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introc++
Intro for c++ from MIT
- 2010-11-23 20:46:39下载
- 积分：1
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Matlab_Basic
把一些Matlab基础教程进行打包，是PDF格式或PPT格式的，希望对初学者有用！(Some basic tutorials for Matlab packaged are the PDF format or PPT format, and I hope it is useful for beginners!)
- 2009-07-10 09:31:27下载
- 积分：1
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VP
说明： 根据ａｂｉｎｉｔ输出的势文件，计算体系的平均势(According abinit potential output file, calculate the average potential of the system)
- 2011-11-01 17:48:12下载
- 积分：1
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apiXC_Hdmi
描述TV HDMI 输入源的信号侦测及Parser过程。(Description HDMI signal input source detection process.)
- 2013-12-13 00:01:03下载
- 积分：1
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bsg
bsg.c - block layer implementation of the sg v4 interface.
- 2015-03-13 10:16:57下载
- 积分：1
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115157686dtc
异步电动机直接转矩控制基本原理及三相交流调速系统仿真(Induction motor direct torque control and the basic principles of three-phase AC Variable Speed System)
- 2010-07-11 16:54:11下载
- 积分：1
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Ev_Model
Electric Vehicle Model
- 2014-12-10 10:01:30下载
- 积分：1
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DM642EDMA
This program uses the timers to trigger EDMA events. These events in turn
* trigger linked EDMA parameter tables to fill a ping pong buffer structure.
* Set a breakpoint on swiProcessFunc(int arg). Then open two memory windows.
* Use ping as the address for one memory window and pong for the other. Then
* run the application. You ll note that the program bounces between the ping
* and pong buffers filling each with a value that comes from the source. The
* source in this case is the SDRAM timer control register and it simulates
* input data(This program uses the timers to trigger EDMA events. These events in turn* trigger linked EDMA parameter tables to fill a ping pong buffer structure.* Set a breakpoint on swiProcessFunc (int arg). Then open two memory windows.* Use ping as the address for one memory window and pong for the other. Then* run the application. You' ll note that the program bounces between the ping* and pong buffers filling each with a value that comes from the source. The* source in this case is the SDRAM timer control register and it simulates* input data)
- 2010-08-07 11:34:58下载
- 积分：1
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speechrecognition
用matlab仿真0到9十个数字的语音识别(Matlab simulation with the 10 numbers 0 to 9 Speech Recognition)
- 2010-01-02 11:54:10下载
- 积分：1
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mofang
魔方还原程序,主函数为Cube,采用CFOP，还原步骤为88步左右，底面十字未优化；采用留个3*3矩阵，前面为froo（数字3），后面为baa（数字-3），左面为leff（数字-2），右面为rigg（数字2），上面为upp（数字0），下面为doo（数字1），(Cube restore process, the main function of Cube, using CFOP, the reduction step is 88 steps or so, the bottom cross unoptimized adopt leave a 3* 3 matrix, in front of froo (figure 3), followed by the baa (figure-3), left the leff (digital-2), the right of rigg (figure 2), at the upp (number 0), following the doo (figure 1),)
- 2013-11-01 14:41:59下载
- 积分：1
