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r2fft
用一个N点复序列的FFT同时计算两个N点实序列离散傅里叶变换(With an N-point complex sequence of N-point FFT calculate two real sequences discrete Fourier transform)
- 2013-05-23 20:11:50下载
- 积分:1
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STT753-2
塔式起重机建模文档,非常棒的商业流程和代码,可以用于ANSYS结构建模学习和工程应用实践练习,其中参数可以用户自己参考修改编写!
此.txt文档为ANSYS的命令流(类似于ForTran)(Tower crane model documentation, great business processes and code that can be used ANSYS structural modeling study and practice of engineering practice, in which parameters can be written to modify the users own reference!)
- 2011-11-26 19:41:58下载
- 积分:1
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g_fitting
使用正交多项式完成数据拟合。程序对读入的gps采样点完成曲线拟合。()
- 2007-08-01 18:25:08下载
- 积分:1
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WilsonMethod
进行结构响应分析的Wilson法,是直接积分的基本算法之一,相信进行有限元动力分析编程的诸位都能用上(structural response analysis of Wilson, which is direct integration one of the basic algorithm, believe that the dynamic finite element analysis program you can use)
- 2006-07-09 11:15:54下载
- 积分:1
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生活垃圾转运站选址lingo 求解编程
二级转运站选址数学规划模型求解的lingo 程序(Lingo Program for Solving Mathematical Programming Model of Secondary Transfer Station Location)
- 2020-06-16 14:20:01下载
- 积分:1
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dit_fft
本程序包括:DFT(离散傅里叶变换),FFT(快速傅里叶变换),IFFT(快速傅里叶逆变换),conv(卷积:圆周卷积、线性卷积),主要应用于信号处理,既可以对实序列做上述操作,也可以对复序列。(This process includes: DFT (Discrete Fourier Transform), FFT (fast Fourier transform), IFFT (inverse fast Fourier transform), conv (convolution: circular convolution, linear convolution), mainly used signal processing, either on the actual sequence to do the operation, can also be complex sequence.)
- 2010-09-22 21:30:23下载
- 积分:1
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Explain-of-SAP5
解释有限元程序SAP5的原理,结构,算例的一本书籍(The finite element program SAP5 explain the principles, structure, examples)
- 2017-03-30 19:48:51下载
- 积分:1
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statecom
说明: STATCOM STATIC COMPENSATOR
- 2019-11-20 18:13:47下载
- 积分:1
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jvzhenlianxi
实现构建矩阵,矩阵的相加,相减,清零,以及转置,还有相乘的功能。(To build matrix, the matrix addition, subtraction, reset, and transposed, and the function of the multiplication.
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- 2012-08-22 09:29:57下载
- 积分:1
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rates
基于CPLEX半连续变量求解发电机组单时段经济调度问题,给定负荷需求,求每台发电机输出P=0或Pmin<=P<=Pmax,使发电成本最小。(Semi-continuous variable based on CPLEX to solve a single period of economic generator scheduling problem, given the load demand, seeking each generator output P = 0 or Pmin <= P <= Pmax, the minimum cost of power generation.)
- 2011-07-04 22:16:33下载
- 积分:1