-
BER_OFDM_BPSK
BER for OFDM using BPSK with 52 subcarriers using 64 IFFT.
- 2010-08-18 22:17:33下载
- 积分:1
-
system_identification_qpsk
随机产生一QPSK调制信号,通过一随机产生的低通滤波器,然后使用LMS算法实现了系统哦你的辨识。(Randomly generate a QPSK modulation signal generated by a random low-pass filter, and then use the LMS algorithm of the system Oh, your identification.)
- 2010-12-13 23:32:53下载
- 积分:1
-
OFDMqpsk
简单的OFDM仿真,采用QPSK调制,有串并转换,保护间隔,fft/ifft等(Simple OFDM simulation using QPSK modulation, there are serial to parallel conversion, guard interval, fft/ifft etc.)
- 2020-07-03 00:00:01下载
- 积分:1
-
texture1
本人编写的用直方图提取纹理特征的源码,希望对大家有用(I am prepared to use the histogram of the source texture feature extraction, in the hope that useful)
- 2007-11-28 10:57:53下载
- 积分:1
-
LMS
很好的介绍了自适应滤波器的一些相关算法,lms算法,rls算法等等(Very good introduction to some of the adaptive filter correlation algorithm, lms algorithm, rls algorithm and so on)
- 2009-03-17 11:03:05下载
- 积分:1
-
mseq-modulation
基于m序列的DS-CDMA通信系统仿真的调制解调模型(M-sequence-based DS-CDMA communication system simulation model of modulation and demodulation)
- 2013-08-25 20:45:26下载
- 积分:1
-
c3
说明: modelling of two cstrs in series. chemical
- 2015-01-25 13:50:23下载
- 积分:1
-
动力失稳
说明: matlab转子系统仿真程序,可用于计算动力失稳,判断系统稳定性(Matlab rotor system simulation program)
- 2020-10-19 10:38:59下载
- 积分:1
-
FPGA-Prototyping-By-Verilog-Examples.pdf
Book that explains the proper use of verily code and the basics of writing. Examples are also provided
- 2014-10-26 04:37:14下载
- 积分:1
-
fractal-use
分形的练习一
①Koch曲线
用复数的方法来迭代Koch曲线
clear i 防止i被重新赋值
A=[0 1] 初始A是连接(0,0)与(1,0)的线段
t=exp(i*pi/3)
n=2 n是迭代次数
for j=0:n
A=A/3 a=ones(1,2*4^j)
A=[A (t*A+a/3) (A/t+(1/2+sqrt(3)/6*i)*a) A+2/3*a]
end
plot(real(A),imag(A))
axis([0 1 -0.1 0.8])
②Sierpinski三角形
A=[0 1 0.5 0 0 1] 初始化A
n=3 迭代次数
for i=1:n
A=A/2 b=zeros(1,3^i) c=ones(1,3^i)/2
A=[A A+[c b] A+[c/2 c]]
end
for i=1:3^n
patch(A(1,3*i-2:3*i),A(2,3*i-2:3*i), b ) patch填充函数
end
(Fractal
Exercise One
The ① Koch curve
Plural iteration Koch curve
clear i to prevent i is reassigned
A = [0 1] initial A is a connection (0,0) and (1,0) of the segments
t = exp (i* pi/3)
n = 2 n is the number of iterations
for j = 0: n
A = A/3 a = ones (1,2* 4 ^ j)
A = [A (t* A+ a/3) (A/t+ (1/2+ sqrt (3)/6* i)* a) A+2/3* a]
end
plot (real (A), imag (A))
axis ([0 1-0.1 0.8])
② Sierpinski triangle
A = [0 1 0.5 0 0 1] initialized A
n = 3 the number of iterations.
for i = 1: n
A = A/2 b = zeros (1,3 ^ i) c = ones (1,3 ^ i)/2
A = [A A+ [c b] A+ [c/2 c]]
end
for i = 1:3 ^ n
patch (A (1,3* i-2: 3* i), A (2,3* i-2: 3* i), b ) patch filled function
end)
- 2013-03-02 10:03:09下载
- 积分:1