在跳频通信中，用来控制载波频率跳变的地址码序列称为跳频序列^[1].对跳频序列的研究主要包括2个方面^[2-3]：一方面是寻找跳频序列各种参数受到的理论限制；另一方面是构造满足理论界限的跳频序列集.文献[3-7]给出了跳频序列有关一维汉明相关函数的理论界.

跳频序列偶适用于发送端和接收端使用不同地址码的跳频通信系统，这类通信系统采用失配滤波实现相关接收^[8-9].目前只有跳频序列偶包含时延变量的一维汉明相关函数的理论界^[10].在实际系统中，跳频信号除存在传输时延外，还可能发生频率偏移.特别是像雷达等系统在高速运动的情况下，传输信号的频率可能发生偏移，跳频序列的频隙可能移位至其他频隙，导致与其他跳频序列发生碰撞，影响通信质量.因此有必要在传统汉明相关函数中引入频移变量，建立跳频序列偶的时频二维移位汉明相关函数的概念及其相关理论界.给出了跳频序列偶时频二维移位汉明相关函数的概念，证明了二维相关跳频序列偶的唯一性，推导了跳频序列偶的二维移位汉明相关函数的理论界.

1 跳频序列偶二维移位汉明相关函数的概念

跳频序列偶的时频二维移位汉明相关函数的概念定义如下.

定义1  设频率集F是一个加法群，(x, y)是F上周期为L的跳频序列x=(x(0), x(1), …, x(L－1))和y=(y(0), y(1), …, y(L－1))组成的序列偶，(x′, y′)是F上周期为L的跳频序列x′=(x′(0), x′(1), …, x′(L－1))和y′=(y′(0), y′(1), …, y′(L－1))组成的序列偶.称

$ {H_{\left( {x,y} \right)}}\left( {\tau ,d} \right) = \sum\limits_{j = 0}^{L - 1} {h\left[ {x\left( j \right),y\left( {j + \tau } \right) + d} \right]} $

(1)

为(x, y)的时频二维周期汉明自相关函数.称

$ \begin{array}{*{20}{c}} {{H_{\left( {x,y} \right)\left( {x',y'} \right)}}\left( {\tau ,d} \right) = {H_{\left( {x,y'} \right)}}\left( {\tau ,d} \right) = }\\ {\sum\limits_{j = 0}^{L - 1} {h\left[ {x\left( j \right),y'\left( {j + \tau } \right) + d} \right]} } \end{array} $

(2)

为(x, y)和(x′, y′)的时频二维周期汉明互相关函数.其中，h[·]为汉明函数，即当u=v时，h[u, v]=1，当u≠v时，h[u, v]=0，τ表示时延，d表示频移，0≤τ≤L－1，d∈F，且j+τ≡(j+τ) mod L，j=0, 1, …, L－1.

当d=0时，二维周期汉明相关函数退化成一维周期汉明相关函数.因此二维周期汉明相关函数是一维周期汉明相关函数的推广.

$ {H_{\rm{m}}}\left( {x,y} \right) = \max \left\{ {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)\left| {\left( {\tau ,d} \right) \ne \left( {0,0} \right)} \right.} \right\} $

(3)

表示跳频序列偶(x, y)的异相二维汉明自相关函数最大值.

$ {H_{\rm{m}}}\left[ {\left( {x,y} \right)\left( {x',y'} \right)} \right] = \max \left\{ {{H_{\left( {x,y} \right)\left( {x',y'} \right)}}\left( {\tau ,d} \right)} \right\} $

(4)

表示跳频序列偶(x, y)和(x′, y′)的二维汉明互相关函数最大值.

$ \bar H\left( {x,y} \right) = \frac{1}{{Lq - 1}}\sum\limits_{\left( {\tau ,d} \right) \ne \left( {0,0} \right)} {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)} $

(5)

表示跳频序列偶(x, y)的平均二维周期汉明自相关函数值.其中：L为序列长度，q为频隙数目.

$ \bar H\left[ {\left( {x,y} \right)\left( {x',y'} \right)} \right] = \frac{1}{{Lq}}\sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {{H_{\left( {x,y} \right)\left( {x',y'} \right)}}\left( {\tau ,d} \right)} } $

(6)

表示跳频序列偶(x, y)和(x′, y′)的平均二维周期汉明互相关函数值.其中：L为序列长度，q为频隙数目.

$ \begin{array}{*{20}{c}} {{H_{{\rm{am}}}}\left( S \right) = }\\ {\max \left\{ {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)\left| {\left( {x,y} \right) \in S,\left( {\tau ,d} \right) \ne \left( {0,0} \right)} \right.} \right\}} \end{array} $

(7)

$ \begin{array}{*{20}{c}} {{H_{{\rm{cm}}}}\left( S \right) = }\\ {\max \left\{ {{H_{\left( {x,y'} \right)}}\left( {\tau ,d} \right)\left| {\left( {x,y} \right)} \right.,\left( {x',y'} \right) \in S,x \ne y'} \right\}} \end{array} $

(8)

$ {H_{\rm{m}}}\left( S \right) = \max \left\{ {{H_{{\rm{am}}}}\left( S \right),{H_{{\rm{cm}}}}\left( S \right)} \right\} $

(9)

分别表示跳频序列偶集S的最大异相二维周期汉明自相关、最大二维周期汉明互相关和最大二维周期汉明相关.

$ {S_{\rm{a}}}\left( S \right) = \sum\limits_{\left( {x,y} \right) \in S} {\sum\limits_{\left( {\tau ,d} \right) \ne \left( {0,0} \right)} {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)} } $

(10)

$ {S_{\rm{c}}}\left( S \right) = \frac{1}{2}\sum\limits_{\begin{array}{*{20}{c}} {\left( {x,y} \right) \in S}\\ {\left( {x',y'} \right) \in S}\\ {x \ne y} \end{array}} {\sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {{H_{\left( {x,y} \right)\left( {x',y'} \right)}}\left( {\tau ,d} \right)} } } $

(11)

分别表示跳频序列偶集S的二维周期汉明自相关碰撞总数和二维周期汉明互相关碰撞总数.

$ {A_{\rm{a}}}\left( S \right) = \frac{{{S_{\rm{a}}}\left( S \right)}}{{M\left( {Lq - 1} \right)}} $

(12)

$ {A_{\rm{c}}}\left( S \right) = \frac{{2{S_{\rm{c}}}\left( S \right)}}{{M\left( {M - 1} \right)Lq}} $

(13)

分别表示跳频序列偶集S的平均二维周期汉明自相关和平均二维周期汉明互相关.

2 二维相关跳频序列偶的唯一性

定理1  设频率集F={f₀, f₁, …, f_q－1}是一个加法群，x=(x(0), x(1), …, x(L－1))，y=(y(0), y(1), …, y(L－1))和y′=(y′(0), y′(1), …, y′(L－1))为F上长度为L的跳频序列，若跳频序列偶(x, y)和(x, y′)的二维周期汉明自相关函数相等，即H_{(x, y)}(τ, d)= H_{(x, y′)}(τ, d)(0≤τ≤L－1，d∈F)，则y=y′.

证明  设

$ {f_x}\left( {a,b} \right) = \sum\limits_{i = 0}^{q - 1} {\sum\limits_{j = 0}^{L - 1} {{x_{ij}}{a^i}{b^j}} } ,{f_y}\left( {a,b} \right) = \sum\limits_{i = 0}^{q - 1} {\sum\limits_{j = 0}^{L - 1} {{y_{ij}}{a^i}{b^j}} } $

是以a、b为变量的二元多项式函数.其中，

$ {x_{ij}} = \left\{ \begin{array}{l} 1,\;\;\;\;i = x\left( j \right)\\ 0,\;\;\;\;i \ne x\left( j \right) \end{array} \right. $

$ {y_{\left( {i - d} \right)\left( {j + \tau } \right)}} = \left\{ \begin{array}{l} 1,\;\;\;\;i = y\left( {j + \tau } \right) + d\\ 0,\;\;\;\;i \ne y\left( {j + \tau } \right) + d \end{array} \right. $

则

$ \begin{array}{*{20}{c}} {{f_x}\left( {a,b} \right) = {f_y}\left( {{a^{ - 1}},{b^{ - 1}}} \right) = }\\ {\sum\limits_{i = 0}^{q - 1} {\sum\limits_{j = 0}^{L - 1} {{x_{ij}}{a^i}{b^j}} } \sum\limits_{k = 0}^{q - 1} {\sum\limits_{h = 0}^{L - 1} {{y_{kh}}{a^{ - k}}{b^{ - h}}} } = }\\ {\sum\limits_{i = 0}^{q - 1} {\sum\limits_{j = 0}^{L - 1} {\sum\limits_{k = 0}^{q - 1} {\sum\limits_{h = 0}^{L - 1} {{x_{ij}}{y_{kh}}{a^{i - k}}{b^{j - h}}} } } } } \end{array} $

令i－k=d，j－h=－τ，则

$ \begin{array}{*{20}{c}} {{f_x}\left( {a,b} \right){f_y}\left( {{a^{ - 1}},{b^{ - 1}}} \right) = }\\ {\sum\limits_{i = 0}^{q - 1} {\sum\limits_{j = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {\sum\limits_{\tau = 0}^{L - 1} {{x_{ij}}{y_{\left( {i - d} \right)\left( {j + \tau } \right)}}{a^d}{b^{ - \tau }}} } } } } \end{array} $

其中

$ \begin{array}{*{20}{c}} {\sum\limits_{d = 0}^{q - 1} {\sum\limits_{\tau = 0}^{L - 1} {{x_{ij}}{y_{\left( {i - d} \right)\left( {j + \tau } \right)}}} } = \sum\limits_{i = x\left( j \right) = y\left( {j + \tau } \right) + d} {\sum\limits_{j = 0}^{L - 1} {1 \times 1} } = }\\ {\sum\limits_{x\left( j \right) = y\left( {j + \tau } \right) + d} 1 = \sum\limits_{j = 0}^{L - 1} {h\left[ {x\left( j \right),y\left( {j + \tau } \right) + d} \right]} = }\\ {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)} \end{array} $

因此

$ {f_x}\left( {a,b} \right){f_y}\left( {{a^{ - 1}},{b^{ - 1}}} \right) = \sum\limits_{d = 0}^{q - 1} {\sum\limits_{\tau = 0}^{L - 1} {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right){a^d}{b^{ - \tau }}} } $

设$ {{f}_{{{y}'}}}\left( a, b \right)=\sum\limits_{i=0}^{q-1}{\sum\limits_{j=0}^{L-1}{y_{ij}^{\prime }{{a}^{i}}{{b}^{j}}}} $，同理可得

$ {f_x}\left( {a,b} \right){f_{y'}}\left( {{a^{ - 1}},{b^{ - 1}}} \right) = \sum\limits_{d = 0}^{q - 1} {\sum\limits_{\tau = 0}^{L - 1} {{H_{\left( {x,y'} \right)}}\left( {\tau ,d} \right){a^d}{b^{ - \tau }}} } $

因为H_{(x, y)}(τ, d)=H_{(x, y′)}(τ, d)，所以

$ {f_x}\left( {a,b} \right){f_y}\left( {{a^{ - 1}},{b^{ - 1}}} \right) = {f_x}\left( {a,b} \right){f_{y'}}\left( {{a^{ - 1}},{b^{ - 1}}} \right) $

即f_x(a, b)[f_y(a^－1, b^－1)－f_y′(a^－1, b^－1)]=0

上式两边同时乘以f_y(a^－1, b^－1)，得

$ \begin{array}{*{20}{c}} {{f_x}\left( {a,b} \right){f_y}\left( {{a^{ - 1}},{b^{ - 1}}} \right)\left[ {{f_y}\left( {{a^{ - 1}},{b^{ - 1}}} \right) - } \right.}\\ {\left. {{f_{y'}}\left( {{a^{ - 1}},{b^{ - 1}}} \right)} \right] = 0} \end{array} $

因为f_x(a, b)f_y(a^－1, b^－1)不等于0，则有

$ {f_y}\left( {{a^{ - 1}},{b^{ - 1}}} \right) = {f_{y'}}\left( {{a^{ - 1}},{b^{ - 1}}} \right) $

所以y=y′.

3 跳频序列偶二维周期汉明相关函数的理论界

引理1  设S是由频率集F={f₀, f₁, …, f_q－1}上M个周期为L的二维相关跳频序列偶组成的集合，对于任意频率f_i∈F，(x, y)∈S，用u_x(f_i)=$ \sum\limits_{j=0}^{L-1} $h(x(j), f_i)表示频率f_i在x中出现的次数，则

$ \sum\limits_{x,y \in S} {\sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)} } } = \sum\limits_{x,y \in S} {\sum\limits_{f \in F} {{u_x}\left( f \right){u_{{y^{\left( \tau \right)}} + d}}\left( f \right)} } $

(14)

证明  由于

$ \begin{array}{*{20}{c}} {\sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)} } = }\\ {\sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {\sum\limits_{j = 0}^{L - 1} {h\left( {y\left( {j + \tau } \right) + d,x\left( j \right)} \right)} } } = }\\ {\sum\limits_{j = 0}^{L - 1} {{u_{{y^{\left( \tau \right)}} + d}}\left( {x\left( j \right)} \right)} = \sum\limits_{i = 0}^{q - 1} {{u_x}\left( {{f_i}} \right){u_{{y^{\left( \tau \right)}} + d}}\left( {{f_i}} \right)} = }\\ {\sum\limits_{f \in F} {{u_x}\left( f \right){u_{{y^{\left( \tau \right)}} + d}}\left( f \right)} } \end{array} $

其中y^(τ)+d=(y(0+τ)+d, y(1+τ)+d, …, y(L－1+τ)+d)，u_x(f_i)表示频率f_i在跳频序列x中出现的次数，u_y^(τ)+d(f_i)表示频率f_i在跳频序列y^(τ)+d中出现的次数.所以得

$ \sum\limits_{x,y \in S} {\sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)} } } = \sum\limits_{x,y \in S} {\sum\limits_{f \in F} {{u_x}\left( f \right){u_{{y^{\left( \tau \right)}} + d}}\left( f \right)} } $

定理2  对于频率集合F={f₀, f₁, …, f_q－1}上的(L, q)二维相关跳频序列偶(x, y)，其异相二维自相关函数最大值H_m(x, y)满足

$ {H_m}\left( {x,y} \right) \ge \frac{{\sum\limits_{i = 0}^{q - 1} {{e_i}{d_i}} - L + d\left( {x,y} \right)}}{{Lq - 1}} $

(15)

其中：d_i、e_i(i=0, 1, …, q－1)分别表示第i个频率f_i在跳频序列x和y的一个周期中出现的次数，d(x, y)表示跳频序列x和y之间的汉明距离.

证明  由式(5)和式(6)知

$ \begin{array}{*{20}{c}} {\sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)} } = }\\ {L - d\left( {x,y} \right) + \sum\limits_{\left( {\tau ,d} \right) \ne \left( {0,0} \right)} {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)} = }\\ {L - d\left( {x,y} \right) + \left( {Lq - 1} \right)\bar H\left( {x,y} \right)} \end{array} $

由引理1知

$ \begin{array}{*{20}{c}} {\sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)} } = \sum\limits_{i = 0}^{q - 1} {{u_x}\left( {{f_i}} \right){u_{{y^{\left( \tau \right)}} + d}}\left( {{f_i}} \right)} = }\\ {\sum\limits_{f \in F} {{u_x}\left( f \right){u_{{y^{\left( \tau \right)}} + d}}\left( f \right)} } \end{array} $

已知频率集合F={f₀, f₁, …, f_q－1}，d_i=u_x(f_i)，e_i=u_y^(τ)+d(f_i)，0≤i≤q－1，则有

$ \begin{array}{*{20}{c}} {\sum\limits_{f \in F} {{u_x}\left( f \right){u_{{y^{\left( \tau \right)}} + d}}\left( f \right)} = \sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)} } = }\\ {L - d\left( {x,y} \right) + \left( {Lq - 1} \right)\bar H\left( {x,y} \right)} \end{array} $

即

$ \begin{array}{*{20}{c}} {\bar H\left( {x,y} \right) = \frac{{\sum\limits_{f \in F} {{u_x}\left( f \right){u_{{y^{\left( \tau \right)}} + d}}\left( f \right) - L + d\left( {x,y} \right)} }}{{Lq - 1}} = }\\ {\frac{{\sum\limits_{i = 1}^{q - 1} {{e_i}{d_i} - L + d\left( {x,y} \right)} }}{{Lq - 1}}} \end{array} $

$ {H_m}\left( {x,y} \right) \ge \bar H\left( {x,y} \right) = \frac{{\sum\limits_{i = 1}^{q - 1} {{e_i}{d_i} - L + d\left( {x,y} \right)} }}{{Lq - 1}} $

定理3  设跳频序列偶(x, y)和(x′, y′)分别是频率集合F={f₀, f₁, …, f_q－1}上2个(L, q)二维相关跳频序列偶，并设a_i、b_i、c_i、d_i(i=0, 1, …, q－1)分别表示第i个频率f_i在二维相关跳频序列x、y、x′、y′的一个周期中出现的次数，且所有f_i的排列使得a_i和c分别形成递减数列，即a₀≥a₁≥…≥a_q－1，c₀≥c₁≥ …≥c_q－1，则二维相关跳频序列偶(x, y)与(x′, y′)之间的二维汉明互相关函数最大值H_m[(x, y)(x′, y′)]满足

$ \begin{array}{*{20}{c}} {{H_m}\left[ {\left( {x,y} \right)\left( {x',y'} \right)} \right] \ge }\\ {\frac{1}{{3Lq - 2}}\left[ {\sum\limits_{i = 0}^{q - 1} {\left( {{a_i}{b_i} + {c_i}{d_i} + {a_i}{d_i}} \right) - 2\left( {L - {d_{\min }}} \right)} } \right]} \end{array} $

(16)

且满足下述条件时，式(16)右边达到最小：

1) b₀≤b₁≤…≤b_q-1，d₀≤d₁≤…≤d_q－1，b_i、d_i形成递增数列；

2) d_min=min{d(x, y), d(x′, y′)}.

证明  设

$ \begin{array}{*{20}{c}} {E\left[ {\left( {x,y} \right)\left( {x',y'} \right)} \right] = }\\ {\frac{1}{{3Lq - 2}}\left[ {\sum\limits_{\left( {\tau ,d} \right) \ne \left( {0,0} \right)} {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)} + } \right.}\\ {\left. {\sum\limits_{\left( {\tau ,d} \right) \ne \left( {0,0} \right)} {{H_{\left( {x',y'} \right)}}\left( {\tau ,d} \right)} + \sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {{H_{\left( {x,y'} \right)}}\left( {\tau ,d} \right)} } } \right]} \end{array} $

由式(5)和式(6)知

$ \begin{array}{*{20}{c}} {E\left[ {\left( {x,y} \right)\left( {x',y'} \right)} \right] = }\\ {\frac{1}{{3Lq - 2}}\left[ {\left( {Lq - 1} \right)\bar H\left( {x,y} \right) + } \right.}\\ {\left( {Lq - 1} \right)\bar H\left( {x',y'} \right) + Lq\bar H\left[ {\left( {x,y} \right)\left( {x',y'} \right)} \right]} \end{array} $

由式(5)、式(6)、式(14)知

$ \begin{array}{*{20}{c}} {\left( {Lq - 1} \right)\bar H\left( {x,y} \right) = \sum\limits_{f \in F} {{u_x}\left( f \right){u_{{y^{\left( \tau \right)}} + d}}\left( f \right)} - }\\ {\left( {L - d\left( {x,y} \right)} \right)\left( {Lq - 1} \right)\bar H\left( {x',y'} \right) = }\\ {\sum\limits_{f \in F} {{u_{x'}}\left( f \right){u_{{{y'}^{\left( \tau \right)}} + d}}\left( f \right)} - \left( {L - d\left( {x',y'} \right)} \right)}\\ {Lq\bar H\left[ {\left( {x,y} \right)\left( {x',y'} \right)} \right] = \sum\limits_{f \in F} {{u_x}\left( f \right){u_{{{y'}^{\left( \tau \right)}} + d}}\left( f \right)} } \end{array} $

由于，a_i=u_x(f_i)，b_i=u_y^(τ)+d(f_i)，c_i=u_x′(f_i)，d_i=u_y′^(τ)+d(f_i)，0≤i≤q－1，有

$ \begin{array}{*{20}{c}} {E\left[ {\left( {x,y} \right)\left( {x',y'} \right)} \right] = }\\ {\frac{1}{{3Lq - 2}}\left[ {\sum\limits_{f \in F} {{u_x}\left( f \right){u_{{y^{\left( \tau \right)}} + d}}\left( f \right)} - } \right.}\\ {\left( {L - d\left( {x,y} \right)} \right) + \sum\limits_{f \in F} {{u_{x'}}\left( f \right){u_{{{y'}^{\left( \tau \right)}} + d}}\left( f \right)} - }\\ {\left( {L - d\left( {x',y'} \right)} \right) + \sum\limits_{f \in F} {{u_x}\left( f \right){u_{{{y'}^{\left( \tau \right)}} + d}}\left( f \right)} = }\\ {\frac{1}{{3Lq - 2}}\left[ {\sum\limits_{i = 0}^{q - 1} {\left( {{a_i}{b_i} + {c_i}{d_i} + {a_i}{d_i}} \right) - \left( {L - d\left( {x,y} \right)} \right) - } } \right.}\\ {\left. {\left( {L - d\left( {x',y'} \right)} \right)} \right] \ge }\\ {\frac{1}{{3Lq - 2}}\left[ {\sum\limits_{i = 0}^{q - 1} {\left( {{a_i}{b_i} + {c_i}{d_i} + {a_i}{d_i}} \right) - 2\left( {L - {d_{\min }}} \right)} } \right]} \end{array} $

故得

$ \begin{array}{*{20}{c}} {{H_m}\left[ {\left( {x,y} \right)\left( {x',y'} \right)} \right] \ge E\left[ {\left( {x,y} \right)\left( {x',y'} \right)} \right] \ge }\\ {\frac{1}{{3Lq - 2}}\left[ {\sum\limits_{i = 0}^{q - 1} {\left( {{a_i}{b_i} + {c_i}{d_i} + {a_i}{d_i}} \right) - 2\left( {L - {d_{\min }}} \right)} } \right]} \end{array} $

满足条件1)、2)时，由切比雪夫不等式可知，等式右边达到最小.

4 跳频序列偶集最大二维周期汉明相关函数的理论界

定理4  设S是由频率集F={f₀, f₁, …, f_q－1}上M个(L, q)二维相关跳频序列偶组成的集合，(x, y)∈S，(x′, y′)∈S，τ=0, 1, …, L－1，d=0, 1, …, q－1.令函数

$ {H_{{\rm{CA}}}}\left( S \right) = \sum\limits_{\begin{array}{*{20}{c}} {\left( {x,y} \right) \in S}\\ {\left( {x',y'} \right) \in S} \end{array}} {\sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {{H_{\left( {x',y'} \right)}}\left( {\tau ,d} \right)} } } $

(17)

则

$ \begin{array}{*{20}{c}} {{H_{{\rm{CA}}}}\left( S \right) \le M\left( {L - {d_{\min }}} \right) + M\left( {Lq - 1} \right){H_{{\rm{am}}}}\left( S \right) + }\\ {M\left( {M - 1} \right)Lq{H_{{\rm{cm}}}}\left( S \right)} \end{array} $

其中

$ {d_{\min }} = \min \left\{ {d\left( {x,y} \right)\left| {\left( {x,y} \right) \in S} \right.} \right\} $

$ \begin{array}{*{20}{c}} {{H_{{\rm{am}}}}\left( S \right) = }\\ {\max \left\{ {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)\left| {\left( {x,y} \right) \in S} \right.,\left( {\tau ,d} \right) \ne \left( {0,0} \right)} \right\}} \end{array} $

$ \begin{array}{*{20}{c}} {{H_{{\rm{cm}}}}\left( S \right) = }\\ {\max \left\{ {{H_{\left( {x,y'} \right)}}\left( {\tau ,d} \right)\left| {\left( {x,y} \right)} \right.,\left( {x',y'} \right) \in S,x \ne y'} \right\}} \end{array} $

证明

$ \begin{array}{*{20}{c}} {{H_{{\rm{CA}}}}\left( S \right) = \sum\limits_{\left( {x,y} \right) \in S} {H\left( {x,y} \right)\left( {0,0} \right)} + }\\ {\sum\limits_{\left( {x,y} \right) \in S} {\sum\limits_{\left( {\tau ,d} \right) \ne \left( {0,0} \right)} {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)} } + }\\ {\sum\limits_{\begin{array}{*{20}{c}} {\left( {x,y} \right) \in S}\\ {\left( {x',y'} \right) \in S}\\ {x \ne y'} \end{array}} {\sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {{H_{\left( {x',y'} \right)}}\left( {\tau ,d} \right)} } } \le }\\ {M\left( {L - {d_{\min }}} \right)M\left( {Lq - 1} \right){H_{{\rm{am}}}}\left( S \right) + }\\ {M\left( {M - 1} \right)Lq{H_{{\rm{cm}}}}\left( S \right)} \end{array} $

定理5  设S是由频率集F={f₀, f₁, …, f_q－1}上M个(L, q)二维相关跳频序列偶组成的集合，令$ {{H}_{\text{CA}}}\left( S \right)=\sum\limits_{\begin{smallmatrix} \ \left( x, y \right)\in S \\ \left( x\prime, y\prime \right)\in S \end{smallmatrix}}{\sum\limits_{\tau =0}^{L-1}{\sum\limits_{d=0}^{q-1}{{{H}_{\left( x, y \right)(x\prime, y\prime )}}\left( \tau, d \right)}}} $，$ \left( x, y \right)\in S, \left( x\prime, y\prime \right)\in S $，则

$ {H_{{\rm{CA}}}}\left( S \right) \ge \frac{{{L^2}{M^2}}}{q} $

(18)

证明  对于任意整数i=0, 1, …, q－1，令g_i=$ \sum\limits_{x\in S} $u_x(f_i)，由定义1知

$ \begin{matrix} {{H}_{\left( x,y \right)\left( {x}',{y}' \right)}}\left( \tau ,d \right)={{H}_{\left( x,{y}' \right)}}\left( \tau ,d \right)= \\ \sum\limits_{j=0}^{L-1}{h\left[ x\left( j \right),{y}'\left( j+\tau \right)+d \right]} \\ \end{matrix} $

那么

$ \begin{array}{*{20}{c}} {\sum\limits_{j = 0}^{L - 1} {{u_{y'\left( \tau \right) + d}}\left( {x\left( j \right)} \right)} = \sum\limits_{j = 0}^{L - 1} {h\left( {y' + x\left( j \right)} \right)} = }\\ {\sum\limits_{j = 0}^{L - 1} {\sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {h\left[ {x\left( j \right),y'\left( {j + \tau } \right) + d} \right]} } } = }\\ {\sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {{H_{\left( {x,y} \right)\left( {x',y'} \right)}}\left( {\tau ,d} \right)} } } \end{array} $

由引理1得

$ \begin{array}{*{20}{c}} {{H_{{\rm{CA}}}}\left( S \right) = \sum\limits_{\begin{array}{*{20}{c}} {\left( {x,y} \right) \in S}\\ {\left( {x',y'} \right) \in S} \end{array}} {\sum\limits_{j = 0}^{L - 1} {{u_{{{y'}^{\left( \tau \right)}} + d}}\left( {x\left( j \right)} \right)} } = }\\ {\sum\limits_{\begin{array}{*{20}{c}} {\left( {x,y} \right) \in S}\\ {\left( {x',y'} \right) \in S} \end{array}} {\sum\limits_{i = 0}^{q - 1} {{u_x}\left( {{f_j}} \right){u_{{{y'}^{\left( \tau \right)}} + d}}\left( {{f_i}} \right)} } = \sum\limits_{i = 0}^{q - 1} {{{\left( {\sum\limits_{\left( {x,y} \right) \in S} {{u_x}\left( {{f_i}} \right)} } \right)}^2}} } \end{array} $

即$ {{H}_{\text{CA}}}\left( S \right)=\sum\limits_{i=0}^{q-1}{g_{i}^{2}} $，并且

$ \begin{array}{*{20}{c}} {{g_i} \ge 0,i = 0,1, \cdots ,q - 1}\\ {\sum\limits_{i = 0}^{q - 1} {{g_i}} = \sum\limits_{\left( {x,y} \right) \in S} {\sum\limits_{i = 0}^{q - 1} {{u_x}\left( {{f_i}} \right)} } = \sum\limits_{\left( {x,y} \right) \in S} {L = LM} } \end{array} $

利用拉格朗日法，求得在上述条件下$ \sum\limits_{i=0}^{q-1}{g_{i}^{2}} $的最小值是$ \frac{{{L}^{2}}{{M}^{2}}}{q} $，即证得$ {{H}_{\text{CA}}}\left( S \right)\ge \frac{{{L}^{2}}{{M}^{2}}}{q} $.

推论1  设S是由频率集F={f₀, f₁, …, f_q－1}上M个(L, q)二维相关跳频序列偶组成的集合，τ=0, 1, …, L－1，d=0, 1, …, q－1，则

$ \begin{array}{*{20}{c}} {q\left( {Lq - 1} \right){H_{{\rm{am}}}}\left( S \right) + q\left( {M - 1} \right)Lq{H_{{\rm{cm}}}}\left( S \right) \ge }\\ {{L^2}M - q\left( {L - {d_{\min }}} \right)} \end{array} $

(19)

其中

$ \begin{array}{l} {d_{\min }} = \min \left\{ {d\left( {x,y} \right)\left| {\left( {x,y} \right) \in S} \right.} \right\}\\ {H_{\rm{m}}}\left( S \right) = \max \left\{ {{H_{{\rm{am}}}}\left( S \right),{H_{{\rm{cm}}}}\left( S \right)} \right\} \end{array} $

证明  由定理4和定理5，得

$ \begin{array}{*{20}{c}} {M\left( {L - {d_{\min }}} \right) + M\left( {Lq - 1} \right){H_{{\rm{am}}}}\left( S \right) + }\\ {M\left( {M - 1} \right)Lq{H_{{\rm{cm}}}}\left( S \right) \ge \frac{{{L^2}{M^2}}}{q}}\\ {q\left( {L - {d_{\min }}} \right) + q\left( {Lq - 1} \right){H_{{\rm{am}}}}\left( S \right) + }\\ {q\left( {M - 1} \right)Lq{H_{{\rm{cm}}}}\left( S \right) \ge {L^2}M}\\ {q\left( {Lq - 1} \right){H_{{\rm{am}}}}\left( S \right) + q\left( {M - 1} \right)Lq{H_{{\rm{cm}}}}\left( S \right) \ge }\\ {{L^2}M - q\left( {L - {d_{\min }}} \right)} \end{array} $

推论2  设S是由频率集F={f₀, f₁, …, f_q－1}上M个(L, q)二维相关跳频序列偶组成的集合，令H_m=max{H_am, H_cm}，可得

$ {H_{\rm{m}}}\left( S \right) \ge \frac{{{L^2}M - q\left( {L - {d_{\min }}} \right)}}{{q\left( {LqM - 1} \right)}} $

(20)

证明  由推论1得

$ \begin{array}{*{20}{c}} {q\left( {Lq - 1} \right){H_{{\rm{am}}}}\left( S \right) + q\left( {M - 1} \right)Lq{H_{{\rm{cm}}}}\left( S \right) \ge }\\ {{L^2}M - q\left( {L - {d_{\min }}} \right)}\\ {\left[ {q\left( {Lq - 1} \right) + q\left( {M - 1} \right)Lq} \right]{H_{\rm{m}}}\left( S \right) \ge }\\ {{L^2}M - q\left( {L - {d_{\min }}} \right)}\\ {{H_{\rm{m}}}\left( S \right) \ge \frac{{{L^2}M - q\left( {L - {d_{\min }}} \right)}}{{q\left( {LqM - 1} \right)}}} \end{array} $

5 跳频序列偶集平均二维周期汉明相关函数的理论界

定理6  设S是由频率集F={f₀, f₁, …, f_q－1}上M个(L, q)二维相关跳频序列偶组成的集合，则

$ \begin{array}{*{20}{c}} {\frac{{{A_{\rm{a}}}\left( S \right)}}{{Lq\left( {M - 1} \right)}}\frac{{{A_{\rm{c}}}\left( S \right)}}{{Lq - 1}} - \frac{{{d_{\min }}}}{{Lq\left( {Lq - 1} \right)\left( {M - 1} \right)}} \ge }\\ {\frac{{LM - q}}{{{q^2}\left( {Lq - 1} \right)\left( {M - 1} \right)}}} \end{array} $

(21)

其中：A_a(S)和A_c(S)分别表示跳频序列偶集S的平均二维周期汉明自相关和平均二维周期汉明互相关，d_min=min{d(x, y)|(x, y)∈S}.

证明

$ \begin{array}{*{20}{c}} {{H_{{\rm{CA}}}}\left( S \right) = \sum\limits_{\begin{array}{*{20}{c}} {\left( {x,y} \right) \in S}\\ {\left( {x',y'} \right) \in S} \end{array}} {\sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {{H_{\left( {x,y} \right)\left( {x',y'} \right)}}\left( {\tau ,d} \right)} } } = }\\ {\sum\limits_{\left( {x,y} \right) \in S} {{H_{\left( {x,y} \right)}}\left( {0,0} \right)} + \sum\limits_{\left( {x,y} \right) \in S} {\sum\limits_{\left( {\tau ,d} \right) \ne \left( {0,0} \right)} {{H_{\left( {x,y} \right)}}\left( {\tau ,d} \right)} } + }\\ {\sum\limits_{\begin{array}{*{20}{c}} {\left( {x,y} \right) \in S}\\ {\left( {x',y'} \right) \in S}\\ {x \ne y'} \end{array}} {\sum\limits_{\tau = 0}^{L - 1} {\sum\limits_{d = 0}^{q - 1} {{H_{\left( {x',y'} \right)}}\left( {\tau ,d} \right)} } } \le }\\ {M\left( {L - {d_{\min }}} \right) + {S_{\rm{a}}}\left( S \right) + 2{S_{\rm{c}}}\left( S \right)} \end{array} $

由定理5知，$ {{H}_{\text{CA}}}\left( S \right)\ge \frac{{{L}^{2}}{{M}^{2}}}{q} $.故得

$ M\left( {L - {d_{\min }}} \right) + {S_{\rm{a}}}\left( S \right) + 2{S_{\rm{c}}}\left( S \right) \ge \frac{{{L^2}{M^2}}}{q} $

根据式(10)~式(13)中A_a(S)和A_c(S)分别与S_a(S)和S_c(S)的关系知

$ \begin{array}{*{20}{c}} {M\left( {L - {d_{\min }}} \right) + M\left( {Lq - 1} \right){A_{\rm{a}}}\left( S \right) + }\\ {M\left( {M - 1} \right)Lq{A_{\rm{c}}}\left( S \right) \ge \frac{{{L^2}{M^2}}}{q}}\\ {M\left( {Lq - 1} \right){A_{\rm{a}}}\left( S \right) + M\left( {M - 1} \right)Lq{A_{\rm{c}}}\left( S \right) - M{d_{\min }} \ge }\\ {\frac{{{L^2}{M^2}}}{q} - LM}\\ {\frac{{{A_{\rm{a}}}\left( S \right)}}{{Lq\left( {M - 1} \right)}} + \frac{{{A_{\rm{c}}}\left( S \right)}}{{Lq - 1}} - \frac{{{d_{\min }}}}{{Lq\left( {Lq - 1} \right)\left( {M - 1} \right)}} \ge }\\ {\frac{{LM - q}}{{{q^2}\left( {Lq - 1} \right)\left( {M - 1} \right)}}} \end{array} $

6 结束语

给出了跳频序列偶的时频二维移位汉明相关函数的概念.证明了二维相关跳频序列偶的唯一性，保证了这类信号在实际应用中的唯一接收.导出了由二维相关跳频序列偶的汉明相关函数、序列偶个数、频隙个数和序列长度构成的理论界，对于构造满足理论界的二维相关跳频序列偶集具有重要指导意义.