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pt
说明: 括发送端,信道和接收端代码),还有OFDM的说明 文件列表(点击判断是否您需要的文件,如果是垃圾请在下面评价投诉): (... (Including transmitter, channel and receiver-side code), as well as OFDM-documentation list (click on the file you need to determine whether, if the rating is spam complaints below): (...)
- 2009-09-11 16:49:36下载
- 积分:1
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dzcl
海量工业数据处理使用,包括了稳态测试,剔除坏值,计算标准差和平均值等(Massive use of industrial data processing, including the steady-state test, remove bad values, calculate the standard deviation and mean, etc.)
- 2013-08-23 19:15:20下载
- 积分:1
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PID_MATLAB_simulation
PID的matlab仿真,简单实用,强烈建议下载(simulation of PID algrithm)
- 2012-10-22 22:30:52下载
- 积分:1
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BoxDimension_TS
求混沌时间序列的盒子维(box dimension)覆盖法(Seeking chaotic time series box dimensions (box dimension) covering method)
- 2015-03-18 16:33:47下载
- 积分:1
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5
一道程序编译顺序的考题,涉及到函数调用的先后顺序及运算符号的优先级等问题。下面我展开给你讲。
C的程序编译总是从main函数开始的,这道题的重点在“fun((int)fun(a+c,b),a-c)) ”语句。
系统首先要确定最外层 fun()函数的实参,第一个参数的确定需要递归调用fun()函数(不妨称其为内层函数)。内层函数的两个参数分别为x=a+b=2+8=10、y=b=5,执行函数体x+y=10+5=15,于是得外层函数的参数x=15。其另一个参数y=a-c=2-b=-6,再次执行函数体,得最终返回值x+y=15+(-6)=9。 (Compiling together the sequence of test procedures, involving the sequence of function calls and operator symbols, such as the priority problem. Now I give you to start speaking. Procedures for C compiler always start from the main function and at这道题the focus of " fun ((int) fun (a+ c, b), ac)) " statement. System must first determine the most outer layer of fun () function of real parameters, the first parameters of recursive calls required fun () function (may be called the inner function). Inner function separately for the two parameters x = a+ b = 2+8 = 10, y = b = 5, to execute the function body x+ y = 10+5 = 15, then the outer function parameters were x = 15 . Its another parameter y = ac = 2-b =- 6, once again to execute the function body may eventually return the value of x+ y = 15+ (-6) = 9.)
- 2009-03-15 15:36:23下载
- 积分:1
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resistor
可以計算resistor 電阻值的一個小program(should calculate the resistance of an electric resistor~)
- 2010-12-05 12:16:53下载
- 积分:1
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Codeforces
Solution for some problems in Codeforces
- 2014-01-09 23:02:19下载
- 积分:1
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source
This derivation of the normalised least mean square algorithm is based on Farhang-
Boroujeny 1999, pp.172-175, and Diniz 1997, pp 150-3. To derive the NLMS algorithm
we consider the standard LMS recursion, for which we select a variable step size
parameter, μ(n). This parameter is selected so that the error value , e+(n), will be
- 2011-01-12 17:04:12下载
- 积分:1
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pcm
此程序为将a律13折线的PCM信号,进行采样量化输出,和其中的量化精度可达0.01(This procedure will be a law broken line 13 of the PCM signal sample quantization output, and the quantitative accuracy of 0.01)
- 2007-08-20 09:32:29下载
- 积分:1
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dct88
基于dct的数字水印算法(A watermark sourcecode based on DCT.)
- 2005-01-21 16:59:36下载
- 积分:1