimator Then we compare the performance versus M for different SNR. Here we simulate two cases: SNR = 0 dB (low SNR) and 40 dB (high SNR). The frequency offset fe = 0.2. The simulation results are shown in Figure 16. We can see that for SNR = 0 dB, estimator (12) will exceed Bolcskei estimator when M > 850; while for SNR = 40 dB, estimator (12) works above the SNR threshold and is always superior to Bolcskei estimator. This implies that estimator (12) is better than Bolcskei estimator for large enough M . In fact from Figure16(b), we can see clearly that mse level of estimator (12) decreases much faster than that of Bolcskei estimator with increasing M . Simulation 9: Comparison of estimator (13) to Ciblat estimator In [6], Ciblat and Serpedin designed a blind CFO estimation algorithm based on conjugate cyclostationarity of the received sequence. However, the authors made a little small mistake on the model of OFDM/OQAM, thus their model is equal to adding a weighting factor j m to the mth subchannel. If there is no weighting, it can be proved that Ciblat and Serpedin’s estimator doesn’t work either, which is same to Bolcskei estimator [4] and our estimator. Simulation shows that Ciblat and Serpedin’s estimator works poorly for weighting method 1, i.e. adding some null-subchannels. Thus we compare only estimator (13) to Ciblat and Serpedin’s. We set fe = 0.2 and M = 256. The simulation results are shown in Figure 17. We can see that for the same weighting factors, estimator (13) exceeds Ciblat and Serpedin’s 13

0.02 −2

10 0

−4

10

Estimator (19) with L=1: 0dB Estimator (19) with L=1: 40dB Bolcskei estimator: 0dB Bolcskei estimator: 40dB

−0.02 −6

mse

bias

10 Estimator (19) with L=1: 0dB Estimator (19) with L=1: 40dB Bolcskei estimator: 0dB Bolcskei estimator: 40dB

−0.04

−0.06

−10

10

−12

−0.08

−0.1

−8

10

10

−14

500

1000

1500

2000

2500

3000

3500

10

4000

500

1000

1500

M

2000

2500

3000

3500

4000

M

(a) Bias versus M

(b) Mse versus M

Figure 16: The comparison of performance versus M between estimator (26) and Bolcskei estimator

estimator.√From Figure √ 5, we can see that estimator (13) has lower SNR threshold and mse. Especially for w1 = 2/2, w2 = 6/2, estimator (13)’s SNR threshold is about 6dB while Ciblat and Serpedin’s doesn’t work well even for every high SNR 60dB. For this case, we should increase M to make Ciblat estimator work. −1

10

Estimator (20): w =√2/2, w =√6/2 1

2

Estimator (20): w1=1/2, w2=√7/2

−2

10

Ciblat estimator: w =√2/2, w =√6/2 1

2

Ciblat estimator: w =1/2, w =√7/2 1

−3

2

10

−4

mse

10

−5

10

−6

10

−7

10

−8

10

−9

10

0

10

20

30 SNR (dB)

40

50

60

Figure 17: The comparison of mse versus SNR of estimator (13) to Ciblat estimator

A

Proof of

PN −1 m=0

Am,k (τ, fe ) = 0

P∞ Proof. First by defining Pm,k (f ) = s=−∞ pm,k [s] e−jπsf (note here Pm,k (f ) is normalized with respect to 1/T instead of the sampling rate 2/T , thus Pm,k (f ) should be periodic to f with a period 2 instead of 1), we can use the decimator formula (see formular 4.1.13 in [16]) to get ∞ X

def

P 1m,k (f ) =

pm,k [2n] e−j2πnf =

n=−∞

¢ 1¡ Pm,k (f ) + Pm,k (f − 1) . 2

(24)

It is more difficult to get the frequency response of the decimated filter pm,k [2n + 1]. We can view P∞ pm,k [2n+1] as the 2 times decimated version of pm,k [s+1], while s=−∞ pm,k [s+1] e−jπsf = ejπf Pm,k (f ), then we have def

P 2m,k (f ) =

∞ X

pm,k [2n + 1] e−j2πnf =

n=−∞

14

¢ 1 jπf ¡ Pm,k (f ) − Pm,k (f − 1) . e 2

(25)

R 0.5 P∞ We can use the discrete form of Parseval’s relation, i.e. g1 [n] g2 [n] = −0.5 G1 (f ) G2 (−f ) df , n=−∞ P∞ P∞ where G1 (f ) = n=−∞ g1 [n] e−j2πf n and G2 (f ) = n=−∞ g2 [n] e−j2πf n . For the case of τ = 2 q, i.e. τ is even, by using (24) and (25), we can rewrite (8) as ∞ X

Am,k (τ, fe ) =

(pm,k [2n] pm,k [2n + 2q] − pm,k [2n + 1] pm,k [2n + 2q + 1])

n=−∞ Z 0.5

(P 1m,k (f ) P 1m,k (−f ) − P 2m,k (f ) P 2m,k (−f )) e−j2πf q df

=

−0.5 Z 0.5

1 = 2

(Pm,k (f ) Pm,k (1 − f ) + Pm,k (−f ) Pm,k (f − 1)) e−jπf τ df.

(26)

−0.5

Similarly, for τ = 2q + 1, we have Am,k (τ, fe ) =

∞ X

(pm,k [2n] pm,k [2n + 2q + 1] − pm,k [2n + 1] pm,k [2n + 2q + 2])

n=−∞ Z 0.5 ³

=

´ P 1m,k (f ) P 2m,k (−f ) e−j2πf q − P 2m,k (f ) P 1m,k (−f ) e−j2πf (q+1) df

−0.5 Z 0.5

1 = 2

(−Pm,k (f ) Pm,k (1 − f ) + Pm,k (−f ) Pm,k (f − 1)) e−jπf τ df.

(27)

−0.5

By combining (26) and (27), we have Am,k (τ, fe ) =

1 2

1 = 2 =

1 2

Z

³

0.5

Pm,k (f ) Pm,k (1 − f ) ejπf τ + Pm,k (−f ) Pm,k (f − 1) ejπ(f +1)τ

−0.5 Z 0.5

Pm,k (f ) Pm,k (1 − f ) e −0.5 Z 1

jπf τ

1 df + 2

Z

1.5

´ df

Pm,k (f ) Pm,k (1 − f ) ejπf τ df

0.5

Pm,k (f ) Pm,k (1 − f ) ejπf τ df.

(28)

−1

Then we define Tm,k (f ) = Pm,k (f ) Pm,k (1 − f )

(29)

that is the integral kernel of (28). From the definition of pm,k [l] shown in (4) and the relationship of pm,k [s] = pm,k [ N2 s], we can write the spectrum of pm,k [s] as π

Pm,k (f ) = ej 2 (m−k)

∞ X

G(f − (m − k) − 2n) G(f + fe − 2n),

(30)

n=−∞

where G(f ) is the spectrum of square root raised cosine pulse with a roll off factor α. Note that G(f ) is 1+α band limited to [− 1+α 2 , 2 ] (normalized with respect to 1/T ), then Pm,k (f ) is just a periodic extension of G(f − (m − k)) G(f + fe ). Substituting (30) into (29), we have " ∞ # X m−k Tm,k (f ) = (−1) G(f − (m − k) − 2n1 ) G(f + fe − 2n1 ) " ×

n1 =−∞ ∞ X

#

G(−f − (m − k) − 2n2 − 1) G(−f + fe − 2n2 − 1)

n2 =−∞

= (−1)m−k

∞ X

∞ X £ G(f − (m − k) − 2n1 ) G(f + fe − 2n1 )

n1 =−∞ n2 =−∞

¤ × G(f + (m − k) + 2n2 + 1) G(f + 2n2 + 1 − fe ) . 15

(31)

Without the loss of generality, we may assume 0 ≤ fe < 0.5. Then Pm,k (f ) is nonzero only if m ∈ {k − 2, k − 1, k, k + 1}. First we check the case of m = k − 2. Substituting m = k − 2 into (31), we have ∞ X

Tk−2,k (f ) =

∞ X

G(f − 2n1 + 2) G(f + fe − 2n1 ) G(f + 2n2 − 1) G(f + 2n2 + 1 − fe ).

(32)

n1 =−∞ n2 =−∞

Since G(f ) is nonzero only if f ∈ [−1, 1], it can be easily verified that the products of G(f − 2n1 + 2) G(f + 2n2 − 1) and G(f + fe − 2n1 ) G(f + 2n2 + 1 − fe ) are nonzero only if 1 < 2 (n1 + n2 ) < 5 and −3 + 2fe < 2 (n1 + n2 ) < 1 + 2fe respectively. Since 0 ≤ fe < 0.5, and n1 and n2 is integer, the two conditions can’t be satisfied in the same time, we can conclude that Tk−2,k (f ) = 0. This implies that Ak−2,k (τ, fe ) = 0. Then we try to analyze the case of m = k: ∞ X

Tk,k (f ) =

∞ X

G(f − 2n1 ) G(f + fe − 2n1 ) G(f + 2n2 + 1) G(f + 2n2 + 1 − fe )

n1 =−∞ n2 =−∞

=

=

−n X1

∞ X

(a)

G(f − 2n1 ) G(f + fe − 2n1 ) G(f + 2n2 + 1) G(f + 2n2 + 1 − fe )

n1 =−∞ n2 =−n1 −1 ∞ X

£ G(f − 2n) G(f + fe − 2n) G(f − 2n − 1) G(f − 2n − 1 − fe )

n=−∞

¤ + G(f − 2n) G(f + fe − 2n) G(f − 2n + 1) G(f − 2n + 1 − fe ) ,

(33)

where (a) follows from the fact that the product of G(f − 2n1 ) G(f + fe − 2n1 ) G(f + 2n2 + 1) G(f + 2n2 + 1 − fe ) is nonzero only if n2 ∈ {−n1 − 1, −n1 }. For the case of m = k − 1: Tk−1,k (f ) = − (a)

= −

= −

∞ X

∞ X

G(f − 2n1 + 1) G(f + fe − 2n1 ) G(f + 2n2 ) G(f + 2n2 + 1 − fe )

n1 =−∞ n2 =−∞ ∞ X

G(f − 2n + 1) G(f + fe − 2n) G(f − 2n) G(f − 2n + 1 − fe )

n=−∞ ∞ X

G(f − 2n) G(f + fe − 2n) G(f − 2n + 1) G(f − 2n + 1 − fe ),

(34)

n=−∞

where (a) follows from the fact that the product of G(f − 2n1 + 1) G(f + fe − 2n1 ) G(f + 2n2 ) G(f + 2n2 + 1 − fe ) is nonzero only if n2 = −n1 . At last for the case of m = k + 1: ∞ X

Tk+1,k (f ) = − =− =−

∞ X

G(f − 2n1 − 1) G(f + fe − 2n1 ) G(f + 2n2 + 2) G(f + 2n2 + 1 − fe )

n1 =−∞ n2 =−∞ ∞ X

G(f − 2n − 1) G(f + fe − 2n) G(f − 2n) G(f − 2n − 1 − fe )

n=−∞ ∞ X

G(f − 2n) G(f + fe − 2n) G(f − 2n − 1) G(f − 2n − 1 − fe ).

(35)

n=−∞

Thus from (33), (34) and (35), we can get that N −1 X

Tm,k (f ) =

m=0

which implies that

PN −1 m=0

k+1 X m=k−1

Am,k (τ, fe ) = 0. 16

Tm,k (f ) = 0

(36)

B

Derivation of asymptotical variance

Since ek [s] satisfies the mixing condition (14), we can use the Lemma 1 introduced by Ciblat [15], which can be written as: Lemma 1. Let

then

M −1 X

1

def

(K)

sM,k (f ) =

M K+1

sK ek [s] e−j2πf s ,

s=0

° ° ° a.s. ° (K) ∀ K ∈ N, sup °sM,k (f )° −→ 0, as M → ∞.

(37)

f ∈[0,1]

Here a.s. stands for ’almost sure’. Then we have the below theorems: a.s. Theorem 1. For ek [s, τ ] = yk [s, τ ] − cˆk [s, τ ] satisfies mixing condition (14), we have fˆe,M − fe −→ 0 a.s. and M (fˆe,M − fe ) −→ 0 as M → ∞.

Proof. Recalling that yk [s, τ ] = cˆk [s, τ ] + ek [s, τ ], we immediately have yk [s] = rk (fe ) ej2π(fe +1/2)s + ek [s] and can write the objective function in (12) as ° °2 L ° M −1 ³ ° ´ X ° 1 X ° JM (f ) = rkl (fe ) ej2π(fe +1/2)s + ekl [s] e−j2πf s ° ° °M ° s=0 l=1 °2 ° ! Ã M −1 L ° ° X 1 X j2π(fe +1/2−f )s ° ° (0) e + sM,kl (f )° . = °rkl (fe ) ° ° M s=0

l=1

1 M

PM −1

(0)

a.s.

ej2π(fe +1/2−f )s → δ[fe + 1/2 − f ]; and by Lemma 1, sM,kl (f ) −→ 0. Then a.s. arg maxf ∈(0,1) JM (f ) −→ fe + 1/2, fˆe,M − fe −→ 0. Also based on Lemma 3 shown in [15], M (fˆe,M − P a.s. M −1 j2π(fe +1/2−f )s 1 fe ) −→ 0 as M → ∞; otherwise M will not converge to δ[fe + 1/2 − f ]. s=0 e As M → ∞,

s=0 a.s.

Theorem 2. M 3/2 (fˆe,M − fe ) is asymptotically Gaussian. Proof. Since fˆe,M = arg maxf ∈(0,1) JM (f ) − 21 , we have ¯ dJM (f ) ¯¯ = 0. df ¯f =fˆe,M + 1 2

Then using the first order Taylor expansion of

dJM (f ) df

around fe + 12 , we have

¯ ¯ ¯ 2 ¯ dJM (f ) ¯¯ dJM (f ) ¯¯ ˆe,M − fe ) d JM (f ) ¯ = + ( f , ¯ ¯ ¯ 2 df df df f =fˆe,M + 1 f =fe + 1 f =fξ + 1 2

where fξ lies between fe and fˆe,M . Then we have

2

2

M 3/2 (fˆe,M − fe ) = −A−1 M BM ,

(38)

where ¯ 1 d2 JM (f ) ¯¯ M2 df 2 ¯f =fξ + 1 2 ¯ 1 dJM (f ) ¯¯ =√ . df ¯f =fe + 1 M

AM = BM

2

17

(39)

Now we try to calculate AM for M → ∞. First we define Ykl (f ) = PL 2 JM (f ) = l=1 kYkl (f )k and

1 M

PM −1 s=0

ykl [s] e−j2πf s , then

ð ¡ ¢ ° ¾! ½ L L 2 X ° ∂Ykl (f ) °2 d2 JM (f ) X d2 YkHl (f ) Ykl (f ) ∂ Y (f ) k l H ° ° = =2 . ° ∂f ° + Re Ykl (f ) ∂f 2 df 2 df 2 l=1

l=1

It can be easily calculated that AM

L 2 X = 2 M l=1

∂ K Ykl (f ) ∂f K

=

1 M

PM −1 s=0

(−j2πs)K ykl [s] e−j2πf s . Then we have

ð ° ½ ¾!¯¯ 2 ° ∂Ykl (f ) °2 ∂ Y (f ) ¯ k l H ° ° ¯ ° ∂f ° + Re Ykl (f ) ∂f 2 ¯

f =fξ + 21

°2 ° M −1 L ° X 2 ° ° 1 X −j2πf s ° = (−j2πs) ykl [s] e ° ° 2 ° M ° M s=0 l=1 f =fξ + 12 à ! !¯¯ H à L M −1 M −1  ¯ X X X 2 1 1 ¯ + Re ykl [s] e−j2πf s (−j2πs)2 ykl [s] e−j2πf s 2  M ¯¯ M M s=0 s=0 l=1

. f =fξ + 21

1

Recalling that ykl [s] = rkl (fe ) ej2π(fe + 2 )s + ekl [s], then as M → ∞, fξ → fe , and based on Lemma 1, the contribution of ekl [s] disappear, we have à a.s.

AM −→ 2 → −

M −1 2π X s M 2 s=0

2π 3

2

L X

!2

L X

à 2

krkl (fe )k − 2

l=1

M −1 4π 2 X 2 s M 3 s=0

!

L X

2

krkl (fe )k

l=1

2

krkl (fe )k .

(40)

l=1

Now we will show that BM is asymptotically Gaussian distributed as M → ∞. First we have BM

¯ ½ ¾¯ L X 2 1 dJM (f ) ¯¯ ∂Ykl (f ) ¯¯ H √ Re Ykl (f ) = =√ ¯ df ¯f =fe + 1 ∂f M M f =fe + 21 l=1 2  à !H à !¯¯ L M −1 M −1  ¯ X X 1 X 4π ¯ √ =− Re j ykl [s] e−j2πf s s ykl [s] e−j2πf s  ¯¯ M s=0 M M s=0 l=1

f =fe + 1

2 Ã !H Ã ! M −1 M −1 L M −1   X X X X 1 1 4π 1 4π r (f ) k e √l = s+ √ s ekl [s] e−j2π(fe + 2 )s Im rkl (fe ) + ekl [s] e−j2π(fe + 2 )s   M s=0 M M s=0 M M s=0 l=1  !H ÃM −1 L −1  4π rH (f ) M X X X 1 1 2π (M − 1) e kl √ √ Im = rkl (fe ) s ekl [s] e−j2π(fe + 2 )s + ekl [s] e−j2π(fe + 2 )s  M M M M s=0 s=0 l=1 Ã M −1 !H Ã ! M −1  X 1 1 1 X 4π √ + ekl [s] e−j2π(fe + 2 )s s ekl [s] e−j2π(fe + 2 )s  M s=0 M M s=0

=

L X l=1

½ ¾ 1 2 π (M − 1) (0)H 1 1 (1) 1 (1) (0)H Im 4π rH (f ) E (f + ) + E (f + ) r (f ) + 4 π s (f + ) E (f + ) , kl e e kl e M,kl e M,kl e M,kl e 2 M 2 2 M,kl 2 (41)

where (K) EM,k (f )

=

1 √

M −1 X

MK M

s=0

18

sK ek [s] e−j2πf s .

(42)

(K)

It can be proved that EM,k (fe + 12 ) is asymptotically zero-mean Gaussian distributed (confer [15] for a.s.

(0)H

more details). As M → ∞, we have sM,k (fe + 12 ) −→ 0 according Lemma 1, and we can rewrite (41) as a.s.

BM −→ j 2 π

L X

M −1 a.s. M −→

1. Therefore

RH kl EM,kl ,

(43)

l=1

where h Rk =

rH k (fe ) 2

−rH k (fe )

−

rT k (fe ) 2

rTk (fe )

iH

h iT (1)T (0)H (1)H 1 1 1 1 EM,k = E(0)T . (f + ) E ) E ) E ) (f + (f + (f + e e e e M,k M,k M,k M,k 2 2 2 2

(44)

a.s.

Then BM −→ N (0, σB2 ), as M → ∞, where σB2 is asymptotic covariance of BM . Thus we can conclude a.s. 9 σ2  . that M 3/2 (fˆe,M − fe ) −→ N (0, σ 2 ) with covariance σ 2 = 4 PL B 2 2 4π l=1 krkl (fe )k Base on Theorem 2, the mse of fˆe,M can be immediately written as γ =

2 9 σB

 . 2 2 krkl (fe )k L Then the only work left is to calculate σB2 . For sparsely distributed null-subchannel indices {kl }l=1 , ekl [s] are independent since the shaping pulse f [l] and g[l] are bandlimited to [−1/T, 1/T ]. Then based on (43), we have ∗ σB2 = E [BM BM ] = 4 π2

L X

4 π4

M3

P

L l=1

£ ¤ H RH kl E EM,kl EM,kl Rkl .

(45)

l=1

h i The matrix E EM,k EH M,k can be expressed as  PM,k (0, 0) P £ ¤  M,k (1, 0) E EM,k EH M,k =  e ∗ PM,k (0, 0) e ∗ (1, 0) P M,k

PM,k (0, 1) PM,k (1, 1) e ∗ (0, 1) P M,k e ∗ (1, 1) P M,k

e M,k (0, 0) P e M,k (1, 0) P P∗M,k (0, 0) P∗M,k (1, 0)

 e M,k (0, 1) P e M,k (1, 1) P  , P∗M,k (0, 1) P∗M,k (1, 1)

(46)

where · PM,k (K1 , K2 ) = E

(K ) EM,k1 (fe

¸ 1 (K2 )H 1 + ) EM,k (fe + ) 2 2

1

M −1 M −1 X X

1

M −1 M −1 X X

£ ¤ −j2π(fe + 1 )(s1 −s2 ) 2 sK1 sK2 E ek [s1 ] eH k [s2 ] e M K1 +K2 +1 s =0 s =0 1 2 1 2 · ¸ 1 1 (K ) (K2 )T 1 e PM,k (K1 , K2 ) = E EM,k (fe + ) EM,k (fe + ) 2 2 =

=

M K1 +K2 +1

£ ¤ −j2π(fe + 1 )(s1 +s2 ) T 1 K2 2 sK . 1 s2 E ek [s1 ] ek [s2 ] e

s1 =0 s2 =0

ee [λ, τ1 , τ2 ] = E [ek [s + λ, τ1 ] ek [s, τ2 ]] e−j4πfe s . We further define Rek [λ, τ1 , τ2 ] = E [ek [s + λ, τ1 ] e∗k [s, τ2 ]] and R k ee [λ, τ1 , τ2 ] are not a function of s, i.e. wide sense stationary, It will be shown later that Rek [λ, τ1 , τ2 ], R k so we omit the time instant s. As M → ∞, by using the mixing condition, we can write the entries of

19

e M,k (K1 , K2 ) as PM,k (K1 , K2 ) and P £ ¤ lim PM,k (K1 , K2 ) τ1 ,τ2 = lim

M −1 M −1 X X

1

M →∞ M K1 +K2 +1

M →∞

M →∞ M K1 +K2 +1

à =

∞ X

s1 =0 s2 =0 s1 X

M −1 X

1

= lim

1

−j2π(fe + 2 )(s1 −s2 ) 1 K2 sK 1 s2 Rek [s1 − s2 , τ1 , τ2 ] e

1

K2 1 sK Rek [λ, τ1 , τ2 ] e−j2π(fe + 2 )λ 1 (s1 − λ)

s1 =0 λ=s1 −M +1

!Ã

−j2π(fe + 12 )λ

lim

Rek [λ, τ1 , τ2 ] e

M →∞

λ=−∞

£ ¤ e M,k (K1 , K2 ) lim P τ

1 ,τ2

M →∞

1 M K1 +K2 +1

M −1 X

! 1 +K2 sK 1

s1 =0

1 1 Se (fe + , τ1 , τ2 ) = K1 + K2 + 1 k 2 M −1 −1 XM X 1 j2π(2fe +1)s2 −j2π(fe + 12 )(s1 +s2 ) 1 K2 e = lim sK e 1 s2 Rek [s1 − s2 , τ1 , τ2 ] e M →∞ M K1 +K2 +1 s =0 s =0 1

= lim

M →∞

à =

1 M K1 +K2 +1

∞ X

2

M −1 X

s1 X

s1 =0 λ=s1 −M +1 −j2π(fe + 12 )λ

ee [λ, τ1 , τ2 ] e R k

1

K2 e 1 sK Rek [λ, τ1 , τ2 ] e−j2π(fe + 2 )λ 1 (s1 − λ)

!Ã lim

M →∞

λ=−∞

1 M K1 +K2 +1

M −1 X

! 1 +K2 sK 1

s1 =0

1 1 = Seek (fe + , τ1 , τ2 ), K1 + K2 + 1 2

(47)

where def

Sek (f, τ1 , τ2 ) =

def Seek (f, τ1 , τ2 ) =

∞ X λ=−∞ ∞ X

Rek [λ, τ1 , τ2 ] e−j2πf λ ee [λ, τ1 , τ2 ] e−j2πf λ . R k

(48)

λ=−∞

Using (47), we immediately have that PM,kl (0, 1) = PM,kl (1, 0) = 12 PM,kl (0, 0), PM,kl (1, 1) = 1 1 e 1 e e e e 3 PM,kl (0, 0) and PM,kl (0, 1) = PM,kl (1, 0) = 2 PM,kl (0, 0), PM,kl (1, 1) = 3 PM,kl (0, 0) as M → ∞. Then by substituting (46) into (45), we have σB2 =

L n o 2 π2 X H ∗ e Re rH kl (fe ) PM,kl (0, 0) rkl (fe ) − rkl (fe ) PM,kl (0, 0) rkl (fe ) . 3 i=1

(49)

e M,k (0, 0) to Pk , P e k respectively, and substituting the By shortening the denotements PM,k (0, 0), P expression of σB2 into γ, we then have (49). While the work is still not finished since Sek (f, τ1 , τ2 ), Seek (f, τ1 , τ2 ) are not known. First we need to ee [λ, τ1 , τ2 ]. calculate Rek [λ, τ1 , τ2 ] and R k A. calculation of cross correlation function of ek [s, τ ] Recalling that ek [s, τ ] = yk [s, τ ] − c˜k [s, τ ] = bk [s + τ ] bk [s] − rk (τ, fe ) ej2π(fe +1/2)s , we have Rek [λ, τ1 , τ2 ] = E [ek [s + λ, τ1 ] e∗k [s, τ2 ]] ´∗ i ´³ h³ bk [s + τ2 ] bk [s] − rk (τ2 , fe ) ej2π(fe +1/2)s = E bk [s + λ + τ1 ] bk [s + λ] − rk (τ1 , fe ) ej2π(fe +1/2)(s+λ) £ ¤ = E bk [s + λ + τ1 ] bk [s + λ] b∗k [s + τ2 ] b∗k [s] − rk∗ (τ2 , fe ) E [bk [s + λ + τ1 ] bk [s + λ]] e−j2π(fe +1/2)s − rk (τ1 , fe ) E [b∗k [s + τ2 ] b∗k [s]] ej2π(fe +1/2)(s+λ) + rk (τ1 , fe ) rk∗ (τ2 , fe ) ej2π(fe +1/2)λ £ ¤ = E bk [s + λ + τ1 ] bk [s + λ] b∗k [s + τ2 ] b∗k [s] − rk∗ (τ2 , fe ) c˜k [s + λ, τ1 ] e−j2π(fe +1/2)s − rk (τ1 , fe ) c˜∗k [s, τ2 ] ej2π(fe +1/2)(s+λ) + rk (τ1 , fe ) rk∗ (τ2 , fe ) ej2π(fe +1/2)λ £ ¤ = E bk [s + λ + τ1 ] bk [s + λ] b∗k [s + τ2 ] b∗k [s] − rk (τ1 , fe ) rk∗ (τ2 , fe ) ej2π(fe +1/2)λ . 20

(50)

Then based on (6), we can continue the first term in the right-hand side of (50) as ¤ £ E bk [s + λ + τ1 ] bk [s + λ] b∗k [s + τ2 ] b∗k [s] "Ã N −1 X jπfe (s+λ+τ1 ) =E e wm µm m=0

×

³

∞ X

aR m [n] pm,k [s

+ λ + τ1 − 2n] + j

(−1)(m−k) aIm [n] pm,k [s

´

!

+ λ + τ1 − 2n − 1] + νk [s + λ + τ1 ]

n=−∞

à ×

e

jπfe (s+λ)

wm µm

e−jπfe (s+τ2 )

∞ ³ X

aR m [n] pm,k [s

+ λ − 2n] + j

(−1)(m−k) aIm [n] pm,k [s

´ + λ − 2n − 1] + νk [s + λ]

!

n=−∞

m=0

à ×

N −1 X

N −1 X

∞ ³ X

wm µ∗m

∗ (m−k) I aR am [n] p∗m,k [s + τ2 − 2n − 1] m [n] pm,k [s + τ2 − 2n] − j (−1)

´

n=−∞

m=0

+ νk∗ [s + τ2 ]) !# Ã N −1 ∞ ³ ´ X X −jπfe s ∗ R ∗ (m−k) I ∗ ∗ × e wm µm am [n] pm,k [s − 2n] − j (−1) am [n] pm,k [s − 2n − 1] + νk [s] n=−∞

m=0

N −1 X

= ejπfe (2λ+τ1 −τ2 )

wm1 wm2 wm3 wm4 µm1 µm2 µ∗m3 µ∗m4

m1 ,m2 ,m3 ,m4 =0

×

∞ X

E

h³ ´ (m1 −k) I aR am1 [n1 ] pm1 ,k [s + λ + τ1 − 2n1 − 1] m1 [n1 ] pm1 ,k [s + λ + τ1 − 2n1 ] + j (−1)

n1 ,n2 ,n3 ,n4 =−∞

³ ´ (m2 −k) I × aR [n ] p [s + λ − 2n ] + j (−1) a [n ] p [s + λ − 2n − 1] 2 m ,k 2 2 m ,k 2 m2 m2 2 2 ³ ´ R ∗ (m3 −k) I ∗ × am3 [n3 ] pm3 ,k [s + τ2 − 2n3 ] − j (−1) am3 [n3 ] pm3 ,k [s + τ2 − 2n3 − 1] ´i ³ (m4 −k) I ∗ am4 [n4 ] p∗m4 ,k [s − 2n4 − 1] × aR m4 [n4 ] pm4 ,k [s − 2n4 ] − j (−1) +

N −1 ∞ X X σa2 jπfe λ 2 e E [νk [s + λ + τ1 ] νk∗ [s + τ2 ]] |wm µm | pm,k [n + λ] p∗m,k [n] 2 n=−∞ m=0

+

N −1 ∞ X X σa2 jπfe (λ−τ2 ) 2 e E [νk [s + λ + τ1 ] νk∗ [s]] |wm µm | pm,k [n + λ − τ2 ] p∗m,k [n] 2 n=−∞ m=0

+

N −1 ∞ X X σa2 jπfe (λ+τ1 ) 2 e E [νk [s + λ] νk∗ [s + τ2 ]] |wm µm | pm,k [n + λ + τ1 ] p∗m,k [n] 2 n=−∞ m=0

+

N −1 ∞ X X σa2 jπfe (λ+τ1 −τ2 ) 2 e E [νk [s + λ] νk∗ [s]] |wm µm | pm,k [n + λ + τ1 − τ2 ] p∗m,k [n] 2 n=−∞ m=0

+ E [νk [s + λ + τ1 ] νk [s + λ] νk∗ [s + τ2 ] νk∗ [s]] .

(51)

21

Then we can continue the summation in the first term in the right-hand side of (51) as N −1 X

wm1 wm2 wm3 wm4 µm1 µm2 µ∗m3 µ∗m4

m1 ,m2 ,m3 ,m4 =0 ∞ X

×

E

h³ ´ (m1 −k) I aR [n ] p [s + λ + τ − 2n ] + j (−1) a [n ] p [s + λ + τ − 2n − 1] 1 m ,k 1 1 1 m ,k 1 1 m1 m1 1 1

n1 ,n2 ,n3 ,n4 =−∞

³

´ (m2 −k) I × aR am2 [n2 ] pm2 ,k [s + λ − 2n2 − 1] m2 [n2 ] pm2 ,k [s + λ − 2n2 ] + j (−1) ³ ´ ∗ (m3 −k) I × aR am3 [n3 ] p∗m3 ,k [s + τ2 − 2n3 − 1] m3 [n3 ] pm3 ,k [s + τ2 − 2n3 ] − j (−1) ³ ´i ∗ (m4 −k) I × aR am4 [n4 ] p∗m4 ,k [s − 2n4 − 1] m4 [n4 ] pm4 ,k [s − 2n4 ] − j (−1) ¶ −1 µ h ∞ X ¡ ¢4 i 3 σa4 NX 4 = E aR [n] − |w µ | pm,k [n + λ + τ1 ] pm,k [n + λ] p∗m,k [n + τ2 ] p∗m,k [n] m m k 4 n=−∞ m=0 "N −1 # ∞ X σa4 X 2 (wm µm ) + (pm,k [2n + λ + τ1 ] pm,k [2n + λ] − pm,k [2n + λ + τ1 + 1] pm,k [2n + λ + 1]) 4 m=0 n=−∞ "N −1 # ∞ X X ¡ ∗ ¢ ∗ 2 ∗ ∗ ∗ × (wm µm ) pm,k [2n + τ2 ] pm,k [2n] − pm,k [2n + τ2 + 1] pm,k [2n + 1] m=0

+

σa4 4

σ4 + a 4

"N −1 X

n=−∞ 2

|wm µm |

m=0

# "N −1 X

n=−∞

m=0

"N −1 X

∞ X ¡ ¢ pm,k [n + λ + τ1 ] p∗m,k [n]

2

|wm µm |

2

|wm µm |

∞ X ¡

pm,k [n + λ −

τ2 ] p∗m,k [n]

# ¢

n=−∞

m=0

# " −1 # ∞ ∞ X X ¡ ¢ NX ¡ ¢ 2 ∗ ∗ pm,k [n + λ + τ1 − τ2 ] pm,k [n] |wm µm | pm,k [n + λ] pm,k [n] .

n=−∞

m=0

n=−∞

(52) Since νk [s] is the filtered and N/2 times down-sampled noise, and f [l] is bandlimited to [−1/T, 1/T ], we have νk [s] =

∞ X

2π

π

ν[l] e−j( N l+ 2 )k f [

l=−∞

=

N 2

∞ X s1 =−∞

ν[

N s − l] 2

π N N s1 ] e−j(πs1 + 2 )k f [ (s − s1 )]. 2 2

(53)

Then we can rewrite the last term in the right-hand side of (51) as E [νk [s + λ + τ1 ] νk [s + λ] νk∗ [s + τ2 ] νk∗ [s]] !Ã ∞ ! µ ¶4 "Ã X ∞ X π π N N N N N = ν[ s2 ] e−j(πs2 + 2 )k f [ (s + λ − s2 )] E ν[ s1 ] e−j(πs1 + 2 )k f [ (s + λ + τ1 − s1 )] 2 2 2 2 2 s2 =−∞ s1 =−∞ !∗ Ã ∞ !∗ # Ã ∞ X X π π N N N N ν[ s4 ] e−j(πs4 + 2 )k f [ (s − s4 ))] × ν[ s3 ] e−j(πs3 + 2 )k f [ (s + τ2 − s3 )] 2 2 2 2 s4 =−∞ s3 =−∞ µ ¶ ∞ ³ h i ´ N 4 X N N N N 4 = E |ν[l]| − 2 σν4 f [ s] f [ (s + τ2 )] f [ (s + λ)] f [ (s + λ + τ1 )] 2 2 2 2 2 s=−∞ ! Ã ! µ ¶4 "Ã X ∞ ∞ X N N N N N f [ s] f [ (s + λ)] + σν4 f [ s] f [ (s + λ + τ1 − τ2 )] 2 2 2 2 2 s=−∞ s=−∞ Ã ∞ !Ã ∞ !# X X N N N N + f [ s] f [ (s + λ + τ1 )] f [ s] f [ (s + λ − τ2 )] . (54) 2 2 2 2 s=−∞ s=−∞ 22

We assume that f [l] and g[l] are identical real-valued symmetric pulses. Reminding that ν[l] is zeromean white Gaussian noise with correlation function σν2 δ[τ ], and the real and imaginary parts of ν[l] are (d) i.i.d., we can write the correlation function of filtered noise νk [s] as h i (d) cνk [τ ] = E νk [s + τ ] νk∗ [s] !Ã ∞ !∗ # "Ã ∞ X X 2π π 2π π N N ν[l1 ] e−j( N l1 + 2 )k f [ (s + τ ) − l1 ] ν[l2 ] e−j( N l2 + 2 )k f [ s − l2 ] =E 2 2 l1 =−∞

= σν2

∞ X l=−∞

(d)

f [l +

l2 =−∞

N (d) τ ] f [l] = σν2 pt [τ ], 2

(55)

def

where pt [s] = (f [l] ∗ f [−l])|l= N s = (f [l] ∗ g[l])|l= N s is N/2 times down-sampled filter of the overall 2 2 response of the cascade of transmitter and h receiver i filters. 4 Similarly, the fourth order moment E |ν[l]| can be written as h i h i 4 4 4 2 2 E |ν[l]| = E (Re {ν[l]}) + (Im {ν[l]}) + 2 (Re {ν[l]}) (Im {ν[l]}) = 2 σν4 .

23

(56)

Then by substituting the results of (51), (52), (54), (55) and (56) into (50), we have ½µ h ¶ ¡ R ¢4 i 3 σa4 jπfe (2λ+τ1 −τ2 ) Rek [λ, τ1 , τ2 ] = e E ak [n] − 4 ×

N −1 X

|wm µm |

m=0

4

∞ X

pm,k [n + λ + τ1 ] pm,k [n + λ] p∗m,k [n + τ2 ] p∗m,k [n]

n=−∞

# "N −1 ∞ X σa4 X 2 (wm µm ) (pm,k [2n + λ + τ1 ] pm,k [2n + λ] − pm,k [2n + λ + τ1 + 1] pm,k [2n + λ + 1]) + 4 m=0 n=−∞ "N −1 # ∞ X X ¡ ∗ ¢ ∗ 2 ∗ ∗ ∗ × (wm µm ) pm,k [2n + τ2 ] pm,k [2n] − pm,k [2n + τ2 + 1] pm,k [2n + 1] m=0

n=−∞

"N −1 # " −1 # ∞ ∞ X X ¡ ¢ NX ¡ ¢ σa4 X 2 2 ∗ ∗ + |wm µm | pm,k [n + λ + τ1 ] pm,k [n] |wm µm | pm,k [n + λ − τ2 ] pm,k [n] 4 m=0 n=−∞ n=−∞ m=0 "N −1 # "N −1 #) ∞ ∞ X X X ¡ ¢ ¡ ¢ σa4 X 2 2 ∗ ∗ + |wm µm | pm,k [n + λ + τ1 − τ2 ] pm,k [n] |wm µm | pm,k [n + λ] pm,k [n] 4 m=0 n=−∞ n=−∞ m=0 +

N −1 ∞ X X σa2 σν2 jπfe λ (d) 2 e pt [λ + τ1 − τ2 ] |wm µm | pm,k [n + λ] p∗m,k [n] 2 n=−∞ m=0

+

N −1 ∞ X X σa2 σν2 jπfe (λ−τ2 ) (d) 2 e pt [λ + τ1 ] |wm µm | pm,k [n + λ − τ2 ] p∗m,k [n] 2 n=−∞ m=0

N −1 ∞ X X σa2 σν2 jπfe (λ+τ1 ) (d) 2 e pt [λ − τ2 ] |wm µm | + pm,k [n + λ + τ1 ] p∗m,k [n] 2 n=−∞ m=0 N −1 ∞ X σa2 σν2 jπfe (λ+τ1 −τ2 ) (d) X 2 e pt [λ] |wm µm | pm,k [n + λ + τ1 − τ2 ] p∗m,k [n] 2 n=−∞ m=0 !Ã ∞ ! µ ¶4 "Ã X ∞ X N N N N N 4 f [ s] f [ (s + λ + τ1 − τ2 )] f [ s] f [ (s + λ)] + σν 2 2 2 2 2 s=−∞ s=−∞ !Ã ∞ !# Ã ∞ X X N N N N f [ s] f [ (s + λ − τ2 )] + f [ s] f [ (s + λ + τ1 )] 2 2 2 2 s=−∞ s=−∞

+

− rk (τ1 , fe ) rk∗ (τ2 , fe ) ej2π(fe +1/2)λ (57) We note that Rek [λ, τ1 , τ2 ] is not a function of s. This confirms that ek [s, τ ] is stationary. We can further simplify the expression. Based on (8), we have ∞ X

(pm,k [2n + λ + τ1 ] pm,k [2n + λ] − pm,k [2n + λ + τ1 + 1] pm,k [2n + λ + 1]) = (−1)λ Am,k (τ1 , fe )

n=−∞ ∞ X ¡ ∗ ¢ pm,k [2n + τ2 ] p∗m,k [2n] − p∗m,k [2n + τ2 + 1] p∗m,k [2n + 1] = A∗m,k (τ2 , fe ). n=−∞

24

(58)

Then based on the definition of rk (τ, fe ) shown in (10), we have # "N −1 ∞ X σa4 jπfe (2λ+τ1 −τ2 ) X 2 e (wm µm ) (pm,k [2n + λ + τ ] pm,k [2n + λ] − pm,k [2n + λ + τ + 1] pm,k [2n + λ + 1]) 4 n=−∞ m=0 "N −1 # ∞ X X ¡ ∗ ¢ ∗ 2 ∗ ∗ ∗ × (wm µm ) pm,k [2n + τ ] pm,k [2n] − pm,k [2n + τ + 1] pm,k [2n + 1] n=−∞

m=0

σ4 = a (−1)λ ej2πfe λ 4

Ã

jπfe τ1

e

= rk (τ1 , fe ) rk∗ (τ2 , fe ) e

N −1 X

!Ã 2

(wm µm ) Am,k (τ1 , fe )

m=0 j2π(fe +1/2)λ

e

jπfe τ2

N −1 X

!∗ 2

(wm µm ) Am,k (τ2 , fe )

m=0

.

Thus we can simplify Rek [λ, τ1 , τ2 ] as ½µ h ¶ ¡ ¢4 i 3 σa4 [n] Rek [λ, τ1 , τ2 ] = ejπfe (2λ+τ1 −τ2 ) E aR − k 4 ×

N −1 X m=0

|wm µm |

4

∞ X

pm,k [n + λ + τ1 ] pm,k [n + λ] p∗m,k [n + τ2 ] p∗m,k [n]

n=−∞

"N −1 # " −1 # ∞ ∞ X X ¡ ¢ NX ¡ ¢ σa4 X 2 2 ∗ ∗ |wm µm | pm,k [n + λ + τ1 ] pm,k [n] |wm µm | pm,k [n + λ − τ2 ] pm,k [n] + 4 m=0 n=−∞ n=−∞ m=0 "N −1 # "N −1 #) ∞ ∞ X X X ¡ ¢ ¡ ¢ σa4 X 2 2 ∗ ∗ + |wm µm | pm,k [n + λ + τ1 − τ2 ] pm,k [n] |wm µm | pm,k [n + λ] pm,k [n] 4 m=0 n=−∞ n=−∞ m=0 +

N −1 ∞ X X σa2 jπfe λ (d) 2 e pt [λ + τ1 − τ2 ] |wm µm | pm,k [n + λ] p∗m,k [n] 2 n=−∞ m=0

+

N −1 ∞ X X σa2 jπfe (λ−τ2 ) (d) 2 e pt [λ + τ1 ] |wm µm | pm,k [n + λ − τ2 ] p∗m,k [n] 2 n=−∞ m=0

+

N −1 ∞ X X σa2 jπfe (λ+τ1 ) (d) 2 e pt [λ − τ2 ] |wm µm | pm,k [n + λ + τ1 ] p∗m,k [n] 2 n=−∞ m=0

N −1 ∞ X σa2 jπfe (λ+τ1 −τ2 ) (d) X 2 + e pt [λ] |wm µm | pm,k [n + λ + τ1 − τ2 ] p∗m,k [n] 2 n=−∞ m=0 !Ã ∞ ! µ ¶4 "Ã X ∞ X N N N N N 4 f [ s] f [ (s + λ + τ1 − τ2 )] f [ s] f [ (s + λ)] + σν 2 2 2 2 2 s=−∞ s=−∞ Ã ∞ !Ã ∞ !# X X N N N N + f [ s] f [ (s + λ + τ1 )] f [ s] f [ (s + λ − τ2 )] . 2 2 2 2 s=−∞ s=−∞

B. calculation of conjugate cross correlation function of ek [s, τ ]

25

(59)

ee (λ, τ1 , τ2 ) = e−j4πfe s E [ek [s + λ, τ1 ] ek [s, τ2 ]], we have First based on the definition of R k ee [λ, τ1 , τ2 ] R k = e−j4πfe s E [ek [s + λ, τ1 ] ek [s, τ2 ]] h³ ´³ ´i = e−j4πfe s E bk [s + λ + τ1 ] bk [s + λ] − rk (τ1 , fe ) ej2π(fe +1/2)(s+λ) bk [s + τ2 ] bk [s] − rk (τ2 , fe ) ej2π(fe +1/2)s h i £ ¤ = e−j4πfe s E bk [s + λ + τ1 ] bk [s + λ] bk [s + τ2 ] bk [s] − rk (τ2 , fe ) E bk [s + λ + τ1 ] bk [s + λ] e−j2π(fe +1/2)s h i − rk (τ1 , fe ) E bk [s + τ2 ] bk [s] ej2π(fe +1/2)(λ−s) + rk (τ1 , fe ) rk (τ2 , fe ) ej2π(fe +1/2)λ £ ¤ = e−j4πfe s E bk [s + λ + τ1 ] bk [s + λ] bk [s + τ2 ] bk [s] − rk (τ2 , fe ) c˜k [s + λ, τ1 ] e−j2π(fe +1/2)s − rk (τ1 , fe ) c˜k [s, τ2 ] ej2π(fe +1/2)(λ−s) + rk (τ1 , fe ) rk (τ2 , fe ) ej2π(fe +1/2)λ £ ¤ = e−j4πfe s E bk [s + λ + τ1 ] bk [s + λ] bk [s + τ2 ] bk [s] − rk (τ1 , fe ) rk (τ2 , fe ) ej2π(fe +1/2)λ .

(60)

Then based on (6), we can continue the first term in the right-hand side of (60) as £ ¤ e−j4πfe s E bk [s + λ + τ1 ] bk [s + λ] bk [s + τ2 ] bk [s] "Ã N −1 X −j4πfe s =e E ejπfe (s+λ+τ1 ) wm µm m=0

×

∞ ³ X

aR m [n] pm,k [s

+ λ + τ1 − 2n] + j

(−1)(m−k) aIm [n] pm,k [s

!

´

+ λ + τ1 − 2n − 1] + νk [s + λ + τ1 ]

n=−∞

à ×

ejπfe (s+λ)

ejπfe (s+τ2 )

×

N −1 X

e

jπfe s

N −1 X

´ (m−k) I aR am [n] pm,k [s + λ − 2n − 1] + νk [s + λ] m [n] pm,k [s + λ − 2n] + j (−1)

!

n=−∞ ∞ ³ ´ X (m−k) I am [n] pm,k [s + τ2 − 2n − 1] + νk [s + τ2 ] aR m [n] pm,k [s + τ2 − 2n] + j (−1)

wm µm

n=−∞

m=0

Ã

∞ ³ X

wm µm

m=0

à ×

N −1 X

wm µm

∞ X

³

aR m [n] pm,k [s

− 2n] + j

(−1)(m−k) aIm [n] pm,k [s

´

!

!#

− 2n − 1] + νk [s]

n=−∞

m=0

N −1 X

= ejπfe (2λ+τ1 +τ2 )

wm1 wm2 wm3 wm4 µm1 µm2 µm3 µm4

m1 ,m2 ,m3 ,m4 =0

×

∞ X

E

h³ ´ (m1 −k) I aR [n ] p [s + λ + τ − 2n ] + j (−1) a [n ] p [s + λ + τ − 2n − 1] 1 m ,k 1 1 1 m ,k 1 1 m1 m1 1 1

n1 ,n2 ,n3 ,n4 =−∞

³ ´ (m2 −k) I × aR ] p [s + λ − 2n ] + j (−1) a ] p [s + λ − 2n − 1] [n [n 2 2 m2 2 m2 ,k m2 2 m2 ,k ³ ´ (m3 −k) I × aR am3 [n3 ] pm3 ,k [s + τ2 − 2n3 − 1] m3 [n3 ] pm3 ,k [s + τ2 − 2n3 ] + j (−1) ³ ´i (m4 −k) I × aR [n ] p [s − 2n ] + j (−1) a [n ] p [s − 2n − 1] 4 4 m4 4 m4 ,k m4 4 m4 ,k + e−j4πfe s E [νk [s + λ + τ1 ] νk [s + λ] νk [s + τ2 ] νk [s]] .

26

(61)

The summation in the first term of (61) can be continued as N −1 X

wm1 wm2 wm3 wm4 µm1 µm2 µm3 µm4

m1 ,m2 ,m3 ,m4 =0 ∞ X

×

E

h³ ´ (m1 −k) I aR [n ] p [s + λ + τ − 2n ] + j (−1) a [n ] p [s + λ + τ − 2n − 1] 1 m ,k 1 1 1 m ,k 1 1 m1 m1 1 1

n1 ,n2 ,n3 ,n4 =−∞

³

´ (m2 −k) I × aR am2 [n2 ] pm2 ,k [s + λ − 2n2 − 1] m2 [n2 ] pm2 ,k [s + λ − 2n2 ] + j (−1) ³ ´ (m3 −k) I × aR am3 [n3 ] pm3 ,k [s + τ2 − 2n3 − 1] m3 [n3 ] pm3 ,k [s + τ2 − 2n3 ] + j (−1) ³ ´i (m4 −k) I × aR am4 [n4 ] pm4 ,k [s − 2n4 − 1] m4 [n4 ] pm4 ,k [s − 2n4 ] + j (−1) ¶ −1 µ h ∞ X ¡ ¢4 i 3 σa4 NX 4 = E aR [n] − (w µ ) pm,k [n + λ + τ1 ] pm,k [n + λ] pm,k [n + τ2 ] pm,k [n] m m k 4 n=−∞ m=0 "N −1 # ∞ X σa4 X 2 + (wm µm ) (pm,k [2n + λ + τ1 ] pm,k [2n + λ] − pm,k [2n + λ + τ1 + 1] pm,k [2n + λ + 1]) 4 m=0 n=−∞ "N −1 # ∞ X X 2 × (wm µm ) (pm,k [2n + τ2 ] pm,k [2n] − pm,k [2n + τ2 + 1] pm,k [2n + 1]) m=0

σa4

"N −1 X

n=−∞ 2

#

∞ X

(wm µm ) (pm,k [2n + λ + τ1 ] pm,k [2n + τ2 ] − pm,k [2n + λ + τ1 + 1] pm,k [2n + τ2 + 1]) 4 m=0 n=−∞ "N −1 # ∞ X X 2 (wm µm ) × (pm,k [2n + λ] pm,k [2n] − pm,k [2n + λ + 1] pm,k [2n + 1])

+

m=0

σa4

"N −1 X

n=−∞ 2

#

∞ X

(wm µm ) (pm,k [2n + λ + τ1 ] pm,k [2n] − pm,k [2n + λ + τ1 + 1] pm,k [2n + 1]) 4 m=0 n=−∞ "N −1 # ∞ X X 2 × (wm µm ) (pm,k [2n + λ] pm,k [2n + τ2 ] − pm,k [2n + λ + 1] pm,k [2n + τ2 + 1]) . +

m=0

(62)

n=−∞

Now we try to calculate the second term in the right-hand side of (61) as E [νk [s + λ + τ1 ] νk [s + λ] νk [s + τ2 ] νk [s]] ¸ · N N N N = E νk [ (s + λ + τ1 )] νk [ (s + λ)] νk [ (s + τ2 )] νk [ s] 2 2 2 2 "Ã ∞ !Ã ∞ ! X X π π 2π N N l + )k l + )k −j( 2π −j( 1 2 N 2 N 2 =E f [ (s + λ + τ1 ) − l1 ] f [ (s + λ − l2 ] ν[l1 ] e ν[l2 ] e 2 2 l1 =−∞ l2 =−∞ Ã ∞ !Ã ∞ !# X X π 2π π N N −j( 2π l + )k −j( l + )k 3 4 N 2 N 2 × ν[l3 ] e f [ (s + τ2 ) − l3 ] ν[l4 ] e f [ s − l4 ] 2 2 l3 =−∞

l4 =−∞

∞ h i X 2π π N N N 4 = E (ν[l]) f [l] f [l + τ2 ] f [l + λ] f [l + (λ + τ1 )]e−j 4( N l+ 2 )k 2 2 2 l=−∞ µ h ∞ i ³ h i´2 ¶ X 2π π N N N 4 2 = 2 E Re {ν[l]} − 6 E Re {ν[l]} f [l] f [l + τ2 ] f [l + λ] f [l + (λ + τ1 )]e−j 4( N l+ 2 )k 2 2 2 l=−∞

= 0.

(63)

27

Then substituting (62) and (63) into (61), then into (60), we have ee [λ, τ1 , τ2 ] R k =e

(µ

jπfe (2λ+τ1 +τ2 )

h¡

E

aR k [n]

¢4 i

−

3 σa4 4

¶ NX −1

4

(wm µm )

∞ X

pm,k [n + λ + τ1 ] pm,k [n + λ] pm,k [n + τ2 ] pm,k [n]

n=−∞

m=0

# "N −1 ∞ X σa4 X 2 (wm µm ) (pm,k [2n + λ + τ1 ] pm,k [2n + λ] − pm,k [2n + λ + τ1 + 1] pm,k [2n + λ + 1]) + 4 m=0 n=−∞ "N −1 # ∞ X X 2 × (wm µm ) (pm,k [2n + τ2 ] pm,k [2n] − pm,k [2n + τ2 + 1] pm,k [2n + 1]) m=0

n=−∞

m=0

n=−∞

"N −1 # ∞ X σa4 X 2 + (wm µm ) (pm,k [2n + λ + τ1 ] pm,k [2n + τ2 ] − pm,k [2n + λ + τ1 + 1] pm,k [2n + τ2 + 1]) 4 m=0 n=−∞ "N −1 # ∞ X X 2 × (wm µm ) (pm,k [2n + λ] pm,k [2n] − pm,k [2n + λ + 1] pm,k [2n + 1]) σa4

"N −1 X

2

#

∞ X

(wm µm ) (pm,k [2n + λ + τ1 ] pm,k [2n] − pm,k [2n + λ + τ1 + 1] pm,k [2n + 1]) 4 m=0 n=−∞ "N −1 #) ∞ X X 2 × (wm µm ) (pm,k [2n + λ] pm,k [2n + τ2 ] − pm,k [2n + λ + 1] pm,k [2n + τ2 + 1]) +

n=−∞

m=0

− rk (τ1 , fe ) rk (τ2 , fe ) ej2π(fe +1/2)λ .

(64)

ee [λ, τ1 , τ2 ] is independent to time instant s. In addition, different from Re [λ, τ1 , τ2 ], We can see that R k k it is also unrelated to noise ν[l]. Similarly, based on (58) and the definition of rk (τ, fe ), we can simplify ee [λ, τ1 , τ2 ] as the expression of R k ½µ h ¶ ¡ R ¢4 i 3 σa4 ee [λ, τ1 , τ2 ] = ejπfe (2λ+τ1 +τ2 ) [n] − R E a k k 4 ×

N −1 X

4

(wm µm )

m=0

∞ X

pm,k [n + λ + τ1 ] pm,k [n + λ] pm,k [n + τ2 ] pm,k [n]

n=−∞

"N −1 # ∞ X σa4 X 2 (wm µm ) + (pm,k [2n + λ + τ1 ] pm,k [2n + τ2 ] − pm,k [2n + λ + τ1 + 1] pm,k [2n + τ2 + 1]) 4 m=0 n=−∞ "N −1 # ∞ X X 2 × (wm µm ) (pm,k [2n + λ] pm,k [2n] − pm,k [2n + λ + 1] pm,k [2n + 1]) m=0

σa4

"N −1 X

n=−∞ 2

#

∞ X

(pm,k [2n + λ + τ1 ] pm,k [2n] − pm,k [2n + λ + τ1 + 1] pm,k [2n + 1]) (wm µm ) 4 m=0 n=−∞ "N −1 #) ∞ X X 2 × (pm,k [2n + λ] pm,k [2n + τ2 ] − pm,k [2n + λ + 1] pm,k [2n + τ2 + 1]) . (wm µm )

+

m=0

n=−∞

C. Calculation of Sek (f, τ1 , τ2 ) and Seek (f, τ1 , τ2 ) ee [λ, τ1 , τ2 ] shown in (64), we are Having got the expressions for Rek [λ, τ1 , τ2 ] shown in (57) and R k now ready to calculate Sek (f, τ1 , τ2 ) and Seek (f, τ1 , τ2 ) based on (48). First we consider Sek (f, τ1 , τ2 ). It is more convenient to rewrite the summations in (57) in frequency domain. By recalling that Pm,k (f ) =

28

(65)

P∞ s=−∞

pm,k [s] e−jπf s , and using Parseval’s relation, we have

∞ X

pm,k [n + λ + τ1 ] pm,k [n + λ] p∗m,k [n + τ2 ] p∗m,k [n]

n=−∞

1 = 8 =

1 8

Z

1

−1 Z 1

µ Z ejπf λ µZ

−1

1

Pm,k (f1 ) Pm,k (f − f1 ) e

jπf1 τ1

−1

¶∗

1

df1

Pm,k (f2 ) Pm,k (f − f2 ) e

−1 1

Z

−1

¶ µZ

1

−1

jπf2 τ2

df2

df

¶

∗ ∗ Pm,k (f1 ) Pm,k (f − f1 ) Pm,k (f2 ) Pm,k (f − f2 ) ejπ(f1 τ1 −f2 τ2 ) df1 df2

ejπf λ df,

(66)

and ∞ X

pm,k [n + λ] p∗m,k [n] =

n=−∞ ∞ X

pm,k [n + λ +

n=−∞ ∞ X

τ1 ] p∗m,k [n]

1 = 2

pm,k [n + λ − τ2 ] p∗m,k [n] =

1 2

pm,k [n + λ + τ1 − τ2 ] p∗m,k [n] =

1 2

n=−∞ ∞ X n=−∞

Similarly, by noting that G(f ) = µ µ

µ

N 2

¶2 X ∞

¶2 X ∞ N 2

µ

1 2

s=−∞

f[

s=−∞

¶2 X ∞ N

f[

f[

N 2

P∞ s=−∞

Z

1

2

|Pm,k (f )| ejπf λ df

−1

Z

1

³

2

|Pm,k (f )| ejπf τ1

´

ejπf λ df

−1

Z

1

³

2

|Pm,k (f )| e−jπf τ2

´

ejπf λ df

−1

Z

1

³

´ 2 |Pm,k (f )| ejπf (τ1 −τ2 ) ejπf λ df.

(67)

−1

f [ N2 s] e−jπf s , we have

N N 1 s] f [ (s + λ)] = 2 2 2

N N 1 s] f [ (s + λ + τ1 )] = 2 2 2 N N 1 s] f [ (s + λ − τ2 )] = 2 2 2

Z

1

G2 (f ) ejπf λ df

−1

Z

1

¡ 2 ¢ G (f ) ejπf τ1 ejπf λ df

−1

Z

1

¡ 2 ¢ G (f ) e−jπf τ2 ejπf λ df

2 −1 s=−∞ ¶2 X Z 1³ ∞ ´ N N 1 N f [ s] f [ (s + λ + τ1 − τ2 )] = G2 (f ) ejπf (τ1 −τ2 ) ejπf λ df. 2 2 2 2 −1 s=−∞

29

(68)

Then we have Sek (f, τ1 , τ2 ) =

µ h ¶ −1 ¡ ¢4 i 3 σa4 NX 1 jπfe (τ1 −τ2 ) 4 e E aR − |wm µm | [n] k 4 4 m=0 Z 1Z 1 ∗ ∗ × Pm,k (f1 ) Pm,k (2(f − fe ) − f1 ) Pm,k (f2 ) Pm,k (2(f − fe ) − f2 ) ejπ(f1 τ1 −f2 τ2 ) df1 df2 −1

+

σa4 8

−1

Z

jπ(fe (τ1 +τ2 )−2f τ2 )

1

e

−1

ÃN −1 X

ÃN −1 X m=0

2

|wm µm | |Pm,k (2f1 )|

2

! ÃN −1 X

! ÃN −1 X 2

! 2

|wm µm | |Pm,k (2(f − fe − f1 ))|

2

ej2πf1 (τ1 +τ2 ) df1

m=0

! σa4 jπfe (τ1 −τ2 ) 2 2 2 + e |wm µm | |Pm,k (2f1 )| |wm µm | |Pm,k (2(f − fe − f1 ))| ej2πf1 (τ1 −τ2 ) df1 8 −1 m=0 m=0 ÃN −1 ! 2 2 Z 1 X σ σ 2 2 |wm µm | |Pm,k (2(f − f1 ) − fe )| G2 (2f1 ) ej2πf1 (τ1 −τ2 ) df1 + a ν 4 −1 m=0 ! Z ÃN −1 σa2 σν2 −j2πf τ2 1 X 2 2 + e |wm µm | |Pm,k (2(f − f1 ) − fe )| G2 (2f1 ) ej2πf1 (τ1 +τ2 ) df1 4 −1 m=0 ! Z ÃN −1 σa2 σν2 j2πf τ1 1 X 2 2 + e |wm µm | |Pm,k (2(f − f1 ) − fe )| G2 (2f1 ) e−j2πf1 (τ1 +τ2 ) df1 4 −1 m=0 ! Z ÃN −1 2 2 σa σν j2πf (τ1 −τ2 ) 1 X 2 2 e |wm µm | |Pm,k (2(f − f1 ) − fe )| G2 (2f1 ) e−j2πf1 (τ1 −τ2 ) df1 + 4 −1 m=0 Z ´ σ 4 1 ³ j2πf1 (τ1 −τ2 ) + ν e + ej2πf τ1 e−j2πf1 (τ1 +τ2 ) G2 (2f1 ) G2 (2(f − f1 )) df1 . 2 −1 Z

1

Then we immediately have 1 Sek (fe + , τ1 , τ2 ) 2 µ h ¶ ¡ ¢4 i 3 σa4 1 jπfe (τ1 −τ2 ) = e E aR [n] − k 4 4 µZ ¶ µZ N −1 1 X 4 × |wm µm | Pm,k (f ) Pm,k (1 − f ) ejπf τ1 df m=0

−1

−1

¶∗ Pm,k (f ) Pm,k (1 − f ) ejπf τ2 df

! 1 σa4 2 2 2 τ2 jπfe (τ1 −τ2 ) + (−1) e |wm µm | |Pm,k (f )| |wm µm | |Pm,k (1 − f )| ejπf (τ1 +τ2 ) df 8 −1 m=0 m=0 ! ÃN −1 ! Z 1 ÃNX −1 4 X σa jπfe (τ1 −τ2 ) 2 2 2 2 e |wm µm | |Pm,k (f )| + |wm µm | |Pm,k (1 − f )| ejπf (τ1 −τ2 ) df 8 −1 m=0 m=0 ÃN −1 ! 2 2 Z 1 X σ σ 2 2 + a ν |wm µm | |Pm,k (1 + fe − f )| G2 (f ) ejπf (τ1 −τ2 ) df 4 −1 m=0 ! Z ÃN −1 2 2 σa σν −jπ(2fe +1)τ2 1 X 2 2 e |wm µm | |Pm,k (1 + fe − f )| G2 (f ) ejπf (τ1 +τ2 ) df + 4 −1 m=0 ! Z ÃN −1 σa2 σν2 jπ(2fe +1)τ1 1 X 2 2 e |wm µm | |Pm,k (1 + fe − f )| G2 (f ) e−jπf (τ1 +τ2 ) df + 4 −1 m=0 ! Z ÃN −1 σa2 σν2 jπ(2fe +1)(τ1 −τ2 ) 1 X 2 2 + e |wm µm | |Pm,k (1 + fe − f )| G2 (f ) e−jπf (τ1 −τ2 ) df 4 −1 m=0 ´ ³ 4 Z 1 σ (70) G2 (f ) G2 (2fe + 1 − f ) ejπf (τ1 −τ2 ) + ejπ(2fe +1)τ1 e−jπf (τ1 +τ2 ) df. + ν 2 −1 30 Z

ÃN −1 X

1

! ÃN −1 X 2

(69)

Now we will calculate Seek (f, τ1 , τ2 ). First we have ∞ X n=−∞ Z 1

1 = 8

pm,k [n + λ + τ1 ] pm,k [n + λ] pm,k [n + τ2 ] pm,k [n] µZ

1

Z

¶

1

jπ(f1 τ1 +f2 τ2 )

Pm,k (f1 ) Pm,k (f − f1 ) Pm,k (f2 ) Pm,k (−f − f2 ) e −1

−1

df2 df1

ejπf λ df.(71)

−1

Also based on the definition of Am,k (τ, fe ) in (8), we have ∞ X

(pm,k [2n + λ + τ1 ] pm,k [2n + τ2 ] − pm,k [2n + λ + τ1 + 1] pm,k [2n + τ2 + 1]) = (−1)τ2 Am,k (λ + τ1 − τ2 , fe )

n=−∞ ∞ X

(pm,k [2n + λ] pm,k [2n] − pm,k [2n + λ + 1] pm,k [2n + 1]) = Am,k (λ, fe )

n=−∞ ∞ X n=−∞ ∞ X

(pm,k [2n + λ + τ1 ] pm,k [2n] − pm,k [2n + λ + τ1 + 1] pm,k [2n + 1]) = Am,k (λ + τ1 , fe ) (pm,k [2n + λ] pm,k [2n + τ2 ] − pm,k [2n + λ + 1] pm,k [2n + τ2 + 1]) = (−1)τ2 Am,k (λ − τ2 , fe ).

n=−∞

(72) Thus by using (28), we have µ h ¶ −1 ¡ ¢4 i 3 σa4 NX ejπfe (τ1 +τ2 ) 4 Seek (f, τ1 , τ2 ) = E aR [n] − (wm µm ) k 4 4 m=0 Z 1Z 1 × Pm,k (f1 ) Pm,k (2(f − fe ) − f1 ) Pm,k (f2 ) Pm,k (−2(f − fe ) − f2 ) ejπ(f1 τ1 +f2 τ2 ) df2 df1 −1

−1 τ2

+ (−1) e ×

ÃN −1 X

jπfe (τ1 +τ2 )

σa4 8

1

ÃN −1 X

−1

m=0

Z

! 2

(wm µm ) Pm,k (2f1 ) Pm,k (1 − 2f1 ) !

2

m=0

+ (−1)τ2 ×

ÃN −1 X

ej2πf1 (τ1 −τ2 ) df1

(wm µm ) Pm,k (2(f − fe − f1 )) Pm,k (1 − 2(f − fe − f1 )) σa4 −j2π(f −f e)τ2 jπfe (τ1 +τ2 ) e e 8

1

ÃN −1 X

−1

m=0

Z

! 2

(wm µm ) Pm,k (2f1 ) Pm,k (1 − 2f1 ) !

2

(wm µm ) Pm,k (2(f − fe − f1 )) Pm,k (1 − 2(f − fe − f1 ))

ej2πf1 (τ1 +τ2 ) df1 .

(73)

m=0

Then we have 1 Seek (fe + , τ1 , τ2 ) 2 µ h ¶ ¡ ¢4 i 3 σa4 1 jπfe (τ1 +τ2 ) = e E aR − [n] k 4 4 µZ ¶ µZ N −1 1 X 4 × (wm µm ) Pm,k (f ) Pm,k (1 − f ) ejπf τ1 df m=0

−1

1

¶ Pm,k (f ) Pm,k (1 − f ) ejπf τ2 df

−1

ÃN −1 X

!2 1 4 σ 2 (wm µm ) Pm,k (f ) Pm,k (1 − f ) ejπf (τ1 −τ2 ) df + (−1)τ2 a ejπfe (τ1 +τ2 ) 8 −1 m=0 !2 Z ÃN −1 σa4 jπfe (τ1 +τ2 ) 1 X 2 + e (wm µm ) Pm,k (f ) Pm,k (1 − f ) ejπf (τ1 +τ2 ) df. 8 −1 m=0 Z

31

(74)

Since Pm,k (f ) can be deferred from the shaping pulse G(f ) and channel response, based on (70) and (74), we can now calculate Sek (fe + 1/2, τ1 , τ2 ) and Seek (fe + 1/2, τ1 , τ2 ).

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