Transmit Antenna Combination Optimization for Generalized Spatial Modulation Systems

Generalized spatial modulation (GSM), where both the transmit antenna combination (TAC) index and the amplitude phase modulation (APM) symbols convey information, is a novel low-complexity and high efficiency multiple-input multiple-output (MIMO) technique. In conventional GSM (C-GSM), the legitimate TACs are selected randomly to transmit the APM symbols. However, the number of the TACs must be a power of two, hence the excess TACs are discarded, resulting in wasting some resource. To address these issues, in this paper, an optimal TAC set-aided enhanced GSM (E-GSM) scheme is proposed, where the optimal TAC set is selected with the aid of the channel state information (CSI) by maximizing the minimum Euclidean distance. Furthermore, a hybrid mapping GSM (HM-GSM) scheme operating without CSI knowledge is investigated, where the TAC selection and bit-to-TAC mapping are both taken into consideration for optimizing the average Hamming distance. Finally, an enhanced high throughput GSM (E-HT-GSM) scheme is developed, which makes full use of all the TACs. This scheme is capable of achieving an extra one bit transmission rate per time slot. Moreover, the rotated phase is employed and optimized for the reused TACs. Our simulation results show that the proposed E-GSM system and HM-GSM system are capable of outperforming the C-GSM system. Furthermore, the E-HT-GSM system is capable of obtaining one extra bit transmission rate per time slot compared with the C-GSM system.

AbbreviationExpansion

List of Acronyms
ABEP	Average Bit Error Probability
AMP	Amplitude Phase Modulation
AHD	Average Hamming Distance
BER	Bit Error Rate
C-GSM	Conventional Generalized Spatial Modulation
CS	Compressive Sensing
CSI	Channel State Information
ED	Euclidean Distances
E-GSM	Enhanced Generalized Spatial Modulation
E-HT-GSM	Enhanced High Throughput GSM
GSM	Generalized Spatial Modulation
HD	Hamming Distance
HM-GSM	Hybrid Mapping based GSM
ML	Maximum Likelihood
MED	Minimum Euclidean Distance
MIMO	Multiple Input Multiple Output
PEP	Pairwise Error Probability
RF	Radio Frequency
SM	Spatial Modulation
TA	Transmit Antennas
TAC	Transmit Antenna Combination
List of Symbols
$B$	Transmission rate
$B_{1}$	The number of bits that one TAC index carried
$B_{2}$	The number of bits that APM symbols carried
$C_{N_{t}}^{N_{u}}$	The total number of TACs
$C_{\text {C-GSM}}$	The complexity of C-GSM system
$C_{\text {E-GSM}}^{\text {Op}}$	The complexity of optimal TAC selection
$D({\bar \varsigma })$	The value of AHD
$F({\bar \varsigma })$	The value of PEP
$\mathbf {H}$	The channel matrix
$\mathbb {I}_{q}$	The q-th TAC set
$\mathbb {I}_{\text {all}}$	The TAC set including all the TACs
$I_{j}$	The j-th TAC
$M$	The modulation order of APM
${N_{t}}$	The total number of TAs
${N_{u}}$	The number of activated TAs
$N_{\text {all}}$	The total number of TACs
$N$	The number of one legitimate TAc set
$N_{\text {left}}$	The number of discarded TACs
$N_{r}$	The number of receiver antennas
$\mathbf {n}$	The noise matrix
$P_{b}$	The ABEP of GSM system
$\mathbf {y}$	The receive signal

SECTION I. Introduction

Generalized Spatial Modulation (GSM) [1], [2] is a novel low-complexity high-efficiency scheme relying on a reduced number of Radio Frequency (RF) chain for Multi-Input Multi-Output (MIMO) transmission. In the GSM scheme, multiple Transmit Antennas (TAs) are activated symbol in each interal, hence GSM may be flexibly reconfigured between Spatial Modulation (SM) [3]–​[5] and Spatial Multiplexing [6]. Specifically, in the GSM scheme, the information bits are conveyed by the index of the activated Transmit Antenna Combination (TAC) as well as by the Amplitude Phase Modulation (APM) symbols. Consequently, the GSM scheme attains a high bandwidth efficiency, despite using a low number of RF chains at the transmitter. Several independent studies have proved that SM and GSM are both promising candidates in the future wireless communications [7]–​[11].

In GSM systems, ${N_{u}}$ out of ${N_{t}}$ (${N_{u}}<{N_{t}}$ ) TAs are activated for data transmission, so that $N_{\text {all}}=C_{N_{t}}^{N_{u}}$ legitimate TACs are available. Among them, $N={2^{\lfloor {\log _{2}({C_{N_{t}}^{N_{u}}})} \rfloor }}$ TACs are chosen randomly for encoding the information bits, where $\lfloor \bullet \rfloor$ denotes the floor operation and $C_{x}^{y}$ represents the binary coefficient. The remaining $N_{\text {left}}=(N_{\text {all}}-N)$ TACs are discarded, hence resulting in wasting some resource. Therefore, how to choose the optimal TAC set and exploiting the remaining TACs become challenging research problems.

Early contributions on GSM have been mainly focused on the low-complexity receiver design [12]–​[19], on the performance and achieve rate analysis [20]–​[22], and on their applications for specific communication scenarios [23]. Specifically, Xiao et al. [12] proposed a low-complexity near- Maximum Likelihood (ML) detector. The detectors of [13]–​[16] exhibited a considerably reduced complexity, since the sparsity of the activated antennas was exploited based on the classic Compressive Sensing (CS) algorithms. The threshold-aided CS detector of [16] is capable of striking a better trade-off between the performance and complexity than other detection schemes. Moreover, the detector of [17] employed both forward error correction codes and turbo equalization to achieve the highest possible coding gain. Additionally, the low-complexity GSM conceived receiver design for dispersive channels have been investigated in [18] and [19]. On the other hand, the capacity of GSM system was analyzed in [20]–​[22] and the achievable rate was quantified in [21]. Since its high capacity and energy efficiency, GSM is a promising candidate for millimeter-wave communications [23]. However, the above research did not consider the Channel Sate Information (CSI) at the transmitter.

Recently, some transceiver designs of SM-based systems were developed in [24]–​[32], as shown in Table 1. Moreover, the CSI based link adaptation, antenna selection-aided distributed/cooperative protocols and precoding were investigated in [33]–​[40] for SM, as shown in Table 2. However, to the best of our knowledge how to select the TAC of GSM has not been investigated to date. Against the above background, the major contributions of this paper are summarized as follows:

1. An Enhanced GSM (E-GSM) system is proposed, where the TAC set is selected by exploiting the knowledge of the CSI based on maximizing the Minimum Euclidean Distance (MED). A low-complexity optimal TAC selection algorithm is developed for the E-GSM system. Our simulation results show that the proposed E-GSM scheme is capable of providing considerable performance gain over the Conventional GSM (C-GSM) system.

2. A Hybrid Mapping based GSM (HM-GSM) system operating without CSI knowledge is developed, where the TAC selection and bit-to-TAC mapping are both taken into consideration for improving the Average Bit Error Probability (ABEP).

3. Then an Enhanced Throughput GSM (E-HT-GSM) system is proposed, which conveys an extra bit. Specifically, the proposed E-HT-GSM system makes full use of all the TACs.

TABLE 1 Major Contributions on SM Transceiver Design Without CSI Knowledge

TABLE 2 Major Contributions on SM Transceiver Design Operating CSI Knowledge

The remainder of this paper is organized as follows. Section II gives a rudimentary introduction of the C-GSM system. In Section III, the optimal-TAC-set based E-GSM with CSI is introduced and a low-complexity TAC selection algorithm is proposed. In Section IV, a hybrid bit-to-TAC mapping method dispensing with CSI knowledge is proposed for the HM-GSM system to achieve a better ABEP. In Section V, an E-HT-GSM scheme is proposed, where all the TACs are exploited and the phase optimization as well as its performance analysis is offered in Section V. Section VI presents the simulation results, while Section VI concludes this paper. For reader’s convenience, the full table of contents is include here.

Notation: $\left \|{ \cdot }\right \|_{F}$ denotes the Frobenious norm of a matrix; $\left |{ \cdot }\right |$ represents the magnitude of a complex quantity; ${\left ({\cdot }\right)^{T}}$ and ${\left ({\cdot }\right)^{H}}$ stand for the transpose and the Hermitian transpose of a vector/matrix, respectively.

SECTION II. Conventional GSM

In C-GSM systems, ${N_{u}}$ out of ${N_{t}}$ (${N_{u}}<{N_{t}}$ ) TAs are activated for data transmission, so that $C_{N_{t}}^{N_{u}}$ legitimate TACs are available, and $N={2^{\lfloor {\log _{2}({C_{N_{t}}^{N_{u}}})} \rfloor }}$ TACs are exploited for encoding the information bits. In each time slot, a vector of $B$ information bits is partitioned in two parts, where ${B_{1}}=\lfloor {\log _{2}({C_{N_{t}}^{N_{u}}})} \rfloor$ bits are used to select a TAC $I_{i}$ ($i \in (1,N)$ ), and ${B_{2}}={N_{u}}{\log _{2}}(M)$ bits are used to modulate $N_{u}~M$ -APM symbols. Hence, the C-GSM transmit vector can be represented as \begin{align*} {\mathbf {x}}_{(I_{i},\mathbf {s})}= {\left [{ {\ldots,0,{s_{i_{1}}},0,\ldots,0,{s_{i_{2}}},0\ldots,0,{s_{i_{N_{u}}}},0,\ldots } }\right]^{T}}, \\\tag{1}{}\end{align*} where $(i_{1},\ldots,i_{N_{u}})$ indicates the activated TA indices.

Let ${\mathbf {H}} \in {\mathbb {C}^{N_{r} \times {N_{t}}}}$ and ${{\mathbf {n}}} \in {\mathbb {C}^{N_{r} \times 1}}$ be the MIMO channel matrix and noise matrix, whose entries are complex-valued Gaussian distributed, yielding $\mathcal {CN}(0,1)$ and $\mathcal {CN}(0,{\sigma ^{2}})$ , respectively. The received signal ${{\mathbf {y}}} \in {\mathbb {C}^{N_{r} \times 1}}$ can be formulated as \begin{equation*} {\mathbf {y}} = {\mathbf {H}}{{\mathbf {x}}_{(I_{i},{\mathbf {s}}_{i})}} + {\mathbf {n}} = \sum \limits _{t = {i_{1}}}^{{i_{N_{u}}}} {{{\mathbf {h}}_{t}}{s_{t}}} + {\mathbf {n}} = {{\mathbf {H}}_{I_{i}}}{\mathbf {s}}_{i} + {\mathbf {n}},\tag{2}\end{equation*} where ${{\mathbf {H}}_{I_{i}}} = ({{{\mathbf {h}}_{i_{1}}},{{\mathbf {h}}_{i_{2}}},\ldots,{{\mathbf {h}}_{{i_{N_{u}}}}}})$ is the sub-matrix of $\mathbf {H}$ with ${N_{u}}$ columns, and ${\mathbf {s}}_{i} = (s_{i_{1}},\ldots,s_{i_{N_{u}}})^{T}$ is the transmit symbol vector corresponding to the TAC $I_{i}=(i_{1},\ldots,i_{N_{u}})$ .

It follows from Eq. (2) that, the optimal ML-based demodulator can be formulated as \begin{equation*} {(\hat I,{\hat {\mathbf {s}}})_{\text {ML}}} = \mathop {\arg \min }\limits _{I \in \mathbb {I},{\mathbf {s}} \in \mathbb {S} } \left \|{ {{\mathbf {y}} - {{\mathbf {H}}_{I}}{\mathbf {s}}} }\right \|_{F}^{2},\tag{3}\end{equation*} where $\mathbb {I}=[I_{1}, I_{2},\ldots,I_{N}]$ is the set of TAC, and $\mathbb {S}$ is the set of $N_{u}$ -element symbol vector.

SECTION III. Transmit Antenna Combination Selection for the GSM System Relying on Channel State Information

In the C-GSM system, $N$ legitimate TACs are selected randomly from $C_{N_{t}}^{N_{u}}$ ones, whose performance can be further improved by selecting the optimal TAC set. In this section, an optimal TAC set based E-GSM system is proposed, where $N$ legitimate TACs are obtained utilizing the CSI instead of selecting them randomly based on maximizing the MED of GSM system. Then, a low-complexity optimal algorithm is introduced.

A. Optimal TAC Set Based E-GSM

For the C-GSM system having $N_{t}$ TAs and $N_{u}$ activated TAs, there are a total of $N_{\text {all}}=C_{N_{t}}^{N_{u}}$ TACs, and $N$ out of $N_{\text {all}}$ TACs are selected randomly. Hence, there are $C_{N_{\text {all}}}^{N}$ possible TAC sets $\mathbb {I}_{q}$ , $q=(1,\ldots,C_{N_{\text {all}}}^{N})$ . In the C-GSM system, a specific TAC set is selected randomly from $\{\mathbb {I}_{1},\ldots,\mathbb {I}_{C_{N_{\text {all}}}^{N}}\}$ . In this section, the optimal TAC set having the largest MED is selected from the $C_{N_{\text {all}}}^{N}$ TAC sets, as shown in Fig. 1. More specifically, for each TAC set $\mathbb {I}_{q}$ , the MED of the specific channel matrix $\mathbf {H}$ between the two GSM symbols ${\mathbf {x}_{m}}$ and ${\mathbf {x}_{n}}$ can be defined as [31]\begin{align*} d_{\min }^{q}=&~\min \limits _{{{\mathbf {x}}_{m}} \ne {{\mathbf {x}}_{n}}} ||{\mathbf {H}}{{\mathbf {x}}_{m}} - {\mathbf {H}}{{\mathbf {x}}_{n}}||_{F}^{2} \\=&~\min \limits _{I_{m},{I_{n}} \in {\mathbb {I}_{q}},({I_{m}},{{\mathbf {s}}_{m}}) \ne ({I_{n}},{{\mathbf {s}}_{n}})} ||{{\mathbf {H}}_{I_{m}}}{{\mathbf {s}}_{m}} - {{\mathbf {H}}_{I_{n}}}{{\mathbf {s}}_{n}}||_{F}^{2}.\tag{4}\end{align*} where $I_{m}=(m_{1},m_{2},\ldots,m_{N_{u}})$ and $I_{n}=(n_{1},n_{2},\ldots,n_{N_{u}})$ denote two TACs. For $C_{N_{\text {all}}}^{N}$ TAC sets we have $C_{N_{\text {all}}}^{N}$ MEDs and the optimal TAC set index having the largest MED can be obtained as \begin{equation*} \hat q = \arg \max \limits _{q \in \{1,2,\ldots,C_{N_{\text {all}}}^{N}\}} (d_{\min }^{q}).\tag{5}\end{equation*} Then the optimal TAC set for the specific channel matrix is $\mathbb {I}_{\hat q}$ . In the optimal TAC set selection, we have to calculate $C_{N_{\text {all}}}^{N}$ MEDs. For each MED, $NM^{N_{u}}$ Euclidean distances have to be calculated to get the minimum one. Hence, a total of $C_{N_{\text {all}}}^{N}NM^{N_{u}}$ Euclidean Distances (ED) should be computed for the optimal TAC set section. In fact, there are too many MEDs that are repeatly calculated. Hence, the optimal TAC set selection can be simplified by computing each ED only once. Assuming that the set $\mathbb {I}_{\text {all}}$ contains all the TACs as $\mathbb {I}_{\text {all}}=[I_{1},\ldots,I_{N},\ldots,I_{N_{\text {all}}}]$ , each TAC corresponding to $M^{N_{u}}$ GSM symbols, the set $\mathbb {X}_{\text {all}}$ containing $N_{\text {all}}M^{N_{u}}$ possible GSM vectors is expressed as \begin{equation*} \mathbb {X}_{\text {all}}=[\mathbf {x}_{1},\ldots,\mathbf {x}_{NM^{N_{u}}},\ldots,\mathbf {x}_{N_{\text {all}}M^{N_{u}}}],\tag{6}\end{equation*} Then, the ED of a specific channel matrix $\mathbf {H}$ between the two different GSM symbols ${\mathbf {x}_{m}} \in \mathbb {X}_{\text {all}}$ and ${\mathbf {x}_{n}} \in \mathbb {X}_{\text {all}}$ can be defined as \begin{equation*} d_{m,n}=||{\mathbf {H}}{{\mathbf {x}}_{m}} - {\mathbf {H}}{{\mathbf {x}}_{n}}||_{F}^{2}, m\ne n.\tag{7}\end{equation*} There are a total of $({N_{\text {all}}M^{N_{u}}}-1)^{2}\,\,\{d_{m,n}\}_{m\ne n=1}^{{N_{\text {all}}M^{N_{u}}}}$ values and we only have to calculate $({N_{\text {all}}M^{N_{u}}}-1)^{2}/2$ values due to the symmetry of the GSM symbols ($d_{m,n}=d_{n,m}$ ). The simplified optimal TAC set selection works follows.

• Step 1:

  Obtain the $({N_{\text {all}}M^{N_{u}}}-1)^{2}$ values of $d_{m,n}$ according to (7), which is shown in Table 3, where we have $N_{M}=M^{N_{u}}$ and $d_{m,n}=d_{n,m}$ .

• Step 2:

  Obtain the legitimate TAC sets $\mathbb {I}_{q}$ , $q=(1,\ldots,C_{N_{\text {all}}}^{N})$ .

• Step 3:

  For each TAC set $\mathbb {I}_{q}$ , find the $d_{\min }^{q}$ based on Table 3. Specifically, if we have $\mathbb {I}_{q}=[I_{1},\ldots,I_{N}]$ , we find the MED $d_{\min }^{q}$ from the values of $d_{m,n}$ corresponding to $\mathbb {I}_{q}$ based on Table 3.

• Step 4:

  Get the optimal TAC set index by (5).

TABLE 3 Values of $d_{m,n}$

FIGURE 1.

System model of the proposed E-GSM scheme.

Although the simplified optimal TAC set can be obtained by only calculating $({{N_{\text {all}}}M^{N_{u}}}-1)^{2}/2$ EDs, its complexity is still excessive for large values of $N_{t}$ and $N_{u}$ . Next, a simplified optimal TAC set selection algorithm is introduced.

B. Low-Complexity Optimal TAC Selection Based E-GSM

In this section, low-complexity optimal TAC selection algorithms are developed for $M=1$ and $M>1$ respectively. Specifically, for the case of $M=1$ , the activated antenna indices transmit the symbol ‘1,’ which is also termed as Generalized Space Shift Keying (GSSK) scheme [41].

1) TAC Design for $M=1$

For the case of $M=1$ , the activated TAC transmits the symbol ‘1,’ so that the ED matrix can be defined as \begin{equation*} {\mathbf {W}} = \left [{ {\begin{array}{*{20}{c}} {{w_{11}}}&\quad {{w_{12}}}&\quad \cdots &\quad {{w_{1{N_{{\mathrm {all}}}}}}}\\ {{w_{21}}}&\quad {{w_{22}}}&\quad \cdots &\quad {{w_{2{N_{{\mathrm {all}}}}}}}\\ \vdots &\quad \vdots &\quad \ddots &\quad \vdots \\ {{w_{{N_{{\mathrm {all}}}}1}}}&\quad {{w_{{N_{{\mathrm {all}}}}2}}}&\quad \cdots &\quad {{w_{{N_{{\mathrm {all}}}}{N_{{\mathrm {all}}}}}}} \end{array}} }\right]\!,\tag{8}\end{equation*} where the element $w_{mn}$ , $m=(1,\ldots,{N_{{\mathrm {all}}}})$ , $n=(1,\ldots,{N_{{\mathrm {all}}}})$ denotes the ED between two GSM symbols as \begin{align*} w_{mn}=&~||{{\mathbf {H}}_{I_{m}}} - {{\mathbf {H}}_{I_{n}}})||_{F}^{2} \\=&~\left \|{ {({{\mathbf {h}}_{m_{1}}} + {{\mathbf {h}}_{m_{2}}} + \cdots {{\mathbf {h}}_{{m_{N_{u}}}}}) - ({{\mathbf {h}}_{n_{1}}} + {{\mathbf {h}}_{n_{2}}} + \cdots {{\mathbf {h}}_{{n_{N_{u}}}}})} }\right \|_{F}^{2} \\=&~\! \begin{cases} 0,&\quad \!\!\!\!\! {\mathrm {if~}}{I_{m}} = {I_{n}}\\ ||{{\mathbf {H}}_{D_{m}}} - {{\mathbf {H}}_{D_{n}}})||_{F}^{2},&\quad \!\!\!\!\! {\mathrm {if~}}{I_{m}} \ne {I_{n}}, \end{cases}\tag{9}\end{align*} where ${D_{m}}$ and ${D_{n}}$ represent the different elements between $I_{m}$ and $I_{n}$ . Assuming that the identical element set between $I_{m}$ and $I_{n}$ is $\Lambda = {\mathrm {intersect}}({I_{m}},{I_{n}})$ , the values of ${D_{m}}$ and ${D_{n}}$ are obtained by \begin{equation*} {D_{m}} = \mathrm {setdiff}({I_{m}}, \Lambda), {D_{n}} = {\mathrm {setdiff}}({I_{n}}, \Lambda),\tag{10}\end{equation*} where $\text {intersect}(A,B)$ denotes the function returning the identical elements between $A$ and $B$ , while setdiff(A, B) denotes the function returning different elements between $A$ and $B$ . As a result, (5) can be reformulated as \begin{align*} d_{\min }^{q}=&~\min \limits _{{{\mathbf {x}}_{m}} \ne {{\mathbf {x}}_{n}}} ||{\mathbf {H}}{{\mathbf {x}}_{m}} - {\mathbf {H}}{{\mathbf {x}}_{n}}||_{F}^{2} \\=&~\min \limits _{\forall {D_{m,}}{D_{n}}} ||{{\mathbf {H}}_{D_{m}}} - {{\mathbf {H}}_{D_{n}}})||_{F}^{2}.\tag{11}\end{align*} For $q=(1,\ldots,C_{N_{\text {all}}}^{N})$ , there are a lot of repeated calculations of $||{{\mathbf {H}}_{D_{m}}} - {{\mathbf {H}}_{D_{n}}})||_{F}^{2}$ . To reduce the complexity of finding the optimal set based TAC selection, only the different values of $||{{\mathbf {H}}_{D_{m}}} - {{\mathbf {H}}_{D_{n}}})||_{F}^{2}$ are calculated.

For the generalized cases of $I_{m}=(m_{1},m_{2},\ldots,m_{N_{u}})$ and $I_{n}=(n_{1},n_{2},\ldots,n_{N_{u}})$ , assuming that the size of $\Lambda$ is $\beta$ , the number of elements in $D_{m}$ and $D_{n}$ are $N_{u}- \beta$ . Then the number of the identical values of $||{{\mathbf {H}}_{D_{m}}} - {{\mathbf {H}}_{D_{n}}})||_{F}^{2}$ is \begin{equation*} {N_{{\mathrm {same}}}} = 2(C_{N_{t} - 2({N_{u}} - \beta)}^\beta).\tag{12}\end{equation*} As a result, the number of different nonzero values of $||{{\mathbf {H}}_{D_{m}}} - {{\mathbf {H}}_{D_{n}}})||_{F}^{2}$ in the ED matrix $\mathbf {W}$ is \begin{equation*} {N_{ED}} = \sum \limits _{\beta = 0}^{N_{u} - 1} {\frac {{C_{N_{t}}^\beta C_{N_{t} - \beta }^{N_{u} - \beta }C_{N_{t} - \beta - {N_{u}}}^{N_{u} - \beta }}}{2}}.\tag{13}\end{equation*} For the generalized cases, there are a total of $N_{\text {all}}$ TACs and $(N_{\text {all}}-N)$ TACs have to be removed from $\mathbb {I}_{\text {all}}$ . After obtaining the ${N_{ED}}$ different EDs $\mathbf {w}=[w_{1}, \ldots, w_{N_{ED}}]$ , the TAC selection for $M=1$ operates as follows.

• Step 1:

  Obtain the ED matrix $\mathbf {W}$ based on ${N_{ED}}$ different EDs $\mathbf {w}=[w_{1}, \ldots, w_{N_{ED}}]$ .

• Step 2:

  Sort the $N_{ED}$ EDs in an ascending order as \begin{equation*} [v_{1},v_{2},\ldots v_{N_{ED}}]=\text {sort}(\mathbf {w},`ascend').\tag{14}\end{equation*}

• Step 3:

  Remove all the values of $v_{1}~v_{2}$ , $\ldots$ , in the sequence from the matrix $\mathbf {W}$ until we removed $N_{\text {left}}$ TACs defining $\mathbb {I}_{\text {Re}}$ .

• Step 4:

  Obtain the optimal TAC set as $\mathbb {I}_{o}=\mathbb {I}_{\text {all}}\backslash \mathbb {I}_{\text {Re}}$ .

Considering $N_{t}=4\,\,N_{u}=2$ for example, the EDs between any two GSM symbols are presented in Table 4 and we have \begin{align*} {w_{12}}=&~{w_{21}} = {w_{56}} = {w_{65}} = ||{{\mathbf {h}}_{2}}-{{\mathbf {h}}_{3}}||_{F}^{2}; \\ {w_{13}}=&~{w_{31}} = {w_{46}} = {w_{64}} = ||{{\mathbf {h}}_{2}}-{{\mathbf {h}}_{4}}||_{F}^{2}; \\ {w_{14}}=&~{w_{41}} = {w_{36}} = {w_{63}} = ||{{\mathbf {h}}_{1}}-{{\mathbf {h}}_{3}}||_{F}^{2}; \\ {w_{15}}=&~{w_{51}} = {w_{26}} = {w_{62}} = ||{{\mathbf {h}}_{1}}-{{\mathbf {h}}_{4}}||_{F}^{2}; \\ {w_{16}}=&~{w_{61}} = ||({{\mathbf {h}}_{1}} + {{\mathbf {h}}_{2}}) - ({{\mathbf {h}}_{3}} + {{\mathbf {h}}_{4}})||_{F}^{2}; \\ {w_{23}}=&~{w_{32}} = {w_{45}} = {w_{54}} = ||{{\mathbf {h}}_{3}}-{{\mathbf {h}}_{4}}||_{F}^{2}; \\ {w_{24}}=&~{w_{42}} = {w_{35}} = {w_{53}} = ||{{\mathbf {h}}_{1}}-{{\mathbf {h}}_{2}}||_{F}^{2}; \\ {w_{25}}=&~{w_{52}} = ||({{\mathbf {h}}_{1}} + {{\mathbf {h}}_{3}}) - ({{\mathbf {h}}_{2}} + {{\mathbf {h}}_{4}})||_{F}^{2}; \\ {w_{34}}=&~{w_{43}} = ||({{\mathbf {h}}_{1}} + {{\mathbf {h}}_{4}}) - ({{\mathbf {h}}_{2}} + {{\mathbf {h}}_{3}})||_{F}^{2}.\tag{15}\end{align*} As a result, there are only a total of 9 EDs $\mathbf {w}=[w_{12}, w_{13}, w_{14}, w_{15}, w_{16}, w_{23}, w_{24}, w_{25}, w_{34}]$ to be computed for TAC selection. If the minimum value of $\mathbf {w}$ is $w_{12}$ , the removed TAC set $\mathbb {I}_{\text {Re}}$ may represent any of the combinations $\{(I_{1}, I_{5}), (I_{1},I_{6}), (I_{2}, I_{5}), (I_{2},I_{6})\}$ .

TABLE 4 ED Matrix $\mathbf {W}$ of $N_{t}=4, M=1$

2) TAC Design for $M>1$

For the case of $M>1$ , the elements in $\mathbf {W}$ of (8) can be defined as \begin{align*} {w_{mn}}=&~\min \limits _{\forall {{\mathbf {s}}_{m}},{{\mathbf {s}}_{n}}} ||{{\mathbf {H}}_{I_{m}}}{{\mathbf {s}}_{m}} - {{\mathbf {H}}_{I_{n}}}{{\mathbf {s}}_{n}}||_{F}^{2} \\=&~\! {\begin{cases} \min \limits _{\forall {{\mathbf {s}}_{m}},{{\mathbf {s}}_{n}}} ||{{\mathbf {H}}_{I_{m}}}({{\mathbf {s}}_{m}} - {{\mathbf {s}}_{n}})||_{F}^{2},&\quad {\mathrm {if~}}{I_{m}} = {I_{n}},{{\mathbf {s}}_{m}} \ne {{\mathbf {s}}_{n}}\\ \min \limits _{\forall {{\mathbf {s}}_{m}},{{\mathbf {s}}_{n}}} ||{{\mathbf {H}}_\Lambda }({\mathbf {s}}_{m}^\Lambda - {\mathbf {s}}_{n}^\Lambda) \\ \quad~ \quad + {{\mathbf {H}}_{D_{m}}}\bar {{\mathbf {s}}}_{m}^\Lambda \!-\! {{\mathbf {H}}_{D_{n}}}\bar {{\mathbf { s}}}_{n}^\Lambda ||_{F}^{2},&\quad {\mathrm {if~}}{I_{m}} \ne {I_{n}} \end{cases}}\!\!\! \\\tag{16}\end{align*} where ${\mathbf {s}}_{m}^\Lambda$ and ${\mathbf {s}}_{n}^\Lambda$ represent the specific subsets of ${{\mathbf {s}}_{m}}$ and ${{\mathbf {s}}_{n}}$ corresponding to the $\Lambda$ index set and $\bar {{\mathbf {s}}}_{m}^\Lambda = {\mathbf {s}}_{m}^{}\backslash {\mathbf {s}}_{m}^\Lambda,\bar {{\mathbf {s}}}_{n}^\Lambda = {\mathbf {s}}_{n}^{}\backslash {\mathbf {s}}_{n}^\Lambda$ . Due to the associated symmetry, each $w_{mn}$ can be obtained by computing $[(1 + {M^{N_{u}}}){M^{N_{u}}}]/2$ EDs in (16). Since we have $w_{mn}=w_{nm}$ , the ED matrix $\mathbf {W}$ of (8) can be obtained by computing $N_{ED}^{M} = [({N_{all}} + 1){N_{all}}]/2$ different values ${w_{mn}}$ . After obtaining the ED matrix $\mathbf {W}$ , the TAC selection can be completed by the Steps.1–4 of the TAC design of $M=1$ .

Considering $N_{t}=4\,\,N_{u}=2$ for example, the ED matrix $\mathbf {W}$ is presented in Table 5, where $w_{mn}$ can be obtained by computing 10 EDs. Observe from Eq. (15) and Tables 4–​5 that when the number of common minimum values of ED matrix $\mathbf {W}$ satisfied that $N_{\text {same}}/2>N_{\text {left}}$ , the minimum value will still exist after removing $N_{\text {left}}$ TACs. In this case, the value of (11) is the same as that of the conventional TAC set.

TABLE 5 ED Matrix $\mathbf {W}$ of $N_{t}=4, M$ > 1

C. Complexity Analysis of E-GSM Systems

In this section, the complexity orders of the ML-aided C-GSM and of the proposed E-GSM are compared in terms of the numbers of real-valued multiplications and additions. For the specific matrices of $\mathbf {A}\in \mathbb {C}^{m \times n}$ , $\mathbf {B}\in \mathbb {C}^{n \times p}$ , $\mathbf {c}\in \mathbb {C}^{n \times 1}$ and $\mathbf {d}\in \mathbb {C}^{n \times 1}$ , the operations of ${\mathbf {A}}{\mathbf {B}}$ , $\|{\mathbf {c}}\|_{F}^{2}$ and ${\mathbf {c}} \pm {\mathbf {d}}$ require $8mnp-2mp$ , $4n-1$ , and $2n$ Floating-point operations (Flops), respectively. Accordingly, the complexity order of the C-GSM relying on the ML detector becomes \begin{equation*} C_{\text {C-GSM}}=(8N_{r}N_{u} +4N_{r}-1)NM^{N_{u}},\tag{17}\end{equation*} since the operation ${\mathrm {||}}{\mathbf {y}} - {{\mathbf {H}}_{I}}{\mathbf {s}}|{|_{F}^{2}}$ requires $C_{F}=(8N_{r}N_{u} +4N_{r}-1)$ Flops, and this operation is computed $2^{B}=NM^{N_{u}}$ times;

The complexity order of the low-complexity optimal TAC set based E-GSM system with ML detector is \begin{align*} C_{{\mathrm {E - GSM}}}^{{\mathrm {Op}}} \!= \!\! \begin{cases} {C_{{\mathrm {C - GSM}}}} \!+\! {C_{F}}N_{ED}^{},&\quad \text {if} ~ M = 1\\ {C_{{\mathrm {C - GSM}}}} \!+\! {C_{F}}N_{ED}^{M}\displaystyle \frac {{(1 \!+\! {M^{N_{u}}}){M^{N_{u}}}}}{2},&\quad \text {else} \end{cases}\!\!\!\!\! \\\tag{18}\end{align*} since $C_{N_{t}}^{N_{u}}(M^{N_{u}}-1)^{2}/2$ EDs are calculated in the TAC set selection.

SECTION IV. Transmit Antenna Combination Selection for the GSM System Operating Without Channel State Information

In this selection, the ABEP assisted TAC selection conceived for the HM-GSM system operating without CSI knowledge is investigated. Firstly, the ABEP analysis of the GSM system is carried out. Then, its TAC selection is developed. Finally, a novel bit-to-TAC mapping approach is proposed for further improving the GSM system’s performance.

A. ABEP Analysis of the GSM System

In this section, the ABEP performance of the GSM system is derived. Let us denote the transmit and receive signal of GSM by ${{\mathbf {x}}_{i}}$ and ${{\mathbf {x}}_{j}}$ , respectively. Then the ABEP upper bound is given by \begin{equation*} {P_{b}} = \sum \limits _{\forall {{\mathbf {x}}_{i}}} {\sum \limits _{\forall {{\mathbf {x}}_{j}} \ne {{\mathbf {x}}_{i}}} {\frac {{{d_{{{\mathbf {x}}_{i}},{{\mathbf {x}}_{j}}}}{\bar P_{e}}({{\mathbf {x}}_{i}} \to {{\mathbf {x}}_{j}})}}{{B{2^{B}}}}} },\tag{19}\end{equation*} where ${\bar P_{e}}({{\mathbf {x}}_{i}} \to {{\mathbf {x}}_{j}})$ is the Pairwise Error Probability (PEP) and ${d_{{{\mathbf {x}}_{i}},{{\mathbf {x}}_{j}}}}$ is the number of bit errors associated with the corresponding PEP event. According to [24], ${\bar P_{e}}({{\mathbf {x}}_{i}} \to {{\mathbf {x}}_{j}})$ may be expressed as \begin{align*} {\bar P_{e}}({{\mathbf {x}}_{i}} \to {{\mathbf {x}}_{j}}) = F(\bar \zeta) = \gamma (\bar \varsigma)\sum \limits _{k = 0}^{N_{r} - 1} {C_{N_{r} - 1 + k}^{k}} {[1 - \gamma \left ({{\bar \varsigma } }\right)]^{k}}, \\\tag{20}\end{align*} where $\gamma \left ({{ \bar \varsigma } }\right) = \frac {1}{2}\left ({{1 - \sqrt {\frac {\bar \varsigma /2}{{1 + {\bar \varsigma } /2}}} } }\right)$ and ${\bar \varsigma }$ is the mean value of $\varsigma = ||{\mathbf {H}}({{\mathbf {x}}_{i}} - {{\mathbf {x}}_{j}})||_{F}^{2}/2{\sigma ^{2}}$ with $N_{r}=1$ . For the GSM system, assuming that the antenna indices of the transmit signal ${{\mathbf {x}}_{i}}$ and the estimated ${\mathbf {x}}_{j}$ are $(l_{1},l_{2},\ldots,l_{N_{u}})$ and $(\hat l_{1},\hat l_{2},\ldots,\hat l_{N_{u}})$ , respectively, and the corresponding symbol vectors are $\mathbf {s}=[s_{1}, s_{2},\ldots,s_{N_{u}}]^{T}$ and $\hat {\mathbf {s}}=[\hat s_{1}, \hat s_{2},\ldots,\hat s_{N_{u}}]^{T}$ . Then the value of ${\bar \varsigma }$ for the GSM system is given by \begin{align*} {\bar \varsigma }= \! \begin{cases} \displaystyle \frac {{{{\left |{ {s_{1} - {\hat s_{1}}} }\right |}^{2}} + \cdots + {{\left |{ {{s_{N_{u}}} - {\hat s_{N_{u}}}} }\right |}^{2}}}}{{2{\sigma ^{2}}}},\quad {\text {if }m=N_{u}} \\[-2pt] \displaystyle \frac {{{{\left |{ {s_{1} - {\hat s_{1}}} }\right |}^{2}} + \cdots + {{\left |{ {s_{m} - {\hat s_{m}}} }\right |}^{2}} + 2({N_{u}} - m)}}{{2{\sigma ^{2}}}},\\[-2pt] &\hspace {-5.2pc}\quad {\text {if}}~ 0 <m< N_{u} \\[-2pt] \displaystyle \frac {{2{N_{u}}}}{{2{\sigma ^{2}}}},&\hspace {-5.4pc}\quad {\text {if}} ~ m=0,\\[-2pt] \end{cases} \\[-2pt]\tag{21}\end{align*} where $m$ is the number of identical antenna indices between $(l_{1},l_{2},\ldots,l_{N_{u}})$ and $(\hat l_{1},\hat l_{2},\ldots,\hat l_{N_{u}})$ . Based on the values of ${\bar \varsigma }$ obtained in Eq. (21), the ABEP of the GSM system can be evaluated using (19).

Observe from (21) that there are lots of identical ${\bar \varsigma }$ values and the number of different values of ${\bar \varsigma }$ is finite for a fixed constellation size $M$ , $N_{r}$ and SNR variance $\sigma ^{2}$ . Hence, the above ABEP expressions can be further simplified. Assuming that the GSM systems have $n$ different values ${\bar \varsigma }$ as ${\bar \varsigma }_{1}$ ,$\ldots$ ,${\bar \varsigma }_{n}$ , the corresponding PEP values $F({\bar \varsigma _{1}})$ ,$\ldots$ ,$F({\bar \varsigma _{n}})$ can be obtained from (20), so that (19) can be represented as \begin{align*} {P_{b}}=&~\frac {{\sum \limits _{t = 1}^{\lambda _{1}} {d_{\bar \varsigma _{1}}^{t}F({\bar \varsigma _{1}})} + \sum \limits _{t = 1}^{\lambda _{2}} {d_{\bar \varsigma _{2}}^{t}F({\bar \varsigma _{2}})} + \cdots + \sum \limits _{t = 1}^{\lambda _{n}} {d_{\bar \varsigma _{2}}^{t}F({\bar \varsigma _{n}})} }}{{B{2^{B}}}} \\=&~D({\bar \varsigma _{1}})F({\bar \varsigma _{1}}) + D({\bar \varsigma _{2}})F({\bar \varsigma _{2}}) + \cdots + D({\bar \varsigma _{n}})F({\bar \varsigma _{n}}) \\\tag{22}{}\end{align*} with \begin{align*}&D({\bar \varsigma _{p}}) = {{\sum \limits _{t = 1}^{\lambda _{p}} {d_{\bar \varsigma _{p}}^{t}} }}/{{B{2^{B}}}},\quad p = 1,\ldots,n, \\[-8pt]&\sum \limits _{t = 1}^{\lambda _{1}} {d_{\bar \varsigma _{1}}^{t}} + \cdots + \sum \limits _{t = 1}^{\lambda _{n}} {d_{\bar \varsigma _{n}}^{t}} = \sum \limits _{i = 1}^{2^{B}} {\sum \limits _{j = 1}^{2^{B}} {d\left ({{{{\mathbf {x}}_{i}},{{\mathbf {x}_{j}}}} }\right)} } = {2^{B}}\sum \limits _{u = 1}^{B} {C_{B}^{u}u},\!\!\! \\\tag{23}{}\end{align*} where ${\lambda _{p}}~p\in \{1,\ldots,n\}$ is the total number of candidate ${\bar \zeta }_{p}$ for all the ${{\mathbf {x}}_{i}}$ and ${{\mathbf {x}}_{j}}$ , while $d_{\bar \varsigma _{p}}^{t}$ is the corresponding Hamming Distance (HD), and $D({\bar \varsigma _{p}})$ is the Average HD (AHD) for the value ${\bar \zeta }_{p}$ .

Next, AHD based TAC selection is developed. Since only the TACs are selected, we employ $M=1$ for our GSM system for simplicity, so that Eq. (21) can be further simplified to \begin{equation*} {\bar \varsigma }= \! \begin{cases} 0,&\quad {\text {if }m=N_{u}} \\ \displaystyle \frac {2({N_{u}} - m)}{{2{\sigma ^{2}}}},&\quad {\text {if}}~ 0 <m< N_{u} \\ \displaystyle \frac {{2{N_{u}}}}{{2{\sigma ^{2}}}},&\quad {\text {if}} ~ m=0. \\ \end{cases}\tag{24}\end{equation*} As a result, there are a total of $N_{u}$ different values ${\bar \varsigma }$ as ${\bar \varsigma }_{1}$ ,$\ldots$ ,${\bar \varsigma }_{N_{u}}$ as \begin{align*} {\bar \varsigma _{1}}=&~\frac {1}{\sigma ^{2}},\quad m = {N_{u}} - 1 \\ {\bar \varsigma _{2}}=&~\frac {2}{\sigma ^{2}},\quad m = {N_{u}} - 2 \\&~\vdots \\ { \bar \varsigma _{N_{u}}}=&~\frac {N_{u}}{\sigma ^{2}},\quad m = 0.\tag{25}\end{align*} Hence, the ABEP of the GSM system associated with $M=1$ is dominated by the values of ${\bar \varsigma _{n}}$ and $F\left ({{\bar \varsigma _{n}} }\right)$ . The relationship between ${\bar \varsigma _{n}}$ and $F\left ({{\bar \varsigma _{n}} }\right)$ is presented graphically in Fig. 2. Observe from Fig. 2 that we have \begin{equation*} F\left ({{\bar \varsigma _{1}} }\right) >>F\left ({{\bar \varsigma _{2}} }\right) > \cdots > F\left ({{{\bar \varsigma _{N_{u}}}} }\right).\tag{26}\end{equation*} where $x>>y$ denotes the value of $x$ is much larger than that of $y$ . Hence, the associated optimization principle can be invoked for designing a GSM system having a small value of $D({\bar \varsigma _{1}})$ . According to (23), the value of $D({\bar \varsigma _{1}})$ is dominated by two parts: the total number ${\lambda _{1}}$ of candidate ${\bar \zeta }_{1}\,\,(m=N_{u}-1)$ and its corresponding HD of $d_{\bar \varsigma _{1}}^{t}\,\,(t=1,\ldots,{\lambda _{1}})$ . Next, the TAC selection aims for attaining a small value of ${\lambda _{1}}$ and $d_{\bar \varsigma _{1}}^{t}\,\,(t=1,\ldots,{\lambda _{1}})$ , which are introduced as follows.

FIGURE 2.

The values of $F\left ({{\bar \varsigma } }\right)$ .

B. TAC Selection

For each TAC $I_{j}, j\in [1,{N_{\text {all}}}]$ , the set ${\psi _{j}} = {{\mathbb {I}}_{{\mathrm {all}}}}\backslash {I_{j}}$ consists of $N_{u}$ parts $\psi _{j}^{N_{u}-1}$ , $\psi _{j}^{N_{u}-2}$ and $\psi _{j}^{0}$ , where $\forall {I_{r}} \in \psi _{j}^{m}\,\,m=0,\ldots,N_{u}-1$ indicates that there are $m$ common indices between ${I_{r}}$ and ${I_{j}}$ , while $A\backslash B$ represents removing $B$ from $A$ . Next, the TAC selections invoked for specific antenna setups are introduced as follows.

1) TAC Selection for $N_{t}=4, N_{u}=2$

Fig. 3 portrays the TAC selection for the case of $N_{t}=4$ and $N_{u}=2$ . In this scenario, we have ${\mathbb {I}}_{\text {all}}=[(1,2),(1,3),(1,4),(2,3),(2,4)]$ . For any two TACs from ${\mathbb {I}}_{\text {all}}$ , the two TACs are connected by the curves seen in Fig. 3, if there are $N_{u}-1$ common indices between the two TACs. As a result, the TAC selection problem has been transformed as to how we remove TACs having the lowest number of curves. For the case of $N_{t}=4$ and $N_{u}=2$ , we have \begin{align*} \underbrace {(1,2)}_{I_{1}} \to \psi _{1}^{1}=&~\{ \underbrace {(1,3)}_{I_{2}},\underbrace {(1,4)}_{I_{3}},\underbrace {(2,3)}_{I_{4}},\underbrace {(2,4)}_{I_{5}}\}; \psi _{1}^{0} = \{ \underbrace {(3,4)}_{I_{6}}\} \\ \underbrace {(1,3)}_{I_{2}} \to \psi _{2}^{1}=&~\{ \underbrace {(1,2)}_{I_{1}},\underbrace {(1,4)}_{I_{3}},\underbrace {(2,3)}_{I_{4}},\underbrace {(3,4)}_{I_{6}}\}; \psi _{2}^{0} = \{ \underbrace {(2,4)}_{I_{5}} \\ \underbrace {(1,4)}_{I_{3}} \to \psi _{3}^{1}=&~\{ \underbrace {(1,2)}_{I_{1}},\underbrace {(1,3)}_{I_{2}},\underbrace {(2,4)}_{I_{5}},\underbrace {(3,4)}_{I_{6}}\}; \psi _{3}^{0} = \{ \underbrace {(2,3)}_{I_{4}}. \\\tag{27}{}\end{align*}

Observe from Fig. 3 that by removing $({I_{1}},\psi _{1}^{0})$ , $({I_{2}},\psi _{2}^{0})$ and $({I_{3}},\psi _{3}^{0})$ , we can obtain a smaller value of ${\lambda _{1}}$ .

FIGURE 3.

The TAC selection for HM-GSM system having $N_{t}=4$ and $N_{u}=2$ .

2) TAC Selection for $N_{t}=5, N_{u}=2$

Fig. 4 portrays the TAC selection for the case of $N_{t}=5$ and $N_{u}=2$ . For the case of $N_{t}=5$ and $N_{u}=2$ , we have ${\mathbb {I}}_{\text {all}}=[(1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4),(3,5), (4,5)]$ . $N=8$ TACs should be selected from the set ${\mathbb {I}}_{\text {all}}$ . For the case of $N_{t}=5$ and $N_{u}=2$ , we have \begin{equation*} \underbrace {(1,2)}_{I_{1}} \to \psi _{1}^{0} = \{ \underbrace {(3,4)}_{I_{8}},\underbrace {(3,5)}_{I_{9}},\underbrace {(4,5)}_{{I_{10}}}\}.\tag{28}\end{equation*} Observe from Fig. 4 that by removing $({I_{1}},{I_{8}})$ , $({I_{1}},{I_{9}})$ and $({I_{1}},{I_{10}})$ , we arrive at at a reduced value of ${\lambda _{1}}$ .\begin{align*} \psi _{1}^{1}=&~\{ {(1,3),(1,4),(1,5),(2,3),(2,4),(2,5)}\} \\ \psi _{1}^{0}=&~\{ {(3,4),(3,5),(4,5)}\}.\tag{29}\end{align*} Then we can get the TAC set as \begin{align*} {\mathbb {I}_{o}}=\{ {\underbrace {(1,3),(1,4),(1,5),(2,3),(2,4),(2,5)}_{\psi _{1}^{1}},\underbrace {(3,4),(3,5)}_{{\mathcal {I}_{{\mathrm {left}}}}}}\}.\!\!\! \\\tag{30}{}\end{align*}

FIGURE 4.

The TAC selection for HM-GSM system having $N_{t}=5$ and $N_{u}=2$ .

3) Generalized TAC Selection for $N_{u}=2$

Based on the above TAC selection examples, the generalized TAC selection design is introduced as follows.

• Step 1:

  For the specific TAC $I_{j},j\in [1,{N_{\text {all}}}]$ , obtain the corresponding TAC set $\psi _{j}^{N_{u}-1}$ , $\psi _{j}^{N_{u}-2},\ldots,\psi _{j}^{0}$ .

• Step 2:

  Remove ${N_{\text {all}}}-N+1$ TACs ${\mathcal {I}_{j}}$ from the TAC set ${\psi _{j}^{N_{u}-2},\ldots,\psi _{j}^{0}}$ and obtain the ${\mathcal{ I}}_{\text {left}}=\{\psi _{j}^{N_{u}-2},\ldots,\psi _{j}^{0}\} \backslash {\mathcal {I}_{j}}$

• Step 3:

  Obtain the final TAC set ${\mathbb {I}}_{o}=\{\psi _{j}^{N_{u}-1}, {\mathcal{ I}}_{\text {left}} \}$ .

Due to the associated symmetry, the TAC selection is the same for each $I_{j}$ . For simplicity, we introduce the TAC selection process for $I_{1}=[1,2,\ldots,N_{u}]$ in detail as follows.

For the case of $N_{u}=2$ , the TAC set can be expressed as \begin{equation*} {{\mathbb {I}}}_{{N_{{\mathrm {all}}}}} = \left \{{\! \begin{array}{c} \underbrace {(1,2),(1,3) \cdots (1,{N_{t}})}_{N_{t} - 1}\\ \underbrace {(2,3),(2,4) \cdots (2,{N_{t}})}_{N_{t} - 2}\\ \underbrace {(3,4),(3,5) \cdots (3,{N_{t}})}_{N_{t} - 3}\\ {} \vdots \\ \underbrace {({N_{t}} - 1,{N_{t}})}_{1} \end{array} }\right \}.\tag{31}\end{equation*}

Firstly, for the TAC $I_{1}=(1,2)$ , we can obtain the corresponding set $\psi _{1}^{ 1}$ and $\psi _{1}^{0}$ as \begin{align*} \psi _{1}^{ 1}=&~\left \{{\! \begin{array}{c} \underbrace {((1,3), \cdots (1,{N_{t}})}_{N_{t} - 2}\\ \underbrace {(2,3), \cdots (2,{N_{t}})}_{N_{t} - 2} \end{array} }\right \},\quad \! \psi _{1}^{0} = \left \{{\! \begin{array}{c} \underbrace {(3,4), \cdots (3,{N_{t}})}_{N_{t} - 3}\\ {} \vdots \\ \underbrace {({N_{t}} - 1,{N_{t}})}_{1} \end{array} }\right \}\!. \\\tag{32}{}\end{align*}

Secondly, select $N-(N_{t}-2)^{2}$ TACs from $\psi _{1}^{0}$ as ${\mathcal{I}}_{\text {left}} \subset \psi _{1}^{0}$ and obtain the final TAC set as ${\mathbb {I}_{o}}=\{\psi _{1}^{ 1}, {\mathcal{I}}_{\text {left}}\}$ .

4) Generalized TAC Selection for $N_{u}>2$

For the case of $N_{u}>2$ , there are a total of $C_{N_{t}}^{N_{u}}$ TACs. For the TAC $I_{1}=({1,2,\ldots,N_{u}})$ , to obtain the set $\psi _{1}^{m}$ , $m=0,\ldots,N_{u}-1$ , we first choose the $m$ indices from $I_{1}=(1,2,\ldots,N_{u})$ and choose the left $N_{u}-m$ indices from $(N_{u}+1,\ldots,N_{t})$ , hence there are a total number of $C_{N_{u}}^{m}C_{N_{t}-N_{u}}^{N_{u}-m}$ TACs in the $\psi _{1}^{m}$ . The set $\psi _{1}^{N_{u} - 1}$ , $\psi _{1}^{N_{u} - 2}$ ,$\ldots$ ,$\psi _{1}^{0}$ can be obtain as follows \begin{align*} \psi _{1}^{N_{u} - 1}=&~\left \{{\! \begin{array}{c} \underbrace {(1,2,\ldots,({N_{u}} - 1),{N_{u}} + 1), \cdots (1,2,\ldots,{N_{t}})}_{N_{t} - {N_{u}}}\\[-2pt] \underbrace {(1,3\ldots,{N_{u}},{N_{u}} + 1), \cdots (1,3,\ldots,{N_{u}},{N_{t}})}_{N_{t} - {N_{u}}}\\[-2pt] \vdots \\[-2pt] \underbrace {(2,3\ldots,{N_{u}},{N_{u}} + 1), \cdots (2,3,\ldots,{N_{u}},{N_{t}})}_{N_{t} - {N_{u}}} \end{array} }\right \} \\[-2pt] \psi _{1}^{N_{u} - 2}=&~\left \{{\! \begin{array}{c} \underbrace {(1,2,\ldots,({N_{u}} - 1),{N_{u}} + 1), \cdots (1,2,\ldots,{N_{t}})}_{N_{t} - {N_{u}}}\\[-2pt] {} \vdots \\[-2pt] \underbrace {(2,3\ldots,{N_{u}},{N_{u}} + 1), \cdots (2,3,\ldots,{N_{u}},{N_{t}})}_{N_{t} - {N_{u}}} \end{array} }\right \} \\[-2pt]&\qquad \qquad \quad \qquad \qquad \qquad {} \vdots \\[-3pt] \psi _{1}^{0}=&~\left \{{\! \begin{array}{c} \underbrace {({N_{u}} + 1, \cdots 2{N_{u}}), \cdots ({N_{u}} + 1, \cdots {N_{t}})}_{N_{t} - 2{N_{u}} + 1}\\[-2pt] {}~ \vdots \\[-2pt] {}\underbrace {({N_{t}} - {N_{u}} + 1, \cdots,{N_{t}})}_{N_{t} - {N_{u}}} \end{array} }\right \}.\tag{33}\end{align*}

Based on the generalized TAC selection principle, the final TAC set can be obtained as \begin{equation*} {\mathbb {I}_{o}} = \{ \psi _{1}^{N_{u} - 1},{\mathcal {I}_{{\mathrm {left}}}}\}\tag{34}\end{equation*} where ${\mathcal {I}_{{\mathrm {left}}}}$ is the first $N-(C_{N_{u}}^{1}C_{N_{t}-N_{u}}^{1})$ values of the set $\psi _{1}^{ \ne ({N_{u}}-1)}=\{ \psi _{1}^{N_{u}-2}, \cdots \psi _{1}^{0}\}$

C. Bit-to-TAC Mapping

After obtaining the TAC set ${\mathbb {I}_{o}}$ , bit-to-TAC mapping is the key issue to obtain small HD values of $d_{\bar \varsigma _{1}}^{t}$ . The possible HDs between any two block of bits with length of $B$ can be expressed as \begin{equation*} \mathbb {H} = (\underbrace {1,\ldots,1}_{C_{B}^{1}},\underbrace {2,\ldots,2}_{C_{B}^{2}},\ldots,\underbrace {u,\ldots,u}_{C_{B}^{u}},\ldots,\underbrace B_{C_{B}^{B}}).\tag{35}\end{equation*} For each TAC $I_{o,i}\in {\mathbb {I}_{o}}$ , $i=(1,2,\ldots,N)$ , there are ${\lambda _{1}^{i}}$ TACs termed as $\psi _{o,i}^{N_{u}-1}$ having $N_{u}-1$ common indices as $I_{o,i}$ , resulting in ${\lambda _{1}^{i}}$ HDs as $d_{\bar \varsigma _{1}}^{1},d_{\bar \varsigma _{1}}^{2},\ldots,d_{\bar \varsigma _{1}}^{\lambda _{1}^{i}}$ . Assuming that ${{\mathbf {H}_{\lambda _{1}^{i}}}}=(d_{\bar \varsigma _{1}}^{1},d_{\bar \varsigma _{1}}^{2},\ldots,d_{\bar \varsigma _{1}}^{\lambda _{1}^{i}})$ denotes a set having ${\lambda _{1}^{i}}$ elements of $\mathbb {H}$ , $D_{\bar \varsigma _{1}}^{i} = \sum {{\mathbf {H}}_{\lambda _{1}^{i}}}$ represents the total number of common HDs between $I_{o,i}$ and $\psi _{o,i}^{N_{u}-1}$ , we have \begin{align*} {\lambda _{1}}=&~\lambda _{1}^{1} + \lambda _{1}^{2} + \cdots + \lambda _{1}^{N}, \\ D_{\bar \varsigma _{1}}^{i}=&~\sum {{{\mathbf {H}}_{\lambda _{1}^{i}}}} = \sum \limits _{t = 1}^{\lambda _{1}^{i}} {d_{\bar \varsigma _{1}}^{t}}, \\ \sum \limits _{t = 1}^{\lambda _{1}} {d_{\bar \varsigma _{1}}^{t}}=&~\sum \limits _{i = 1}^{N} {D_{\bar \varsigma _{1}}^{i}}.\tag{36}\end{align*} Then the value of $D({\bar \varsigma _{1}})$ can be represented as \begin{equation*} D({\bar \varsigma _{1}}) = \frac {{\sum \limits _{i = 1}^{N} {D_{\bar \varsigma _{1}}^{i}} }}{{{B^{2B}}}}.\tag{37}\end{equation*} As a result, the bit-to-TAC mapping has to satisfy \begin{align*}&{\min \limits _{{{\mathbf {H}}_{\lambda _{1}^{i}}}}~ [D({\bar \varsigma _{1}})] } \\&s.t.~{\lambda _{1}} = \lambda _{1}^{1} + \lambda _{1}^{2} + \cdots + \lambda _{1}^{N}, \\&\hphantom {s.t.~} D_{\bar \varsigma _{1}}^{i} = \sum {{{\mathbf {H}}_{\lambda _{1}^{i}}}},{{\mathbf {H}}_{\lambda _{1}^{i}}} \in \mathbb {H}\tag{38}\end{align*} where ${\lambda _{1}}$ can be obtained in Section IV-B. Next, the bit-to-TAC mapping of $N_{t}=4, N_{u}=2$ and $N_{t}=5, N_{u}=2$ is introduced first, then it is extended to the generalized cases.

1) Bit-to-TAC Mapping for $N_{t}=4, N_{u}=2$

Fig. 5 presents the bit-to-TAC mapping for the case of $N_{t}=4$ and $N_{u}=2$ . In this case, we have $B=2$ and \begin{align*} \underbrace {(1,3)}_{{I_{o,1}}} \to \psi _{o,1}^{1}=&~\left \{{ {(1,4),(2,3)} }\right \} \\ \underbrace {(1,4)}_{{I_{o,2}}} \to \psi _{o,2}^{1}=&~\left \{{ {(1,4),(2,4)} }\right \} \\ \underbrace {(2,3)}_{{I_{o,3}}} \to \psi _{o,3}^{1}=&~\left \{{ {(1,3),(2,4)} }\right \} \\ \underbrace {(2,4)}_{{I_{o,4}}} \to \psi _{o,4}^{1}=&~\left \{{ {(1,4),(2,3)} }\right \}.\tag{39}\end{align*}

FIGURE 5.

Bit-to-TAC mapping for HM-GSM system having $N_{t}=4$ and $N_{u}=2$ .

According to (35)–​(38), the HD set ${\mathbf {H}_{\lambda _{1}^{i}}}$ can be obtained by \begin{align*} \underbrace {(1,3)}_{{I_{o,1}}} \to \lambda _{1}^{1}=&~2 \to {\mathbf {H}_{\lambda _{1}^{1}}} = (1,1) \\ \underbrace {(1,4)}_{{I_{o,2}}} \to \lambda _{1}^{2}=&~2 \to {\mathbf {H}_{\lambda _{1}^{2}}} = (1,1) \\ \underbrace {(2,3)}_{{I_{o,3}}} \to \lambda _{1}^{3}=&~2 \to {\mathbf {H}_{\lambda _{1}^{3}}} = (1,1) \\ \underbrace {(2,4)}_{{I_{o,4}}} \to \lambda _{1}^{4}=&~2 \to {\mathbf {H}_{\lambda _{1}^{4}}} = (1,1).\tag{40}\end{align*} As seen from Fig. 5, the bit-to-TAC mapping can be obtained as \begin{equation*} 00 \to (1,3),01 \to (1,4),10 \to (2,3),11 \to (2,4).\tag{41}\end{equation*}

2) Bit-to-TAC Mapping for $N_{t}=5, N_{u}=2$

Fig. 6 presents the bit-to-TAC mapping for the case of $N_{t}=5$ and $N_{u}=2$ . For the scenario of $N_{t}=5$ and $N_{u}=2$ , we have $B=3$ and \begin{align*} \underbrace {(1,4)}_{{I_{o,1}}} \to \psi _{o,1}^{1}=&~\left \{{ {(1,3),(1,5),(2,4),(3,4)} }\right \} \\ \underbrace {(1,5)}_{{I_{o,2}}} \to \psi _{o,2}^{1}=&~\left \{{ {(1,3),(1,4),(2,5),(3,5)} }\right \} \\ \underbrace {(2,4)}_{{I_{o,3}}} \to \psi _{o,3}^{1}=&~\left \{{ {(1,4),(2,3),(2,5),(3,4)} }\right \} \\ \underbrace {(2,5)}_{{I_{o,4}}} \to \psi _{o,4}^{1}=&~\left \{{ {(1,5),(2,3),(2,4),(3,5)} }\right \} \\ \underbrace {(2,3)}_{{I_{o,5}}} \to \psi _{o,5}^{1}=&~\left \{{ {(1,3),(2,4),(2,5),(3,4),(3,5)} }\right \} \\ \underbrace {(1,3)}_{{I_{o,6}}} \to \psi _{o,6}^{1}=&~\left \{{ {(1,4),(1,5),(2,3),(3,4),(3,5)} }\right \} \\ \underbrace {(3,4)}_{{I_{o,7}}} \to \psi _{o,7}^{1}=&~\left \{{ {(1,3),(1,4),(2,3),(2,4),(3,5)} }\right \} \\ \underbrace {(3,5)}_{{I_{o,8}}} \to \psi _{o,8}^{1}=&~\left \{{ {(1,3),(1,5),(2,3),(2,5),(3,4)} }\right \}.\tag{42}\end{align*} According to (36)–​(38), the set ${\mathbf {H}_{\lambda _{1}^{i}}}$ for each TAC ${I_{o,i}}$ can be expressed as \begin{align*} \underbrace {(1,4)}_{{I_{o,1}}} \to \lambda _{1}^{1}=&~4 \to {\mathbf {H}_{\lambda _{1}^{1}}} = (1,1,1,2) \\ \underbrace {(1,5)}_{{I_{o,2}}} \to \lambda _{1}^{2}=&~4 \to {\mathbf {H}_{\lambda _{1}^{2}}} = (1,1,1,2) \\ \underbrace {(2,4)}_{{I_{o,3}}} \to \lambda _{1}^{3}=&~4 \to {\mathbf {H}_{\lambda _{1}^{3}}} = (1,1,1,2) \\ \underbrace {(2,5)}_{{I_{o,4}}} \to \lambda _{1}^{4}=&~4 \to {\mathbf {H}_{\lambda _{1}^{4}}} = (1,1,1,2) \\ \underbrace {(1,3)}_{{I_{o,5}}} \to \lambda _{1}^{5}=&~5 \to {\mathbf {H}_{\lambda _{1}^{5}}} = (1,1,1,2,2) \\ \underbrace {(2,3)}_{{I_{o,6}}} \to \lambda _{1}^{6}=&~5 \to {\mathbf {H}_{\lambda _{1}^{6}}} = (1,1,1,2,2) \\ \underbrace {(3,4)}_{{I_{o,7}}} \to \lambda _{1}^{7}=&~5 \to {\mathbf {H}_{\lambda _{1}^{7}}} = (1,1,1,2,2) \\ \underbrace {(3,5)}_{{I_{o,8}}} \to \lambda _{1}^{8}=&~5 \to {\mathbf {H}_{\lambda _{1}^{8}}} = (1,1,1,2,2).\tag{43}\end{align*}

FIGURE 6.

Bit-to-TAC mapping for HM-GSM system having $N_{t}=5$ and $N_{u}=2$ .

Hence, the general bit-to-TAC mapping principle is based on the values of ${\mathbf {H}_{\lambda _{1}^{i}}}$ in (43), which is detailed as follows.

Initialize $(1,3) \to 000,(1,4) \to 001$ .

• Step 1:

  The bit-to-TAC mapping begins from $\psi _{o,1}^{1}$ . Since the TAC mapping should satisfy Eq. (43), we have $(1,5) \in \{ 010,011,100,101\}$ .

• Step 2:

  If $(1,5) \to 010$ , we have ${\mathbf {H}_{\lambda _{1}^{1}}}=(1,2)$ and ${\mathbf {H}_{\lambda _{1}^{2}}}=(1,2)$ . According to Eq. (43), we have $(2,4),(3,4) \in \{ 101,011\}$ and $(2,5),(3,5) \in \{ 011,110\}$ . The bits set {011, 101, 110} should be mapped to four TACs (2, 4), (3, 4), (2, 5), (3, 5), resulting in mapping errors. Hence, $'010'$ cannot be mapped to (1, 5).

• Step 3:

  If $(1,5) \to 011$ , we have ${\mathbf {H}_{\lambda _{1}^{1}}}=(1,1)$ and ${\mathbf {H}_{\lambda _{1}^{2}}}=(1,2)$ . According to (42)–​(43), we have $(2,5),(3,5) \in \{ 010,111\}$ and the following expressions can be formulated \begin{align*} \underbrace {(1,4)}_{001} \to \psi _{o,1}^{1}=&~\{ {\underbrace {(1,3)}_{000 \to 1},\underbrace {(1,5)}_{011 \to 1},(2,4),(3,4)} \} \\ \underbrace {(1,5)}_{011} \to \psi _{o,2}^{1}=&~\{ {\underbrace {(1,3)}_{000 \to 2},\underbrace {(1,4)}_{001 \to 1},\underbrace {(2,5)}_{010/111},\underbrace {(3,5)}_{010/111}} \}.\tag{44}\end{align*}

• Step 4:

  The bit-to-TAC mapping continues from the TAC (2, 5). If $(2,5)\to 010$ , we have \begin{equation*} \underbrace {(2,5)}_{010} \to \psi _{o,4}^{1} = \{ \underbrace {(1,5)}_{011 \to 1},\underbrace {(2,3)}_{110 \to 1},\underbrace {(2,4)}_{000 \to 1},\underbrace {(3,5)}_{111 \to 2}\}.\tag{45}\end{equation*} Then we should have $(2,4)\to 000$ in order to satisfy Eq. (43), which is the same as $(1,3)\to 000$ . Hence, (2, 5) cannot be mapped to 010 and the bit-to-TAC mapping starts from $(2,5)\to 111$ as \begin{equation*} \underbrace {(2,5)}_{111} \to \psi _{o,4}^{1} = \{ \underbrace {(1,5)}_{011 \to 1},\underbrace {(2,3)}_{110/101},\underbrace {(2,4)}_{110/101},\underbrace {(3,5)}_{010 \to 2}\}.\tag{46}\end{equation*}

• Step 5:

  According to (46), the bit-to-TAC mapping continues from the TAC (2, 4). If $(2,4)\to 110$ , the HD between (2, 4) and (1, 4) is 3, so that we have $(2,4)\to 101$ and the following expressions can be formulated \begin{align*} \underbrace {(2,5)}_{111} \to \psi _{o,4}^{1}=&~\{ \underbrace {(1,5)}_{011 \to 1},\underbrace {(2,3)}_{110 \to 1},\underbrace {(2,4)}_{101 \to 1},\underbrace {(3,5)}_{010 \to 2}\} \\ \underbrace {(1,4)}_{001} \to \psi _{o,1}^{1}=&~\{ {\underbrace {(1,3)}_{000 \to 1},\underbrace {(1,5)}_{011 \to 1},\underbrace {(2,4)}_{101 \to 1},\underbrace {(3,4)}_{100 \to 2}} \} \\ \underbrace {(2,4)}_{101} \to \psi _{o,3}^{1}=&~\{ \underbrace {(1,4)}_{001 \to 1},\underbrace {(2,3)}_{110 \to 2},\underbrace {(2,5)}_{111 \to 1},\underbrace {(3,4)}_{100 \to 1}\}.\tag{47}\end{align*} Finally, as shown in Fig. 6, the bit-to-TAC mapping for the case of $N_{t}=5$ and $N_{u}=2$ is given by \begin{align*} 000\to&~(1,3),001 \to (1,4),010 \to (3,5),011 \to (1,5) \\ 100\to&~(3,4),101 \to (2,4),110 \to (2,3),111 \to (2,5). \\\tag{48}\end{align*}

3) Bit-to-TAC Mapping for Generalized Cases

Based on the bit-to-TAC mapping of the above specific examples, the generalized bit-to-TAC mapping is formulated as follows.

• Step 1:

  Obtain the TAC set ${\mathbb {I}_{o}}$ and the HD set $\mathbb {H}$ according to (34) and (38).

• Step 2:

  For each TAC $I_{o,i}\in {\mathbb {I}_{o}}$ , obtain the set $\psi _{o,i}^{N_{u}-1}$ and the value ${\lambda _{1}^{i}}$ according to (34).

• Step 3:

  Obtain the HD set ${{\mathbf {H}_{\lambda _{1}^{i}}}}$ according to (38).

• Step 4:

  Complete the bit-to-TAC mapping based on $I_{o,i}\in {\mathbb {I}_{o}}$ , $\psi _{o,i}^{N_{u}-1}$ , ${{\mathbf {H}_{\lambda _{1}^{i}}}}$ with $D_{\bar \varsigma _{1}}^{i}$ .

Assuming that $\mathbb {B}=\{\mathbf {b}_{1},\ldots,\mathbf {b}_{N}\}$ is the set of information bits, the bit-to-TAC mapping can be detailed as in Algorithm 1. As shown in Algorithm 1, there are lots of mapping options, which have the smallest value $d_{\lambda _{1}^{i}}^{\text {min}}$ . In fact, we only have to find a single appropriate mapping, hence the complexity of the bit-to-TAC mapping is determined by ${\mathbb {I}_{o}}$ and by the specific technique we select for mapping. Taking $N_{t}=6$ and $N_{u}=2$ as an example, we have ${\mathbb {I}_{o}}=\{(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6)\}$ , then the bit-to-TAC mapping can be found without any extra complexity as \begin{align*} (1,3)\to&~000,(1,4) \to 001,(1,5) \to 010,(1,6) \to 011 \\ (2,3)\to&~100,(2,4) \to 101,(2,5) \to 110,(2,6) \to 111. \\\tag{49}\end{align*}

Algorithm 1 Bit-to-TAC Mapping Based the Obtained TAC Set $\mathbb{I}_{o}$

Input:

${\mathbb {I}_{o}}$ , $\psi _{o,i}^{N_{u}-1}$ , ${\lambda _{1}^{i}}~{{\mathbf {H}_{\lambda _{1}^{i}}}}$ , $i=1,\ldots,N$ , $D_{\bar \varsigma _{1}}^{i}$ , $\mathbf {b}_{o,1}=\mathbf {b}_{1}$ , $\mathbf {b}_{o,2}=\mathbf {b}_{2}\,\,D_{\text {max}}^{i}=\max ({{{\mathbf {H}_{\lambda _{1}^{i}}}}})\,\,{{\mathcal {H}_{\lambda _{1}^{i}}}}=\phi$ .

Output:

$\hat {\mathbf {b}}$ .

1:

For the TAC ${I}_{o,3}$ , obtain the possible bits set as ${\mathbb {B}}_{o,3}$ with number of $n_{o,3}$ ;

2:

for $t_{3} \in (1,n_{o,3})$ do

3:

$\mathbf {b}_{o,3}={\mathbb {B}}_{o,3}^{t_{3}}$ , ${\mathbb {B}}_{o,3}^{t_{3}}$ is the $t_{3}$ -th element of ${\mathbb {B}}_{o,3}$ .

4:

Update ${{\mathcal {H}_{\lambda _{1}^{1}}}}$ , ${{\mathcal {H}_{\lambda _{1}^{2}}}}$ , ${{\mathcal {H}_{\lambda _{1}^{3}}}}$ , where ${{\mathcal {H}_{\lambda _{1}^{i}}}}$ denotes the HDs between two TACs inside the set $\psi _{o,i}^{N_{u}-1}$ .

5:

if $\max ({{\mathcal {H}_{\lambda _{1}^{i}}}})>D_{\text {max}}^{i}~\|~\sum {{{\mathcal {H}_{\lambda _{1}^{i}}}}}>D_{\bar \varsigma _{1}}^{i}$ then

6:

break;

7:

else

8:

$\mathbf {b}_{o,3}={\mathbb {B}}_{o,3}^{t_{3}}$

9:

end if

10:

For the TAC ${I}_{o,4}$ , obtain the possible bits set as ${\mathbb {B}}_{o,4}$ with number of $n_{o,4}$ ;

11:

for $t_{4} \in (1,n_{o,4})$ do

12:

$\mathbf {b}_{o,4}={\mathbb {B}}_{o,4}^{t_{4}}$ .

13:

Update ${{\mathcal {H}_{\lambda _{1}^{1}}}}$ , ${{\mathcal {H}_{\lambda _{1}^{2}}}}$ , ${{\mathcal {H}_{\lambda _{1}^{3}}}}$ and ${{\mathcal {H}_{\lambda _{1}^{4}}}}$ .

14:

if $\max ({{\mathcal {H}_{\lambda _{1}^{i}}}})>D_{\text {max}}^{i} ~ \| ~\sum {{{\mathcal {H}_{\lambda _{1}^{i}}}}}>D_{\bar \varsigma _{1}}^{i}$ then

15:

break;

16:

else

17:

$\mathbf {b}_{o,4}={\mathbb {B}}_{o,4}^{t_{4}}$

18:

end if

19:

end for

20:

Get the possible bits set ${\mathbb {B}}_{o,5}$ with number of $n_{o,5}$ ;

21:

for $t_{5} \in (1,n_{o,5})$ do

22:

$\ldots$

23:

end for

24:

$\ldots$

25:

Get the possible bits set ${\mathbb {B}}_{o,N}$ with number of $n_{o,N}$ ;

26:

for $t_{N} \in (1,n_{o,N})$ do

27:

$\ldots$

28:

end for

29:

end for

30:

$\hat {\mathbf {b}}=\{\mathbf {b}_{1}, \mathbf {b}_{2},{\mathbb {B}}_{o,3}^{t_{3}},{\mathbb {B}}_{o,3}^{t_{4}},\ldots,{\mathbb {B}}_{o,N}^{t_{N}}\}$

In order to provide further insights, we compare the AHD of the proposed HM-GSM system to that of the C-GSM system in Table 4. Observe from Table 4 that the proposed HM-GSM scheme is capable of attaining a significantly lower AHD $D({\bar \varsigma _{1}})$ than the C-GSM system.

SECTION V. Enhanced High Throughput GSM

In the above section, we have analyzed how to choose an optimal TAC set from the entire TAC set space, which is capable of providing a performance gain over the C-GSM system. In this section we discuss, how to exploiting the remaining TACs to increase the throughput.

A. Proposed E-HT-GSM System

In the C-GSM system, only $N=2^{\lfloor \log _{2}(C_{N_{t}}^{N_{u}})\rfloor }$ TACs represent antenna index, while the remaining $C_{N_{t}}^{N_{u}}-N$ TACs are wasted. Here we propose reusing the remaining $(2^{\log _{2}(N)+1}-C_{N_{t}}^{N_{u}})$ TACs to convey an extra bit per time slot. The detailed process of the proposed E-HT-GSM is as follows.

• Step 1:

  Determine the number of bits that the conventional TAC index conveys as \begin{equation*} R_{c}={\lfloor \log _{2}(C_{N_{t}}^{N_{u}})\rfloor }\tag{50}\end{equation*}

• Step 2:

  Extend the number of bits conveyed to \begin{equation*} R_{p}={\lfloor \log _{2}(C_{N_{t}}^{N_{u}})\rfloor }+1.\tag{51}\end{equation*} Then the total number of TAC required by the proposed E-HT-GSM scheme becomes $N^{'}=2^{R_{p}}$ .

• Step 3:

  Naturally, we can only arrange for $R_{p}$ bits be conveyed, if $(N^{'}-C_{N_{t}}^{N_{u}})$ TACs are reused randomly and distinguishing rotated phase $\theta$ is employed for the reused TACs. The extended TAC set becomes $\mathbb {I}_{\text {high}}=[\mathbb {I}_{\text {all}}, \mathbb {I}_{\text {repeat}}]$ .

• Step 4:

  GSM mapping utilizing the extended TAC set $\mathbb {I}_{\text {high}}$ .

Let us consider $N_{t}=4$ for example, then the set of legitimate TACs for $N_{t}=4$ is expressed as \begin{equation*} \mathbb {I}_{\text {all}}=\left \{{ {\left [{ {\begin{array}{*{20}{c}} 1 \\ 1 \\ 0 \\ 0 \end{array}} }\right],\left [{ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \\ 0 \end{array}} }\right],\left [{ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 0 \\ 1 \end{array}} }\right],\left [{ {\begin{array}{*{20}{c}} 0 \\ 1 \\ 1 \\ 0 \end{array}} }\right],\left [{ {\begin{array}{*{20}{c}} 0 \\ 1 \\ 0 \\ 1 \end{array}} }\right],\left [{ {\begin{array}{*{20}{c}} 0 \\ 0 \\ 1 \\ 1 \end{array}} }\right]} }\right \}\!.\tag{52}\end{equation*}

When reusing the two TACs $[{1 1 0 0}]^{T}$ and $[{0 0 1 1}]^{T}$ , the repeated TACs are given by \begin{equation*} {\mathbb {I}}{_{{\mathrm {repeat}}}}=\left \{{ {\left [{ {\begin{array}{*{20}{c}} 1\\ 1\\ 0\\ 0 \end{array}} }\right]{e^{j\theta }}\left [{ {\begin{array}{*{20}{c}} 0\\ 0\\ 1\\ 1 \end{array}} }\right]{e^{j\theta }}} }\right \}\!.\tag{53}\end{equation*}

According to (20) and (21), the extended TAC set is given by \begin{align*} {\mathbb {I}_{\text {high}}}=&~\Biggl \{{ {\left [{ {\begin{array}{*{20}{c}} 1 \\ 1 \\ 0 \\ 0 \end{array}} }\right],\left [{ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \\ 0 \end{array}} }\right],\left [{ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 0 \\ 1 \end{array}} }\right],\left [{ {\begin{array}{*{20}{c}} 0 \\ 1 \\ 1 \\ 0 \end{array}} }\right]\!,} } \\&\qquad ~\qquad { {\left [{ {\begin{array}{*{20}{c}} 0 \\ 1 \\ 0 \\ 1 \end{array}} }\right],\left [{ {\begin{array}{*{20}{c}} 0 \\ 0 \\ 1 \\ 1 \end{array}} }\right],\left [{ {\begin{array}{*{20}{c}} {{e^{j\theta }}} \\ {{e^{j\theta }}} \\ 0 \\ 0 \end{array}} }\right],\left [{ {\begin{array}{*{20}{c}} 0 \\ 0 \\ {{e^{j\theta }}} \\ {{e^{j\theta }}} \end{array}} }\right]} }\Biggr \}\!.\tag{54}\end{align*} At the receiver, the ML detector and a range of low-complexity detectors designed for C-GSM systems can also be applied to the proposed E-HT-GSM systems.

B. The Optimization of $\theta$

In this section, the value of $\theta$ is optimized by maximizing the Minimum Distance (MD) [42] associated with $\theta$ between the pair of E-HT-GSM symbols. Assuming that $\mathbf {x}_{i}$ and $\mathbf {x}_{j}$ are the transmit and estimated E-HT-GSM symbols, respectively, the MD between $\mathbf {x}_{i}$ and $\mathbf {x}_{j}$ can be expressed as \begin{equation*} \delta _{\min }(\mathbf {x}_{i},\mathbf {x}_{j})=\min _{\mathbf {x}_{i},\mathbf {x}_{j}}\det (\mathbf {x}_{i}-\mathbf {x}_{j})(\mathbf {x}_{i}-\mathbf {x}_{j})^{H}.\tag{55}\end{equation*} The optimization principle is based on maximizing the value of $\delta _{\min }(\mathbf {x}_{i},\mathbf {x}_{j})$ . For simplicity, let us consider $N_{u}=2$ for example. Specifically, the calculation is based on four scenarios:

1. Both $\mathbf {x}_{i}$ and $\mathbf {x}_{j}$ are independent of $\theta$ . In this case, the value of $\delta _{\min }(\mathbf {x}_{i},\mathbf {x}_{j})$ is independent of $\theta$ as well;

2. Both $\mathbf {x}_{i}$ and $\mathbf {x}_{j}$ are associated with the $\theta$ . In this case, the calculation is the same as the scenario 1);

3. $\mathbf {x}_{i}$ is independent of $\theta$ and $\mathbf {x}_{j}$ is associated with $\theta$ ;

4. $\mathbf {x}_{i}$ is associated with $\theta$ and $\mathbf {x}_{j}$ is independent of $\theta$ .

The calculation process of scenario 3) and scenario 4) is the same due to their symmetry. As a result, we can focus on scenario 3) for our analysis. Assuming that the TAC and symbol vectors of $\mathbf {x}_{i}$ and $\mathbf {x}_{j}$ are given as $(l_{1},l_{2})$ , $\mathbf {s}=[s_{1}, s_{2}]^{T}$ , and $(\hat l_{1},\hat l_{2})$ , $\hat {\mathbf { s}}=[\hat s_{1}, \hat s_{2}]^{T}$ , respectively, the value of $\delta _{\min }(\mathbf {x}_{i},\mathbf {x}_{j})$ can be calculated for three independent cases.

• Case 1:

  All the activate antenna indices of $\mathbf {x}_{i}$ and $\mathbf {x}_{j}$ are the same. Assuming $l_{1}=\hat l_{1}=1$ , $l_{2}=\hat l_{2}=2$ , $\mathbf x_{i}=\left[{\begin{matrix}s_{1}&s_{2}&\mathbf {0}_{2\times (N_{t}-2)}\\ \end{matrix}}\right]^{T}$ , and $\mathbf x_{j}=\left[{\begin{matrix}\hat s_{1}&\hat s_{2}&\mathbf {0}_{2\times (N_{t}-2)} \end{matrix}}\right]^{T}e^{j\theta }$ for example, the MD between $\mathbf {x}_{i}$ and $\mathbf {x}_{j}$ obeys \begin{align*} \delta _{\min }(\mathbf x_{i},\mathbf x_{j})=&~\min _{\mathbf x_{i},\mathbf x_{j}}\det (\mathbf x_{i}-\mathbf x_{j})(\mathbf x_{i}-\mathbf x_{j})^{H} \\=&~\min _{s_{1},\hat s_{1},s_{2},\hat s_{2}}\det \left({\begin{matrix}s_{1}-\hat {s}_{1}e^{j\theta }&s_{2}-\hat {s}_{2}e^{j\theta }&\mathbf {0}_{2\times (N_{t}-2)} \end{matrix}}\right) \\&\times \left({\begin{matrix}s_{1}-\hat {s}_{1}e^{j\theta }&s_{2}-\hat {s}_{2}e^{j\theta }&\mathbf {0}_{2\times (N_{t}-2)} \end{matrix}}\right)^{H} \\=&~\min _{s_{1},\hat s_{1},s_{2},\hat s_{2}}\det (||s_{1}||^{2}+||s_{2}||^{2}+||\hat {s}_{1}||^{2}+||\hat {s}_{2}||^{2} \\&-\,(s_{1}\hat {s}_{1}^{*}+s_{2}\hat {s}_{2}^{*})e^{j\theta }-(s_{1}^{*}\hat {s}_{1}+s_{2}^{*}\hat {s}_{2})e^{j\theta }).\tag{56}\end{align*}

• Case 2:

  Only a single activated antenna index of $\mathbf {x}_{i}$ and $\mathbf {x}_{j}$ is the same. Assuming $s1=1,\hat l_{1}=2$ , $l_{2}=1,\hat l_{2}=3$ , $\mathbf x_{i}=\left[{\begin{matrix}s_{1}&s_{2}&\mathbf {0}_{2\times (N_{t}-2)}\\ \end{matrix}}\right]^{T}$ , and $\mathbf x_{j}=\left[{\begin{matrix}\hat s_{1}&0&\hat s_{2}&\mathbf {0}_{2\times (N_{t}-3)} \end{matrix}}\right]^{T}e^{j\theta }$ for example, the MD between $\mathbf {x}_{i}$ and $\mathbf {x}_{j}$ obeys \begin{align*} \delta _{\min }(\mathbf x_{i},\mathbf x_{j})=&~\min _{\mathbf x_{i},\mathbf x_{j}}\det (\mathbf x_{i}-\mathbf x_{j})(\mathbf x_{i}-\mathbf x_{j})^{H} \\=&~\min _{s_{1},\hat s_{1},s_{2},\hat s_{2}}\det \left({\begin{matrix}s_{1}-\hat {s}_{1}e^{j\theta }&s_{2}&-\hat {s}_{2}e^{j\theta }&\mathbf {0}_{2\times (N_{t}-2)} \end{matrix}}\right) \\&\times \left({\begin{matrix}s_{1}-\hat {s}_{1}e^{j\theta }&s_{2}&-\hat {s}_{2}e^{j\theta }&\mathbf {0}_{2\times (N_{t}-3)} \end{matrix}}\right)^{H} \\=&~\min _{s_{1},\hat s_{1},s_{2},\hat s_{2}}\det (||s_{1}||^{2}+||s_{2}||^{2}+||\hat {s}_{1}||^{2}+||\hat {s}_{2}||^{2} \\&-\,s_{1}\hat {s}_{1}^{*}e^{j\theta }+s_{1}^{*}\hat {s}_{1}e^{j\theta }).\tag{57}\end{align*}

• Case 3:

  None of the activated antenna indices are the same. Considering $l_{1}=1,\hat l_{1}=2$ , $l_{2}=3,\hat l_{2}=4$ , $\mathbf x_{i}=\left ({\begin{matrix}s_{1}&s_{2}&\mathbf {0}_{2\times (N_{t}-2)}\\ \end{matrix}}\right)$ , and $\mathbf x_{j}=\left ({\begin{matrix} 0&0&\hat s_{1}&\hat s_{2}&\mathbf {0}_{2\times (N_{t}-4)} \end{matrix}}\right)e^{j\theta }$ for example, the MD between $\mathbf {x}_{i}$ and $\mathbf {x}_{j}$ obeys \begin{align*}&\hspace {-2pc}\delta _{\min }(\mathbf x_{i},\mathbf x_{j}) \\=&~\min _{\mathbf x_{i},\mathbf x_{j}}\det (\mathbf x_{i}-\mathbf x_{j})(\mathbf x_{i}-\mathbf x_{j})^{H} \\=&~\min _{s_{1},\hat s_{1},s_{2},\hat s_{2}}\det \left({\begin{matrix}s_{1}&\quad s_{2}&\quad -\hat {s}_{1}e^{j\theta }&-\hat {s}_{2}e^{j\theta }&\quad \mathbf {0}_{2\times (N_{t}-4)} \end{matrix}}\right) \\&\times \left({\begin{matrix}s_{1}&\quad s_{2}&\quad -\hat {s}_{1}e^{j\theta }&\quad -\hat {s}_{2}e^{j\theta }&\quad \mathbf {0}_{2\times (N_{t}-4)} \end{matrix}}\right)^{H} \\=&~\min _{s_{1},\hat s_{1},s_{2},\hat s_{2}}\det (||s_{1}||^{2}+||s_{2}||^{2}+||\hat {s}_{1}||^{2}+||\hat {s}_{2}||^{2}).\tag{58}\end{align*} As a result, the values of $\delta _{\min }(\mathbf x_{i},\mathbf x_{j})$ for the three cases can be summarized as (59), as shown at the bottom of the next page.

  According to (59), we have\begin{align*} \delta _{\min }(\mathbf {x}_{i},\mathbf {x}_{j})=&~\min _{x_{ij},\hat {x}_{ij}}\det (||s_{1}||^{2}+||s_{2}||^{2}+||\hat {s}_{1}||^{2}+||\hat {s}_{2}||^{2} \\&-\,(s_{1}\hat {s}_{1}^{*}+s_{2}\hat {s}_{2}^{*})e^{j\theta }-(s_{1}^{*}\hat {s}_{1}+s_{2}^{*}\hat {s}_{2})e^{j\theta }).\tag{60}\end{align*} To provide further insights, Fig. 7 shows the value of $\delta _{\min }(\mathbf {x}_{i}, \mathbf {x}_{j})$ in (60) for BPSK, QPSK, 16 QAM and 64-QAM in conjunction with different $\theta$ . If $\theta \in [0, \pi /2]$ , as seen from Fig. 7, $\theta$ can be optimized by maximizing the value of $\delta _{\min }(\mathbf {x}_{i}, \mathbf {x}_{j})$ in (60) as \begin{align*} \theta =\! {\begin{cases} \displaystyle \frac {\pi }{2},&\quad {\mathrm {BPSK}}\\[6pt] \displaystyle \frac {\pi }{4},&\quad {\mathrm {QPSK}}\\[6pt] \displaystyle \frac {\pi }{7} {~\text {or }} \left ({{\displaystyle \frac {\pi }{2} - \displaystyle \frac {\pi }{7}} }\right),&\quad {\mathrm { 16QAM}}\\[6pt] \displaystyle \frac {\pi }{16} {~\text {or }} \left ({{\displaystyle \frac {\pi }{2} - \displaystyle \frac {\pi }{16}} }\right),&\quad {\mathrm {64QAM}}. \end{cases}}\tag{61}\end{align*}

FIGURE 7.

The values of $\delta _{\min } (\mathbf {x}_{i},\mathbf {x}_{j})$ for different modulation orders.

SECTION VI. Simulation Results

In this section, the performances of the proposed E-GSM, HM-GSM and E-HT-GSM schemes are presented and compared under different antenna configurations. In all the simulation results, perfect channel state information is assumed and the values of the specific $\theta$ are selected based on Eq. (59). ML detectors are employed at the receiver for Figs. 8–​14. Moreover, the analytical ABEP performances of the proposed systems are added as benchmarkers, which can be obtained by (19). As seen from Fig. 8–​14, the upper bound becomes very tight upon increasing the SNR values, which is helpful for evaluating the BER performances of the proposed systems.

FIGURE 8.

Performance comparison of the proposed schemes and of the conventional GSM systems using $N_{t}=4, N_{r}=4, N_{u}=2$ and $M=1$ at 2 bits/symbol.

FIGURE 9.

Performance comparison of the proposed schemes and of the conventional GSM systems using $N_{t}=4, N_{r}=4, N_{u}=2$ and $M=2$ at 3 bits/symbol.

FIGURE 10.

Performance comparison of the proposed schemes and of the conventional GSM systems using $N_{t}=4, N_{r}=4, N_{u}=2$ and $M=4$ at 4 bits/symbol.

FIGURE 11.

Complexity comparison of the proposed schemes and of the conventional GSM system using $N_{t}=4$ , $N_{r}=4$ and $N_{u}=2$ .

FIGURE 12.

Performance comparison of the proposed schemes and of the conventional GSM systems using $N_{t}=5, N_{r}=4, N_{u}=3$ .

FIGURE 13.

Performance comparison of the proposed schemes and of the conventional GSM systems using $N_{t}=6, N_{r}=4, N_{u}=2$ .

FIGURE 14.

Performance comparisons of the QPSK-aided E-HT-GSM systems and of the QPSK-aided C-GSM systems using $N_{t}=4, N_{r}=4, N_{u}=2$ and $N_{t}=8, N_{r}=8, N_{u}=3$ .

Specifically, Figs. 8–​10 compare the performances of both the proposed E-GSM, and of the HM-GSM systems to the C-GSM system using $N_{t}=4, N_{r}=4, N_{u}=2$ at different values of $M$ . Fig. 11 compares the complexity of the proposed E-GSM and HM-GSM systems to that of the C-GSM system for the same setups as in Figs. 8–​10. The mapping principles of the HM-GSM and C-GSM system are shown in Table 6. Observe from Figs. 8–​11 that the proposed E-GSM system is capable of outperforming the C-GSM system by 5 dB, 3.5 dB and 2.5 dB for $M=1$ , $M=2$ and $M=4$ respectively at the BER=10⁻⁵ at an acceptable extra complexity. The proposed HM-GSM scheme is capable of outperforming C-GSM system by 1 dB, 0.6 dB and 0.4 dB for $M=1$ , $M=2$ and $M=4$ at the BER=10⁻⁵ at the same complexity. Moreover, observe from Fig. 10 that the proposed E-GSM and HM-GSM systems also perform better than the scheme in [2].

TABLE 6 AHD Comparisons Between the Proposed HM-GSM and C-GSM Systems

In order to provide further insights, Figs. 12–​13 compare the performances of the proposed E-GSM, HM-GSM to the C-GSM system using $N_{t}=5, N_{r}=4, N_{u}=3$ and $N_{t}=6, N_{r}=4, N_{u}=2$ , respectively. Observe from Fig. 12 that the performance advantage of the E-GSM over the C-GSM system becomes modest. This is because for the case of $N_{t}=5, N_{u}=3$ , we have $N_{\text {left}}=2$ , while $N_{\text {same}}/2=3$ for $\beta =2$ , $N_{\text {same}}/2=1$ for $\beta =1$ , $N_{\text {same}}/2=1$ for $\beta =0$ remain valid according to (12). When the number of common minimum values in $\mathbf {W}$ satisfies $N_{\text {same}}/2>N_{\text {left}}=2$ , the performance of the optimal TAC set is the same as that of the conventional TAC set. For the case of $N_{t}=6, N_{r}=4, N_{u}=2$ , $N_{\text {left}}=7$ , $N_{\text {same}}/2=4$ for $\beta =1$ and $N_{\text {same}}/2=1$ for $\beta =0$ . As a result, $N_{\text {same}}/2<N_{\text {left}}$ always remains true and the proposed E-GSM system using $N_{t}=6, N_{r}=4, N_{u}=2$ is capable of providing significant performance gain over the C-GSM system, as observed in Fig. 13.

Finally, Fig. 14 compares the performance of the proposed E-HT-GSM system to that of the C-GSM system at different transmission rates. Specifically, for the C-GSM system, we employ QPSK modulation for the cases of $N_{t}=4, N_{r}=4, N_{u}=2$ and $N_{t}=8, N_{r}=8, N_{u}=3$ to obtain the normalized throughput of 6 bits/symbol and 11 bits/symbol. For the proposed E-HT-GSM systems relying on the above setups, we achieve an extra bit of throughput, yielding 7 bits/symbol and 12 bits/symbol, respectively. Observe from Fig. 14 that the proposed E-HT-GSM system is capable of increasing the throughput by one bit per channel use at a 0.5 dB performance loss at BER = 10⁻⁴ for both the above mentioned setups.

SECTION VII. Conclusions

The problem of TAC set optimization has been investigated. Firstly, a low-complexity TAC selection algorithm relying on CSI knowledge was designed for our E-GSM systems. Then, hybrid bit-to-TAC mapping based TAC optimization operating without CSI knowledge was designed for the GSM system. The proposed E-GSM system and HM-GSM system are capable of outperforming the C-GSM system at a negligible extra complexity, while the proposed E-HT-GSM system conceived is capable of increasing the throughput per time slot by one bit at a negligible performance loss.