Introduction
Modern multiple-input multiple-output (MIMO) technologies, including massive (cellular) MIMO and distributed (cell-free) MIMO, are designed to operate in time-division duplexing (TDD) with half-duplex access points (APs) [? , ? , ? ]. Conventionally, these systems operate in static TDD, where the fraction of time allocated to uplink (UL) respectively downlink (DL) transmissions is fixed and aligned among all APs. Contrarily, in dynamic TDD operation, each individual AP can adjust its UL and DL allocation in every coherence interval. This allows dynamic TDD systems to cater asymmetric and time-varying UL and DL traffic by adjusting the allocation based on the instantaneous data demands [? ]. However, dynamic TDD introduces cross-link interference among APs and among user equipment (UEs), and this interference must be managed to maximize the benefits of dynamic TDD [? ].
Several studies have shown advantages of dynamic TDD over static TDD, predominantly in the special case where each AP is restricted to operate in either UL or DL throughout the entire frame, i.e., the APs do not switch between UL and DL within a frame [? , ? ]. This operation mode is often referred to as network-assisted full-duplex and will be considered in this work. Further, dynamic TDD has been discussed and is featured in the 3GPP standards [? , ? ], which shows its potential for implementation in practical systems.
The use of half-duplex APs makes dynamic TDD a cost-effective flexible duplexing mode. It can be enabled in existing MIMO systems, in contrast to full-duplex schemes that require additional complex and costly hardware for self-interference cancellation [? , ? , ? , ? ]. In fact, networks with half-duplex APs and dynamic TDD operation can outperform corresponding full-duplex networks, both in terms of spectral and energy efficiency [? , ? , 15 ]. For full-duplex to compete with dynamic TDD, the self-interference within the full-duplex APs must be highly mitigated, enforcing strict hardware requirements.
Although dynamic TDD systems do not suffer from self-interference, the cross-link interference remains. Specifically, the AP-AP interference (AAI), which occurs when neighboring APs operate in opposite transmission directions on the same time-frequency resources, can significantly limit the UL data rates in dynamic TDD unless mitigated [? , ? ].
State-of-the-Art in AP-AP Interference Mitigation
Existing methods for AAI mitigation primarily rely on strict coordination of the AP scheduling, beamforming, or power control. One exception is [16 ] , wherein the authors show that the AAI can be perfectly mitigated in dynamic TDD massive MIMO systems when the number of AP antennas per UE tends to infinity. However, several hundreds of antennas per AP are required to get close to this limit, which is not always feasible in practical systems.
More specifically, the papers [? , ? , 17 , 18 , 19 ] optimize the system throughput by scheduling the transmission modes (UL or DL) of the APs in each coherence interval. In [17 ] , the AP mode selection is optimized in order to maximize the sum UL/DL SE, whereas [? ] and [? ] formulate similar problems that also include AP power control in the optimization problem. Further, [19 ] maximizes the geometric mean of the per-UE spectral efficiency to achieve fairness, including the DL beamforming design in the optimization problem.
The existing works closest to ours consider coordinated AP-AP beamforming methods where the transmit precoding at the
Parts of this paper were presented at the 2024 International Conference on Acoustics, Speech, and Signal Processing (ICASSP), see [? ].
M. Andersson, J. Åslund and E. G. Larsson are with the Department of Electrical Engineering (ISY), Linköping University, 581 83 Linköping, Sweden. Email: {martin.andersson, jan.aslund, erik.g.larsson}@liu.se.
T. T. Vu is with the School of Computer, Data and Mathematical Sciences, Western Sydney University, Australia. Email: t.vu4@westernsydney.edu.au.
P. Frenger is with Ericsson Research, 583 30 Linköping, Sweden. Email: pal.frenger@ericsson.com.
This work was partially supported by the Wallenberg AI, Autonomous Systems and Software Program (WASP) funded by the Knut and Alice Wallenberg Foundation, and partially supported by ELLIIT. The work of T. T. Vu was partially supported by the Australian Research Council Discovery Early Career Researcher Award (DECRA) under Project DE260100816. DL APs or the receive combining at the UL APs is designed to suppress AAI [20 , 21 , 22 , 23 ] . For example, [23 ] applies zero-forcing receive combining at the UL APs to nullify the AAI from the DL APs, while [21 ] designs the beamforming at the DL APs such that the AAI at the UL APs is constrained.
An important limitation of the existing methods for AAI mitigation, specifically those based on coordinated beamforming cited above, is their reliance on precise AP-AP channel knowledge; they require frequent estimation of these channels. This introduces a vast overhead for transmission of pilot sequences between APs for AP-AP channel estimation. During this pilot phase, the DL APs must halt their data transmission (reducing the DL pre-log factor) and the UL UEs must be silent to avoid corrupting the AP-AP channel estimates at the UL APs (reducing the UL pre-log factor), which heavily limits performance and scalability. Furthermore, most of these methods require adaptation of the DL transmissions (e.g., through DL beamforming design) in order to alleviate the AAI, wasting degrees of freedom on interference mitigation and impairing the DL data rates.
Contributions of the Paper
In this paper, we develop an alternative strategy by exploiting that the UL APs know the data signals transmitted by the DL APs. This lets us use the DL transmit signals as implicit pilots to avoid an overhead for AP-AP channel estimation. However, these DL transmit signals are rank-deficient, and hence cannot be used for AP-AP channel estimation within a single coherence interval. But provided that the AP-AP channels are slowly time-varying (compared to the UE-AP channels)1
More specifically, we formulate the AAI mitigation problem as a joint estimation of the UL data and the AP-AP channel over multiple coherence intervals during which the AP-AP channel is constant. Mathematically, this gives rise to a coupled (two-sided) least-squares problem of the form
min { X t } , H ∑ t ∈ T ∥ Y t − G t X t − H V t S t ∥ 2 (1) \min_{\{X_t\},H}\sum_{t\in\mathcal{T}}\left\|Y_t-G_tX_t-HV_tS_t\right\|^2 \tag*{(1)} { X t } , H min t ∈ T ∑ ∥ Y t − G t X t − H V t S t ∥ 2 (1) where { X t } \{X_t\} { X t } H \mathbf{H} H { Y t } \{Y_t\} { Y t } { G t } \{G_t\} { G t } { V t } \{V_t\} { V t } { S t } \{S_t\} { S t }
Solving (1) is straightforward assuming that a unique solution exists. However, to the best of our knowledge, the existence of a unique solution to problems of the form (1) has not been analyzed in the literature, not even from a purely mathematical perspective. Some results exist on the (unique) solvability of matrix equations of the form Y = G X + H V S \mathbf{Y}=\mathbf{G}\mathbf{X}+\mathbf{H}\mathbf{V}\mathbf{S} Y = GX + HVS X \mathbf{X} X H \mathbf{H} H ∣ T ∣ = 1 |\mathcal{T}|=1 ∣ T ∣ = 1 [24 , 25 , 26 ] . However, these results do not generalize to equation systems that define the solution of (1) (with a summation over t t t
[27 ] develops iterative algorithms to solve systems of coupled matrix equations, but it assumes that a unique solution of (1) exists and does not characterize detailed conditions for the existence of a unique solution. Problem (1) can also be expressed on vector form using the Khatri-Rao product [28 ] , and one possible approach could be to bound the rank of the resulting effective regression matrix, using results from multilinear decomposition theory. Specifically, [29 ] and [30 ] give k k k k ′ k' k ′
Our technical contributions are:
We formulate the AAI mitigation problem as in (1), by leveraging the slow time-variation of the AP-AP channels and knowledge of the DL transmit signals. Further, we calculate the mean-square error (MSE) and derive an achievable spectral efficiency (SE) for the UL data symbol estimates given by the solution to (1). Numerical simulations show that our proposed method enables effective AAI mitigation and significantly outperforms baseline receivers both in terms of MSE and SE. When the AP-AP channel varies very slowly over time, the performance of our proposed technique approaches that of a genie-aided baseline with perfect AAI mitigation.
For cases where (1) does not have a unique solution, we propose a modified (uniquely solvable) least-squares problem by silencing a subset of the UL data symbols in each coherence interval.
As a methodological contribution to least-squares estimation, we investigate the existence of a unique solution to (1) by analyzing the rank of its regression matrix. We provide the following specific results (we refer to Section IV for the formal statements):
Theorem 1: A necessary condition in terms of ∣ T ∣ |\mathcal{T}| ∣ T ∣
Theorem 2: The necessary condition for identifiability in Theorem 1 is sufficient in the case where there are equally many UL and DL AP antennas, and the number of UL/DL UEs present in the network is half of the number of AP antennas.
Theorem 3: In the case ∣ T ∣ = 1 |\mathcal{T}|=1 ∣ T ∣ = 1 never identifiable.
This paper is a comprehensive extension of our conference paper [? ], which was limited to single-slot processing (i.e., ∣ T ∣ = 1 |\mathcal{T}|=1 ∣ T ∣ = 1 V t S t \mathbf{V}_t\mathbf{S}_t V t S t
Notation
We denote scalars by regular lowercase symbols. Bold symbols in lowercase and uppercase are column vectors and matrices, respectively. The set of complex numbers is denoted by C \mathbb{C} C ( ⋅ ) ∗ (\cdot)^* ( ⋅ ) ∗ ( ⋅ ) T (\cdot)^T ( ⋅ ) T ( ⋅ ) H (\cdot)^H ( ⋅ ) H ⊗ \otimes ⊗ vec ( ⋅ ) \operatorname{vec}(\cdot) vec ( ⋅ ) Tr ( ⋅ ) \operatorname{Tr}(\cdot) Tr ( ⋅ ) det ( ⋅ ) \det(\cdot) det ( ⋅ ) ( ⋅ ) − 1 (\cdot)^{-1} ( ⋅ ) − 1 N ( ⋅ ) \mathcal{N}(\cdot) N ( ⋅ ) dim ( ⋅ ) \dim(\cdot) dim ( ⋅ ) Cov ( ⋅ ) \operatorname{Cov}(\cdot) Cov ( ⋅ ) Bias ( ⋅ ) \operatorname{Bias}(\cdot) Bias ( ⋅ ) MSE ( ⋅ ) \operatorname{MSE}(\cdot) MSE ( ⋅ ) C N ( ⋅ , ⋅ ) \mathcal{CN}(\cdot,\cdot) C N ( ⋅ , ⋅ ) ∥ ⋅ ∥ \lVert\cdot\rVert ∥ ⋅ ∥ [ ⋅ ] i : j , k : l [\cdot]_{i:j,k:l} [ ⋅ ] i : j , k : l i , … , j i,\ldots,j i , … , j k , … , l k,\ldots,l k , … , l
System Model
We consider the UL data transmission in a MIMO system with dynamic TDD operation. Specifically, one AP equipped with M M M K K K N N N J J J 2 K < M K<M K < M J < N J<N J < N Figure 1 .
Figure 1. Illustration of the considered MIMO system operating in dynamic TDD with AP-AP interference.
Remark 1. For clarity of presentation, we consider the case with co-located AP antennas and single-antenna UEs. However, our system model is general and covers the cases of arbitrarily distributed AP antennas and multi-antenna UEs.
In the sequel, we refer to the cross-link interference from the DL AP to the UL AP as AP-AP interference (AAI). Our aim is to detect the data transmitted by the UL UEs to the UL AP, in the presence of the AAI caused by the DL AP.
We assume a block-fading model where all channels stay constant throughout each coherence interval. Suppose that the APs are static in space and the UEs are mobile, such that the AP-AP channel varies much more slowly over time (i.e., has a longer coherence time) than the UE-AP channels. Assuming that the coherence bandwidths of all channels are of the same order, the AP-AP channel will have a larger coherence interval length than the UE-AP channels. Concretely, let τ \tau τ T T T t ∈ T ≜ { 1 , … , T } t \in\mathcal{T} \triangleq\{1,\ldots,T\} t ∈ T ≜ { 1 , … , T }
We define the collective channel matrix from the K K K t ∈ T t \in\mathcal{T} t ∈ T
G t = [ g 1 , t ⋯ g K , t ] ∈ C M × K , (2) \mathbf{G}_t = \left[\begin{array}{ccc} \mathbf{g}_{1,t} & \cdots& \mathbf{g}_{K,t} \end{array}\right] \in\mathbb{C}^{M \times K}, \tag*{(2)} G t = [ g 1 , t ⋯ g K , t ] ∈ C M × K , (2) where g k , t ∈ C M × 1 \mathbf{g}_{k,t} \in\mathbb{C}^{M \times1} g k , t ∈ C M × 1 k k k
H = [ h 1 ⋯ h N ] ∈ C M × N (3) \mathbf{H} = \left[\begin{array}{ccc} \mathbf{h}_1 & \cdots& \mathbf{h}_N \end{array}\right] \in\mathbb{C}^{M \times N} \tag*{(3)} H = [ h 1 ⋯ h N ] ∈ C M × N (3) be the interference channel from the DL AP to the UL AP, which is constant during all t ∈ T t \in\mathcal{T} t ∈ T h n ∈ C M × 1 \mathbf{h}_n \in\mathbb{C}^{M \times1} h n ∈ C M × 1 n n n Figure 2 .
Figure 2. The UE-AP and AP-AP channels during a sequence of T T T
Further, let X t ∈ C K × τ \mathbf{X}_t \in\mathbb{C}^{K \times\tau} X t ∈ C K × τ K K K t t t S t ∈ C J × τ \mathbf{S}_t \in\mathbb{C}^{J \times\tau} S t ∈ C J × τ J J J J J J V t ∈ C N × J \mathbf{V}_t \in\mathbb{C}^{N \times J} V t ∈ C N × J V t S t ∈ C N × τ \mathbf{V}_t\mathbf{S}_t \in\mathbb{C}^{N \times\tau} V t S t ∈ C N × τ
The signal received by the UL AP in each coherence interval t ∈ T t \in\mathcal{T} t ∈ T
Y t = G t X t + H V t S t + W t ∈ C M × τ , (4) \mathbf{Y}_t = \mathbf{G}_t\mathbf{X}_t + \mathbf{H}\mathbf{V}_t\mathbf{S}_t + \mathbf{W}_t \in\mathbb{C}^{M \times\tau}, \tag*{(4)} Y t = G t X t + H V t S t + W t ∈ C M × τ , (4) where W t \mathbf{W}_t W t { G t } \{\mathbf{G}_t\} { G t } { S t } \{\mathbf{S}_t\} { S t }
Assumption 1. The UL AP knows the UE-AP channels { G t } \{\mathbf{G}_t\} { G t } { S t } \{\mathbf{S}_t\} { S t } { V t } \{\mathbf{V}_t\} { V t } H \mathbf{H} H
In practice the UE-AP channels { G t } \{\mathbf{G}_t\} { G t } { X t } \{\mathbf{X}_t\} { X t } { G t } \{\mathbf{G}_t\} { G t } { S t } \{\mathbf{S}_t\} { S t } { V t } \{\mathbf{V}_t\} { V t }
Here, we define the UL data detection3 H \mathbf{H} H { X t } \{\mathbf{X}_t\} { X t } H \mathbf{H} H { V t S t } \{\mathbf{V}_t\mathbf{S}_t\} { V t S t }
min { X t } , H ∑ t ∈ T ∥ Y t − G t X t − H V t S t ∥ 2 , (5) \underset{\{\mathbf{X}_t\},\mathbf{H}}{\min}\ \sum_{t \in\mathcal{T}}\left\lVert\mathbf{Y}_t-\mathbf{G}_t\mathbf{X}_t-\mathbf{H}\mathbf{V}_t\mathbf{S}_t\right\rVert^2, \tag*{(5)} { X t } , H min t ∈ T ∑ ∥ Y t − G t X t − H V t S t ∥ 2 , (5) which is a joint estimation of the UL data matrices { X t } \{\mathbf{X}_t\} { X t } H \mathbf{H} H T T T t ∈ T t \in\mathcal{T} t ∈ T H \mathbf{H} H Γ \mathbf{\Gamma} Γ Δ \mathbf{\Delta} Δ Λ \mathbf{\Lambda} Λ
vec ( Γ Δ Λ ) = ( Λ T ⊗ Γ ) vec ( Δ ) . (6) \operatorname{vec}(\mathbf{\Gamma}\mathbf{\Delta}\mathbf{\Lambda}) = (\mathbf{\Lambda}^{T} \otimes\mathbf{\Gamma})\operatorname{vec}(\mathbf{\Delta}). \tag*{(6)} vec ( ΓΔΛ ) = ( Λ T ⊗ Γ ) vec ( Δ ) . (6) Let y t ≜ vec ( Y t ) \mathbf{y}_t \triangleq\operatorname{vec}(\mathbf{Y}_t) y t ≜ vec ( Y t ) x t ≜ vec ( X t ) \mathbf{x}_t \triangleq\operatorname{vec}(\mathbf{X}_t) x t ≜ vec ( X t ) h ≜ vec ( H ) \mathbf{h} \triangleq\operatorname{vec}(\mathbf{H}) h ≜ vec ( H ) w t ≜ vec ( W t ) \mathbf{w}_t \triangleq\operatorname{vec}(\mathbf{W}_t) w t ≜ vec ( W t ) Y t \mathbf{Y}_t Y t
y t ≜ [ ( I τ ⊗ G t ) ( S t T V t T ⊗ I M ) ] [ x t h ] + w t . (7) \mathbf{y}_t \triangleq\left[(\mathbf{I}_{\tau} \otimes\mathbf{G}_t)\quad(\mathbf{S}_t^{T}\mathbf{V}_t^{T} \otimes\mathbf{I}_M)\right]
\begin{bmatrix}
\mathbf{x}_t\\
\mathbf{h}
\end{bmatrix}
+\mathbf{w}_t. \tag*{(7)} y t ≜ [ ( I τ ⊗ G t ) ( S t T V t T ⊗ I M ) ] [ x t h ] + w t . (7) To formulate the joint estimation problem over all t ∈ T t \in\mathcal{T} t ∈ T { y t } \{y_t\} { y t }
y ≜ [ y 1 ⋮ y T ] = A z + w ∈ C T M τ × 1 , (8) y \triangleq
\begin{bmatrix}
y_1\\
\vdots\\
y_T
\end{bmatrix}
= A z + w \in\mathbb{C}^{TM\tau\times1}, \tag*{(8)} y ≜ y 1 ⋮ y T = A z + w ∈ C T M τ × 1 , (8) where the regression matrix A ∈ C T M τ × ( T K τ + M N ) A \in\mathbb{C}^{TM\tau\times(TK\tau+MN)} A ∈ C T M τ × ( T K τ + M N )
A ≜ [ I τ ⊗ G 1 0 S 1 T V 1 T ⊗ I M ⋮ ⋱ ⋮ 0 I τ ⊗ G T S T T V T T ⊗ I M ] , (9) A \triangleq
\begin{bmatrix}
I_\tau\otimes G_1 & 0 & S_1^{T}V_1^{T} \otimes I_M\\
\vdots& \ddots& \vdots\\
0 & I_\tau\otimes G_T & S_T^{T}V_T^{T} \otimes I_M
\end{bmatrix}, \tag*{(9)} A ≜ I τ ⊗ G 1 ⋮ 0 0 ⋱ I τ ⊗ G T S 1 T V 1 T ⊗ I M ⋮ S T T V T T ⊗ I M , (9) and
z ≜ [ x h ] , x ≜ [ x 1 ⋮ x T ] , w ≜ [ w 1 ⋮ w T ] . (10) z \triangleq
\begin{bmatrix}
x\\
h
\end{bmatrix},
\quad
x \triangleq
\begin{bmatrix}
x_1\\
\vdots\\
x_T
\end{bmatrix},
\quad
w \triangleq
\begin{bmatrix}
w_1\\
\vdots\\
w_T
\end{bmatrix}. \tag*{(10)} z ≜ [ x h ] , x ≜ x 1 ⋮ x T , w ≜ w 1 ⋮ w T . (10) Now, the original least-squares problem (5) can be expressed on the equivalent vector form
min z ∥ y − A z ∥ 2 . (11) \min_z \ \lVert y-Az\rVert^2. \tag*{(11)} z min ∥ y − A z ∥ 2 . (11) If A A A
z ^ ≜ [ x ^ h ^ ] = arg min z ∥ y − A z ∥ 2 = ( A H A ) − 1 A H y . (12) \hat{z} \triangleq
\begin{bmatrix}
\hat{x}\\
\hat{h}
\end{bmatrix}
= \operatorname*{arg\,min}_z \lVert y-Az\rVert^2
= (A^{H}A)^{-1}A^{H}y. \tag*{(12)} z ^ ≜ [ x ^ h ^ ] = z arg min ∥ y − A z ∥ 2 = ( A H A ) − 1 A H y . (12) However, in general A A A 4 A A A x x x A A A A A A T T T A A A T = 1 T=1 T = 1
The rest of the paper is structured as follows. Section IV is concerned with the existence of a unique solution to (11). In Section V we analyze the statistics of the unique solution (12), and develop remedies for the rank-deficient case. Finally, in Section VII, we evaluate our proposed methods numerically.
Uniqueness of the Solution to (11)
Finding conditions for uniqueness of the solution to (11) is vital. If the least-squares problem lacks a unique solution, the estimation problem is unidentifiable and infinitely many sets of data matrices { X t } \{X_t\} { X t } H H H
The solution (12) is unique if A A A A A A
Assumption 2. The coherence interval length τ \tau τ A A A
T M τ ≥ T K τ + M N ⟺ τ ≥ M N T ( M − K ) . (13) TM\tau\geq TK\tau+MN
\quad\Longleftrightarrow\quad
\tau\geq\frac{MN}{T(M-K)}. \tag*{(13)} T M τ ≥ T K τ + M N ⟺ τ ≥ T ( M − K ) M N . (13) This is a reasonable assumption which will be fulfilled in practice; see Section VI-D.
However, A A A A A A A A A A A A
Analyzing the Rank of A
Our further analysis will rely on one critical assumption.
Assumption 3. All elements of { G t } \{G_t\} { G t } { V t } \{V_t\} { V t } { S t } \{S_t\} { S t }
From Assumption 3, it follows that all { G t } \{\mathbf{G}_t\} { G t } { V t } \{\mathbf{V}_t\} { V t } { S t } \{\mathbf{S}_t\} { S t } { V t S t } \{\mathbf{V}_t\mathbf{S}_t\} { V t S t }
rank ( V t S t ) = min { rank ( V t ) , rank ( S t ) } = J . (14) \operatorname{rank}(\mathbf{V}_t\mathbf{S}_t)=\min\{\operatorname{rank}(\mathbf{V}_t),\operatorname{rank}(\mathbf{S}_t)\}=J. \tag*{(14)} rank ( V t S t ) = min { rank ( V t ) , rank ( S t )} = J . (14) Next, we calculate a lower bound on the rank of A \mathbf{A} A T T T A \mathbf{A} A
Theorem 1. *The rank of A \mathbf{A} A 5
rank ( A ) = T K τ + R , (15) \operatorname{rank}(\mathbf{A})=TK\tau+R, \tag*{(15)} rank ( A ) = T K τ + R , (15) for a non-negative constant R ≤ M N R\le MN R ≤ M N N ( A ) \mathcal{N}(\mathbf{A}) N ( A ) A \mathbf{A} A
dim ( N ( A ) ) = T K τ + M N − rank ( A ) = M N − R . (16) \dim(\mathcal{N}(\mathbf{A}))=TK\tau+MN-\operatorname{rank}(\mathbf{A})=MN-R. \tag*{(16)} dim ( N ( A )) = T K τ + M N − rank ( A ) = M N − R . (16) Furthermore, for the matrix A \mathbf{A} A
T ≥ M N ( M − K ) J . (17) T\ge\frac{MN}{(M-K)J}. \tag*{(17)} T ≥ ( M − K ) J M N . (17) Proof. See Appendix A.
Note that (17) is a necessary but not sufficient condition for full-rank A \mathbf{A} A A \mathbf{A} A
Henceforth, we will limit our analysis to two cases:
Case 1 (M = N M=N M = N K = J K=J K = J T T T T T T A \mathbf{A} A
Case 2 (T = 1 T=1 T = 1 M M M N N N K K K J J J T = 1 T=1 T = 1 A \mathbf{A} A
Case 1 (M = N M=N M = N K = J K=J K = J T T T
Calculating the exact rank of A \mathbf{A} A M = N M=N M = N K = J K=J K = J
Theorem 2. When M = N = 2 K = 2 J M=N=2K=2J M = N = 2 K = 2 J A \mathbf{A} A
T ≥ 4. (18) T\ge4. \tag*{(18)} T ≥ 4. (18) Proof. See Appendix B.
Note that the condition (18) is consistent with (17) by inserting M = N = 2 K = 2 J M=N=2K=2J M = N = 2 K = 2 J M = N M=N M = N K = J K=J K = J
Conjecture 1. With M = N M=N M = N K = J K=J K = J A \mathbf{A} A
T ≥ M 2 ( M − K ) K . (19) T\ge\frac{M^2}{(M-K)K}. \tag*{(19)} T ≥ ( M − K ) K M 2 . (19) The interpretation of Conjecture 1 is that A \mathbf{A} A T T T
Case 2 (T = 1 T=1 T = 1 M M M N N N K K K J J J
With T = 1 T=1 T = 1 A \mathbf{A} A
A = [ I τ ⊗ G 1 S 1 T V 1 T ⊗ I M ] ∈ C M τ × ( K τ + M N ) . (20) \mathbf{A}=\left[\mathbf{I}_{\tau}\otimes\mathbf{G}_1\mathbf{S}_1^{T}\mathbf{V}_1^{T}\otimes\mathbf{I}_M\right]\in\mathbb{C}^{M\tau\times(K\tau+MN)}. \tag*{(20)} A = [ I τ ⊗ G 1 S 1 T V 1 T ⊗ I M ] ∈ C M τ × ( K τ + M N ) . (20) We derive a closed-form expression for the rank of A \mathbf{A} A
Theorem 3. With T = 1 T=1 T = 1 A \mathbf{A} A (20) is:
rank ( A ) = K τ + ( M − K ) J . (21) \operatorname{rank}(\mathbf{A})=K\tau+(M-K)J. \tag*{(21)} rank ( A ) = K τ + ( M − K ) J . (21) Hence, since A \mathbf{A} A
dim ( N ( A ) ) = K τ + M N − rank ( A ) = M ( N − J ) + K J . (22) \dim(\mathcal{N}(\mathbf{A}))=K\tau+MN-\operatorname{rank}(\mathbf{A})
=M(N-J)+KJ. \tag*{(22)} dim ( N ( A )) = K τ + M N − rank ( A ) = M ( N − J ) + K J . (22) Proof. See Appendix C.
Theorem 3 states that A \mathbf{A} A never has full rank when T = 1 T=1 T = 1 T = 1 T=1 T = 1
Joint Data Detection and AAI Mitigation
In this section, we present methods for joint UL data detection and AAI mitigation. First, we calculate the covariance matrix of x ^ \hat{\mathbf{x}} x ^ A \mathbf{A} A A \mathbf{A} A
Throughout, we assume that the elements of all W t \mathbf{W}_t W t C N ( 0 , 1 ) \mathcal{CN}(0,1) C N ( 0 , 1 ) H \mathbf{H} H
Proposed Method for Full-Rank A \mathbf{A} A
When A \mathbf{A} A { x ^ t } \{\hat{\mathbf{x}}_t\} { x ^ t }
Proposition 1. The covariance matrix Cov ( x ^ t ) \operatorname{Cov}(\hat{\mathbf{x}}_t) Cov ( x ^ t ) x ^ t \hat{\mathbf{x}}_t x ^ t t t t t t t K τ × K τ K\tau\times K\tau K τ × K τ ( A H A ) − 1 (\mathbf{A}^{H}\mathbf{A})^{-1} ( A H A ) − 1 x ^ \hat{\mathbf{x}} x ^
Proof. See Appendix D.
Note that the covariance matrix (23) contains two terms. The first term arises due to the receiver noise at the UL AP. The second term comes from the imperfect AAI mitigation and depends on the covariance matrix of the AP-AP channel estimate in (24). This is intuitive; a good AP-AP channel estimate has a small covariance matrix, which reduces the uncertainty in the data estimates.
Proposed Method for Rank-Deficient A \mathbf{A} A T = 1 T=1 T = 1
As alluded to in Section III, we can reduce the dimension of the nullspace of A \mathbf{A} A T = 1 T=1 T = 1 6
For simplicity, we drop the index t t t A \mathbf{A} A effective AP-AP channel F ≜ H V ∈ C M × J \mathbf{F}\triangleq\mathbf{H}\mathbf{V}\in\mathbb{C}^{M\times J} F ≜ HV ∈ C M × J H \mathbf{H} H
Cov ( x ^ t ) = I τ ⊗ ( G t H G t ) − 1 + ( S t T V t T ⊗ ( G t H G t ) − 1 G t H ) E ( V t ∗ S t ∗ ⊗ G t ( G t H G t ) − 1 ) , (23) \operatorname{Cov}(\hat{\mathbf{x}}_t)=\mathbf{I}_{\tau}\otimes(\mathbf{G}_t^{H}\mathbf{G}_t)^{-1}+(\mathbf{S}_t^{T}\mathbf{V}_t^{T}\otimes(\mathbf{G}_t^{H}\mathbf{G}_t)^{-1}\mathbf{G}_t^{H})\operatorname{E}(\mathbf{V}_t^{*}\mathbf{S}_t^{*}\otimes\mathbf{G}_t(\mathbf{G}_t^{H}\mathbf{G}_t)^{-1}), \tag*{(23)} Cov ( x ^ t ) = I τ ⊗ ( G t H G t ) − 1 + ( S t T V t T ⊗ ( G t H G t ) − 1 G t H ) E ( V t ∗ S t ∗ ⊗ G t ( G t H G t ) − 1 ) , (23) E ≜ Cov ( h ^ ) = ( ∑ t ∈ T V t ∗ S t ∗ S t T V t T ⊗ ( I M − G t ( G t H G t ) − 1 G t H ) ) − 1 . (24) \mathbf{E}\triangleq\operatorname{Cov}(\hat{\mathbf{h}})=\left(\sum_{t\in\mathcal{T}}\mathbf{V}_t^{*}\mathbf{S}_t^{*}\mathbf{S}_t^{T}\mathbf{V}_t^{T}\otimes(\mathbf{I}_{M}-\mathbf{G}_t(\mathbf{G}_t^{H}\mathbf{G}_t)^{-1}\mathbf{G}_t^{H})\right)^{-1}. \tag*{(24)} E ≜ Cov ( h ^ ) = ( t ∈ T ∑ V t ∗ S t ∗ S t T V t T ⊗ ( I M − G t ( G t H G t ) − 1 G t H ) ) − 1 . (24) reduces the number of unknown channel parameters from M N MN M N M J MJ M J A \mathbf{A} A
Y = G X + H V S + W = G X + F S + W , (25) \mathbf{Y}=\mathbf{G}\mathbf{X}+\mathbf{H}\mathbf{V}\mathbf{S}+\mathbf{W}=\mathbf{G}\mathbf{X}+\mathbf{F}\mathbf{S}+\mathbf{W}, \tag*{(25)} Y = GX + HVS + W = GX + FS + W , (25) and the corresponding least-squares problem becomes
min X , F ∥ Y − G X − F S ∥ 2 . (26) \min_{\mathbf{X},\mathbf{F}}\left\|\mathbf{Y}-\mathbf{G}\mathbf{X}-\mathbf{F}\mathbf{S}\right\|^{2}. \tag*{(26)} X , F min ∥ Y − GX − FS ∥ 2 . (26) Define f ≜ vec ( F ) \mathbf{f}\triangleq\operatorname{vec}(\mathbf{F}) f ≜ vec ( F )
y = A ′ z ′ + w , (27) \mathbf{y}=\mathbf{A}'\mathbf{z}'+\mathbf{w}, \tag*{(27)} y = A ′ z ′ + w , (27) where A ′ ∈ C M τ × ( K τ + M J ) \mathbf{A}'\in\mathbb{C}^{M\tau\times(K\tau+MJ)} A ′ ∈ C M τ × ( K τ + M J ) z ′ ∈ C ( K τ + M J ) × 1 \mathbf{z}'\in\mathbb{C}^{(K\tau+MJ)\times1} z ′ ∈ C ( K τ + M J ) × 1
A ′ = [ I τ ⊗ G S T ⊗ I M ] , z ′ = [ x f ] . (28) \mathbf{A}'=\left[\mathbf{I}_{\tau}\otimes\mathbf{G}\quad\mathbf{S}^{T}\otimes\mathbf{I}_{M}\right],\qquad
\mathbf{z}'=\begin{bmatrix}\mathbf{x}\\\mathbf{f}\end{bmatrix}. \tag*{(28)} A ′ = [ I τ ⊗ G S T ⊗ I M ] , z ′ = [ x f ] . (28) By following the same idea as in the proof of Theorem 3 it follows immediately that dim ( N ( A ′ ) ) = K J \dim(\mathcal{N}(\mathbf{A}'))=KJ dim ( N ( A ′ )) = K J V \mathbf{V} V F \mathbf{F} F M ( N − J ) + K J M(N-J)+KJ M ( N − J ) + K J K J KJ K J
We will show that (26) becomes uniquely solvable if the K K K J J J J J J
X = [ 0 K × J X ˉ ] , (29) \mathbf{X}=\begin{bmatrix}\mathbf{0}_{K\times J}&\bar{\mathbf{X}}\end{bmatrix}, \tag*{(29)} X = [ 0 K × J X ˉ ] , (29) where X ˉ ∈ C K × ( τ − J ) \bar{\mathbf{X}}\in\mathbb{C}^{K\times(\tau-J)} X ˉ ∈ C K × ( τ − J ) τ − J \tau-J τ − J
Y = G [ 0 K × J X ˉ ] + F S + W . (30) \mathbf{Y}=\mathbf{G}\begin{bmatrix}\mathbf{0}_{K\times J}&\bar{\mathbf{X}}\end{bmatrix}+\mathbf{F}\mathbf{S}+\mathbf{W}. \tag*{(30)} Y = G [ 0 K × J X ˉ ] + FS + W . (30) Define x ˉ ≜ vec ( X ˉ ) \bar{\mathbf{x}}\triangleq\operatorname{vec}(\bar{\mathbf{X}}) x ˉ ≜ vec ( X ˉ )
y = [ I τ ⊗ G S T ⊗ I M ] [ 0 K J × 1 x ˉ f ] + w = A ˉ z ˉ + w , (31) \mathbf{y}=\left[\mathbf{I}_{\tau}\otimes\mathbf{G}\quad\mathbf{S}^{T}\otimes\mathbf{I}_{M}\right]
\begin{bmatrix}\mathbf{0}_{KJ\times1}\\\bar{\mathbf{x}}\\\mathbf{f}\end{bmatrix}
+\mathbf{w}=\bar{\mathbf{A}}\bar{\mathbf{z}}+\mathbf{w}, \tag*{(31)} y = [ I τ ⊗ G S T ⊗ I M ] 0 K J × 1 x ˉ f + w = A ˉ z ˉ + w , (31) where A ˉ ∈ C M τ × ( K ( τ − J ) + M J ) \bar{\mathbf{A}}\in\mathbb{C}^{M\tau\times(K(\tau-J)+MJ)} A ˉ ∈ C M τ × ( K ( τ − J ) + M J ) z ˉ ∈ C ( K ( τ − J ) + M J ) × 1 \bar{\mathbf{z}}\in\mathbb{C}^{(K(\tau-J)+MJ)\times1} z ˉ ∈ C ( K ( τ − J ) + M J ) × 1
A ˉ = [ 0 M J × K ( τ − J ) S T ⊗ I M I τ − J ⊗ G ] , z ˉ = [ x ˉ f ] . (32) \bar{\mathbf{A}}=
\left[
\begin{array}{c|c}
\mathbf{0}_{MJ\times K(\tau-J)} & \mathbf{S}^{T}\otimes\mathbf{I}_{M}\\
\mathbf{I}_{\tau-J}\otimes\mathbf{G} &
\end{array}
\right],\qquad
\bar{\mathbf{z}}=\begin{bmatrix}\bar{\mathbf{x}}\\\mathbf{f}\end{bmatrix}. \tag*{(32)} A ˉ = [ 0 M J × K ( τ − J ) I τ − J ⊗ G S T ⊗ I M ] , z ˉ = [ x ˉ f ] . (32) Note that A ˉ \bar{\mathbf{A}} A ˉ K J KJ K J A ′ \mathbf{A}' A ′ A ˉ \bar{\mathbf{A}} A ˉ
Theorem 4. The matrix A ˉ \bar{\mathbf{A}} A ˉ
rank ( A ˉ ) = K ( τ − J ) + M J . (33) \operatorname{rank}(\bar{\mathbf{A}})=K(\tau-J)+MJ. \tag*{(33)} rank ( A ˉ ) = K ( τ − J ) + M J . (33) Proof. See Appendix E.
We present two algorithms to estimate x ˉ \bar{\mathbf{x}} x ˉ y \mathbf{y} y
1) Algorithm 1 (Joint Estimation): Here, in the same spirit as the joint estimation in (12), we directly obtain x ˉ ^ \hat{\bar{\mathbf{x}}} x ˉ ^
z ˉ ^ ≜ [ x ˉ ^ f ^ ] = argmin z ˉ ∥ y − A ˉ z ˉ ∥ 2 = ( A ˉ H A ˉ ) − 1 A ˉ H y . (34) \hat{\bar{\mathbf{z}}}\triangleq
\begin{bmatrix}\hat{\bar{\mathbf{x}}}\\\hat{\mathbf{f}}\end{bmatrix}
=\mathop{\operatorname{argmin}}_{\bar{\mathbf{z}}}\left\|\mathbf{y}-\bar{\mathbf{A}}\bar{\mathbf{z}}\right\|^{2}
=(\bar{\mathbf{A}}^{H}\bar{\mathbf{A}})^{-1}\bar{\mathbf{A}}^{H}\mathbf{y}. \tag*{(34)} z ˉ ^ ≜ [ x ˉ ^ f ^ ] = argmin z ˉ y − A ˉ z ˉ 2 = ( A ˉ H A ˉ ) − 1 A ˉ H y . (34) We calculate the covariance matrix of the data estimate x ˉ ^ \hat{\bar{\mathbf{x}}} x ˉ ^
Proposition 2. The covariance matrix Cov ( x ˉ ^ ) \operatorname{Cov}(\hat{\bar{\mathbf{x}}}) Cov ( x ˉ ^ ) K ( τ − J ) × K ( τ − J ) K(\tau-J)\times K(\tau-J) K ( τ − J ) × K ( τ − J ) ( A ˉ H A ˉ ) − 1 (\bar{\mathbf{A}}^{H}\bar{\mathbf{A}})^{-1} ( A ˉ H A ˉ ) − 1 z ˉ ^ \hat{\bar{\mathbf{z}}} z ˉ ^ Cov ( x ˉ ^ ) \operatorname{Cov}(\hat{\bar{\mathbf{x}}}) Cov ( x ˉ ^ ) S ˉ \bar{\mathbf{S}} S ˉ J J J S \mathbf{S} S
Proof. The derivation follows the same steps as the proof of Proposition 1 and is thus omitted.
2) Algorithm 2 (Channel Estimation Followed by Data Detection): In Algorithm 1, the first J J J F \mathbf{F} F J J J J J J Y \mathbf{Y} Y
Y 1 : J ≜ G [ 0 K × J X ˉ ] 1 : J + F S 1 : J + W 1 : J = F S 1 : J + W 1 : J . (35) \mathbf{Y}_{1:J}\triangleq\mathbf{G}\left[\mathbf{0}_{K\times J}\ \bar{\mathbf{X}}\right]_{1:J}+\mathbf{F}\mathbf{S}_{1:J}+\mathbf{W}_{1:J}
=\mathbf{F}\mathbf{S}_{1:J}+\mathbf{W}_{1:J}. \tag*{(35)} Y 1 : J ≜ G [ 0 K × J X ˉ ] 1 : J + F S 1 : J + W 1 : J = F S 1 : J + W 1 : J . (35) where ( ⋅ ) 1 : J (\cdot)_{1:J} ( ⋅ ) 1 : J J J J F \mathbf{F} F
F ^ = Y 1 : J S 1 : J H ( S 1 : J S 1 : J H ) − 1 . (36) \hat{\mathbf{F}}=\mathbf{Y}_{1:J}\mathbf{S}_{1:J}^{H}(\mathbf{S}_{1:J}\mathbf{S}_{1:J}^{H})^{-1}. \tag*{(36)} F ^ = Y 1 : J S 1 : J H ( S 1 : J S 1 : J H ) − 1 . (36) Now partition Y \mathbf{Y} Y Y = [ Y 1 : J Y ˉ ] \mathbf{Y}=[\mathbf{Y}_{1:J}\ \bar{\mathbf{Y}}] Y = [ Y 1 : J Y ˉ ] Y ˉ \bar{\mathbf{Y}} Y ˉ F ^ S ˉ \hat{\mathbf{F}}\bar{\mathbf{S}} F ^ S ˉ
Y ˉ − F ^ S ˉ = G X ˉ + ( F − F ^ ) S ˉ + W ˉ . (37) \bar{\mathbf{Y}}-\hat{\mathbf{F}}\bar{\mathbf{S}}=\mathbf{G}\bar{\mathbf{X}}+(\mathbf{F}-\hat{\mathbf{F}})\bar{\mathbf{S}}+\bar{\mathbf{W}}. \tag*{(37)} Y ˉ − F ^ S ˉ = G X ˉ + ( F − F ^ ) S ˉ + W ˉ . (37) The least-squares estimate of X ˉ \bar{\mathbf{X}} X ˉ
X ˉ ^ = ( G H G ) − 1 G H ( Y ˉ − F ^ S ˉ ) , (38) \hat{\bar{\mathbf{X}}}=(\mathbf{G}^{H}\mathbf{G})^{-1}\mathbf{G}^{H}(\bar{\mathbf{Y}}-\hat{\mathbf{F}}\bar{\mathbf{S}}), \tag*{(38)} X ˉ ^ = ( G H G ) − 1 G H ( Y ˉ − F ^ S ˉ ) , (38) and on vector form, x ˉ ^ = vec ( X ˉ ^ ) \hat{\bar{\mathbf{x}}}=\operatorname{vec}(\hat{\bar{\mathbf{X}}}) x ˉ ^ = vec ( X ˉ ^ )
Despite the seemingly different nature of the two algorithms (joint estimation respectively channel estimation followed by data detection), we show that they have the same solution.
Theorem 5. The solutions x ˉ ^ \hat{\bar{\mathbf{x}}} x ˉ ^
Proof. See Appendix F.
The Proposed Methods are Independent of the AAI Power
Notably, the proposed methods (both for the full-rank case and the rank-deficient case) are independent of the DL SNR, i.e., they are independent of the interference power.
We can observe this by looking at the covariance matrices of the data estimates in (23) and (35). The first term in the covariance matrices depends only on { G t } \{\mathbf{G}_t\} { G t } { S t } \{\mathbf{S}_t\} { S t } E \mathbf{E} E { S t } \{\mathbf{S}_t\} { S t } S t = ρ d S ˉ t \mathbf{S}_t=\sqrt{\rho_d}\bar{\mathbf{S}}_t S t = ρ d S ˉ t ρ d \rho_d ρ d
Cov ( x ^ ) = I τ − J ⊗ ( G H G ) − 1 + ( S ˉ T ⊗ ( G H G ) − 1 G H ) E ˉ ( S ˉ ∗ ⊗ G ( G H G ) − 1 ) (39) \operatorname{Cov}(\hat{\mathbf{x}})=\mathbf{I}_{\tau-J}\otimes(\mathbf{G}^{H}\mathbf{G})^{-1}+(\bar{\mathbf{S}}^{T}\otimes(\mathbf{G}^{H}\mathbf{G})^{-1}\mathbf{G}^{H})\bar{\mathbf{E}}(\bar{\mathbf{S}}^{*}\otimes\mathbf{G}(\mathbf{G}^{H}\mathbf{G})^{-1}) \tag*{(39)} Cov ( x ^ ) = I τ − J ⊗ ( G H G ) − 1 + ( S ˉ T ⊗ ( G H G ) − 1 G H ) E ˉ ( S ˉ ∗ ⊗ G ( G H G ) − 1 ) (39) E ˉ ≜ Cov ( f ^ ) = ( ( S ˉ ∗ S ˉ T ⊗ I M ) − ( S ˉ ∗ S ˉ T ⊗ G ( G H G ) − 1 G H ) ) − 1 . (40) \bar{\mathbf{E}}\triangleq\operatorname{Cov}(\hat{\mathbf{f}})=\left((\bar{\mathbf{S}}^{*}\bar{\mathbf{S}}^{T}\otimes\mathbf{I}_{M})-(\bar{\mathbf{S}}^{*}\bar{\mathbf{S}}^{T}\otimes\mathbf{G}(\mathbf{G}^{H}\mathbf{G})^{-1}\mathbf{G}^{H})\right)^{-1}. \tag*{(40)} E ˉ ≜ Cov ( f ^ ) = ( ( S ˉ ∗ S ˉ T ⊗ I M ) − ( S ˉ ∗ S ˉ T ⊗ G ( G H G ) − 1 G H ) ) − 1 . (40) SNR and the elements of S ~ t \widetilde{\mathbf{S}}_{t} S t E = E ~ / ρ d \mathbf{E}=\widetilde{\mathbf{E}}/\rho_{d} E = E / ρ d
E ~ = ( ∑ t ∈ T V t ∗ S ~ t ∗ S ~ t T V t T ⊗ ( I M − G t ( G t H G t ) − 1 G t H ) ) − 1 . \widetilde{\mathbf{E}}=\left(\sum_{t\in\mathcal{T}}\mathbf{V}_{t}^{*}\widetilde{\mathbf{S}}_{t}^{*}\widetilde{\mathbf{S}}_{t}^{T}\mathbf{V}_{t}^{T}\otimes(\mathbf{I}_{M}-\mathbf{G}_{t}(\mathbf{G}_{t}^{H}\mathbf{G}_{t})^{-1}\mathbf{G}_{t}^{H})\right)^{-1}. E = ( t ∈ T ∑ V t ∗ S t ∗ S t T V t T ⊗ ( I M − G t ( G t H G t ) − 1 G t H ) ) − 1 . Now consider the second term of (23) :
( S ~ t T V t T ⊗ ( G t H G t ) − 1 G t H ) E ( V t ∗ S ~ t ∗ ⊗ G t ( G t H G t ) − 1 ) = ( ρ d S ~ t T V t T ⊗ ( G t H G t ) − 1 G t H ) E ~ / ρ d \begin{aligned}
(\widetilde{\mathbf{S}}_{t}^{T}\mathbf{V}_{t}^{T}\otimes(\mathbf{G}_{t}^{H}\mathbf{G}_{t})^{-1}\mathbf{G}_{t}^{H})\mathbf{E}(\mathbf{V}_{t}^{*}\widetilde{\mathbf{S}}_{t}^{*}\otimes\mathbf{G}_{t}(\mathbf{G}_{t}^{H}\mathbf{G}_{t})^{-1}) \\
&=(\sqrt{\rho_{d}}\widetilde{\mathbf{S}}_{t}^{T}\mathbf{V}_{t}^{T}\otimes(\mathbf{G}_{t}^{H}\mathbf{G}_{t})^{-1}\mathbf{G}_{t}^{H})\widetilde{\mathbf{E}}/\rho_{d}
\end{aligned} ( S t T V t T ⊗ ( G t H G t ) − 1 G t H ) E ( V t ∗ S t ∗ ⊗ G t ( G t H G t ) − 1 ) = ( ρ d S t T V t T ⊗ ( G t H G t ) − 1 G t H ) E / ρ d × ( V t ∗ ρ d S ~ t ∗ ⊗ G t ( G t H G t ) − 1 ) = ( S ~ t T V t T ⊗ ( G t H G t ) − 1 G t H ) E ~ ( V t ∗ S ~ t ∗ ⊗ G t ( G t H G t ) − 1 ) , (41) \begin{aligned}
\qquad\times(\mathbf{V}_{t}^{*}\sqrt{\rho_{d}}\widetilde{\mathbf{S}}_{t}^{*}\otimes\mathbf{G}_{t}(\mathbf{G}_{t}^{H}\mathbf{G}_{t})^{-1}) \\
&=(\widetilde{\mathbf{S}}_{t}^{T}\mathbf{V}_{t}^{T}\otimes(\mathbf{G}_{t}^{H}\mathbf{G}_{t})^{-1}\mathbf{G}_{t}^{H})\widetilde{\mathbf{E}}(\mathbf{V}_{t}^{*}\widetilde{\mathbf{S}}_{t}^{*}\otimes\mathbf{G}_{t}(\mathbf{G}_{t}^{H}\mathbf{G}_{t})^{-1}), \tag*{(41)}
\end{aligned} × ( V t ∗ ρ d S t ∗ ⊗ G t ( G t H G t ) − 1 ) = ( S t T V t T ⊗ ( G t H G t ) − 1 G t H ) E ( V t ∗ S t ∗ ⊗ G t ( G t H G t ) − 1 ) , (41) where the final expression is independent of ρ d \rho_{d} ρ d (23) , and hence the full covariance matrix, is independent of the DL SNR. (The motivation for (35) is analogous; we omit the details for brevity.) The intuition is that the channel estimate h ^ \hat{\mathbf{h}} h ^ f ^ \hat{\mathbf{f}} f ^
Baseline Methods
We present two baselines to compare our proposed methods against: a genie-aided detector with perfect AAI mitigation, and a worst-case detector without AAI mitigation.
Genie-Aided Baseline (Perfect AAI Mitigation)
We assume that the AP-AP channel H \mathbf{H} H
Y ˙ t ≜ Y t − H V t S t = G t X t + W t ∈ C M × τ . (42) \dot{\mathbf{Y}}_{t}\triangleq\mathbf{Y}_{t}-\mathbf{H}\mathbf{V}_{t}\mathbf{S}_{t}=\mathbf{G}_{t}\mathbf{X}_{t}+\mathbf{W}_{t}\in\mathbb{C}^{M\times\tau}. \tag*{(42)} Y ˙ t ≜ Y t − H V t S t = G t X t + W t ∈ C M × τ . (42) Let y ˙ t = vec ( Y ˙ t ) ∀ t ∈ T \dot{\mathbf{y}}_{t}=\operatorname{vec}(\dot{\mathbf{Y}}_{t})\ \forall t\in\mathcal{T} y ˙ t = vec ( Y ˙ t ) ∀ t ∈ T
y ˙ ≜ [ y ˙ 1 ⋮ y ˙ T ] = A ˙ x + w ∈ C T M τ × 1 , (43) \dot{\mathbf{y}}\triangleq
\begin{bmatrix}
\dot{\mathbf{y}}_{1}\\
\vdots\\
\dot{\mathbf{y}}_{T}
\end{bmatrix}
=\dot{\mathbf{A}}\mathbf{x}+\mathbf{w}\in\mathbb{C}^{TM\tau\times1}, \tag*{(43)} y ˙ ≜ y ˙ 1 ⋮ y ˙ T = A ˙ x + w ∈ C T M τ × 1 , (43) where A ˙ ∈ C T M τ × T K τ \dot{\mathbf{A}}\in\mathbb{C}^{TM\tau\times TK\tau} A ˙ ∈ C T M τ × T K τ
A ˙ = [ I τ ⊗ G 1 0 ⋮ ⋱ 0 I τ ⊗ G T ] . (44) \dot{\mathbf{A}}=
\begin{bmatrix}
\mathbf{I}_{\tau}\otimes\mathbf{G}_{1} & \mathbf{0}\\
\vdots& \ddots\\
\mathbf{0} & \mathbf{I}_{\tau}\otimes\mathbf{G}_{T}
\end{bmatrix}. \tag*{(44)} A ˙ = I τ ⊗ G 1 ⋮ 0 0 ⋱ I τ ⊗ G T . (44) The matrix A ˙ \dot{\mathbf{A}} A ˙ x \mathbf{x} x (43) is
x ^ g ≜ arg min x ∥ y ˙ − A ˙ x ∥ 2 = ( A ˙ H A ˙ ) − 1 A ˙ H y ˙ = ( A ˙ H A ˙ ) − 1 A ˙ H ( A ˙ x + w ) = x + w ˙ , (45) \begin{aligned}
\hat{\mathbf{x}}_{g}\triangleq\underset{\mathbf{x}}{\arg\min}\|\dot{\mathbf{y}}-\dot{\mathbf{A}}\mathbf{x}\|^{2}=(\dot{\mathbf{A}}^{H}\dot{\mathbf{A}})^{-1}\dot{\mathbf{A}}^{H}\dot{\mathbf{y}} \\
&=(\dot{\mathbf{A}}^{H}\dot{\mathbf{A}})^{-1}\dot{\mathbf{A}}^{H}(\dot{\mathbf{A}}\mathbf{x}+\mathbf{w})=\mathbf{x}+\dot{\mathbf{w}}, \tag*{(45)}
\end{aligned} x ^ g ≜ x arg min ∥ y ˙ − A ˙ x ∥ 2 = ( A ˙ H A ˙ ) − 1 A ˙ H y ˙ = ( A ˙ H A ˙ ) − 1 A ˙ H ( A ˙ x + w ) = x + w ˙ , (45) where
w ˙ ≜ ( A ˙ H A ˙ ) − 1 A ˙ H w ∼ C N ( 0 T K τ × 1 , ( A ˙ H A ˙ ) − 1 ) . (46) \dot{\mathbf{w}}\triangleq(\dot{\mathbf{A}}^{H}\dot{\mathbf{A}})^{-1}\dot{\mathbf{A}}^{H}\mathbf{w}\sim\mathcal{CN}(\mathbf{0}_{TK\tau\times1},(\dot{\mathbf{A}}^{H}\dot{\mathbf{A}})^{-1}). \tag*{(46)} w ˙ ≜ ( A ˙ H A ˙ ) − 1 A ˙ H w ∼ C N ( 0 T K τ × 1 , ( A ˙ H A ˙ ) − 1 ) . (46) The covariance of x ^ g \hat{\mathbf{x}}_{g} x ^ g Cov ( x ^ g ) = Cov ( w ˙ ) = ( A ˙ H A ˙ ) − 1 \operatorname{Cov}(\hat{\mathbf{x}}_{g})=\operatorname{Cov}(\dot{\mathbf{w}})=(\dot{\mathbf{A}}^{H}\dot{\mathbf{A}})^{-1} Cov ( x ^ g ) = Cov ( w ˙ ) = ( A ˙ H A ˙ ) − 1
Baseline Without AAI Mitigation
Here, we proceed as if we received the interference-free signal (43) , although we actually received the signal with interference in (8). The estimate with no AAI mitigation, x ^ n \hat{\mathbf{x}}_{n} x ^ n
x ^ n ≜ ( A ˙ H A ˙ ) − 1 A ˙ H y = ( A ˙ H A ˙ ) − 1 A ˙ H ( A z + w ) = ( A ˙ H A ˙ ) − 1 A ˙ H [ S 1 T V 1 T ⊗ I M ⋮ S T T V T T ⊗ I M ] [ x h ] + w ˙ = x + [ S 1 T V 1 T ⊗ ( G 1 H G 1 ) − 1 G 1 H ⋮ S T T V T T ⊗ ( G T H G T ) − 1 G T H ] ⏟ ≜ B h + w ˙ . (47) \begin{aligned}
\hat{\mathbf{x}}_{n}\triangleq(\dot{\mathbf{A}}^{H}\dot{\mathbf{A}})^{-1}\dot{\mathbf{A}}^{H}\mathbf{y}=(\dot{\mathbf{A}}^{H}\dot{\mathbf{A}})^{-1}\dot{\mathbf{A}}^{H}(\mathbf{A}\mathbf{z}+\mathbf{w}) \\
&=(\dot{\mathbf{A}}^{H}\dot{\mathbf{A}})^{-1}\dot{\mathbf{A}}^{H}
\begin{bmatrix}
\mathbf{S}_{1}^{T}\mathbf{V}_{1}^{T}\otimes\mathbf{I}_{M}\\
\vdots\\
\mathbf{S}_{T}^{T}\mathbf{V}_{T}^{T}\otimes\mathbf{I}_{M}
\end{bmatrix}
\begin{bmatrix}
\mathbf{x}\\
\mathbf{h}
\end{bmatrix}
+\dot{\mathbf{w}} \\
&=\mathbf{x}+
\underbrace{
\begin{bmatrix}
\mathbf{S}_{1}^{T}\mathbf{V}_{1}^{T}\otimes(\mathbf{G}_{1}^{H}\mathbf{G}_{1})^{-1}\mathbf{G}_{1}^{H}\\
\vdots\\
\mathbf{S}_{T}^{T}\mathbf{V}_{T}^{T}\otimes(\mathbf{G}_{T}^{H}\mathbf{G}_{T})^{-1}\mathbf{G}_{T}^{H}
\end{bmatrix}
}_{\triangleq\mathbf{B}}
\mathbf{h}+\dot{\mathbf{w}}. \tag*{(47)}
\end{aligned} x ^ n ≜ ( A ˙ H A ˙ ) − 1 A ˙ H y = ( A ˙ H A ˙ ) − 1 A ˙ H ( Az + w ) = ( A ˙ H A ˙ ) − 1 A ˙ H S 1 T V 1 T ⊗ I M ⋮ S T T V T T ⊗ I M [ x h ] + w ˙ = x + ≜ B S 1 T V 1 T ⊗ ( G 1 H G 1 ) − 1 G 1 H ⋮ S T T V T T ⊗ ( G T H G T ) − 1 G T H h + w ˙ . (47) This estimate has the same covariance as the genie-aided estimate, i.e. Cov ( x ^ n ) = Cov ( x ^ g ) \operatorname{Cov}(\hat{\mathbf{x}}_{n})=\operatorname{Cov}(\hat{\mathbf{x}}_{g}) Cov ( x ^ n ) = Cov ( x ^ g )
Bias ( x ^ n ) = B h , (48) \operatorname{Bias}(\hat{\mathbf{x}}_{n})=\mathbf{B}\mathbf{h}, \tag*{(48)} Bias ( x ^ n ) = Bh , (48) where B \mathbf{B} B (47) above.
Practical Considerations
In this section, we will discuss some practical aspects of our models and assumptions.
Feasible Values of T T T
In general, for a wireless channel, the coherence interval length is given by T c B c T_{c}B_{c} T c B c T c T_{c} T c B c B_{c} B c B c B_{c} B c T c T_{c} T c
Let T c , U E − A P T_{c,\mathrm{UE-AP}} T c , UE − AP T c , A P − A P T_{c,\mathrm{AP-AP}} T c , AP − AP τ = T c , U E − A P B c \tau=T_{c,\mathrm{UE-AP}}B_{c} τ = T c , UE − AP B c T τ = T c , A P − A P B c T\tau=T_{c,\mathrm{AP-AP}}B_{c} T τ = T c , AP − AP B c T = T c , A P − A P / T c , U E − A P T=T_{c,\mathrm{AP-AP}}/T_{c,\mathrm{UE-AP}} T = T c , AP − AP / T c , UE − AP ? ], eq:2.3 the coherence time is T c = λ / ( 2 ν ) T_{c}=\lambda/(2\nu) T c = λ / ( 2 ν ) λ \lambda λ ν \nu ν λ = 0.15 \lambda=0.15 λ = 0.15 T c , U E − A P ≈ 90 T_{c,\mathrm{UE-AP}}\approx90 T c , UE − AP ≈ 90 T c , U E − A P ≈ 4.5 T_{c,\mathrm{UE-AP}}\approx4.5 T c , UE − AP ≈ 4.5 T c , A P − A P ≈ 90 T_{c,\mathrm{AP-AP}}\approx90 T c , AP − AP ≈ 90 T c , A P − A P / T c , U E − A P = T ∈ { 1 , … , 20 } T_{c,\mathrm{AP-AP}}/T_{c,\mathrm{UE-AP}}=T\in\{1,\ldots,20\} T c , AP − AP / T c , UE − AP = T ∈ { 1 , … , 20 } T T T
In Table 1 , we investigate whether these values of T T T A \mathbf{A} A T T T T ≥ 10 T\ge10 T ≥ 10 A \mathbf{A} A T T T
Table 1. The necessary values of T T T τ \tau τ A \mathbf{A} A
Latency and Memory Growth
When jointly processing the UL data over T T T T c , A P − A P T_{c,\mathrm{AP-AP}} T c , AP − AP T T T T T T T = 1 T=1 T = 1
Further, the memory demand for storing { Y t } \{\mathbf{Y}_t\} { Y t } { G t } \{\mathbf{G}_t\} { G t } { V t S t } \{\mathbf{V}_t\mathbf{S}_t\} { V t S t } T T T T T T T ( M τ + M K + N τ ) T(M\tau+MK+N\tau) T ( M τ + M K + N τ )
Perfectly Known { G t } \{\mathbf{G}_t\} { G t } { S t } \{\mathbf{S}_t\} { S t } { V t } \{\mathbf{V}_t\} { V t }
In practice, the UE-AP channels { G t } \{\mathbf{G}_t\} { G t } K K K K / τ K/\tau K / τ { G ^ t } \{\hat{\mathbf{G}}_t\} { G ^ t } { G t } \{\mathbf{G}_t\} { G t } A \mathbf{A} A A ˉ \bar{\mathbf{A}} A ˉ A ˙ \dot{\mathbf{A}} A ˙
Further, the assumption of known { V t } \{\mathbf{V}_t\} { V t } { S t } \{\mathbf{S}_t\} { S t } N J + J τ NJ+J\tau N J + J τ { V t S t } \{\mathbf{V}_t\mathbf{S}_t\} { V t S t } N τ N\tau N τ T T T T ⋅ min { J ( N + τ ) , N τ } T\cdot\min\{J(N+\tau),N\tau\} T ⋅ min { J ( N + τ ) , N τ }
For AP-AP transfer, the transport delay (to transfer { V t S t } \{\mathbf{V}_t\mathbf{S}_t\} { V t S t } [31 ] , Table 6.1-1. In a centralized architecture, the transport delay can be reduced further. In any case, the delay is relatively small compared to the latency introduced by the joint processing over T T T T T T
How Large τ \tau τ
In Table 1 , we present the lower bound on τ \tau τ T T T A \mathbf{A} A M = N M=N M = N τ ≥ M 2 / T ( M − K ) \tau\ge M^2/T(M-K) τ ≥ M 2 / T ( M − K )
M 2 T ( M − K ) ≤ ( M − K ) K M 2 M 2 M − K = K . (49) \frac{M^2}{T(M-K)} \le\frac{(M-K)K}{M^2}\frac{M^2}{M-K}=K. \tag*{(49)} T ( M − K ) M 2 ≤ M 2 ( M − K ) K M − K M 2 = K . (49) That is, with M = N M=N M = N τ \tau τ K K K τ ≥ K \tau\ge K τ ≥ K
Computational Complexity
A naïve approach to calculate the least-squares solution in the full-rank case is to directly invert A H A \mathbf{A}^{H}\mathbf{A} A H A O ( ( T K τ + M N ) 3 ) \mathcal{O}((TK\tau+MN)^3) O (( T K τ + M N ) 3 ) T T T A \mathbf{A} A O ( T ⋅ ( ( M N ) 2 + ( M + N ) 2 τ ) + ( M N ) 3 ) \mathcal{O}(T\cdot((MN)^2+(M+N)^2\tau)+(MN)^3) O ( T ⋅ (( M N ) 2 + ( M + N ) 2 τ ) + ( M N ) 3 ) T T T O ( ( M N ) 3 ) \mathcal{O}((MN)^3) O (( M N ) 3 ) M N × M N MN\times MN M N × M N h \mathbf{h} h
Numerical Stability and Robustness
We solve the least-squares problem after a standard column rescaling to unit norm, which yields a well-conditioned system. Let D \mathbf{D} D A \mathbf{A} A
argmin z ∥ y − A z ∥ 2 = argmin z ∥ y − A D − 1 D z ∥ 2 , (50) \underset{\mathbf{z}}{\operatorname{argmin}}\ \lVert\mathbf{y}-\mathbf{A}\mathbf{z}\rVert^2
=\underset{\mathbf{z}}{\operatorname{argmin}}\ \lVert\mathbf{y}-\mathbf{A}\mathbf{D}^{-1}\mathbf{D}\mathbf{z}\rVert^2, \tag*{(50)} z argmin ∥ y − Az ∥ 2 = z argmin ∥ y − A D − 1 Dz ∥ 2 , (50) which can be solved for D z \mathbf{D}\mathbf{z} Dz A D − 1 \mathbf{A}\mathbf{D}^{-1} A D − 1 z ^ \hat{\mathbf{z}} z ^ D z \mathbf{D}\mathbf{z} Dz D − 1 \mathbf{D}^{-1} D − 1
Consider the setup used later in our numerical evaluations (see Section VII-A for details), with M = N = 32 M=N=32 M = N = 32 K = J = 16 K=J=16 K = J = 16 τ = 50 \tau=50 τ = 50 T = 10 T=10 T = 10 A D − 1 \mathbf{A}\mathbf{D}^{-1} A D − 1 ≈ 20 \approx20 ≈ 20
Further evaluations show that our methods are robust to scenarios where the strict independence in Assumption 3 is relaxed. Specifically, when the DL AP applies zero-forcing instead of maximum-ratio precoding (which breaks the independence of the elements in { V t } \{\mathbf{V}_t\} { V t } A \mathbf{A} A A D − 1 \mathbf{A}\mathbf{D}^{-1} A D − 1 ≈ 20 \approx20 ≈ 20 ≈ 30 \approx30 ≈ 30 { G t } \{\mathbf{G}_t\} { G t } A \mathbf{A} A 15 ∘ 15^\circ 1 5 ∘ [32 ] , the rescaled condition number increases to ≈ 100 \approx100 ≈ 100 0 ∘ 0^\circ 0 ∘ { G t } \{\mathbf{G}_t\} { G t } A \mathbf{A} A
These results indicate that the sufficiency condition in Theorem 2 may extend beyond Assumption 3, except in degenerate cases where A \mathbf{A} A 0 ∘ 0^\circ 0 ∘
The Block-Fading Model in our Context
The analysis in this paper, and specifically the signal model in (4), relies on a strict block-fading assumption where all channels are exactly static throughout each coherence interval. This requires the definition of “coherence” to be rather conservative, using a value of τ \tau τ
Consider the example in Section VI-A, with T c , A P − A P ≈ 90 T_{\mathrm{c,AP-AP}} \approx90 T c , AP − AP ≈ 90 T c , U E − A P ∈ [ 4.5 , … , 90 ] T_{\mathrm{c,UE-AP}} \in[4.5,\ldots,90] T c , UE − AP ∈ [ 4.5 , … , 90 ] B c ≈ 300 B_{\mathrm{c}} \approx300 B c ≈ 300 ? ], we get an UE-AP coherence interval length in the range from τ ≈ 1350 \tau\approx1350 τ ≈ 1350 τ ≈ 27000 \tau\approx27000 τ ≈ 27000 τ = 50 \tau= 50 τ = 50 τ \tau τ
However, it should be noted that in many other cases (e.g., ergodic capacity bounding [? ]) one can define larger coherence intervals since the channel responses can be interpolated between coherence intervals , which is not possible for our proposed algorithms.7 [33 ] .
Numerical Results
In this section, we evaluate our proposed methods numerically. First, we calculate the MSEs of the data symbol estimates achieved by our proposed methods, and compare them to those obtained by the two baselines (see Section V-D). Then, we derive achievable SEs for our methods and compare them to the SEs obtained by the baselines and a state-of-the-art zero-forcing based AAI mitigation technique.
Simulation Setup
We consider the setup from Figure 1 with M = N = 32 M=N=32 M = N = 32 K = J = 16 K=J=16 K = J = 16 τ = 50 \tau=50 τ = 50 ρ u \rho_u ρ u ρ d \rho_d ρ d
We assume i.i.d. Rayleigh fading channels and include the UL SNR in the UL channels, i.e., they are generated as
g k , t ∼ C N ( 0 M × 1 , ρ u η k , t β k , t I M ) ∀ k , t (51) \mathbf{g}_{k,t} \sim\mathcal{CN}(\mathbf{0}_{M\times1},\rho_u\eta_{k,t}\beta_{k,t}\mathbf{I}_M)\ \forall k,t \tag*{(51)} g k , t ∼ C N ( 0 M × 1 , ρ u η k , t β k , t I M ) ∀ k , t (51) h n ∼ C N ( 0 M × 1 , α I M ) ∀ n , (52) \mathbf{h}_n \sim\mathcal{CN}(\mathbf{0}_{M\times1},\alpha\mathbf{I}_M)\ \forall n, \tag*{(52)} h n ∼ C N ( 0 M × 1 , α I M ) ∀ n , (52) where β k , t \beta_{k,t} β k , t α \alpha α k k k { η k , t } \{\eta_{k,t}\} { η k , t }
V t = [ v 1 , t ⋯ v J , t ] (53) \mathbf{V}_t=[\mathbf{v}_{1,t}\ \cdots\ \mathbf{v}_{J,t}] \tag*{(53)} V t = [ v 1 , t ⋯ v J , t ] (53) v j , t ∼ C N ( 0 N × 1 , θ j , t γ j , t I N / Ξ ) , (54) \mathbf{v}_{j,t}\sim\mathcal{CN}(\mathbf{0}_{N\times1},\theta_{j,t}\gamma_{j,t}\mathbf{I}_N/\Xi), \tag*{(54)} v j , t ∼ C N ( 0 N × 1 , θ j , t γ j , t I N /Ξ ) , (54) where γ j , t \gamma_{j,t} γ j , t j j j t t t { θ j , t } \{\theta_{j,t}\} { θ j , t } Ξ \Xi Ξ E { Tr ( V t H V t ) } = 1 \mathbb{E}\{\operatorname{Tr}(\mathbf{V}_t^H\mathbf{V}_t)\}=1 E { Tr ( V t H V t )} = 1 { β k , t } \{\beta_{k,t}\} { β k , t } α \alpha α { γ j , t } \{\gamma_{j,t}\} { γ j , t } ? ], with elevations of 1 m and 10 m for the UEs and APs, respectively. The UL and DL power control coefficients are chosen as the ones achieving max-min fairness in a single-cell (static TDD) setup.
We let the DL data symbols, i.e. the elements of { s t } \{\mathbf{s}_t\} { s t } C N ( 0 , ρ d ) \mathcal{CN}(0,\rho_d) C N ( 0 , ρ d )
Evaluation of Mean-Square Errors
Here, we evaluate our proposed methods and compare them to the baselines in terms of their MSEs per data symbol.
1) Calculating MSEs: The MSE of an estimator x ^ \hat{\mathbf{x}} x ^ x \mathbf{x} x MSE ( x ^ ) = E { ∥ x ^ − x ∥ 2 } = Tr ( Cov ( x ^ ) ) + ∥ Bias ( x ^ ) ∥ 2 \operatorname{MSE}(\hat{\mathbf{x}})=\mathbb{E}\{\lVert\hat{\mathbf{x}}-\mathbf{x}\rVert^2\}=\operatorname{Tr}(\operatorname{Cov}(\hat{\mathbf{x}}))+\lVert\operatorname{Bias}(\hat{\mathbf{x}})\rVert^2 MSE ( x ^ ) = E {∥ x ^ − x ∥ 2 } = Tr ( Cov ( x ^ )) + ∥ Bias ( x ^ ) ∥ 2
The bias term is non-zero only in the baseline without AAI mitigation. Normalizing (55) by the dimension of x \mathbf{x} x { V t s t } \{\mathbf{V}_t\mathbf{s}_t\} { V t s t }
MSE ( x ^ g ) = 1 T K ∑ t ∈ T Tr ( ( G t H G t ) − 1 ) . (55) \operatorname{MSE}(\hat{\mathbf{x}}_{\mathrm{g}})=\frac{1}{TK}\sum_{t\in\mathcal{T}}\operatorname{Tr}\left((\mathbf{G}_t^H\mathbf{G}_t)^{-1}\right). \tag*{(55)} MSE ( x ^ g ) = T K 1 t ∈ T ∑ Tr ( ( G t H G t ) − 1 ) . (55) MSE ( x ^ n ) = MSE ( x ^ g ) + 1 T K τ h H ( ∑ t ∈ T V t ∗ S t ∗ S t T V t T ⊗ G t ( G t H G t ) − 2 G t H ) h . (56) \operatorname{MSE}(\hat{\mathbf{x}}_{\mathrm{n}})=\operatorname{MSE}(\hat{\mathbf{x}}_{\mathrm{g}})+\frac{1}{TK\tau}\mathbf{h}^H\left(\sum_{t\in\mathcal{T}}\mathbf{V}_t^*\mathbf{S}_t^*\mathbf{S}_t^T\mathbf{V}_t^T\otimes\mathbf{G}_t(\mathbf{G}_t^H\mathbf{G}_t)^{-2}\mathbf{G}_t^H\right)\mathbf{h}. \tag*{(56)} MSE ( x ^ n ) = MSE ( x ^ g ) + T K τ 1 h H ( t ∈ T ∑ V t ∗ S t ∗ S t T V t T ⊗ G t ( G t H G t ) − 2 G t H ) h . (56) MSE ( x ^ ) = MSE ( x ^ g ) + 1 T K T ∑ t ∈ T Tr ( ( V t ∗ S t ∗ S t T V t T ⊗ G t ( G t H G t ) − 2 G t H ) E ) . (57) \operatorname{MSE}(\hat{\mathbf{x}})=\operatorname{MSE}(\hat{\mathbf{x}}_{\mathrm{g}})+\frac{1}{T K_T}\sum_{t\in\mathcal{T}}\operatorname{Tr}\left(\left(\mathbf{V}_{t}^{*}\mathbf{S}_{t}^{*}\mathbf{S}_{t}^{T}\mathbf{V}_{t}^{T}\otimes\mathbf{G}_{t}(\mathbf{G}_{t}^{H}\mathbf{G}_{t})^{-2}\mathbf{G}_{t}^{H}\right)\mathbf{E}\right). \tag*{(57)} MSE ( x ^ ) = MSE ( x ^ g ) + T K T 1 t ∈ T ∑ Tr ( ( V t ∗ S t ∗ S t T V t T ⊗ G t ( G t H G t ) − 2 G t H ) E ) . (57) MSE ( x ^ ) = MSE ( x ^ g ) ∣ T = 1 + 1 K ( τ − J ) Tr ( ( S ∗ S T ⊗ G ( G H G ) − 2 G H ) E ‾ ) . (58) \operatorname{MSE}(\hat{\mathbf{x}})=\operatorname{MSE}(\hat{\mathbf{x}}_{\mathrm{g}})\big|_{T=1}+\frac{1}{K(\tau-J)}\operatorname{Tr}\left(\left(\mathbf{S}^{*}\mathbf{S}^{T}\otimes\mathbf{G}(\mathbf{G}^{H}\mathbf{G})^{-2}\mathbf{G}^{H}\right)\overline{\mathbf{E}}\right). \tag*{(58)} MSE ( x ^ ) = MSE ( x ^ g ) T = 1 + K ( τ − J ) 1 Tr ( ( S ∗ S T ⊗ G ( G H G ) − 2 G H ) E ) . (58) We observe that the last three MSEs are expressed in terms of the genie-aided baseline MSE plus an additional term arising due to imperfect AP-AP channel estimation. Further, the MSEs of our two proposed methods are independent of the DL SNR ρ d \rho_d ρ d ρ d \rho_d ρ d
Numerical Evaluations: We evaluate our proposed methods, both for full-rank and rank-deficient A A A Figure 3 presents the median MSEs obtained with each of the methods for varying T ∈ { 5 , … , 20 } T\in\{5,\ldots,20\} T ∈ { 5 , … , 20 }
Figure 3. MSEs of the proposed methods and the baselines for different T T T
We observe that the MSE of our proposed method for the full-rank case approaches that of the genie-aided baseline when T T T T = 5 T=5 T = 5 T > 10 T>10 T > 10
Our proposed method for the full-rank case heavily outperforms the baseline without AAI mitigation, even for small values of T T T T = 5 T=5 T = 5 T = 20 T=20 T = 20
Our proposed method for the rank-deficient case gives
an improvement of approximately 4 dB over the baseline without AAI mitigation. This observation indicates that our joint estimation approach in itself achieves a performance gain, but that it is vital to leverage the slow time-variation of the AP-AP channel to reach a performance similar to the genie-aided baseline.
Next, in Figure 4 , we present the median MSEs obtained with the proposed methods and baselines for varying UL and DL SNRs ρ u \rho_u ρ u ρ d \rho_d ρ d
Figure 4. Comparison between the proposed methods and the baselines with different values of the UL and DL SNRs.
When varying the UL SNR, all MSEs decrease linearly (in log-log scale) but the gaps remain equal. This behavior is expected as the UL SNR enters the MSE expressions through { G t } \{\mathbf{G}_t\} { G t }
As mentioned in Section VII-B1 above, the MSEs of our proposed methods are independent of the DL SNR. This behavior can also be observed from Figure 4 (b).
The baseline without AAI mitigation performs better than the proposed method for the rank-deficient case at low DL SNRs. This is because, for decreasing DL SNRs, the gain in mitigating the AAI becomes more limited. We also observe that the MSE without AAI mitigation approaches that of the genie-aided baseline at low DL SNRs, which is reasonable as there is virtually no gain in mitigating the AAI when its power is negligibly small.
Evaluation of Spectral Efficiencies
Next, we derive achievable UL SEs for our proposed methods and benchmark these against an existing AAI mitigation method with zero-forcing combining. The genie-aided method and the method without AAI mitigation will serve as baselines.
1) Proposed Method for Full-Rank A: The least-squares solution in (12) can be written as
z ^ = [ x h ] + ( A H A ) − 1 A H w . (59) \hat{\boldsymbol{z}} = \begin{bmatrix} \boldsymbol{x} \\ \boldsymbol{h} \end{bmatrix} + \left(\boldsymbol{A}^{H}\boldsymbol{A}\right)^{-1}\boldsymbol{A}^{H}\boldsymbol{w}. \tag*{(59)} z ^ = [ x h ] + ( A H A ) − 1 A H w . (59) Thus, for any symbol s ∈ { 1 , … , T K τ } s \in\{1,\ldots,TK\tau\} s ∈ { 1 , … , T K τ } x \boldsymbol{x} x
x ^ s = x s + a s H w , (60) \hat{x}_{s}=x_{s}+\boldsymbol{a}_{s}^{H}\boldsymbol{w}, \tag*{(60)} x ^ s = x s + a s H w , (60) where a s H \boldsymbol{a}_{s}^{H} a s H s s s ( A H A ) − 1 A H (\boldsymbol{A}^{H}\boldsymbol{A})^{-1}\boldsymbol{A}^{H} ( A H A ) − 1 A H { a s } \{\boldsymbol{a}_{s}\} { a s } { G t } \{\boldsymbol{G}_{t}\} { G t } { V t } \{\boldsymbol{V}_{t}\} { V t } { S t } \{\boldsymbol{S}_{t}\} { S t }
E { log 2 ( 1 + 1 Var { a s H w ∣ { G t } , { S t } , { V t } } ) } = E { log 2 ( 1 + 1 ∥ a s ∥ 2 ) } , (61) \begin{aligned}
\mathbb{E}\left\{\log_{2}\left(1+\frac{1}{\operatorname{Var}\{\boldsymbol{a}_{s}^{H}\boldsymbol{w}\mid\{\boldsymbol{G}_{t}\},\{\boldsymbol{S}_{t}\},\{\boldsymbol{V}_{t}\}\}}\right)\right\} \\
&=\mathbb{E}\left\{\log_{2}\left(1+\frac{1}{\|\boldsymbol{a}_{s}\|^{2}}\right)\right\}, \tag*{(61)}
\end{aligned} E { log 2 ( 1 + Var { a s H w ∣ { G t } , { S t } , { V t }} 1 ) } = E { log 2 ( 1 + ∥ a s ∥ 2 1 ) } , (61) where the expectation is taken over { G t } \{\boldsymbol{G}_{t}\} { G t } { S t } \{\boldsymbol{S}_{t}\} { S t } { V t } \{\boldsymbol{V}_{t}\} { V t }
2) Proposed Method for Rank-Deficient A: From (34), the data symbol estimate for any s ∈ { 1 , … , K ( τ − J ) } s \in\{1,\ldots,K(\tau-J)\} s ∈ { 1 , … , K ( τ − J )}
x ^ s = x s + a ˉ s H w , (62) \hat{x}_{s}=x_{s}+\bar{\boldsymbol{a}}_{s}^{H}\boldsymbol{w}, \tag*{(62)} x ^ s = x s + a ˉ s H w , (62) where a ˉ s H \bar{\boldsymbol{a}}_{s}^{H} a ˉ s H s s s ( A ˉ H A ˉ ) − 1 A ˉ H (\bar{\boldsymbol{A}}^{H}\bar{\boldsymbol{A}})^{-1}\bar{\boldsymbol{A}}^{H} ( A ˉ H A ˉ ) − 1 A ˉ H
τ − J τ E { log 2 ( 1 + 1 ∥ a ˉ s ∥ 2 ) } , (63) \frac{\tau-J}{\tau}\mathbb{E}\left\{\log_{2}\left(1+\frac{1}{\|\bar{\boldsymbol{a}}_{s}\|^{2}}\right)\right\}, \tag*{(63)} τ τ − J E { log 2 ( 1 + ∥ a ˉ s ∥ 2 1 ) } , (63) where the pre-log factor accounts for the UL UEs being silent in J J J G \boldsymbol{G} G S \boldsymbol{S} S
3) Genie-Aided Baseline: For any s ∈ { 1 , … , T K τ } s \in\{1,\ldots,TK\tau\} s ∈ { 1 , … , T K τ }
x ^ g , s = x s + a ˙ s H w , (64) \hat{x}_{g,s}=x_{s}+\dot{\boldsymbol{a}}_{s}^{H}\boldsymbol{w}, \tag*{(64)} x ^ g , s = x s + a ˙ s H w , (64) where a ˙ s H \dot{\boldsymbol{a}}_{s}^{H} a ˙ s H s s s ( A ˙ H A ˙ ) − 1 A ˙ H (\dot{\boldsymbol{A}}^{H}\dot{\boldsymbol{A}})^{-1}\dot{\boldsymbol{A}}^{H} ( A ˙ H A ˙ ) − 1 A ˙ H
E { log 2 ( 1 + 1 ∥ a ˙ s ∥ 2 ) } , (65) \mathbb{E}\left\{\log_{2}\left(1+\frac{1}{\|\dot{\boldsymbol{a}}_{s}\|^{2}}\right)\right\}, \tag*{(65)} E { log 2 ( 1 + ∥ a ˙ s ∥ 2 1 ) } , (65) where the expectation is taken over { G t } \{\boldsymbol{G}_{t}\} { G t }
4) Baseline Without AAI Mitigation: From (47), the data symbol estimate for any s ∈ { 1 , … , T K τ } s \in\{1,\ldots,TK\tau\} s ∈ { 1 , … , T K τ }
x ^ n , s = x s + b s H h + a ˙ s H w , (66) \hat{x}_{n,s}=x_{s}+\boldsymbol{b}_{s}^{H}\boldsymbol{h}+\dot{\boldsymbol{a}}_{s}^{H}\boldsymbol{w}, \tag*{(66)} x ^ n , s = x s + b s H h + a ˙ s H w , (66) where b s H \boldsymbol{b}_{s}^{H} b s H s s s B \boldsymbol{B} B
Var { b s H h + a ˙ s H w ∣ { G t } , { S t } , { V t } } = α ∥ b s ∥ 2 + ∥ a ˙ s ∥ 2 . (67) \operatorname{Var}\{\boldsymbol{b}_{s}^{H}\boldsymbol{h}+\dot{\boldsymbol{a}}_{s}^{H}\boldsymbol{w}\mid\{\boldsymbol{G}_{t}\},\{\boldsymbol{S}_{t}\},\{\boldsymbol{V}_{t}\}\}=\alpha\|\boldsymbol{b}_{s}\|^{2}+\|\dot{\boldsymbol{a}}_{s}\|^{2}. \tag*{(67)} Var { b s H h + a ˙ s H w ∣ { G t } , { S t } , { V t }} = α ∥ b s ∥ 2 + ∥ a ˙ s ∥ 2 . (67) With the same reasoning as before, an achievable SE is
E { log 2 ( 1 + 1 α ∥ b s ∥ 2 + ∥ a ˙ s ∥ 2 ) } , (68) \mathbb{E}\left\{\log_{2}\left(1+\frac{1}{\alpha\|\boldsymbol{b}_{s}\|^{2}+\|\dot{\boldsymbol{a}}_{s}\|^{2}}\right)\right\}, \tag*{(68)} E { log 2 ( 1 + α ∥ b s ∥ 2 + ∥ a ˙ s ∥ 2 1 ) } , (68) where the expectation is taken over { G t } \{\boldsymbol{G}_{t}\} { G t } { S t } \{\boldsymbol{S}_{t}\} { S t } { V t } \{\boldsymbol{V}_{t}\} { V t }
5) Zero-Forcing Combining for AAI Mitigation: As a benchmark we consider zero-forcing combining at the UL AP as in [23 ] , which is the existing method closest to ours. This method requires AP-AP channel knowledge and hence an overhead for pilot transmission between the APs. We use M M M H ^ \hat{\boldsymbol{H}} H ^ H \boldsymbol{H} H ? ]. Define the estimation error H ~ = H − H ^ \tilde{\boldsymbol{H}}=\boldsymbol{H}-\hat{\boldsymbol{H}} H ~ = H − H ^ t t t
Y = [ G H ^ V ] [ X S ] + H ~ V S + W . (69) \boldsymbol{Y}=\begin{bmatrix}\boldsymbol{G}&\hat{\boldsymbol{H}}\boldsymbol{V}\end{bmatrix}\begin{bmatrix}\boldsymbol{X}\\\boldsymbol{S}\end{bmatrix}+\tilde{\boldsymbol{H}}\boldsymbol{V}\boldsymbol{S}+\boldsymbol{W}. \tag*{(69)} Y = [ G H ^ V ] [ X S ] + H ~ V S + W . (69) Then, the zero-forcing combining matrix is
Z = ( [ G H ^ V ] H [ G H ^ V ] ) − 1 [ G H ^ V ] H , (70) \boldsymbol{Z}=\left(\begin{bmatrix}\boldsymbol{G}&\hat{\boldsymbol{H}}\boldsymbol{V}\end{bmatrix}^{H}\begin{bmatrix}\boldsymbol{G}&\hat{\boldsymbol{H}}\boldsymbol{V}\end{bmatrix}\right)^{-1}\begin{bmatrix}\boldsymbol{G}&\hat{\boldsymbol{H}}\boldsymbol{V}\end{bmatrix}^{H}, \tag*{(70)} Z = ( [ G H ^ V ] H [ G H ^ V ] ) − 1 [ G H ^ V ] H , (70) which gives
Z Y = [ X S ] + Z ( H ~ V S + W ) . (71) \boldsymbol{Z}\boldsymbol{Y}=\begin{bmatrix}\boldsymbol{X}\\\boldsymbol{S}\end{bmatrix}+\boldsymbol{Z}(\tilde{\boldsymbol{H}}\boldsymbol{V}\boldsymbol{S}+\boldsymbol{W}). \tag*{(71)} Z Y = [ X S ] + Z ( H ~ V S + W ) . (71) Let x s x_{s} x s s ∈ { 1 , … , K τ } s\in\{1,\ldots,K\tau\} s ∈ { 1 , … , K τ } X \boldsymbol{X} X x s x_{s} x s
x ^ s = x s + n s , (72) \hat{x}_{s}=x_{s}+n_{s}, \tag*{(72)} x ^ s = x s + n s , (72) where n s n_{s} n s s s s N \boldsymbol{N} N
τ − M / T τ E { log 2 ( 1 + 1 Var { n s ∣ G , S , V } ) } , (73) \frac{\tau-M/T}{\tau}\mathbb{E}\left\{\log_{2}\left(1+\frac{1}{\operatorname{Var}\{n_{s}\mid\boldsymbol{G},\boldsymbol{S},\boldsymbol{V}\}}\right)\right\}, \tag*{(73)} τ τ − M / T E { log 2 ( 1 + Var { n s ∣ G , S , V } 1 ) } , (73) where the expectation is taken over G \boldsymbol{G} G S \boldsymbol{S} S V \boldsymbol{V} V [23 ] by leveraging the slow time-variation of the AP-AP channel, such that the M M M T T T M / T M/T M / T
6) Numerical Evaluation: In Figure 5 , we evaluate the median average SE obtained by our proposed methods, the two baselines, and the zero-forcing combining technique, for T ∈ { 5 , … , 15 } T\in\{5,\ldots,15\} T ∈ { 5 , … , 15 } K K K
Figure 5. SEs of the proposed methods and the baselines for different T T T
The behavior of the SE is similar to the behavior of the MSE, both for the proposed methods and for the two baselines. Our proposed method for the full-rank case heavily outperforms the proposed method for the rank-deficient case and the baseline without AAI mitigation, and its SE approaches that of the genie-aided baseline when T T T
In terms of SE, the proposed method for the rank-
deficient case still outperforms the baseline without AAI mitigation, even though it suffers from a reduced pre-log factor due to the UL UEs being silent during J J J
The zero-forcing combining benchmark performs worse than the baseline without AAI mitigation. This is because the zero-forcing combining matrix requires M ≥ K + J M \ge K + J M ≥ K + J K K K J J J H V \mathbf{H}\mathbf{V} HV M = K + J M = K + J M = K + J Figure 5 , too many degrees of freedom are spent on AAI mitigation to allow for accurate data detection.
To obtain a fair comparison between our proposed methods and the zero-forcing benchmark, we now evaluate the SE for a varying number of AP antennas M = N ∈ { 16 , … , 32 } M = N \in\{16,\ldots,32\} M = N ∈ { 16 , … , 32 } K = J = 8 K = J = 8 K = J = 8 Figure 6 , we compare the zero-forcing benchmark with T = 1 T = 1 T = 1 T = 10 T = 10 T = 10 T = 10 T = 10 T = 10 8
Figure 6. SEs of the proposed methods and the baselines for different M = N M = N M = N
With T = 10 T = 10 T = 10
Our proposed method for the rank-deficient case outperforms the zero-forcing benchmark with T = 1 T = 1 T = 1
Except for when the number of AP antennas is smallest (where M = K + J M = K + J M = K + J T = 10 T = 10 T = 10 T > 1 T > 1 T > 1
In conclusion, our proposed methods achieve gains over the zero-forcing benchmark in terms of the UL SE in most cases. We end with a remark on the DL performance.
Remark 2. We restrict the simulations to the UL SE. However, our proposed method also improves the DL performance over existing methods, since we overcome the AP-AP channel estimation overhead which increases the pre-log factor of the DL SE. Our proposed method for the full-rank case increases the pre-log factor of the DL SE by M / T M/T M / T
Conclusions and Future Directions
We conclude that slowly time-varying AP-AP channels can be effectively leveraged to enhance interference mitigation in dynamic TDD. To this end, we have proposed a method for AP-AP interference mitigation that uses a least-squares formulation to jointly estimate the UL data and the AP-AP interference channel over multiple coherence intervals, during which the interference channel stays constant. We observed that this problem does not always have a unique solution, and provided an analytical bound on how slowly the AP-AP channel must vary to make the problem uniquely solvable. We solved the formulated least-squares problem in the full-rank case, and proposed a modification of the original problem that has a unique solution in the rank-deficient case.
Numerical simulations showed that our proposed method for the full-rank case significantly outperforms a baseline without interference mitigation and a benchmark zero-forcing receiver, in all relevant cases. Further, it achieves a similar performance as a genie-aided baseline with perfect interference mitigation for very slowly time-varying AP-AP channels.
We must leave the problem of calculating the exact rank of the regression matrix in the general scenario in Theorem 1 as an open question. Other directions for future work could be to develop improved methods for the rank-deficient case, and to extend our proposed method for the full-rank case to overcome the strict block-fading assumption used herein. A further direction is to extend the batch least-squares solution to an online recursive variant that decodes each coherence interval as it arrives, thereby reducing the latency caused by jointly processing multiple coherence intervals.
Proof of Theorem 1
We will split the proof into two parts. The first part proves (15), and the second part derives (17) by building upon the results from part one. The proof of (15) can be simplified by considering the leftmost T M T × T K T TM_T \times TK_T T M T × T K T A \mathbf{A} A
1) Proof of (15): We define the matrix P t = S t T V t T \mathbf{P}_t = \mathbf{S}_t^T \mathbf{V}_t^T P t = S t T V t T p t ( i ) ∈ C 1 × N \mathbf{p}_t^{(i)} \in\mathbb{C}^{1 \times N} p t ( i ) ∈ C 1 × N i ∈ { 1 , … , τ } i \in\{1,\ldots,\tau\} i ∈ { 1 , … , τ } i i i P t \mathbf{P}_t P t p t ( i , j ) p_t^{(i,j)} p t ( i , j ) i ∈ { 1 , … , τ } i \in\{1,\ldots,\tau\} i ∈ { 1 , … , τ } j ∈ { 1 , … , N } j \in\{1,\ldots,N\} j ∈ { 1 , … , N } ( i , j ) (i,j) ( i , j ) P t \mathbf{P}_t P t j j j p t ( i ) \mathbf{p}_t^{(i)} p t ( i )
A = [ G 1 p 1 ( 1 ) ⊗ I M ⋱ ⋮ G t p t ( i ) ⊗ I M ⋱ 0 G T p T ( τ ) ⊗ I M ] . (74) \mathbf{A} =
\begin{bmatrix}
\mathbf{G}_1 & & & \mathbf{p}_1^{(1)} \otimes\mathbf{I}_M \\
& \ddots& & \vdots\\
& & \mathbf{G}_t & \mathbf{p}_t^{(i)} \otimes\mathbf{I}_M \\
& & & \ddots\\
\mathbf{0} & & \mathbf{G}_T & \mathbf{p}_T^{(\tau)} \otimes\mathbf{I}_M
\end{bmatrix}. \tag*{(74)} A = G 1 0 ⋱ G t G T p 1 ( 1 ) ⊗ I M ⋮ p t ( i ) ⊗ I M ⋱ p T ( τ ) ⊗ I M . (74) The matrix A \mathbf{A} A T τ T\tau T τ M M M
[ G t p t ( i ) ⊗ I M ] = [ G t p t ( i , 1 ) I M ⋯ p t ( i , N ) I M ] . (75) \left[\mathbf{G}_t\ \ \mathbf{p}_t^{(i)} \otimes\mathbf{I}_M\right]
=
\left[\mathbf{G}_t\ \ p_t^{(i,1)}\mathbf{I}_M\ \cdots\ p_t^{(i,N)}\mathbf{I}_M\right]. \tag*{(75)} [ G t p t ( i ) ⊗ I M ] = [ G t p t ( i , 1 ) I M ⋯ p t ( i , N ) I M ] . (75) We now perform row operations on each block independently. The columns with only zeros are not affected by row operations; thus we can work with the structure in (76). From Assumption 3, it is possible to perform a sequence of row operations on G t \mathbf{G}_t G t G ˘ t \breve{\mathbf{G}}_t G ˘ t K K K
G t ∼ row ops. G ˘ t ≜ [ × ⋯ × 0 ⋱ ⋮ ⋮ ⋱ × 0 ⋯ 0 ] , K rows M − K rows (76) \mathbf{G}_t \mathrel{\overset{\text{row ops.}}{\sim}} \breve{\mathbf{G}}_t \triangleq
\left[
\begin{array}{cccc}
\times& \cdots& \times\\
0 & \ddots& \vdots\\
\vdots& \ddots& \times\\
0 & \cdots& 0
\end{array}
\right],
\qquad
\begin{array}{l}
K\ \text{rows}\\
M-K\ \text{rows}
\end{array} \tag*{(76)} G t ∼ row ops. G ˘ t ≜ × 0 ⋮ 0 ⋯ ⋱ ⋱ ⋯ × ⋮ × 0 , K rows M − K rows (76) Here, the symbol “× \times × p t ( i ) ⊗ I M \mathbf{p}_t^{(i)} \otimes\mathbf{I}_M p t ( i ) ⊗ I M
p t ( i ) ⊗ I M = [ p t ( i , 1 ) I M ⋯ p t ( i , N ) I M ] ∼ row ops. [ p t ( i , 1 ) Q ˘ t ⋯ p t ( i , N ) Q ˘ t ] = p t ( i ) ⊗ Q ˘ t , (77) \mathbf{p}_t^{(i)} \otimes\mathbf{I}_M
=
\left[p_t^{(i,1)}\mathbf{I}_M\ \cdots\ p_t^{(i,N)}\mathbf{I}_M\right]
\mathrel{\overset{\text{row ops.}}{\sim}}
\left[p_t^{(i,1)}\breve{\mathbf{Q}}_t\ \cdots\ p_t^{(i,N)}\breve{\mathbf{Q}}_t\right]
=
\mathbf{p}_t^{(i)} \otimes\breve{\mathbf{Q}}_t, \tag*{(77)} p t ( i ) ⊗ I M = [ p t ( i , 1 ) I M ⋯ p t ( i , N ) I M ] ∼ row ops. [ p t ( i , 1 ) Q ˘ t ⋯ p t ( i , N ) Q ˘ t ] = p t ( i ) ⊗ Q ˘ t , (77) where Q ˘ t ∈ C M × M \breve{\mathbf{Q}}_t \in\mathbb{C}^{M \times M} Q ˘ t ∈ C M × M
Q ˘ t = [ 1 0 ⋯ ⋯ ⋯ 0 × ⋱ ⋱ ⋮ ⋮ ⋱ 1 0 ⋯ 0 × ⋯ × 1 0 ⋯ ⋮ ⋮ 0 ⋱ ⋮ × ⋯ × 0 ⋯ 1 ] . (78) \breve{\mathbf{Q}}_t =
\begin{bmatrix}
1 & 0 & \cdots& \cdots& \cdots& 0 \\
\times& \ddots& \ddots& & & \vdots\\
\vdots& \ddots& 1 & 0 & \cdots& 0 \\
\times& \cdots& \times& 1 & 0 & \cdots\\
\vdots& & \vdots& 0 & \ddots& \vdots\\
\times& \cdots& \times& 0 & \cdots& 1
\end{bmatrix}. \tag*{(78)} Q ˘ t = 1 × ⋮ × ⋮ × 0 ⋱ ⋱ ⋯ ⋯ ⋯ ⋱ 1 × ⋮ × ⋯ 0 1 0 0 ⋯ ⋯ 0 ⋱ ⋯ 0 ⋮ 0 ⋯ ⋮ 1 . (78) As before, the elements “× \times ×
[ G t p t ( i ) ⊗ I M ] ∼ row ops. [ G ˘ t p t ( i ) ⊗ Q ˘ t ] . (79) \left[\mathbf{G}_t\ \ \mathbf{p}_t^{(i)} \otimes\mathbf{I}_M\right]
\mathrel{\overset{\text{row ops.}}{\sim}}
\left[\breve{\mathbf{G}}_t\ \ \mathbf{p}_t^{(i)} \otimes\breve{\mathbf{Q}}_t\right]. \tag*{(79)} [ G t p t ( i ) ⊗ I M ] ∼ row ops. [ G ˘ t p t ( i ) ⊗ Q ˘ t ] . (79) and the full matrix in (75) becomes
A ∼ row ops. A ˘ ≜ [ I τ ⊗ G ˘ 1 P 1 ⊗ Q ˘ 1 ⋱ ⋮ I τ ⊗ G ˘ T P T ⊗ Q ˘ T ] . (80) \mathbf{A} \mathrel{\overset{\text{row ops.}}{\sim}} \breve{\mathbf{A}} \triangleq
\begin{bmatrix}
\mathbf{I}_\tau\otimes\breve{\mathbf{G}}_1 & & & \mathbf{P}_1 \otimes\breve{\mathbf{Q}}_1 \\
& \ddots& & \vdots\\
& & \mathbf{I}_\tau\otimes\breve{\mathbf{G}}_T & \mathbf{P}_T \otimes\breve{\mathbf{Q}}_T
\end{bmatrix}. \tag*{(80)} A ∼ row ops. A ˘ ≜ I τ ⊗ G ˘ 1 ⋱ I τ ⊗ G ˘ T P 1 ⊗ Q ˘ 1 ⋮ P T ⊗ Q ˘ T . (80) We note that A ˘ \breve{\mathbf{A}} A ˘ T τ T\tau T τ K K K T K τ TK\tau T K τ A \mathbf{A} A T K τ TK\tau T K τ M − K M-K M − K T τ T\tau T τ T τ ( M − K ) T\tau(M-K) T τ ( M − K )
R ≜ [ P 1 ⊗ Q 1 ⋮ P T ⊗ Q T ] ∈ C T τ ( M − K ) × M N , (81) \mathbf{R} \triangleq
\begin{bmatrix}
\mathbf{P}_1 \otimes\mathbf{Q}_1 \\
\vdots\\
\mathbf{P}_T \otimes\mathbf{Q}_T
\end{bmatrix}
\in\mathbb{C}^{T\tau(M-K)\times MN}, \tag*{(81)} R ≜ P 1 ⊗ Q 1 ⋮ P T ⊗ Q T ∈ C T τ ( M − K ) × M N , (81) where Q t ∈ C ( M − K ) × M \mathbf{Q}_t \in\mathbb{C}^{(M-K)\times M} Q t ∈ C ( M − K ) × M M − K M-K M − K Q ˘ t \breve{\mathbf{Q}}_t Q ˘ t Q t \mathbf{Q}_t Q t
Q t ≜ [ Q ‾ t I M − K ] , (82) \mathbf{Q}_t \triangleq\left[\overline{\mathbf{Q}}_t\ \ \mathbf{I}_{M-K}\right], \tag*{(82)} Q t ≜ [ Q t I M − K ] , (82) where Q ‾ t ∈ C ( M − K ) × K \overline{\mathbf{Q}}_t \in\mathbb{C}^{(M-K)\times K} Q t ∈ C ( M − K ) × K Q ˘ t \breve{\mathbf{Q}}_t Q ˘ t R ≜ rank ( R ) R \triangleq\operatorname{rank}(\mathbf{R}) R ≜ rank ( R ) A \mathbf{A} A
rank ( A ) = T K τ + R , (83) \operatorname{rank}(\mathbf{A}) = TK\tau+ R, \tag*{(83)} rank ( A ) = T K τ + R , (83) with R ≤ M N R \le MN R ≤ M N
2) Proof of (17): Now, take a set of row vectors spanning the row space of P t \mathbf{P}_t P t J J J
P ‾ t ≜ [ p t ( 1 ) ⋮ p t ( J ) ] ∈ C J × N . (84) \overline{\mathbf{P}}_t \triangleq
\begin{bmatrix}
\mathbf{p}_t^{(1)} \\
\vdots\\
\mathbf{p}_t^{(J)}
\end{bmatrix}
\in\mathbb{C}^{J \times N}. \tag*{(84)} P t ≜ p t ( 1 ) ⋮ p t ( J ) ∈ C J × N . (84) Then, the row space of P t ⊗ Q t \mathbf{P}_t \otimes\mathbf{Q}_t P t ⊗ Q t P ‾ t ⊗ Q t \overline{\mathbf{P}}_t \otimes\mathbf{Q}_t P t ⊗ Q t R = rank ( R ) = rank ( R ~ ) R = \operatorname{rank}(\mathbf{R}) = \operatorname{rank}(\widetilde{\mathbf{R}}) R = rank ( R ) = rank ( R )
R ~ ≜ [ P ‾ 1 ⊗ Q 1 ⋮ P ‾ T ⊗ Q T ] ∈ C T ( M − K ) J × M N . (85) \widetilde{\mathbf{R}} \triangleq
\begin{bmatrix}
\overline{\mathbf{P}}_1 \otimes\mathbf{Q}_1 \\
\vdots\\
\overline{\mathbf{P}}_T \otimes\mathbf{Q}_T
\end{bmatrix}
\in\mathbb{C}^{T(M-K)J \times MN}. \tag*{(85)} R ≜ P 1 ⊗ Q 1 ⋮ P T ⊗ Q T ∈ C T ( M − K ) J × M N . (85) It follows that
R = rank ( R ~ ) ≤ min { T ( M − K ) J , M N } . (86) R = \operatorname{rank}(\widetilde{\mathbf{R}}) \le\min\{T(M-K)J,MN\}. \tag*{(86)} R = rank ( R ) ≤ min { T ( M − K ) J , M N } . (86) From (16) and (87) we get
dim ( N ( A ) ) = M N − R ≥ M N − min { T ( M − K ) J , M N } . (87) \dim(\mathcal{N}(\mathbf{A})) = MN-R
\ge MN-\min\{T(M-K)J,MN\}. \tag*{(87)} dim ( N ( A )) = M N − R ≥ M N − min { T ( M − K ) J , M N } . (87) Hence, the dimension of the nullspace can only equal zero if R ~ \widetilde{\mathbf{R}} R T ( M − K ) J ≥ M N T(M-K)J \ge MN T ( M − K ) J ≥ M N
Proof of Theorem 2
With M = N = 2 K = 2 J M=N=2K=2J M = N = 2 K = 2 J A \mathbf{A} A R ~ \widetilde{\mathbf{R}} R K K K
A ∈ C 2 T K × ( T K τ + 4 K 2 ) , (88) \mathbf{A} \in\mathbb{C}^{2TK \times(TK\tau+4K^2)}, \tag*{(88)} A ∈ C 2 T K × ( T K τ + 4 K 2 ) , (88) R ~ ∈ C T K 2 × 4 K 2 . (89) \widetilde{\mathbf{R}} \in\mathbb{C}^{TK^2 \times4K^2}. \tag*{(89)} R ∈ C T K 2 × 4 K 2 . (89) By inserting (83) into R ~ \widetilde{\mathbf{R}} R R ‾ \overline{\mathbf{R}} R R ~ \widetilde{\mathbf{R}} R
R ~ ∼ c o l . o p s . R ‾ ≜ [ P ‾ 1 ⊗ Q 1 P ‾ 1 ⊗ I K ⋮ ⋮ P ‾ T ⊗ Q T P ‾ T ⊗ I K ] . (90) \widetilde{\mathbf{R}} \mathrel{\overset{\mathrm{col.\ ops.}}{\sim}} \overline{\mathbf{R}} \triangleq
\begin{bmatrix}
\overline{\mathbf{P}}_1 \otimes\mathbf{Q}_1 & \overline{\mathbf{P}}_1 \otimes\mathbf{I}_K \\
\vdots& \vdots\\
\overline{\mathbf{P}}_T \otimes\mathbf{Q}_T & \overline{\mathbf{P}}_T \otimes\mathbf{I}_K
\end{bmatrix}. \tag*{(90)} R ∼ col. ops. R ≜ P 1 ⊗ Q 1 ⋮ P T ⊗ Q T P 1 ⊗ I K ⋮ P T ⊗ I K . (90) Thus,
R = rank ( R ) = rank ( R ~ ) = rank ( R ‾ ) . (91) R = \operatorname{rank}(\mathbf{R}) = \operatorname{rank}(\widetilde{\mathbf{R}}) = \operatorname{rank}(\overline{\mathbf{R}}). \tag*{(91)} R = rank ( R ) = rank ( R ) = rank ( R ) . (91) Since A \mathbf{A} A rank ( A ) = T K τ + 4 K 2 \operatorname{rank}(\mathbf{A})=TK\tau+4K^2 rank ( A ) = T K τ + 4 K 2 R = rank ( R ‾ ) = 4 K 2 R=\operatorname{rank}(\overline{\mathbf{R}})=4K^2 R = rank ( R ) = 4 K 2 rank ( R ‾ ) ≤ min { T K 2 , 4 K 2 } \operatorname{rank}(\overline{\mathbf{R}}) \le\min\{TK^2,4K^2\} rank ( R ) ≤ min { T K 2 , 4 K 2 } rank ( R ‾ ) = 4 K 2 \operatorname{rank}(\overline{\mathbf{R}})=4K^2 rank ( R ) = 4 K 2 T K 2 ≥ 4 K 2 ⟺ T ≥ 4 TK^2 \ge4K^2 \Longleftrightarrow T \ge4 T K 2 ≥ 4 K 2 ⟺ T ≥ 4
It remains to prove that rank ( R ‾ ) = 4 K 2 \operatorname{rank}(\overline{\mathbf{R}})=4K^2 rank ( R ) = 4 K 2 T ≥ 4 T \ge4 T ≥ 4 R ‾ \overline{\mathbf{R}} R T = 4 T=4 T = 4 R ‾ \overline{\mathbf{R}} R T ≥ 4 T \ge4 T ≥ 4 T = 4 T=4 T = 4 R ‾ \overline{\mathbf{R}} R
R ‾ = [ P ‾ 1 ⊗ Q ‾ 1 P ‾ 1 ⊗ I K P ‾ 2 ⊗ Q ‾ 2 P ‾ 2 ⊗ I K P ‾ 3 ⊗ Q ‾ 3 P ‾ 3 ⊗ I K P ‾ 4 ⊗ Q ‾ 4 P ‾ 4 ⊗ I K ] ∈ C 4 K 2 × 4 K 2 . (92) \overline{\mathbf{R}} =
\begin{bmatrix}
\overline{\mathbf{P}}_1 \otimes\overline{\mathbf{Q}}_1 & \overline{\mathbf{P}}_1 \otimes\mathbf{I}_K \\
\overline{\mathbf{P}}_2 \otimes\overline{\mathbf{Q}}_2 & \overline{\mathbf{P}}_2 \otimes\mathbf{I}_K \\
\overline{\mathbf{P}}_3 \otimes\overline{\mathbf{Q}}_3 & \overline{\mathbf{P}}_3 \otimes\mathbf{I}_K \\
\overline{\mathbf{P}}_4 \otimes\overline{\mathbf{Q}}_4 & \overline{\mathbf{P}}_4 \otimes\mathbf{I}_K
\end{bmatrix}
\in\mathbb{C}^{4K^2 \times4K^2}. \tag*{(92)} R = P 1 ⊗ Q 1 P 2 ⊗ Q 2 P 3 ⊗ Q 3 P 4 ⊗ Q 4 P 1 ⊗ I K P 2 ⊗ I K P 3 ⊗ I K P 4 ⊗ I K ∈ C 4 K 2 × 4 K 2 . (92) Since R ‾ \overline{\mathbf{R}} R R ‾ \overline{\mathbf{R}} R det ( R ‾ ) \det(\overline{\mathbf{R}}) det ( R ) ( 4 K 2 ) ! (4K^2)! ( 4 K 2 )! 4 K 2 4K^2 4 K 2 R ‾ \overline{\mathbf{R}} R [34 ] . This is a polynomial, say p ( ⋅ ) p(\cdot) p ( ⋅ ) { P ‾ t } \{\overline{\mathbf{P}}_t\} { P t } { Q ‾ t } \{\overline{\mathbf{Q}}_t\} { Q t } t ∈ { 1 , … , 4 } t \in\{1,\ldots,4\} t ∈ { 1 , … , 4 }
P = [ P ‾ 1 P ‾ 2 P ‾ 3 P ‾ 4 ] , Q = [ Q ‾ 1 Q ‾ 2 Q ‾ 3 Q ‾ 4 ] . (93) \mathbf{P} =
\begin{bmatrix}
\overline{\mathbf{P}}_1\\
\overline{\mathbf{P}}_2\\
\overline{\mathbf{P}}_3\\
\overline{\mathbf{P}}_4
\end{bmatrix},
\qquad
\mathbf{Q} =
\begin{bmatrix}
\overline{\mathbf{Q}}_1 & \overline{\mathbf{Q}}_2 & \overline{\mathbf{Q}}_3 & \overline{\mathbf{Q}}_4
\end{bmatrix}. \tag*{(93)} P = P 1 P 2 P 3 P 4 , Q = [ Q 1 Q 2 Q 3 Q 4 ] . (93) Then p ( ⋅ ) p(\cdot) p ( ⋅ ) det ( R ‾ ) = p ( P , Q ) \det(\overline{\mathbf{R}})=p(\mathbf{P},\mathbf{Q}) det ( R ) = p ( P , Q )
p ( P , Q ) = f ( P , Q ) det ( [ P ‾ 3 ⊗ I K P ‾ 4 ⊗ I K ] ) + r ( P , Q ) , (94) p(\mathbf{P},\mathbf{Q})=f(\mathbf{P},\mathbf{Q})\det\left(
\begin{bmatrix}
\overline{\mathbf{P}}_3\otimes\mathbf{I}_K\\
\overline{\mathbf{P}}_4\otimes\mathbf{I}_K
\end{bmatrix}
\right)+r(\mathbf{P},\mathbf{Q}), \tag*{(94)} p ( P , Q ) = f ( P , Q ) det ( [ P 3 ⊗ I K P 4 ⊗ I K ] ) + r ( P , Q ) , (94) with
f ( P , Q ) = ∏ k = 1 K ( det ( p ‾ 1 ( k , k ) Q ‾ 1 ) det ( p ‾ 2 ( k , k + K ) Q ‾ 2 ) ) , (95) f(\mathbf{P},\mathbf{Q})=\prod_{k=1}^{K}\left(\det\left(\overline{p}^{(k,k)}_1\overline{\mathbf{Q}}_1\right)\det\left(\overline{p}^{(k,k+K)}_2\overline{\mathbf{Q}}_2\right)\right), \tag*{(95)} f ( P , Q ) = k = 1 ∏ K ( det ( p 1 ( k , k ) Q 1 ) det ( p 2 ( k , k + K ) Q 2 ) ) , (95) where p ‾ t ( i , j ) \overline{p}^{(i,j)}_t p t ( i , j ) ( i , j ) (i,j) ( i , j ) P ‾ t \overline{\mathbf{P}}_t P t r ( P , Q ) r(\mathbf{P},\mathbf{Q}) r ( P , Q ) f ( P , Q ) f(\mathbf{P},\mathbf{Q}) f ( P , Q ) f ( P , Q ) ≢ 0 f(\mathbf{P},\mathbf{Q})\not\equiv0 f ( P , Q ) ≡ 0
det ( [ P ‾ 3 ⊗ I K P ‾ 4 ⊗ I K ] ) = det ( [ P ‾ 3 P ‾ 4 ] ⊗ I K ) = det ( [ P ‾ 3 P ‾ 4 ] ) K det ( I K ) 2 K = det ( [ P ‾ 3 P ‾ 4 ] ) K ≠ 0 , (96) \det\left(
\begin{bmatrix}
\overline{\mathbf{P}}_3\otimes\mathbf{I}_K\\
\overline{\mathbf{P}}_4\otimes\mathbf{I}_K
\end{bmatrix}
\right)
=
\det\left(
\begin{bmatrix}
\overline{\mathbf{P}}_3\\
\overline{\mathbf{P}}_4
\end{bmatrix}
\otimes\mathbf{I}_K
\right)
=
\det\left(
\begin{bmatrix}
\overline{\mathbf{P}}_3\\
\overline{\mathbf{P}}_4
\end{bmatrix}
\right)^K
\det(\mathbf{I}_K)^{2K}
=
\det\left(
\begin{bmatrix}
\overline{\mathbf{P}}_3\\
\overline{\mathbf{P}}_4
\end{bmatrix}
\right)^K
\ne0, \tag*{(96)} det ( [ P 3 ⊗ I K P 4 ⊗ I K ] ) = det ( [ P 3 P 4 ] ⊗ I K ) = det ( [ P 3 P 4 ] ) K det ( I K ) 2 K = det ( [ P 3 P 4 ] ) K = 0 , (96) where in the second equality we used a well-known property for the determinant of a Kronecker product [35 ] . Hence, the product of (96) and (97) in (95) is not the zero polynomial. Further, the same product consists of all monomials in the expansion of the determinant of R ‾ \overline{\mathbf{R}} R 2 K 2K 2 K { p ‾ 1 ( k , k ) Q ‾ 1 } \{\overline{p}^{(k,k)}_1\overline{\mathbf{Q}}_1\} { p 1 ( k , k ) Q 1 } { p ‾ 2 ( k , k + K ) Q ‾ 2 } \{\overline{p}^{(k,k+K)}_2\overline{\mathbf{Q}}_2\} { p 2 ( k , k + K ) Q 2 } 2 K 2 × 2 K 2 2K^2\times2K^2 2 K 2 × 2 K 2 R ‾ \overline{\mathbf{R}} R
It follows that no monomials in the first part of the right-hand side of (95) can be canceled out analytically by any monomials in the second part, since all monomials in the second part contain at least one element from outside the diagonal blocks. Hence, det ( R ‾ ) = p ( P , Q ) ≢ 0 \det(\overline{\mathbf{R}})=p(\mathbf{P},\mathbf{Q})\not\equiv0 det ( R ) = p ( P , Q ) ≡ 0 P ∗ \mathbf{P}^* P ∗ Q ∗ \mathbf{Q}^* Q ∗ p ( P ∗ , Q ∗ ) ≠ 0 p(\mathbf{P}^*,\mathbf{Q}^*)\ne0 p ( P ∗ , Q ∗ ) = 0
Next, note that with M = N = 2 K = 2 J M=N=2K=2J M = N = 2 K = 2 J Q ‾ t = − G t , 2 G t , 1 − 1 \overline{\mathbf{Q}}_t=-\mathbf{G}_{t,2}\mathbf{G}_{t,1}^{-1} Q t = − G t , 2 G t , 1 − 1 G t , 1 \mathbf{G}_{t,1} G t , 1 G t , 2 \mathbf{G}_{t,2} G t , 2 K × K K\times K K × K G t \mathbf{G}_t G t P ‾ t = S ‾ t T V t T \overline{\mathbf{P}}_t=\overline{\mathbf{S}}_t^T\mathbf{V}_t^T P t = S t T V t T S ‾ t T \overline{\mathbf{S}}_t^T S t T K K K S t T \mathbf{S}_t^T S t T
P = [ S ‾ 1 T S ‾ 2 T S ‾ 3 T S ‾ 4 T ] ⏟ ≜ S ‾ [ V 1 T V 2 T V 3 T V 4 T ] ⏟ ≜ V , (97) \mathbf{P}=
\underbrace{\begin{bmatrix}
\overline{\mathbf{S}}_1^T\\
\overline{\mathbf{S}}_2^T\\
\overline{\mathbf{S}}_3^T\\
\overline{\mathbf{S}}_4^T
\end{bmatrix}}_{\triangleq\overline{\mathbf{S}}}
\underbrace{\begin{bmatrix}
\mathbf{V}_1^T\\
\mathbf{V}_2^T\\
\mathbf{V}_3^T\\
\mathbf{V}_4^T
\end{bmatrix}}_{\triangleq\mathbf{V}}, \tag*{(97)} P = ≜ S S 1 T S 2 T S 3 T S 4 T ≜ V V 1 T V 2 T V 3 T V 4 T , (97) Q = − [ G 1 , 2 T G 2 , 2 T G 3 , 2 T G 4 , 2 T ] ⏟ ≜ G 2 [ G 1 , 1 − 1 G 2 , 1 − 1 G 3 , 1 − 1 G 4 , 1 − 1 ] ⏟ ≜ G 1 − 1 ; (98) \mathbf{Q}=-\underbrace{\begin{bmatrix}
\mathbf{G}_{1,2}^T\\
\mathbf{G}_{2,2}^T\\
\mathbf{G}_{3,2}^T\\
\mathbf{G}_{4,2}^T
\end{bmatrix}}_{\triangleq\mathbf{G}_2}
\underbrace{\begin{bmatrix}
\mathbf{G}_{1,1}^{-1} & & & \\
& \mathbf{G}_{2,1}^{-1} & & \\
& & \mathbf{G}_{3,1}^{-1} & \\
& & & \mathbf{G}_{4,1}^{-1}
\end{bmatrix}}_{\triangleq\mathbf{G}_1^{-1}}; \tag*{(98)} Q = − ≜ G 2 G 1 , 2 T G 2 , 2 T G 3 , 2 T G 4 , 2 T ≜ G 1 − 1 G 1 , 1 − 1 G 2 , 1 − 1 G 3 , 1 − 1 G 4 , 1 − 1 ; (98) thus, for the matrices P ∗ , Q ∗ \mathbf{P}^*,\mathbf{Q}^* P ∗ , Q ∗ V ∗ = S ‾ − 1 P ∗ \mathbf{V}^*=\overline{\mathbf{S}}^{-1}\mathbf{P}^* V ∗ = S − 1 P ∗ G 2 ∗ = − Q ∗ G 1 \mathbf{G}_2^*=-\mathbf{Q}^*\mathbf{G}_1 G 2 ∗ = − Q ∗ G 1 q ( ⋅ ) q(\cdot) q ( ⋅ ) p ( ⋅ ) p(\cdot) p ( ⋅ ) q ( V , G 2 ) = p ( P , Q ) q(\mathbf{V},\mathbf{G}_2)=p(\mathbf{P},\mathbf{Q}) q ( V , G 2 ) = p ( P , Q )
det ( R ‾ ) = p ( P , Q ) = p ( S ‾ V , − G 2 G 1 − 1 ) (99) \det(\overline{\mathbf{R}})=p(\mathbf{P},\mathbf{Q})=p(\overline{\mathbf{S}}\mathbf{V},-\mathbf{G}_2\mathbf{G}_1^{-1}) \tag*{(99)} det ( R ) = p ( P , Q ) = p ( S V , − G 2 G 1 − 1 ) (99) = q ( V , G 2 ) ≠ 0 , (100) =q(\mathbf{V},\mathbf{G}_2)\ne0, \tag*{(100)} = q ( V , G 2 ) = 0 , (100) since q ( V ∗ , G 2 ∗ ) = p ( P ∗ , Q ∗ ) ≠ 0 q(\mathbf{V}^*,\mathbf{G}_2^*)=p(\mathbf{P}^*,\mathbf{Q}^*)\ne0 q ( V ∗ , G 2 ∗ ) = p ( P ∗ , Q ∗ ) = 0 q ( V , G 2 ) ≠ 0 q(\mathbf{V},\mathbf{G}_2)\ne0 q ( V , G 2 ) = 0 q ( V , G 2 ) q(\mathbf{V},\mathbf{G}_2) q ( V , G 2 ) V \mathbf{V} V G 2 \mathbf{G}_2 G 2 V ∗ \mathbf{V}^* V ∗ G 2 ∗ \mathbf{G}_2^* G 2 ∗ R ‾ = rank ( R ‾ ) = 4 K 2 \overline{\mathbf{R}}=\operatorname{rank}(\overline{\mathbf{R}})=4K^2 R = rank ( R ) = 4 K 2 A \mathbf{A} A T ≥ 4 T\ge4 T ≥ 4
Proof of Theorem 3
By following the same steps as in the proof of Theorem 1 (with T = 1 T=1 T = 1 rank ( A ) = K τ + rank ( R ) \operatorname{rank}(\mathbf{A})=K\tau+\operatorname{rank}(\mathbf{R}) rank ( A ) = K τ + rank ( R ) R \mathbf{R} R R = S 1 T V 1 T ⊗ Q 1 \mathbf{R}=\mathbf{S}_1^T\mathbf{V}_1^T\otimes\mathbf{Q}_1 R = S 1 T V 1 T ⊗ Q 1 rank ( R ) = rank ( S 1 T V 1 T ) rank ( Q 1 ) = J ( M − K ) \operatorname{rank}(\mathbf{R})=\operatorname{rank}(\mathbf{S}_1^T\mathbf{V}_1^T)\operatorname{rank}(\mathbf{Q}_1)=J(M-K) rank ( R ) = rank ( S 1 T V 1 T ) rank ( Q 1 ) = J ( M − K ) rank ( A ) = K τ + ( M − K ) J \operatorname{rank}(\mathbf{A})=K\tau+(M-K)J rank ( A ) = K τ + ( M − K ) J
Note that this theorem extends [? ], which assumed J = 1 J=1 J = 1 J J J
Proof of Proposition 1
The matrix A H A \mathbf{A}^H\mathbf{A} A H A
A H A = [ B C C H D ] , (101) \mathbf{A}^H\mathbf{A}=
\begin{bmatrix}
\mathbf{B} & \mathbf{C}\\
\mathbf{C}^H & \mathbf{D}
\end{bmatrix}, \tag*{(101)} A H A = [ B C H C D ] , (101) where, with P t = S t T V t T \mathbf{P}_t=\mathbf{S}_t^T\mathbf{V}_t^T P t = S t T V t T
B ≜ [ I T ⊗ G 1 H G 1 0 ⋱ 0 I T ⊗ G T H G T ] , \mathbf{B}\triangleq
\begin{bmatrix}
\mathbf{I}_T\otimes\mathbf{G}_1^H\mathbf{G}_1 & & \mathbf{0}\\
& \ddots& \\
\mathbf{0} & & \mathbf{I}_T\otimes\mathbf{G}_T^H\mathbf{G}_T
\end{bmatrix}, B ≜ I T ⊗ G 1 H G 1 0 ⋱ 0 I T ⊗ G T H G T , C ≜ [ P 1 ⊗ G 1 H ⋮ P T ⊗ G T H ] , D ≜ ∑ t ∈ T ( P t H P t ⊗ I M ) . (102) \mathbf{C}\triangleq
\begin{bmatrix}
\mathbf{P}_1\otimes\mathbf{G}_1^H\\
\vdots\\
\mathbf{P}_T\otimes\mathbf{G}_T^H
\end{bmatrix},
\qquad
\mathbf{D}\triangleq\sum_{t\in\mathcal{T}}(\mathbf{P}_t^H\mathbf{P}_t\otimes\mathbf{I}_M). \tag*{(102)} C ≜ P 1 ⊗ G 1 H ⋮ P T ⊗ G T H , D ≜ t ∈ T ∑ ( P t H P t ⊗ I M ) . (102) Then, by using the inversion lemma for block-partitioned matrices [34 ] , we get
Cov ( x ^ ) = [ ( A H A ) − 1 ] 1 : K τ , 1 : K τ = B − 1 + B − 1 C ( D − C H B − 1 C ) − 1 C H B − 1 . (103) \operatorname{Cov}(\hat{\mathbf{x}})=\left[(\mathbf{A}^H\mathbf{A})^{-1}\right]_{1:K\tau,1:K\tau}
=\mathbf{B}^{-1}+\mathbf{B}^{-1}\mathbf{C}(\mathbf{D}-\mathbf{C}^H\mathbf{B}^{-1}\mathbf{C})^{-1}\mathbf{C}^H\mathbf{B}^{-1}. \tag*{(103)} Cov ( x ^ ) = [ ( A H A ) − 1 ] 1 : K τ , 1 : K τ = B − 1 + B − 1 C ( D − C H B − 1 C ) − 1 C H B − 1 . (103) Define E ~ ≜ B − 1 C ( D − C H B − 1 C ) − 1 C H B − 1 \tilde{\mathbf{E}} \triangleq\mathbf{B}^{-1}\mathbf{C}(\mathbf{D}-\mathbf{C}^{H}\mathbf{B}^{-1}\mathbf{C})^{-1}\mathbf{C}^{H}\mathbf{B}^{-1} E ~ ≜ B − 1 C ( D − C H B − 1 C ) − 1 C H B − 1 T × T T \times T T × T K τ × K τ K_{\tau} \times K_{\tau} K τ × K τ
E ~ = [ E ~ 1 , 1 ⋯ E ~ 1 , T ⋮ ⋱ ⋮ E ~ T , 1 ⋯ E ~ T , T ] . (104) \tilde{\mathbf{E}} =
\begin{bmatrix}
\tilde{\mathbf{E}}_{1,1} & \cdots& \tilde{\mathbf{E}}_{1,T}\\
\vdots& \ddots& \vdots\\
\tilde{\mathbf{E}}_{T,1} & \cdots& \tilde{\mathbf{E}}_{T,T}
\end{bmatrix}. \tag*{(104)} E ~ = E ~ 1 , 1 ⋮ E ~ T , 1 ⋯ ⋱ ⋯ E ~ 1 , T ⋮ E ~ T , T . (104) For any i , j ∈ { 1 , … , T } i,j \in\{1,\ldots,T\} i , j ∈ { 1 , … , T }
E ~ i , j = ( P i ⊗ ( G i H G i ) − 1 G i H ) E ( P j T ⊗ G j ( G j H G j ) − 1 ) , (105) \tilde{\mathbf{E}}_{i,j} = (\mathbf{P}_{i} \otimes(\mathbf{G}_{i}^{H}\mathbf{G}_{i})^{-1}\mathbf{G}_{i}^{H})\mathbf{E}(\mathbf{P}_{j}^{T} \otimes\mathbf{G}_{j}(\mathbf{G}_{j}^{H}\mathbf{G}_{j})^{-1}), \tag*{(105)} E ~ i , j = ( P i ⊗ ( G i H G i ) − 1 G i H ) E ( P j T ⊗ G j ( G j H G j ) − 1 ) , (105) where
E ≜ ( D − C H B − 1 C ) − 1 = ( ∑ t ∈ T P t H P t ⊗ ( I M − G t ( G t H G t ) − 1 G t H ) ) − 1 . (106) \mathbf{E} \triangleq(\mathbf{D}-\mathbf{C}^{H}\mathbf{B}^{-1}\mathbf{C})^{-1}
= \left(\sum_{t\in\mathcal{T}}\mathbf{P}_{t}^{H}\mathbf{P}_{t}\otimes(\mathbf{I}_{M}-\mathbf{G}_{t}(\mathbf{G}_{t}^{H}\mathbf{G}_{t})^{-1}\mathbf{G}_{t}^{H})\right)^{-1}. \tag*{(106)} E ≜ ( D − C H B − 1 C ) − 1 = ( t ∈ T ∑ P t H P t ⊗ ( I M − G t ( G t H G t ) − 1 G t H ) ) − 1 . (106) By extracting the t t t K τ × K τ K_{\tau}\times K_{\tau} K τ × K τ Cov ( x ^ ) \operatorname{Cov}(\hat{\mathbf{x}}) Cov ( x ^ ) P t = S t T V t T \mathbf{P}_{t}=\mathbf{S}_{t}^{T}\mathbf{V}_{t}^{T} P t = S t T V t T
Proof of Theorem 4
To show that A ˉ ∈ C M T × ( K ( τ − J ) + M J ) \bar{\mathbf{A}}\in\mathbb{C}^{M T\times(K(\tau-J)+MJ)} A ˉ ∈ C M T × ( K ( τ − J ) + M J ) K ( τ − J ) + M J K(\tau-J)+MJ K ( τ − J ) + M J A ˉ \bar{\mathbf{A}} A ˉ A ˉ = [ A ˉ 1 A ˉ 2 ] \bar{\mathbf{A}}=[\bar{\mathbf{A}}_{1}\ \bar{\mathbf{A}}_{2}] A ˉ = [ A ˉ 1 A ˉ 2 ]
A ˉ 1 = [ 0 M J × K ( τ − J ) I τ − J ⊗ G ] , A ˉ 2 = S T ⊗ I M . (107) \bar{\mathbf{A}}_{1}=
\begin{bmatrix}
\mathbf{0}_{MJ\times K(\tau-J)}\\
\mathbf{I}_{\tau-J}\otimes\mathbf{G}
\end{bmatrix},\qquad
\bar{\mathbf{A}}_{2}=\mathbf{S}^{T}\otimes\mathbf{I}_{M}. \tag*{(107)} A ˉ 1 = [ 0 M J × K ( τ − J ) I τ − J ⊗ G ] , A ˉ 2 = S T ⊗ I M . (107) From Assumption 3 the left-most block must be full-rank, i.e. rank ( A ˉ 1 ) = K ( τ − J ) \operatorname{rank}(\bar{\mathbf{A}}_{1})=K(\tau-J) rank ( A ˉ 1 ) = K ( τ − J ) M J MJ M J A ˉ 2 \bar{\mathbf{A}}_{2} A ˉ 2 M J MJ M J A ˉ 1 \bar{\mathbf{A}}_{1} A ˉ 1 A ˉ 2 \bar{\mathbf{A}}_{2} A ˉ 2 M J × M J MJ\times MJ M J × M J S \mathbf{S} S rank ( A ˉ 2 ) = rank ( S T ⊗ I M ) = M J \operatorname{rank}(\bar{\mathbf{A}}_{2})=\operatorname{rank}(\mathbf{S}^{T}\otimes\mathbf{I}_{M})=MJ rank ( A ˉ 2 ) = rank ( S T ⊗ I M ) = M J
Proof of Theorem 5
The proof follows the same lines as [? ], which considered the special case of J = 1 J=1 J = 1 Y = [ Y 1 : J Y ˉ ] \mathbf{Y}=[\mathbf{Y}_{1:J}\ \bar{\mathbf{Y}}] Y = [ Y 1 : J Y ˉ ] S = [ S 1 : J S ˉ ] \mathbf{S}=[\mathbf{S}_{1:J}\ \bar{\mathbf{S}}] S = [ S 1 : J S ˉ ]
min X ˉ , F ∥ Y − G [ 0 K × J X ˉ ] − F S ∥ 2 = min X ˉ , F { ∥ Y 1 : J − F S 1 : J ∥ 2 + ∥ Y ˉ − G X ˉ − F S ˉ ∥ 2 } . (108) \min_{\bar{\mathbf{X}},\mathbf{F}}\left\|\mathbf{Y}-\mathbf{G}
\begin{bmatrix}
\mathbf{0}_{K\times J} & \bar{\mathbf{X}}
\end{bmatrix}
-\mathbf{F}\mathbf{S}\right\|^{2}
=
\min_{\bar{\mathbf{X}},\mathbf{F}}\left\{\|\mathbf{Y}_{1:J}-\mathbf{F}\mathbf{S}_{1:J}\|^{2}
+\|\bar{\mathbf{Y}}-\mathbf{G}\bar{\mathbf{X}}-\mathbf{F}\bar{\mathbf{S}}\|^{2}\right\}. \tag*{(108)} X ˉ , F min Y − G [ 0 K × J X ˉ ] − FS 2 = X ˉ , F min { ∥ Y 1 : J − F S 1 : J ∥ 2 + ∥ Y ˉ − G X ˉ − F S ˉ ∥ 2 } . (108) From (109), the solution to X ˉ \bar{\mathbf{X}} X ˉ F \mathbf{F} F
X ˉ ^ = ( G H G ) − 1 G H ( Y ˉ − F S ˉ ) . (109) \hat{\bar{\mathbf{X}}}=(\mathbf{G}^{H}\mathbf{G})^{-1}\mathbf{G}^{H}(\bar{\mathbf{Y}}-\mathbf{F}\bar{\mathbf{S}}). \tag*{(109)} X ˉ ^ = ( G H G ) − 1 G H ( Y ˉ − F S ˉ ) . (109) By inserting (110) into (109), we obtain the remaining problem to minimize over F \mathbf{F} F
min F { ∥ Y 1 : J − F S 1 : J ∥ 2 + ∥ Y ˉ − G ( G H G ) − 1 G H ( Y ˉ − F S ˉ ) − F S ˉ ∥ 2 } = min F { ∥ Y 1 : J − F S 1 : J ∥ 2 + ∥ Π G ⊥ ( Y ˉ − F S ˉ ) ∥ 2 } , (110) \min_{\mathbf{F}}\left\{\|\mathbf{Y}_{1:J}-\mathbf{F}\mathbf{S}_{1:J}\|^{2}
+\|\bar{\mathbf{Y}}-\mathbf{G}(\mathbf{G}^{H}\mathbf{G})^{-1}\mathbf{G}^{H}(\bar{\mathbf{Y}}-\mathbf{F}\bar{\mathbf{S}})-\mathbf{F}\bar{\mathbf{S}}\|^{2}\right\}
=
\min_{\mathbf{F}}\left\{\|\mathbf{Y}_{1:J}-\mathbf{F}\mathbf{S}_{1:J}\|^{2}
+\|\boldsymbol{\Pi}_{\mathbf{G}}^{\perp}(\bar{\mathbf{Y}}-\mathbf{F}\bar{\mathbf{S}})\|^{2}\right\}, \tag*{(110)} F min { ∥ Y 1 : J − F S 1 : J ∥ 2 + ∥ Y ˉ − G ( G H G ) − 1 G H ( Y ˉ − F S ˉ ) − F S ˉ ∥ 2 } = F min { ∥ Y 1 : J − F S 1 : J ∥ 2 + ∥ Π G ⊥ ( Y ˉ − F S ˉ ) ∥ 2 } , (110) where Π G ⊥ ≜ I M − G ( G H G ) − 1 G H \boldsymbol{\Pi}_{\mathbf{G}}^{\perp}\triangleq\mathbf{I}_{M}-\mathbf{G}(\mathbf{G}^{H}\mathbf{G})^{-1}\mathbf{G}^{H} Π G ⊥ ≜ I M − G ( G H G ) − 1 G H G \mathbf{G} G Y 1 : J = Y 1 : J ′ + Y 1 : J ′ ′ \mathbf{Y}_{1:J}=\mathbf{Y}_{1:J}^{\prime}+\mathbf{Y}_{1:J}^{\prime\prime} Y 1 : J = Y 1 : J ′ + Y 1 : J ′′ F = F ′ + F ′ ′ \mathbf{F}=\mathbf{F}^{\prime}+\mathbf{F}^{\prime\prime} F = F ′ + F ′′ Y 1 : J ′ , F ′ ∈ sp ( G ) \mathbf{Y}_{1:J}^{\prime},\mathbf{F}^{\prime}\in\operatorname{sp}(\mathbf{G}) Y 1 : J ′ , F ′ ∈ sp ( G ) Y 1 : J ′ ′ , F ′ ′ ∈ sp ( G ) ⊥ \mathbf{Y}_{1:J}^{\prime\prime},\mathbf{F}^{\prime\prime}\in\operatorname{sp}(\mathbf{G})^{\perp} Y 1 : J ′′ , F ′′ ∈ sp ( G ) ⊥ sp ( G ) \operatorname{sp}(\mathbf{G}) sp ( G ) sp ( G ) ⊥ \operatorname{sp}(\mathbf{G})^{\perp} sp ( G ) ⊥ G \mathbf{G} G
min F { ∥ Y 1 : J ′ + Y 1 : J ′ ′ − F ′ S 1 : J − F ′ ′ S 1 : J ∥ 2 + ∥ Π G ⊥ ( Y ˉ − F ′ S ˉ − F ′ ′ S ˉ ) ∥ 2 } = min F { ∥ Y 1 : J ′ − F ′ S 1 : J ∥ 2 + ∥ Y 1 : J ′ ′ − F ′ ′ S 1 : J ∥ 2 + ∥ Π G ⊥ ( Y ˉ − F ′ ′ S ˉ ) ∥ 2 } . (111) \min_{\mathbf{F}}\left\{\|\mathbf{Y}_{1:J}^{\prime}+\mathbf{Y}_{1:J}^{\prime\prime}-\mathbf{F}^{\prime}\mathbf{S}_{1:J}-\mathbf{F}^{\prime\prime}\mathbf{S}_{1:J}\|^{2}
+\|\boldsymbol{\Pi}_{\mathbf{G}}^{\perp}(\bar{\mathbf{Y}}-\mathbf{F}^{\prime}\bar{\mathbf{S}}-\mathbf{F}^{\prime\prime}\bar{\mathbf{S}})\|^{2}\right\}
=
\min_{\mathbf{F}}\left\{\|\mathbf{Y}_{1:J}^{\prime}-\mathbf{F}^{\prime}\mathbf{S}_{1:J}\|^{2}
+\|\mathbf{Y}_{1:J}^{\prime\prime}-\mathbf{F}^{\prime\prime}\mathbf{S}_{1:J}\|^{2}
+\|\boldsymbol{\Pi}_{\mathbf{G}}^{\perp}(\bar{\mathbf{Y}}-\mathbf{F}^{\prime\prime}\bar{\mathbf{S}})\|^{2}\right\}. \tag*{(111)} F min { ∥ Y 1 : J ′ + Y 1 : J ′′ − F ′ S 1 : J − F ′′ S 1 : J ∥ 2 + ∥ Π G ⊥ ( Y ˉ − F ′ S ˉ − F ′′ S ˉ ) ∥ 2 } = F min { ∥ Y 1 : J ′ − F ′ S 1 : J ∥ 2 + ∥ Y 1 : J ′′ − F ′′ S 1 : J ∥ 2 + ∥ Π G ⊥ ( Y ˉ − F ′′ S ˉ ) ∥ 2 } . (111) Let the solution to (112) be F ^ ⋆ = F ^ ⋆ ′ + F ^ ⋆ ′ ′ \hat{\mathbf{F}}_{\star}=\hat{\mathbf{F}}_{\star}^{\prime}+\hat{\mathbf{F}}_{\star}^{\prime\prime} F ^ ⋆ = F ^ ⋆ ′ + F ^ ⋆ ′′ F ^ ⋆ ′ ∈ sp ( G ) \hat{\mathbf{F}}_{\star}^{\prime}\in\operatorname{sp}(\mathbf{G}) F ^ ⋆ ′ ∈ sp ( G ) F ^ ⋆ ′ ′ ∈ sp ( G ) ⊥ \hat{\mathbf{F}}_{\star}^{\prime\prime}\in\operatorname{sp}(\mathbf{G})^{\perp} F ^ ⋆ ′′ ∈ sp ( G ) ⊥ F ′ ′ \mathbf{F}^{\prime\prime} F ′′
F ^ ⋆ ′ ′ = Y 1 : J ′ ′ S 1 : J H ( S 1 : J S 1 : J H ) − 1 . \hat{\mathbf{F}}_{\star}^{\prime\prime}=\mathbf{Y}_{1:J}^{\prime\prime}\mathbf{S}_{1:J}^{H}(\mathbf{S}_{1:J}\mathbf{S}_{1:J}^{H})^{-1}. F ^ ⋆ ′′ = Y 1 : J ′′ S 1 : J H ( S 1 : J S 1 : J H ) − 1 . Thus, we get
X ˉ ^ = ( G H G ) − 1 G H ( Y ˉ − F ^ ⋆ S ˉ ) = ( G H G ) − 1 G H ( Y ˉ − F ^ ⋆ ′ S ˉ ) = ( G H G ) − 1 G H ( Y ˉ − Y 1 : J ′ S 1 : J H ( S 1 : J S 1 : J H ) − 1 S ˉ ) = ( G H G ) − 1 G H ( Y ˉ − Y 1 : J S 1 : J H ( S 1 : J S 1 : J H ) − 1 S ˉ ) = ( G H G ) − 1 G H ( Y ˉ − F ^ S ˉ ) . (112) \begin{aligned}
\hat{\bar{\mathbf{X}}}=(\mathbf{G}^{H}\mathbf{G})^{-1}\mathbf{G}^{H}(\bar{\mathbf{Y}}-\hat{\mathbf{F}}_{\star}\bar{\mathbf{S}})
=(\mathbf{G}^{H}\mathbf{G})^{-1}\mathbf{G}^{H}(\bar{\mathbf{Y}}-\hat{\mathbf{F}}_{\star}^{\prime}\bar{\mathbf{S}}) \\
&=(\mathbf{G}^{H}\mathbf{G})^{-1}\mathbf{G}^{H}\left(\bar{\mathbf{Y}}-\mathbf{Y}_{1:J}^{\prime}\mathbf{S}_{1:J}^{H}(\mathbf{S}_{1:J}\mathbf{S}_{1:J}^{H})^{-1}\bar{\mathbf{S}}\right) \\
&=(\mathbf{G}^{H}\mathbf{G})^{-1}\mathbf{G}^{H}\left(\bar{\mathbf{Y}}-\mathbf{Y}_{1:J}\mathbf{S}_{1:J}^{H}(\mathbf{S}_{1:J}\mathbf{S}_{1:J}^{H})^{-1}\bar{\mathbf{S}}\right)
=(\mathbf{G}^{H}\mathbf{G})^{-1}\mathbf{G}^{H}(\bar{\mathbf{Y}}-\hat{\mathbf{F}}\bar{\mathbf{S}}). \tag*{(112)}
\end{aligned} X ˉ ^ = ( G H G ) − 1 G H ( Y ˉ − F ^ ⋆ S ˉ ) = ( G H G ) − 1 G H ( Y ˉ − F ^ ⋆ ′ S ˉ ) = ( G H G ) − 1 G H ( Y ˉ − Y 1 : J ′ S 1 : J H ( S 1 : J S 1 : J H ) − 1 S ˉ ) = ( G H G ) − 1 G H ( Y ˉ − Y 1 : J S 1 : J H ( S 1 : J S 1 : J H ) − 1 S ˉ ) = ( G H G ) − 1 G H ( Y ˉ − F ^ S ˉ ) . (112) where in the second and fourth equalities, respectively, we used that F ^ ⋆ ′ ′ , Y 1 : J ′ ′ ∈ sp ( G ) ⊥ \hat{\mathbf{F}}_{\star}^{\prime\prime},\mathbf{Y}_{1:J}^{\prime\prime}\in\operatorname{sp}(\mathbf{G})^{\perp} F ^ ⋆ ′′ , Y 1 : J ′′ ∈ sp ( G ) ⊥ Y 1 : J S 1 : J H ( S 1 : J S 1 : J H ) − 1 \mathbf{Y}_{1:J}\mathbf{S}_{1:J}^{H}(\mathbf{S}_{1:J}\mathbf{S}_{1:J}^{H})^{-1} Y 1 : J S 1 : J H ( S 1 : J S 1 : J H ) − 1 F ^ \hat{\mathbf{F}} F ^
Algorithm and Complexity Analysis for the Full-Rank Case
The unknown vector z = [ x 1 T , … , x T T , h T ] T \mathbf{z}=[\mathbf{x}_{1}^{T},\ldots,\mathbf{x}_{T}^{T},\mathbf{h}^{T}]^{T} z = [ x 1 T , … , x T T , h T ] T A \mathbf{A} A A H A \mathbf{A}^{H}\mathbf{A} A H A 2 × 2 2\times2 2 × 2 B \mathbf{B} B C \mathbf{C} C D \mathbf{D} D B \mathbf{B} B E − 1 = D − C H B − 1 C \mathbf{E}^{-1}=\mathbf{D}-\mathbf{C}^{H}\mathbf{B}^{-1}\mathbf{C} E − 1 = D − C H B − 1 C E \mathbf{E} E
Rather than forming ( A H A ) − 1 (\mathbf{A}^{H}\mathbf{A})^{-1} ( A H A ) − 1 A H A z ^ = A H y \mathbf{A}^{H}\mathbf{A}\hat{\mathbf{z}}=\mathbf{A}^{H}\mathbf{y} A H A z ^ = A H y
B x ^ + C h ^ = ( A H y ) x , (113) \mathbf{B}\hat{\mathbf{x}}+\mathbf{C}\hat{\mathbf{h}}=(\mathbf{A}^{H}\mathbf{y})_{\mathbf{x}}, \tag*{(113)} B x ^ + C h ^ = ( A H y ) x , (113) C H x ^ + D h ^ = ∑ t ∈ T ( P t H ⊗ I M ) y t , (114) \mathbf{C}^{H}\hat{\mathbf{x}}+\mathbf{D}\hat{\mathbf{h}}=\sum_{t\in\mathcal{T}}(\mathbf{P}_{t}^{H}\otimes\mathbf{I}_{M})\mathbf{y}_{t}, \tag*{(114)} C H x ^ + D h ^ = t ∈ T ∑ ( P t H ⊗ I M ) y t , (114) where ( A H y ) x (\mathbf{A}^{H}\mathbf{y})_{\mathbf{x}} ( A H y ) x x \mathbf{x} x A H y \mathbf{A}^{H}\mathbf{y} A H y t t t ( I K ⊗ G t ) y t (\mathbf{I}_{K}\otimes\mathbf{G}_{t})\mathbf{y}_{t} ( I K ⊗ G t ) y t x ^ = B − 1 ( ( A H y ) x − C h ^ ) \hat{\mathbf{x}}=\mathbf{B}^{-1}((\mathbf{A}^{H}\mathbf{y})_{\mathbf{x}}-\mathbf{C}\hat{\mathbf{h}}) x ^ = B − 1 (( A H y ) x − C h ^ ) E − 1 \mathbf{E}^{-1} E − 1
E − 1 h ^ = r ≜ ∑ t ∈ T ( P t H ⊗ ( I M − G t ( G t H G t ) − 1 G t H ) ) y t . (115) \mathbf{E}^{-1}\hat{\mathbf{h}}=\mathbf{r}\triangleq\sum_{t\in\mathcal{T}}\left(\mathbf{P}_{t}^{H}\otimes(\mathbf{I}_{M}-\mathbf{G}_{t}(\mathbf{G}_{t}^{H}\mathbf{G}_{t})^{-1}\mathbf{G}_{t}^{H})\right)\mathbf{y}_{t}. \tag*{(115)} E − 1 h ^ = r ≜ t ∈ T ∑ ( P t H ⊗ ( I M − G t ( G t H G t ) − 1 G t H ) ) y t . (115) Once h ^ \hat{\mathbf{h}} h ^ t t t T T T
x ^ t = ( I K ⊗ ( G t H G t ) − 1 G t H ) ( y t − ( P t ⊗ I M ) h ^ ) . (116) \hat{\mathbf{x}}_{t}=(\mathbf{I}_{K}\otimes(\mathbf{G}_{t}^{H}\mathbf{G}_{t})^{-1}\mathbf{G}_{t}^{H})(\mathbf{y}_{t}-(\mathbf{P}_{t}\otimes\mathbf{I}_{M})\hat{\mathbf{h}}). \tag*{(116)} x ^ t = ( I K ⊗ ( G t H G t ) − 1 G t H ) ( y t − ( P t ⊗ I M ) h ^ ) . (116) (115) is a single M N × M N MN\times MN M N × M N h ^ \hat{\mathbf{h}} h ^ T T T r \mathbf{r} r T T T
We solve (116) without ever forming E \mathbf{E} E E − 1 \mathbf{E}^{-1} E − 1 E − 1 = L L H \mathbf{E}^{-1}=\mathbf{L}\mathbf{L}^{H} E − 1 = L L H L \mathbf{L} L c ≜ L H h ^ \mathbf{c}\triangleq\mathbf{L}^{H}\hat{\mathbf{h}} c ≜ L H h ^ L L H h ^ = r \mathbf{L}\mathbf{L}^{H}\hat{\mathbf{h}}=\mathbf{r} L L H h ^ = r L c = r \mathbf{L}\mathbf{c}=\mathbf{r} Lc = r L H h ^ = c \mathbf{L}^{H}\hat{\mathbf{h}}=\mathbf{c} L H h ^ = c c \mathbf{c} c forward substitution : since L \mathbf{L} L i i i L i i c i = r i − ∑ j < i L i j c j L_{ii}c_{i}=r_{i}-\sum_{j<i}L_{ij}c_{j} L ii c i = r i − ∑ j < i L ij c j i i i 1 1 1 M N MN M N c \mathbf{c} c O ( ( M N ) 2 ) \mathcal{O}((MN)^{2}) O (( M N ) 2 ) h ^ \hat{\mathbf{h}} h ^ backward substitution : since L H \mathbf{L}^{H} L H i i i L i i ∗ h ^ i = c i − ∑ j > i L j i ∗ h ^ j L_{ii}^{*}\hat{h}_{i}=c_{i}-\sum_{j>i}L_{ji}^{*}\hat{h}_{j} L ii ∗ h ^ i = c i − ∑ j > i L j i ∗ h ^ j i i i M N MN M N 1 1 1 h ^ \hat{\mathbf{h}} h ^ O ( ( M N ) 3 ) \mathcal{O}((MN)^{3}) O (( M N ) 3 ) [36 ] , Sec. 4.2.5] and dominates the two triangular solves. Since E − 1 \mathbf{E}^{-1} E − 1 M N × M N MN\times MN M N × M N T T T T T T
The remaining work scales linearly with T T T T T T E − 1 \mathbf{E}^{-1} E − 1 r \mathbf{r} r E − 1 \mathbf{E}^{-1} E − 1 O ( N 2 τ ) O(N^{2}\tau) O ( N 2 τ ) ∑ t P t P t H \sum_{t} \mathbf{P}_{t}\mathbf{P}_{t}^{H} ∑ t P t P t H O ( M 2 K ) O(M^{2}K) O ( M 2 K ) I M − G t ( G t H G t ) − 1 G t H \mathbf{I}_{M}-\mathbf{G}_{t}(\mathbf{G}_{t}^{H}\mathbf{G}_{t})^{-1}\mathbf{G}_{t}^{H} I M − G t ( G t H G t ) − 1 G t H O ( ( M N ) 2 ) O((MN)^{2}) O (( M N ) 2 ) r \mathbf{r} r y t \mathbf{y}_{t} y t Y t ∈ C M × N \mathbf{Y}_{t}\in\mathbb{C}^{M\times N} Y t ∈ C M × N t t t r \mathbf{r} r vec ( ( I M − G t ( G t H G t ) − 1 G t H ) Y t P t ∗ ) \operatorname{vec}((\mathbf{I}_{M}-\mathbf{G}_{t}(\mathbf{G}_{t}^{H}\mathbf{G}_{t})^{-1}\mathbf{G}_{t}^{H})\mathbf{Y}_{t}\mathbf{P}_{t}^{*}) vec (( I M − G t ( G t H G t ) − 1 G t H ) Y t P t ∗ ) O ( M 2 τ + M N τ ) O(M^{2}\tau+MN\tau) O ( M 2 τ + M N τ ) O ( ( M N ) 2 + ( M + N ) 2 τ ) O((MN)^{2}+(M+N)^{2}\tau) O (( M N ) 2 + ( M + N ) 2 τ ) K ≤ M , N K\le M,N K ≤ M , N
Summing over the T T T
O ( T ⋅ ( ( M N ) 2 + ( M + N ) 2 τ ) + ( M N ) 3 ) , (117) O\left(T\cdot\left((MN)^{2}+(M+N)^{2}\tau\right)+(MN)^{3}\right), \tag*{(117)} O ( T ⋅ ( ( M N ) 2 + ( M + N ) 2 τ ) + ( M N ) 3 ) , (117) i.e. linear in T T T O ( ( M N ) 3 ) O((MN)^{3}) O (( M N ) 3 ) z \mathbf{z} z T K τ + M N TK\tau+MN T K τ + M N A H A \mathbf{A}^{H}\mathbf{A} A H A ( T K τ + M N ) × ( T K τ + M N ) (TK\tau+MN)\times(TK\tau+MN) ( T K τ + M N ) × ( T K τ + M N ) O ( ( T K τ + M N ) 3 ) O((TK\tau+MN)^{3}) O (( T K τ + M N ) 3 ) T 3 T^{3} T 3
