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The combinatorial core of the OVSF code assignment problem that arises in UMTS is to assign some nodes of a complete binary tree of height h (the code tree) to n simultaneous connections, such that no two assigned nodes (codes) are on the same root-to-leaf path. Each connection requires a code on a ...

Use this resource to teach yourself Visual C# .NET version 2003, start developing Microsoft .NET–connected applications—one step at a time, master language fundamentals at your own pace and use the learn-by-doing exercises to dig in and code!

马达控制程序的设计（em78） PORT5输出动作的控制，而且外部接有一推动STEP MOTOR的电路。写一程序驱动STEP MOTOR。

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 ...

A new type of cloak is discussed: one that gives all cloaked objects the appearance of a flat conducting sheet. It has the advantage that none of the parameters of the cloak is singular and can in fact be made isotropic. It makes broadband cloaking in the optical frequencies one step closer.

The main MIPS processor of SMP8630 comes with a JTAG interface, allowing:  access to caches and data bus (DRAM) with a bandwidth of about 200kbit/s  examining the processor state whatever the execution mode (monice)  connecting to monice using mdi-server and using a gdb client on the processor to st ...

The following Matlab code converts a Matrix into it a diagonal and off-diagonal component and performs up to 100 iterations of the Jacobi method or until 蔚step < 1e-5

matlab程序，演示了如何画出系统时域和频域的step相应

Login Manager V3.0(LM3.0) is an authentication system which can integrate with any existing website that meets the requirements. LM3.0 provides a gatekeeper where user must be authorized before entering the membership secured areas. Features: 1. Flexibility LM3.0 allows administrator to integrate i ...

迷宫算法（maze） /* Maze * Starting point is m[0][0], need to find a path go to m[9][9]. 0 means OK, * 1 means cannot go there, boundary is 0 and 9, cannot go beyond boundary. * Each step can be made horizontally or vertically for one more grid (diagonal * jump is not allowed). * Your prog ...