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摘要: 由于多输入多输出(MIMO)系统具有波形、空间分集和多路复用等优势，MIMO探通一体化(DFRC)系统通过共享软硬件资源以同时实现目标探测和保密通信功能受到了极大关注。该文针对基于预编码矩阵调制的MIMO探通一体化系统，提出了基于交替方向乘子(ADMM)的一体化信号矩阵设计方法。通过用户和窃听用户参考密码本约束下最大化方向图峰值主瓣旁瓣电平比(PMSR)，保证了探测方向图性能的同时防止通信信息被窃听。针对预编码矩阵通信解调问题，提出了基于交替方向惩罚(ADPM)的排序学习优化解调方法，提升了一体化波形信息解调效率。数值仿真验证了所提设计方法实现探通一体化的有效性，与已有算法相比可实现多用户通信和更高的PMSR。Abstract: Due to several advantages of the Multi-Input Multi-Output (MIMO) system in terms of waveform, space diversity, and multiplexing, the MIMO Dual Function Radar and Communication (DFRC) system, which is responsible for target detection and securing the communication by sharing the software and hardware resources, has attracted great attention. This paper addresses the MIMO DFRC system based on permutation matrix modulation and proposes a DFRC signal matrix design method based on the Alternation Direction Method of Multipliers (ADMM). By maximizing the Peak Mainlobe to Sidelobe level Ratio (PMSR) of the beampattern with the constraints of the reference codebook for both users and eavesdroppers, the system guarantees excellent detection performance along with protecting the communication information from interception. Aiming at the communication demodulation of the permutation matrix, a permutation learning demodulation method based on the Alternating Direction Penalty Method (ADPM) is proposed to improve the demodulation efficiency of the co-use waveform. Numerical simulations verify the effectiveness of the proposed methods to achieve dual function, capable of realizing multiuser communication and deriving higher PMSR compared with the existing counterparts.
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图 1 基于预编码矩阵调制的MIMO探通一体化系统框架示意图
Figure 1. The framework diagram of the MIMO DFRC system based on permutation matrix modulation
图 2 假设
$\theta _1^{{\text{com}}} = - {60^\circ }$ 时发射方向图Figure 2. The transmit beampattern with
$\theta _1^{{\text{com}}} = - {60^\circ }$ 
图 6
${\boldsymbol{s}}\left( {{{65}^\circ }} \right)$ 与${\boldsymbol{s}}\left( { - {{60}^\circ }} \right)$ 元素分布Figure 6. The element distributions of
${\boldsymbol{s}}\left( {{{65}^\circ }} \right)$ and${\boldsymbol{s}}\left( { - {{60}^\circ }} \right)$ 算法1 基于ADMM的一体化信号矩阵设计方法 Alg. 1 DFRC waveform matrix design method based on ADMM 输入：$\left\{ {\boldsymbol{y} }_{i}^{(0)}\right\},{ {\epsilon} }^{(0)},\left\{ {\boldsymbol{z} }_{s}^{(0)}\right\},{ {\eta} }^{(0)},\left\{ {\boldsymbol{x} }_{c}^{(0)}\right\},\left\{ {\boldsymbol{v} }_{h}^{(0)}\right\},\left\{ {\boldsymbol{\mu} }_{i}^{(0)}\right\},\left\{ {\boldsymbol{\iota} }_{s}^{(0)}\right\},\left\{ {\boldsymbol{\xi} }_{c}^{(0)}\right\},\left\{ { {\boldsymbol{\lambda} } }_{h}^{(0)}\right\},{\boldsymbol{n} }^{\left(0\right)},{\boldsymbol{\zeta} }^{\left(0\right)},{\boldsymbol{\rho} },\varDelta ,{\delta }_{1}$； 输出：MIMO一体化系统加权向量$ {{\boldsymbol{\boldsymbol{w}}}^ \star } $； 步骤1. $ t = 0 $； 步骤2. 通过求解以下问题更新$ {\boldsymbol{w}}^{\left(t\text{+}1\right)},\left\{{\boldsymbol{y}}_{i}^{\left(t\text{+}1\right)}\right\},{{\epsilon}}^{\left(t\text{+}1\right)},\left\{{\boldsymbol{z}}_{s}^{\left(t\text{+}1\right)}\right\},{{\eta}}^{\left(t\text{+}1\right)},\left\{{\boldsymbol{x}}_{c}^{\left(t\text{+}1\right)}\right\},\left\{{\boldsymbol{v}}_{h}^{\left(t\text{+}1\right)}\right\}, $$ {{\boldsymbol{\boldsymbol{n}}}^{\left( {t{\text{ + }}1} \right)}} $： ${\boldsymbol{w} }^{(t+1)}:=\mathrm{arg}\;\underset{\boldsymbol{w} }{\mathrm{min} }\;{L}_{ { {\boldsymbol \rho} } }\left({\boldsymbol{w}},\left\{ {\boldsymbol{y} }_{i}^{(t)}\right\},{ {\epsilon} }^{(t)},\left\{ {\boldsymbol{z} }_{s}^{(t)}\right\},{ {\eta} }^{(t)},\left\{ {\boldsymbol{x} }_{c}^{(t)}\right\},\left\{ {\boldsymbol{v} }_{h}^{(t)}\right\},\left\{ {\boldsymbol{\mu} }_{i}^{(t)}\right\},\left\{ {\boldsymbol{\iota} }_{s}^{(t)}\right\},\left\{ {\boldsymbol{\xi} }_{c}^{(t)}\right\},{ {\boldsymbol{\lambda} } }_{h}^{(t)},{\boldsymbol{n} }^{\left(t\right)},{\boldsymbol{\zeta} }^{\left(t\right)}\right)$ (22) $\begin{aligned}& \left\{ {\boldsymbol{y} }_{i}^{(t+1)},{ {\epsilon} }^{(t+1)}\right\}:=\mathrm{arg}\;\underset{ {\boldsymbol{y} }_{i},\epsilon}{\mathrm{min} }\;{L}_{ { {\boldsymbol \rho} } }\left({\boldsymbol{w} }^{(t+1)},\left\{ {\boldsymbol{y} }_{i}\right\},\epsilon,\left\{ {\boldsymbol{z} }_{s}^{(t)}\right\},{ {\eta} }^{(t)}, \left\{ {\boldsymbol{x} }_{c}^{(t)}\right\},\left\{ {\boldsymbol{v} }_{h}^{(t)}\right\},\left\{ {\boldsymbol{\mu} }_{i}^{(t)}\right\},\left\{ {\boldsymbol{\iota} }_{s}^{(t)}\right\},\left\{ {\boldsymbol{\xi} }_{c}^{(t)}\right\},{ {\boldsymbol{\lambda} } }_{h}^{(t)},{\boldsymbol{n} }^{\left(t\right)},{\boldsymbol{\zeta} }^{\left(t\right)}\right)\text{ }\\ & \quad\text{s}\text{.t}\text{. }{\Vert {\boldsymbol{y} }_{i}\Vert }^{2}\ge \epsilon,i=1,2,\cdots ,I; \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\, (23)\end{aligned}$ $\begin{array}{l}\left\{ {\boldsymbol{z} }_{s}^{(t+1)},{ {\eta} }^{(t+1)}\right\}:=\mathrm{arg}\;\underset{ {\boldsymbol{z} }_{s},\eta }{\mathrm{min} }\;{L}_{ { {\boldsymbol \rho} } }\left({\boldsymbol{w} }^{(t+1)},\left\{ {\boldsymbol{y} }_{i}^{(t+1)}\right\},{ {\epsilon} }^{(t+1)},\left\{ {\boldsymbol{z} }_{s}\right\}, \eta ,\left\{ {\boldsymbol{x} }_{c}^{(t)}\right\},\left\{ {\boldsymbol{v} }_{h}^{(t)}\right\},\left\{ {\boldsymbol{\mu} }_{i}^{(t)}\right\},\left\{ {\boldsymbol{\iota} }_{s}^{(t)}\right\},\left\{ {\boldsymbol{\xi} }_{c}^{(t)}\right\},{ {\boldsymbol{\lambda} } }_{h}^{(t)},{\boldsymbol{n} }^{\left(t\right)},{\boldsymbol{\zeta} }^{\left(t\right)}\right)\text{ }\\ \quad\text{s}\text{.t}\text{. }{\Vert {\boldsymbol{z} }_{s}\Vert }^{2}\le \eta ,s=1,2,\cdots ,S; \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\, (24)\end{array}$ $\begin{array}{l}\left\{ {\boldsymbol{x} }_{c}^{(t+1)},{b}_{c}^{(t+1)}\right\}:=\mathrm{arg}\underset{ {\boldsymbol{x} }_{c},{b}_{c} }{\mathrm{min} }{L}_{ { {\boldsymbol \rho} } }\left({\boldsymbol{w} }^{(t+1)},\left\{ {\boldsymbol{y} }_{i}^{(t+1)}\right\},{ {\epsilon} }^{(t+1)},\left\{ {\boldsymbol{z} }_{s}^{(t+1)}\right\}, { {\eta} }^{(t+1)},\left\{ {\boldsymbol{x} }_{c}\right\},\left\{ {\boldsymbol{v} }_{h}^{(t)}\right\},\left\{ {\boldsymbol{\mu} }_{i}^{(t)}\right\},\left\{ {\boldsymbol{\iota} }_{s}^{(t)}\right\},\left\{ {\boldsymbol{\xi} }_{c}^{(t)}\right\},{ {\boldsymbol{\lambda} } }_{h}^{(t)},{\boldsymbol{n} }^{\left(t\right)},{\boldsymbol{\zeta} }^{\left(t\right)}\right)\text{ }\\ \quad\text{s}\text{.t}\text{. }{\boldsymbol{x} }_{c}={b}_{c}{\overline{{\boldsymbol{a}}} }_{1},{l}_{c}\le {b}_{c}\le {u}_{c},c=1,2,\cdots ,C; \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\, (25)\end{array}$ $\begin{aligned}& \left\{ {\boldsymbol{v} }_{h}^{(t+1)},{r}_{h}^{(t+1)} \right\} :=\mathrm{arg}\underset{ {\boldsymbol{v} }_{h},{r}_{h} }{\mathrm{min} }\;{L}_{ { {\boldsymbol \rho} } }\left({\boldsymbol{w} }^{(t+1)},\left\{ {\boldsymbol{y} }_{i}^{(t+1)}\right\},{ {\epsilon} }^{(t+1)},\left\{ {\boldsymbol{z} }_{s}^{(t+1)} \right\}, { {\eta} }^{(t+1)},\left\{ {\boldsymbol{x} }_{c}^{(t+1)} \right\},\left\{ {\boldsymbol{v} }_{h}\right\},\left\{ {\boldsymbol{\mu} }_{i}^{(t)}\right\},\left\{ {\boldsymbol{\iota} }_{s}^{(t)}\right\},\left\{ {\boldsymbol{\xi} }_{c}^{(t)} \right\},{ {\boldsymbol{\lambda} } }_{h}^{(t)},{\boldsymbol{n} }^{\left(t\right)},{\boldsymbol{\zeta} }^{\left(t\right)}\right)\\ & \quad\text{s}\text{.t}\text{. }{\boldsymbol{v} }_{h}={r}_{h} { {\textit{1}}}_{K},h=1,2,\cdots ,E \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\, (26)\end{aligned}$ $\begin{aligned}& {\boldsymbol{n} }^{\left(t\text{+}1\right)}:=\mathrm{arg}\;\underset{\boldsymbol{n} }{\mathrm{min} }\;{L}_{ { {\boldsymbol \rho} } }\left({\boldsymbol{w} }^{(t+1)},\left\{ {\boldsymbol{y} }_{i}^{(t+1)}\right\},{ {\epsilon} }^{(t+1)},\left\{ {\boldsymbol{z} }_{s}^{(t+1)}\right\}, { {\eta} }^{(t+1)},\left\{ {\boldsymbol{x} }_{c}^{(t+1)}\right\},\left\{ {\boldsymbol{v} }_{h}^{(t+1)}\right\},\left\{ {\boldsymbol{\mu} }_{i}^{(t)}\right\},\left\{ {\boldsymbol{\iota} }_{s}^{(t)}\right\},\left\{ {\boldsymbol{\xi} }_{c}^{(t)}\right\},{ {\boldsymbol{\lambda} } }_{h}^{(t)},{\boldsymbol{n} },{\boldsymbol{\zeta} }^{\left(t\right)}\right)\\& \quad\text{s}\text{.t}\text{. }{\Vert {\boldsymbol{n} }\Vert }^{2}=\varDelta \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \quad\;\, (27)\end{aligned}$ 步骤3. 通过下列公式更新$\left\{ {{\boldsymbol{\boldsymbol{\mu}}}_i^{(t + 1)}} \right\},\left\{ {{\boldsymbol{\boldsymbol{\iota}}}_s^{(t + 1)}} \right\},\left\{ {{\boldsymbol{\boldsymbol{\xi}}}_c^{(t + 1)}} \right\},\left\{ {{\boldsymbol{\lambda }}_h^{(t + 1)}} \right\},{{\boldsymbol{\boldsymbol{\zeta}}}^{\left( {t{\text{ + }}1} \right)}}$： $ {\boldsymbol{\boldsymbol{\mu}}}_i^{(t + 1)}: = {\boldsymbol{\boldsymbol{\mu}}}_i^{(t)} + {\boldsymbol{\boldsymbol{y}}}_i^{(t + 1)} - {{\boldsymbol{A}}^{\text{H}}}\left( {{\theta _i}} \right){{\boldsymbol{\boldsymbol{w}}}^{(t + 1)}} $ (28) $ {\boldsymbol{\boldsymbol{\iota}}}_s^{(t + 1)}: = {\boldsymbol{\boldsymbol{\iota}}}_s^{(t)} + {\boldsymbol{\boldsymbol{z}}}_s^{(t + 1)} - {{\boldsymbol{A}}^{\text{H}}}\left( {{\vartheta _s}} \right){{\boldsymbol{\boldsymbol{w}}}^{(t + 1)}} $ (29) $ {\boldsymbol{\boldsymbol{\xi}}}_c^{(t + 1)}: = {\boldsymbol{\boldsymbol{\xi}}}_c^{(t)} + {\boldsymbol{\boldsymbol{x}}}_c^{(t + 1)} - {{\boldsymbol{A}}^{\text{H}}}\left( {\theta _c^{{\text{com }}}} \right){{\boldsymbol{\boldsymbol{w}}}^{(t + 1)}} $ (30) $ {\boldsymbol{\lambda }}_h^{(t + 1)}: = {\boldsymbol{\lambda }}_h^{(t)} + {\boldsymbol{\boldsymbol{v}}}_h^{(t + 1)} - {{\boldsymbol{A}}^{\text{H}}}\left( {\theta _h^{{\text{eav }}}} \right){{\boldsymbol{\boldsymbol{w}}}^{(t + 1)}} $ (31) ${{\boldsymbol{\boldsymbol{\zeta}}}^{\left( {t{\text{ + }}1} \right)}} = {{\boldsymbol{\boldsymbol{\zeta}}}^{\left( t \right)}} + {{\boldsymbol{\boldsymbol{n}}}^{(t + 1)}} - {{\boldsymbol{\boldsymbol{w}}}^{(t + 1)}}$ (32) 步骤4. 如果原始可行性容差$V_1^{(t + 1)} = \displaystyle\sum\limits_{i = 1}^I {{{\left\| {{\boldsymbol{\boldsymbol{y}}}_i^{(t + 1)} - {{\boldsymbol{A}}^{\text{H}}}\left( {{\theta _i}} \right){{\boldsymbol{\boldsymbol{w}}}^{(t + 1)}}} \right\|}^2}} + $$\displaystyle\sum\limits_{s = 1}^S {{{\left\| {{\boldsymbol{\boldsymbol{z}}}_s^{(t + 1)} - {{\boldsymbol{A}}^{\text{H}}}\left( {{\vartheta _s}} \right){{\boldsymbol{\boldsymbol{w}}}^{(t + 1)}}} \right\|}^2}} + $$\displaystyle\sum\limits_{c = 1}^C { { {\left\| { {\boldsymbol{\boldsymbol{x} } }_c^{(t + 1)} - { {\boldsymbol{A} }^{\text{H} } }\left( {\theta _c^{ {\text{com } } } } \right){ {\boldsymbol{\boldsymbol{w} } }^{(t + 1)} } } \right\|}^2} }$
$+\displaystyle\sum\limits_{h = 1}^E { { {\left\| { {\boldsymbol{\boldsymbol{v} } }_h^{(t + 1)} - { {\boldsymbol{A} }^{\text{H} } }\left( {\theta _h^{ {\text{eav } } } } \right){ {\boldsymbol{\boldsymbol{w} } }^{(t + 1)} } } \right\|}^2} }$$+ {\left\| { { {\boldsymbol{\boldsymbol{n} } }^{(t + 1)} } - { {\boldsymbol{\boldsymbol{w} } }^{(t + 1)} } } \right\|^2} \le {\delta _1}$，则输出${{\boldsymbol{\boldsymbol{w}}}^ \star } = {{\boldsymbol{\boldsymbol{w}}}^{\left( {t + 1} \right)}}$；否则$t: = t + 1$，回到步骤2。
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算法2 基于ADPM的排序学习优化解调方法 Alg. 2 The permutation learning demodulation method
based on ADPM输入：${{\boldsymbol{p}}^0},{\boldsymbol{u}}_1^0,{\boldsymbol{u}}_2^0,{\boldsymbol{\kappa }}_1^0,{\boldsymbol{\kappa }}_2^0,{\rho ^0},{\boldsymbol{g}},{\boldsymbol{B}},{\boldsymbol{C}},{\delta _2},{\delta _3}$； 输出：问题(56)的最优解$ {{\boldsymbol{p}}^ \star } $； 步骤1. $ l = 0 $； 步骤2. 通过求解以下问题更新${{\boldsymbol{p}}^{l{\text{ + }}1}},{\boldsymbol{u}}_1^{l{\text{ + }}1},{\boldsymbol{u}}_2^{l{\text{ + }}1}$：
$\begin{aligned} { {\boldsymbol{p} }^{l + 1} }: =& \arg \mathop {\min }\limits_{\boldsymbol{p} } {\mathcal{L}_{ {\rho ^l} } }\left( { {\boldsymbol{p} },{\boldsymbol{u} }_1^l,{\boldsymbol{u} }_2^l,{\boldsymbol{\kappa } }_1^l,{\boldsymbol{\kappa } }_2^l} \right){\text{ } } \\& {\text{s} }{\text{.t} }{\text{. } }{p_i} \in \{ 0,1\} ,i = 1,2, \cdots ,{K^2} \qquad\qquad\qquad\qquad\; (60)\end{aligned}$$\begin{aligned} {\boldsymbol{u} }_1^{l + 1}: =& \arg \mathop {\min }\limits_{ {\boldsymbol{u} }_1^{} } {\mathcal{L}_{ {\rho ^l} } }\left( { { {\boldsymbol{p} }^{l + 1} },{\boldsymbol{u} }_1^{},{\boldsymbol{u} }_2^l,{\boldsymbol{\kappa } }_1^l,{\boldsymbol{\kappa } }_2^l} \right){\text{ } } \\ & {\text{s} }{\text{.t} }{\text{. } }{\boldsymbol{B} }{ {\boldsymbol{u} }_1} = { { { {\textit{1} } } }_K} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\, (61)\end{aligned}$ $\begin{aligned} {\boldsymbol{u} }_2^{l + 1}: =& \arg \mathop {\min }\limits_{ { {\boldsymbol{u} }_2} } {\mathcal{L}_{ {\rho ^l} } }\left( { { {\boldsymbol{p} }^{l + 1} },{\boldsymbol{u} }_1^{l + 1},{\boldsymbol{u} }_2^{},{\boldsymbol{\kappa } }_1^l,{\boldsymbol{\kappa } }_2^l} \right){\text{ } } \\ & {\text{s} }{\text{.t} }{\text{. } }{\boldsymbol{C} }{ {\boldsymbol{u} }_2} = { { { {\textit{1} } } }_K} \qquad\qquad\quad\qquad\qquad\qquad\qquad\quad\; (62)\end{aligned}$ 步骤3. 通过下列公式更新${\boldsymbol{\kappa }}_1^{l + 1},{\boldsymbol{\kappa }}_2^{l + 1},{\rho ^{l + 1}}$： ${\rho }^{l+1}=\left\{\begin{aligned}& {\delta }_{2}{\rho }^{l},\text{ }\Vert {p}^{l+1}-{p}^{l}\Vert =0 \\ & {\rho }^{l},\text{ }其他\text{ } \end{aligned}\right.$ (63) $ {\boldsymbol{\kappa }}_1^{l + 1}: = {\boldsymbol{\kappa }}_1^l + {\rho ^{l + 1}}\left( {{\boldsymbol{u}}_1^{l + 1} - {{\boldsymbol{p}}^{l + 1}}} \right) $ (64) $ {\boldsymbol{\kappa }}_2^{l + 1}: = {\boldsymbol{\kappa }}_2^l + {\rho ^{l + 1}}\left( {{\boldsymbol{u}}_2^{l + 1} - {{\boldsymbol{p}}^{l + 1}}} \right) $ (65) 4. 如果原始可行性容差
${V}_2^{l + 1} = {\left\| { {\boldsymbol{u} }_1^{l + 1} - { {\boldsymbol{p} }^{l + 1} } } \right\|^2} + {\left\| { {\boldsymbol{u} }_2^{l + 1} - { {\boldsymbol{p} }^{l + 1} } } \right\|^2} \le {\delta _3}$，则输出
$ {{\boldsymbol{p}}^ \star } = {{\boldsymbol{p}}^{l + 1}} $；否则$ l: = l + 1 $，回到步骤2。
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表 1 不同算法所得PMSR
Table 1. PMSR derived by different methods
数值 PSK-ADMM QAM-ADMM PSK-TRPIS $K = 4$ 18.7693 18.7715 17.8392 $K = 8$ 18.7693 18.7735 $K = 16$ 18.7535 18.8135
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表 2 不同算法所需时间
Table 2. Computing time required by different methods
数值 PSK-ADMM QAM-ADMM PSK-TRPIS $K = 4$ 8.354 8.434 8.276 $K = 8$ 5.502 5.182 8.281 $K = 16$ 9.221 7.398 8.282
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