Costas arrays have been an interesting combinatorial object for decades because of their optimal aperiodic auto-correlation properties. Meanwhile, it is interesting to find families of Costas arrays or extended arrays with small maximal cross-correlation values, since for applications in multi-user systems, the cross-interferences between different signals should also be small. The objective of this paper is to study several large-size families of Costas arrays or extended arrays, and their values of maximal crosscorrelation are partially bounded for some cases of horizontal shifts $u$ and vertical shifts $v$. Given a prime $p \geq 5$, a large-size family of Costas arrays over $\{1, \ldots, p-1\}$ is investigated, including both the exponential and logarithmic Welch Costas arrays. An upper bound on the maximal cross-correlation of this family for arbitrary $u$ and $v$ is given. We also show that the maximal cross-correlation of the family of power permutations over $\{1, \ldots, p-1\}$ for $u=0$ and $v \neq 0$ is bounded by $\frac{1}{2}+\sqrt{p-1}$. Furthermore, we give the first nontrivial upper bound on the maximal cross-correlation of the larger family including both exponential Welch Costas arrays and power permutations over $\{1, \ldots, p-1\}$ for arbitrary $u$ and $v=0$ that it equals $(p-1) / t$ where $t$ is the smallest prime divisor of $(p-1) / 2$ if p is not a safe prime and is at most $(p-1)^{\frac{1}{2}}+(p-1)^{\frac{1}{4}}+\frac{1}{2}$ otherwise.

翻译：科斯塔斯阵列因其最优的非周期自相关特性，数十年来一直是备受关注的组合对象。同时，寻找具有较小最大互相关值的科斯塔斯阵列族或扩展阵列族也具有重要意义，因为在多用户系统应用中，不同信号之间的交叉干扰也应尽可能小。本文旨在研究若干大规模科斯塔斯阵列族或扩展阵列族，并在某些水平偏移$u$和垂直偏移$v$的情况下，对其最大互相关值进行了部分界定。给定素数$p \geq 5$，本文研究了一个定义在$\{1, \ldots, p-1\}$上的大规模科斯塔斯阵列族，该族同时包含指数型和对数型韦尔奇科斯塔斯阵列。我们给出了该阵列族在任意$u$和$v$下最大互相关的一个上界。我们还证明了当$u=0$且$v \neq 0$时，定义在$\{1, \ldots, p-1\}$上的幂置换族的最大互相关以$\frac{1}{2}+\sqrt{p-1}$为界。此外，对于包含指数型韦尔奇科斯塔斯阵列与定义在$\{1, \ldots, p-1\}$上的幂置换的更大规模阵列族，在任意$u$且$v=0$的情况下，我们首次给出了其最大互相关的非平凡上界：若$p$不是安全素数，则该上界等于$(p-1) / t$，其中$t$为$(p-1) / 2$的最小素因子；否则，该上界至多为$(p-1)^{\frac{1}{2}}+(p-1)^{\frac{1}{4}}+\frac{1}{2}$。