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本文提出在数字移动通信中子带滤波器组处理可以提高不同阵元信号的相关性，从而能改善自适应阵列抑制码间干扰(ISI)和共信道干扰(CCI)的能力．在文[6]的基础上，本文研究了子带滤波器组在宽带自适应天线旁 瓣相消中的应用，对其原理进行了理论分析，提出了有效的子带处理方法．经研究表明，子带滤波器组处理能有效增加主、辅 ...

Memory Management—Algorithms and implementation in C/C++ Introduction Chapter 1 - Memory Management Mechanisms Chapter 2 - Memory Management Policies Chapter 3 - High-Level Services Chapter 4 - Manual Memory Management Chapter 5 - Automatic Memory Management Chapter 6 - Miscellaneous Topics Index ...

用pascal编一个年历具体要求是：设计电子月历的程序 一、任务内容（task） 1．设计的程序应具有以下功能： （1）任意输入某年的某一月份，屏幕应能显示该月的月历； （2）应允许多次从键盘输入某月份,并自行设置结束标志。 （3）如在屏幕上输入2007年12月份的信息,屏幕的显示形式如下： month Sun. Mon. Tue. Wed. Th ...

isual Chat 1.91 Developer Edition - Customize the Visual Chat code regarding your own requirements - Use customchatdev.html for developing / testing - Create .jar and .cab-files containing client-specific .class-files and the images-folder (use zip and cabarc compressing tools) - Adapt the customch ...

用java语言实现的四则运算，采用的是case语句，选择运算符，作相应的运算

this procedure is simulated document management system disk space management, Disk Management, directory management (single-level directory) of a simulation program, as well as university computer courses on operating system design a case, we only reference

本程序仿真一个从卫星到区域数据中心的下行链路（由一组6个载波构成），评估热噪声，交调（IM）失真、由滤波器造成的码间干扰（ISI）以及邻频干扰对下行链路性能的影响。

Free open-source disk encryption software for Windows Vista/XP, Mac OS X, and Linux Main Features: * Creates a virtual encrypted disk within a file and mounts it as a real disk. * Encrypts an entire partition or storage device such as USB flash drive or hard drive. * Encrypts a partition or dr ...

This book uses the Python language to teach pro - gramming concepts and problem -solving skills, without assuming any previous program- ming experience. With easy-to-understand examples, pseudocode, flowcharts, and other tools, the student learns how to design the logic of programs and then implem ...

Instead of finding the longest common subsequence, let us try to determine the length of the LCS. &#1048708 Then tracking back to find the LCS. &#1048708 Consider a1a2…am and b1b2…bn. &#1048708 Case 1: am=bn. The LCS must contain am, we have to find the LCS of a1a2…am-1 and b1b2…bn-1. &#1048708 ...