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Comments on “Performance Analysis of a8to1000. Elements of the data matrixSNwere generated using the QPSK constellation andFis chosen asIM. One Deterministic Channel Estimator for Block hundred independent realizations of channel coefﬁcients and Transmission Systems With Null Guard 10 independent realizations of data blocksSNare used (totally Intervals” 1000 different pairs ofSNandh). Traces ofChhandCCRin (2) are computed for these 1000 realizations and the averages Borching Su,Student Member, IEEE, are reported in Table I. and P. P. Vaidyanathan,Fellow, IEEE tr(C)−tr(C) 2 2hh CR σtr(C)/σ Ntr(Chh)/v CRv(CCR) tr Abstract— In the abovementioned paper a CramerRao bound 8−1.7752− was derived for the performance of a blind channel estimation 12 184.01 1.3373 136.6002 algorithm. In this paper an error in the bound is pointed out 14 6.8590 1.0981 5.2462 and corrected. It is observed here that the performance of the 16 3.5362 0.9760 2.6233 1 said algorithm does not achieve the CramerRao bound.20 1.7197 0.7414 1.3196 100 0.1614 0.1448 0.1147 −2−2 In the above paper [1], important work has been done to1000 1.5149×10 1.4986×100.0109 analyze the algorithm in [2] which solves a blind channel estimation problem. The performance of the algorithm in [2] TABLE I in high SNR region was shown to be as in (33) of [1]. CO FO M PA R I S O NEQ. (33)IN[1]A N DEQ. (2);T H EE N G T HDATA LP E R The CramerRao bound (CRB) of the above mentioned blind M= 12 B L O C KI S estimation problem was shown to be as in (49) of [1]. The coincidence of (33) and (49) led the authors of [1] to claim that the algorithm in [2] is statistically efﬁcient (i.e., achieves We ﬁnd from Table I that there is a signiﬁcant discrepancy the CRB) at high SNR values. However, we have found an between the corrected CRB in (2) and the performance of the error in the derivation of (49), which invalidates this claim. algorithm in [2] (Eq. (33) in [1]), especially whenNis small. Eq. (49) of [1] was derived from (80) in Appendix B of [1]. ∗T Furthermore, whenN <M, the inverse ofS Sin (33) of N N The second equality of (80) is not valid in general since it is [1] does not exist, butCCRin (2) still gives a ﬁnite value. conditioned on the validity of the matrix identity This suggests there might exist algorithms (e.g., see [4]–[6]) H−1H† −1† other than [2] which solve the aforementioned blind estimation (ABA) =A B A(1) problem whenMN <. On the other hand, whenNis large, whereAis a full rank matrix with more columns than rows the difference between traces ofChhandCCRtends to shrink, andBis a square positive deﬁnite matrix. But a simple but it never goes to zero. This observation is accounted for by example shows that this identity is not true in general: set the following lemma, where we use notations from the singular ˜ · ¸value decomposition of theL×LMfullrank matrixV: 1 0 0 1 0 0 £ ¤£ ¤H A=,andB1 1= 0, ˜ (3) 0 1 0V=VU D 01V2, 0 1 2 whereUis a unitary matrix,Dis a diagonal matrix with £ ¤ then the left hand side of (1) isI2whereas the right hand side · ¸ positive diagonal entries, andV:=V1V2is a unitary 1 0 is .matrix.V1andV2are the ﬁrstLand the last(M−1)L 0 2 columns ofV, respectively. A correction to the CRB, however, is easy to make. The corrected CRB can be simply taken as the ﬁrst equality of Lemma 1:IfN≥M, then tr(Chh)≥tr(CCR), with (80) of [1]: equality if and only if h i £ ¤ −1 2∗ ∗T TH ˜ ˜ VI⊗(F SS)V(2)H CCR=σL×NL NF v V1BV2=0(4) 2∗ ∗T T (in the original text [1],σappeared in the denominator, whichwhereB:=IL×L⊗(S FF S)andV1andV2are v N N was presumably a typographical error).deﬁned as in (3). We conduct numerical simulations to compare £ ¡¢¤Proof:Since bothChhandCCRare positive deﬁnite 2†H−T∗T−1−∗ † ˜ ˜ C≈σF(S S)F hh vVIL×L⊗N NV (p.d.), the statement tr(Chh)≥tr(CCR)is equivalent to the statement thatChh−CCRis a positive semideﬁnite matrix. from (33) of [1] and the corrected CRB in (2). The simulation ∗ ∗T T We ﬁrst observe thatBis p.d. sinceS FF Sis p.d. Recall setting basically follows that in [1]: the channel order is chosenN N ˜ the SVD ofVas in (3) whereUandV:= [V1,V2]are asL= 4and the channel coefﬁcients are i.i.d., zeromean, unit unitary matrices andDis a diagonal matrix with positive variance complex Gaussian random variables. The data length H diagonal entries. DeﬁneB2:=V BVwhich is obviously per block isM= 12and the number of blocksNranges from −1 also p.d. Partitinto ionB2andB2 1 Work supported in parts by the NSF grant CCF0428326, ONR grant· ¸· ¸ 0 0 B11B12B B N000140610011, and the Moore Fellowship of the California Institute of−121 11 B2=HandB=0H0, 2 Technology. B B B12B2222 12

0 the same size asD vely, so thatB11andB11have

2

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