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hfm
随机生成十个数,和为1。使用c++语言,在vs2008上运行(Ten randomly generated number, and to 1. Using c++ language, running on vs2008)
- 2010-12-07 22:24:42下载
- 积分:1
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histogram
this program calculates the histogram of an inage
- 2010-05-15 17:49:07下载
- 积分:1
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eigenvalue-problem-of-PDE
以上代码是求解椭圆微分算子的特征值问题的Matlab源代码,采用有限元法的思想来编写的,程序计算结果和理论是相符的。(The above code is to solve the elliptic differential operator eigenvalue problem of the Matlab source code, using the finite element method to write the ideas, procedures, results and theoretical calculation is consistent.)
- 2011-09-06 17:02:49下载
- 积分:1
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signatureNormalization
Signature normalization for signature recognition
- 2015-04-05 16:06:55下载
- 积分:1
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CepSpecPackage
无参数功率谱估计,基于matlab编程语言(Non-parametric power spectrum estimation, based on the matlab programming language)
- 2007-11-30 13:07:47下载
- 积分:1
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bianjiegou
是关于变结构控制的一个简单实例,其中包括控制函数的选取(on variable structure control a simple examples, including the selection control function)
- 2007-04-22 16:32:38下载
- 积分:1
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coilc
Calculate the sensible and latend load of the cooling coil in a VAV Reheat System
- 2011-11-14 05:42:57下载
- 积分:1
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LMS
LMS算法实现自适应滤波
clear close all clc
N=10000 设置仿真长度
信号产生参数设定
a1=-0.195
a1=-1.5955
a2=0.95
R0=[1,a1,a2 a1,1+a2,0 a2,a1,1]
p=[1,0,0]
r=inv(R0)*p 计算理论自相关函数
R=[r(1),r(2) r(2),r(1)] 生成理论自相关矩阵
p1=[r(2),r(3)] 生成互相关
h=inv(R)*p1 计算维纳解
Jmin=r(1)-h *p1 计算维纳解时最小均方误差
u=1/sum(eigs(R)) ( LMS算法实现自适应滤波
clear close all clc
N=10000 设置仿真长度
信号产生参数设定
a1=-0.195
a1=-1.5955
a2=0.95
R0=[1,a1,a2 a1,1+a2,0 a2,a1,1]
p=[1,0,0]
r=inv(R0)*p 计算理论自相关函数
R=[r(1),r(2) r(2),r(1)] 生成理论自相关矩阵
p1=[r(2),r(3)] 生成互相关
h=inv(R)*p1 计算维纳解
Jmin=r(1)-h*p1 计算维纳解时最小均方误差
u=1/sum(eigs(R)) )
- 2021-03-01 22:29:34下载
- 积分:1
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Compress
按照用户输入的压缩率,完成对灰度图的压缩(Compression ratio in accordance with user input to complete the compression of grayscale)
- 2010-10-24 23:11:13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