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kth_order_spectrum_analysis
完整的幅相调制信号可k阶非线性功率谱分析(Complete amplitude and phase modulated signal can be k-order nonlinear power spectrum analysis)
- 2008-12-13 17:01:05下载
- 积分：1
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10
说明： 《Matlab从入门到精通》所有的源程序。(《Matlab from entry to the master》 all of the source.)
- 2009-10-09 01:40:58下载
- 积分：1
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Matlabchengxushili
这是一个关于matlab程序的实例，里面详细介绍了一些函数的使用
(This is a procedure on the matlab example, which detailed the use of some function)
- 2008-05-25 17:52:31下载
- 积分：1
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Simulink_rus
similunk book about rus for advanced simulation and modeling
- 2013-10-07 16:42:12下载
- 积分：1
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elm-kernel
The MATLAB codes ELM with kernels (for both regression and multi-class classification) work linearly similarly to ELM with random hidden nodes. For the sake of convenience, the source codes of ELM with kernels are given separately.
- 2020-12-04 15:59:23下载
- 积分：1
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fhan
说明： ADRC自抗扰算法设计中fhan（）函数(Fhan() function in ADRC algorithm design)
- 2021-03-08 16:29:28下载
- 积分：1
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adboost-demo
adboost算法的一个例子。在Kearns和Valiant在1989年大作中指出了这种算法的可行性。而后，Freund在
1990年以及他和Schapire在 1994-1996年提出了boosting整个算法思路，似乎这种算法走到
了实际应用的开端。然而直到AdaBoost被viola在其人脸识别系统中运用(2001Viola和
Jones)，这种方法才彻底开始暴火.(An example adboost algorithm. Kearns and Valiant pointed at the feasibility of this method in 1989 masterpiece. Then, Freund and Schapire he made in 1990 and in 1994-1996 the idea of boosting the entire algorithm, this algorithm seems to come to the beginning of the practical application. However, until the use of AdaBoost is viola (2001Viola and Jones) in its face recognition systems, this approach was completely start a fire storm.)
- 2013-12-28 22:44:27下载
- 积分：1
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FFTjisuan
利用时域相关定理结合FFT完成相关函数计算源程序，包括时延估计。(Theorems time domain and FFT to complete the correlation function computing source, including delay estimation.)
- 2016-11-29 11:58:41下载
- 积分：1
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base
1.分别在两个图中画出δ(n)和δ(n-2)，其中-10≤n≤10，画出U（n）其中-10≤n≤10。
2.分别在两个图中画出x1(n)=cos(2π×20×0.01×n)，0≤n≤10和x2(n)=(0.8)n，1≤n≤6。
3.由上一题，计算y(n)=x1(n)+x2(n)，并画图。(1. Draw the figure at two δ (n) and δ (n-2), where-10 ≤ n ≤ 10, draw the U (n) where-10 ≤ n ≤ 10. 2. Respectively in the two Draw the figure x1 (n) = cos (2π × 20 × 0.01 × n), 0 ≤ n ≤ 10 and x2 (n) = (0.8) n, 1 ≤ n ≤ 6. 3. from the previous question, calculate y (n) = x1 (n)+ x2 (n), and drawing.)
- 2010-11-26 16:34:06下载
- 积分：1
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hamming_weighted
说明： 均匀线性阵列 波束形成，Hamming加权/Chebychev加权(Uniform Linear Array beamforming ,Hamming/Chebychev Weighted)
- 2009-07-23 21:04:17下载
- 积分：1
