Fast blind channel estimation for space–time block coded MIMO–OFDM systems

Abstract

Fast blind subspace channel estimation using circular property of the channel matrix is investigated for space–time block coded (STBC) multiple-input multiple-output orthogonal frequency division multiplexing (MIMO–OFDM) systems in this paper. The noise subspace computed from the correlation matrix of received signals requires a large number of symbols to converge in the subspace channel estimation. Using the circular property of the channel matrix, we propose both the cyclic repetition method (CRM) and the forward–backward method (FBM) to generate N times of equivalent signals for each STBC–OFDM symbol, respectively, where N is the size of FFT operation. With these equivalent symbols, the proposed CRM, FBM and CRM–FBM (CFBM) channel estimations can perform very well within a few OFDM symbols. The CRM, FBM and CFBM schemes are applicable to the CP-OFDM, ZP-OFDM and VC-OFDM systems, respectively. The identifiability of the subspace channel estimation is investigated that the channel matrix is determined up to two ambiguity matrices. Computer simulations demonstrate that the CRM-based, FBM-based and CFBM-based channel estimations have better performances than the conventional ones.

Appendices

Appendix 1: Proof of Lemma 1

We note that $\mathbf{G}$ and $\mathbf{H}$ have the same rank for $\mathbf{P}_{2N,M_\mathrm{r}}$ and $\mathbf{Q}_{2N,K}$ are orthogonal matrices. Since G is a block circular matrix, it is diagonalizable by DFT/IDFT matrix,

$$\begin{aligned} \mathbf{G}= & {} \mathbf{F}_{N,2M_r }^H diag(\varvec{\Lambda }_G (0),\ldots ,\varvec{\Lambda }_G (N-1))\mathbf{F}_{N,2K} \nonumber \\= & {} \mathbf{F}_{N,2M_r }^H \varvec{\Lambda }_G \mathbf{F}_{N,2K} , \end{aligned}$$

(63)

where $\mathbf{F}_{N,M} =\mathbf{F}_N \otimes \mathbf{I}_M$ and $\varvec{\Lambda }_G (k)$ is the kth frequency response of $\{\mathbf{g}(n)\}$. From (11), $\varvec{\Lambda }_G (k)$ is obtained as

$$\begin{aligned} \varvec{\Lambda }_G (k)= & {} \left( {{\begin{array}{cc} \varvec{\Lambda }_1 (k)&{} \varvec{\Lambda }_2 (k) \\ \varvec{\Lambda }_2^*(k)W_N^{Lk} &{} -\varvec{\Lambda }_1^*(k)W_N^{Lk} \\ \end{array} }} \right) \nonumber \\= & {} \left( {{\begin{array}{cc} \mathbf{I}&{} 0 \\ 0&{} {W_N^{Lk} \mathbf{I}} \\ \end{array} }} \right) \left( {{\begin{array}{cc} \varvec{\Lambda }_1 (k)&{} \varvec{\Lambda }_2 (k) \\ \varvec{\Lambda }_2^*(k)&{} -\varvec{\Lambda }_1^*(k) \\ \end{array} }} \right) \nonumber \\= & {} \left( {{\begin{array}{cc} \mathbf{I}&{} 0 \\ 0&{} W_N^{Lk} \mathbf{I} \\ \end{array} }} \right) \varvec{\Lambda }(k), \end{aligned}$$

(64)

From (62), G and H are of full column rank if and only if $\varvec{\Lambda }(k)$ has full column rank for $k = 0, {\ldots }, N-1$, which proves this lemma.

Appendix 2: Proof of Theorem 2

Since both H and ${\hat{\mathbf{H}}}$ are of full column rank and $span(\mathbf{H}) = span({\hat{\mathbf{H}}})$, we have $\mathbf{H} = {\hat{\mathbf{H}}{} \mathbf{A}}$ where A is a $2KN\times 2KN$ nonsingular matrix. Applying (13) to both H and ${\hat{\mathbf{H}}}, \mathbf{H} = {\hat{\mathbf{H}}{} \mathbf{A}}$ is simplified as $\mathbf{G} = {\hat{\mathbf{G}}{} \mathbf{B}}$ where $\mathbf{B}=\mathbf{Q}_{2N,K}^T \mathbf{AQ}_{2N,K}$. Besides, similar to $\mathbf{G}$ in (63), ${\hat{\mathbf{G}}}$ can be diagonalizable by ${\hat{\mathbf{G}}}=\mathbf{F}_{N,2M_r }^H {\hat{\varvec{\Lambda }}}_G \mathbf{F}_{N,2K}$. Then we have

$$\begin{aligned} \varvec{\Lambda }_G ={\hat{\varvec{\Lambda }}}_G \underbrace{\mathbf{F}_{N,2K} \mathbf{BF}_{N,2K}^H }_{\buildrel \Delta \over = \mathbf{C}}={\hat{\varvec{\Lambda }}}_G \mathbf{C}, \end{aligned}$$

(65)

where C is a $2NK\times 2NK$ matrix. Since $\varvec{\Lambda }_G \hbox { and }{\hat{\varvec{\Lambda }}}_G$ are block diagonal matrices, we can find that C is also a block diagonal matrix with $\mathbf{C}=diag\left( {\mathbf{C}_{0,0} \cdots \mathbf{C}_{N-1,N-1} } \right)$ where $\mathbf{C}_{n,n}$ is a $2K\times 2K$ matrix. Therefore, matrix B becomes

$$\begin{aligned} \mathbf{B}=\mathbf{F}_{N,2K}^H diag\left( {\mathbf{C}_{0,0} \cdots \mathbf{C}_{N-1,N-1} } \right) \mathbf{F}_{N,2K} . \end{aligned}$$

(66)

It is known that a circular matrix can be diagonlizable by the DFT/IDFT matrix, and vice versa. Therefore, matrix B is rewritten as a block circular matrix where the first 2K row vectors are composed by $[\mathbf{B}_{0}{} \mathbf{B}_{1} {\ldots }\mathbf{B}_{N-1}]$ and where $\mathbf{B}_n ,n = 0,\ldots ,N-1$ is a $2K\times 2K$ matrix. Using $\mathbf{G}={\hat{\mathbf{G}}{} \mathbf{B}}$ and B is a block circular matrix, the equations for which the entries in $\mathbf{G}$ are zeros are given by

$$\begin{aligned}&\underbrace{\left[ {{\begin{array}{ccccccc} {{\hat{\mathbf{g}}}(0)} &{} {{\hat{\mathbf{g}}}(1)} &{} \cdots &{} {{\hat{\mathbf{g}}}(L)} &{} &{} &{} \mathbf{0} \\ &{} {{\hat{\mathbf{g}}}(0)} &{} \cdots &{} \cdots &{} {{\hat{\mathbf{g}}}(L)} &{} &{}\\ &{} &{} \ddots &{} \ddots &{} \ddots &{} \ddots &{} \\ \mathbf{0} &{} &{} &{} {{\hat{\mathbf{g}}}(0)} &{} \cdots &{} \cdots &{} {{\hat{\mathbf{g}}}(L)} \\ \end{array} }} \right] }_{\mathbf{T}_{toe} ({\hat{\mathbf{g}}})}\left[ {{\begin{array}{l} {\mathbf{B}_1 } \\ {\mathbf{B}_2 } \\ \vdots \\ {\mathbf{B}_{N-1} } \\ \end{array} }} \right] \nonumber \\&\qquad =\mathbf{T}_{toe} ({\hat{\mathbf{g}}})\left[ {{\begin{array}{l} {\mathbf{B}_1 } \\ {\mathbf{B}_2 } \\ \vdots \\ {\mathbf{B}_{N-1} } \\ \end{array} }} \right] =\mathbf{0}, \end{aligned}$$

(67)

where $\mathbf{T}_{toe} ({\hat{\mathbf{g}}})$ is a $2M_r (N-L-1)\times 2(N-1)K$ block Toeplitz matrix constructing by ${\hat{\mathbf{g}}}=\left[ {{\hat{\mathbf{g}}}(0)\cdots {\hat{\mathbf{g}}}(L)} \right]$. With the assumption that ${\hat{\mathbf{G}}}(z)$ is irreducible, the matrix $\mathbf{T}_{toe} ({\hat{\mathbf{g}}})$ is of full column rank [25] and $\mathbf{B}_i =\mathbf{0}$ for $i = 1,2,\ldots ,N-1$. Then the matrix $\mathbf{B}$ is simplified as $\mathbf{B} = \mathbf{I}_N \otimes \mathbf{B}_0$. It turns out that $\mathbf{G}={\hat{\mathbf{G}}{} \mathbf{B}}$ becomes

$$\begin{aligned} \mathbf{g}(n)={\hat{\mathbf{g}}}(n)\mathbf{B}_0 ,\hbox { }n = 0,\ldots ,L. \end{aligned}$$

(68)

Let $\mathbf{B}_0 =\left[ {{\begin{array}{ll} {\mathbf{b}_1 }&{} {\mathbf{b}_3 } \\ {\mathbf{b}_2 }&{} {\mathbf{b}_4 } \\ \end{array} }} \right]$ where $\mathbf{b}_i \in C^{K\times K}$. Considering $n = k$ and $n = L-k$ in (68), we obtain

$$\begin{aligned} \left( {{\begin{array}{cc} {{\hat{\mathbf{h}}}_1 (k)}&{} {{\hat{\mathbf{h}}}_2 (k)} \\ {{\hat{\mathbf{h}}}_2^*(k)}&{} {-{\hat{\mathbf{h}}}_1^*(k)} \\ \end{array} }} \right) \left[ {{\begin{array}{l} {\mathbf{b}_1 -\mathbf{b}_4^*} \\ {\mathbf{b}_2 +\mathbf{b}_3^*} \\ \end{array} }} \right] =\mathbf{0}. \end{aligned}$$

(69)

Since we assume ${\hat{\mathbf{h}}}(0)$ is of full column rank, we have $\mathbf{b}_4 =\mathbf{b}_1^*$ and $\mathbf{b}_3 =-\mathbf{b}_2^*$. Using these equalities, the proof of Theorem 2 is complete by simplifying (68) into (20).