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OSTBC-BER-PerformanceAnalysi
正交空时分组码误码率的性能分析,可以对在无线传输信道下的源信号进行编码,使其更加适合信道传输(Orthogonal STBC BER Performance Analysis)
- 2011-05-25 12:57:11下载
- 积分:1
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2D_teaching_material
这是2010合工大Robocup机器人足球仿真2D技术培训内部培训资料,我找的好长时间才找到,希望对大家有用!(This is the 2010 co-workers big Robocup robot soccer simulation 2D technical training within the training materials, I am looking for a good long time to find the hope for all of us!)
- 2011-10-06 08:16:58下载
- 积分:1
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FingerPrintProject
Finger Print Code By Matlab
- 2009-10-08 02:21:16下载
- 积分:1
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DTLBQ
说明: 数字低通滤波器设计,MATLAB程序,课程学习报告使用,算法模拟学习很好例子(digital signal)
- 2010-03-28 12:00:19下载
- 积分:1
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hamming_LPF
程序的功能是用海明窗设计低通滤波器,性能指标为wp=0.2pi,ws=0.3pi(The program s function is Hamming window low-pass filter, performance indicators, wp = 0.2pi ws = 0.3pi)
- 2012-05-12 09:48:59下载
- 积分:1
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Array-Pattern-Based-on-FFT
了等距线阵方向图与傅立叶变换的一致性,用F丌算法进行阵列方向图快速计算。通过严格的数学推导,揭示变换序列与可见空问的关系,讨论了不同单元问距情况下的处理方法(In this paper,the consistence of equispaced linear array pattem and foureir transfom is analyzed,so the
linear arrayy pattem can be computed rapidly with FFT a1goritllm.By the use of reUgious mathmatic deducing process,relation between transfotmation sequences and visible space are given,and different handling procedures are discussed in the case of different element spacing. )
- 2011-10-27 22:19:56下载
- 积分:1
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system
自动控制原理中应用仿真软件对系统的频率响应和稳定性的研究(The application of the principle of automatic control system simulation software, the frequency response and stability study)
- 2009-04-28 19:40:00下载
- 积分:1
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fractal
该程序可以实现信号的分形计算,从而求解出所求信号的分形函数,在信号非线性研究中有很重要的作用。(The program can calculate the fractal signal, thus solving the request signals a fractal function, the study of nonlinear signal has a very important role.)
- 2011-09-07 17:11:47下载
- 积分:1
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_FDTD_2D_PML
基于MATLAB的原理应用,此程序可实现2d的吸收边界算法,比较简单(This MATLAB M-file implements the finite-difference time-domain
solution of Maxwell s curl equations over a two-dimensional
Cartesian space lattice comprised of uniform square grid cells.)
- 2011-12-26 19:53:20下载
- 积分:1
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0.618-gold
0.618法求y=sin(x)在[0,2]极大值点(0.618 method y = sin (x) in [0, 2] the maximum points)
- 2015-04-16 16:29:15下载
- 积分:1