A Hybrid Mapping Scheme to Improve Bit Error Rate Performance of Orthogonal Frequency Division Multiplexing with Index Modulation System
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摘要: 为进一步提升索引调制正交频分复用(OFDM-IM)系统在高信噪比(SNR)条件下的误比特率(BER)，该文提出一种混合映射方案。该方案利用高信噪比时索引比特比数据比特错误率更低的特性，混合使用两种不同位数的索引比特分配策略，提高了索引比特的占比。相应地，将子载波激活模式(SAPs)分为超级SAPs和普通SAPs，相比于普通SAPs，超级SAPs对应的索引比特和数据比特分别增加1 bit和减少1 bit。将超级SAP对应的数据比特采用奇偶校验的方式补齐，然后映射为数据符号，提高了系统的分集度，增大了数据符号间的最小欧氏距离。仿真结果表明，相比于以往的映射方案，混合映射方案在误比特率为10–4时可取得1～3 dB的性能增益，有效改善索引调制OFDM系统的误码性能。Abstract: A hybrid mapping scheme is proposed in this paper to improve further the Bit Error Rate (BER) performance of Orthogonal Frequency Division Multiplexing with Index Modulation (OFDM-IM) system for high Signal-to-Noise Ratio (SNR). By taking advantage of the fact that the BER of index bits is lower than that of data bits in the high SNR region, the scheme uses two allocation strategies of index bits with different numbers, which increases the proportion of index bits. Correspondingly, the Subcarrier Activation Patterns (SAPs) are divided into super SAPs and normal SAPs. Compared with normal SAPs, super SAPs have seen increased index bits and decreased data bits by 1 bit respectively. The decreased data bits corresponding to a super SAP are supplemented with a parity-check bit and then mapped into symbols, which improves the diversity order of OFDM-IM system and increases the minimum Euclidean distance between symbols. The simulation results show that, compared with the previous mapping scheme, the hybrid mapping scheme can gain about 1～3 dB when BER is 10–4, and improve effectively the BER performance of OFDM-IM system.
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算法1 组合数法 输入：索引比特序列${{\mathbf{c}}_{\alpha ,0}}$。 输出：SAP $ {{\boldsymbol{P}}_\alpha } $。 步骤1 将${{\mathbf{c}}_{\alpha ,0}}$转换为十进制数$D$。 步骤2 取最大的${J_k}$，满足$C_{ {J_k} }^k \le D$。 步骤3 取最大的${J_{k - 1}}$，满足$C_{ {J_{k - 1} } }^{k - 1} \le D - C_{ {J_k} }^k$。 步骤4 以此类推，直至$C_{ {J_1} }^1 \le D - C_{ {J_k} }^k - C_{ {J_{k - 1} } }^{k - 1} - \cdots - C_{ {J_2} }^2$。 步骤5 将$ J = [{J_k}{J_{k - 1}} \cdots {J_1}] $中元素加1，得到序列${J^*}$。 步骤6 SAP ${{\boldsymbol{P}}_\alpha } = {\text{sort} }({J^*},\;'{\text{ascend} }')$。 
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算法2 格雷码索引映射 输入：索引比特序列${{\mathbf{c}}_{\alpha ,0}}$，SAP矩阵${\mathbf{U}}$。 输出：SAP ${{\boldsymbol{P}}_\alpha }$。 步骤1 将${{\mathbf{c}}_{\alpha ,0}}$看作格雷码序列，转换为自然二进制序列${\mathbf{c} }_{_{\alpha ,0} }^{{\rm{bin}}}$。 步骤2 计算与${\mathbf{c} }_{_{\alpha ,0} }^{{\rm{bin}}}$对应的十进制数$D$。 步骤3 SAP $ {{\boldsymbol{P}}_\alpha } = {\mathbf{U}}(D,:) $。
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算法3 混合映射算法 输入：信息比特序列${{\boldsymbol{c}}_\alpha }$，调制阶数M。 输出：发送符号序列$ {{\boldsymbol{X}}_\alpha } $。 步骤1 将$w$种SAP按照格雷码的方式进行排序，构造$w$行$k$列的SAP矩阵${\boldsymbol{V}}$。 步骤2 将${{\boldsymbol{c}}_{\alpha ,0} }$看作格雷码序列，转换为自然二进制序列，再计算对应的十进制数${u_0}$。 步骤3 若${u_0} < z$，转到步骤4，否则令$u = {u_0} + z$, ${{\boldsymbol{c}}_{\alpha ,{\text{sym} } } } = {{\boldsymbol{c}}_{\alpha ,1} }$，转到步骤5。 步骤4 将${\boldsymbol{c}}_{\alpha ,0}^{\text{*} }$看作格雷码序列，转换为自然二进制序列，再转换为对应的十进制数$u$，对${\boldsymbol{c}}_{\alpha ,1}^{\text{*}}$进行奇(偶)校验，校验位放在${\boldsymbol{c}}_{\alpha ,1}^{\text{*}}$末尾组成
长度为${p_1}$的序列${{\boldsymbol{c}}_{\alpha ,{\text{sym}}}}$。步骤5 选择${\boldsymbol{V}}$的第$u$行作为${{\boldsymbol{P}}_\alpha }$，${{\boldsymbol{c}}_{\alpha ,{\rm{sym}}} }$映射为$k$个$M{\text{ - PSK}}$符号${s_\alpha }$，得到发送符号序列$ {{\boldsymbol{X}}_\alpha } $。
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表 1 (4,2)QPSK两种映射方案的SAP查找表
序号 0 1 2 3 4 5 索引比特 0 0 0 1 1 1 1 0 $ {{\mathbf{P}}_\alpha } $(格雷码) {3,4} {2,3} {1,3} {1,4} 索引比特 0 0 0 0 0 1 0 1 1 0 1 0 1 1 1 0 ${{\mathbf{P}}_\alpha }$(混合) {3,4} {2,3} {2,4} {1,2} {1,3} {1,4}
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表 2 (4,2)QPSK混合映射方案超级SAP对应的符号映射表
序号 0 1 2 3 4 5 6 7 ${\mathbf{c}}_{\alpha ,1}^*$ 0 0 0 0 0 1 0 1 1 0 1 0 1 1 0 1 1 1 1 0 1 1 0 0 偶校验位 0 1 0 1 0 1 0 1 ${{\mathbf{s}}_\alpha }$ +1 +1 +1 –1 +1i –1i +1i +1i –1 +1 –1 –1 –1i –1i –1i +1i
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表 3 不同映射方案索引比特占比及分集度为1和MED的误码情况占比(%)
映射方案 索引比特 分集度为1 MED 组合数法 33.33 40.00 26.67 格雷码索引映射 33.33 40.00 26.67 混合映射 41.67 13.33 13.33
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