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propagationmodel
移动无线系统传播模型概论,里面介绍的模型很全希望对大家有帮助.(Introduction to mobile radio systems propagation model, which describes the model is all we want to help.)
- 2010-10-21 09:02:35下载
- 积分:1
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shixufenxi
主要是用R语言进行时序分析,通过一个带有季节效应的例子给出分析过程(R language is mainly used for timing analysis by example is given with a seasonal effect analysis)
- 2015-04-17 21:54:53下载
- 积分:1
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frft
分数阶傅里叶变换(FrFT)的程序,采用Ozaktas的第二种离散化方法。(Fractional Fourier transform (FrFT) program, using the second method Ozaktas of discrete.)
- 2021-03-16 18:29:21下载
- 积分:1
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simlation-of-mimo
mimo信道、滤波器等实用的仿真还附有对应的说明文件。(simulation of mimo channel filters and so on with reference)
- 2013-12-22 22:25:33下载
- 积分:1
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PCA-K_means
基于PCA降维的k-means算法,附带有检测正确率的源码以及相应的检测数据(PCA dimensionality reduction based on k-means algorithm, source code comes with the correct detection rate and the corresponding test data)
- 2021-01-02 16:18:57下载
- 积分:1
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BFGS
BFGS optimization algorithm
- 2021-02-07 10:49:55下载
- 积分:1
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queure
针对M/PH/1(k)排队系统推导出该排队系统在任意时刻、到达时刻、退去时刻的队列长度状态概率分布、平均队列长度、以及平均等待时间,并编写程序进行数值计算,并对M/E2/1(k), M/M/1(k), M/H2/1(k)排队系统的性能进行数值比较。(against M/PH/1 (k) queuing system derived the queuing system at any time, the arrival time, faded moment of the queue length probability distribution, the average queue length, and the average waiting time, and the preparation procedures for numerical computation, and M/E2/1 (k), M/M/1 (k), M/H2/1 (k) queuing system performance numerical comparison.)
- 2006-08-29 08:23:39下载
- 积分:1
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work
高斯核函数,矩阵自身的高斯核函数计算,需要指定参数(Gaussian kernel function, matrix itself Gaussian kernel function, the need to specify the parameters)
- 2021-04-26 12:08:45下载
- 积分:1
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s_fun_CVT
主要是利用恒压方法实现光伏阵列的最大功率点跟踪(Method is to use constant pressure to achieve the main solar array maximum power point tracking)
- 2011-05-24 15:49:57下载
- 积分:1
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LTE_System_Level_1.4_r570
To solve this instance of the decision problem we must determine whether there is a truth value (TRUE or FALSE) we can assign to each of the variables (x1 through x4) such that the entire expression is TRUE. In this instance, there is such an assignment (x1 = TRUE,x2 = TRUE, x3=TRUE, x4=TRUE), so the answer to this instance is YES. This is one of many possible assignments, with for instance, any set of assignments including x1 = TRUE being sufficient. If there were no such assignment(s), the answer would be NO.
- 2012-04-23 22:00:12下载
- 积分:1