如果多径,还要在接收端引入均衡,均衡有很兆裂多种。
可以看看simulink中自带的例子
饿了。 。 。您的H1确实是瑞利分布,但幅度通道H1啊,我们说“瑞利路”,不意味没御着真正的通道,而是一个复杂的通道,通道幅度瑞利分布,所以你枯物岩H1 = ABS( H1)“这句话应该被删除。
其次,就是我们常说的信道响应,不包含??高斯白噪声,也就蚂铅是说,只H此人。
直接IFFT(H1 )是信道响应。************************************
建模许多通信模块失败的反映,说明这里也不是很清楚。建议向老师
% written by Amir Sarrafzadeh (14Jan2008)% this function generates normalized rayleigh samples based on Inverse DFT
% method as was proposed by David J. Young, and Norman C. Beaulieu
% "The Generation of Correlated Rayleigh Random Variates by Inverse
% Discrete Fourier Transform, "
% Sample Use:
% chan=genRayleighFading(512,ceil(10000/512),1e4,100)
% chan=chan(1:10000)
% where 10000=number of needed samples
% parameters:
% fftsize: size of fft which used
% numBlocks: number of samples/fftsize
% fs: sampling frequency(Hz)
% fd: doppler shift(Hz)
function [ outSignal ] = genRayleighFading( fftSize,numBlocks,fs,fd )
numSamples=fftSize*numBlocks%total number of samples
fM=fd/fs %normalized doppler shift
NfM=fftSize*fM
kM=floor(NfM)%maximum freq of doppler filter in FFT samples
doppFilter=[0,1./sqrt(2*sqrt(1-(((1:kM-1)./NfM).^2))),sqrt((kM/2)*((pi/2)-atan((kM-1)/sqrt(2*kM-1)))),...
zeros(1,fftSize-2*kM-1),sqrt((kM/2)*((pi/2)-atan((kM-1)/sqrt(2*kM-1)))),1./sqrt(2*sqrt(1-(((kM-1:-1:1)./NfM).^2)))].'
sigmaG=sqrt((2*2/(fftSize.^2))*sum(doppFilter.^2))
gSamplesI=randn(numSamples,2) %i.i.d gaussian input samples (in phase)
gSamplesQ=randn(numSamples,2) %i.i.d gaussian input samples (quadrature phase)
gSamplesI=(1/sigmaG)*(gSamplesI(:,1)+1j*gSamplesI(:,2))
gSamplesQ=(1/sigmaG)*(gSamplesQ(:,1)+1j*gSamplesQ(:,2))
%filtering
filterSamples=kron(ones(numBlocks,1),doppFilter)
gSamplesI=gSamplesI.*filterSamples
gSamplesQ=gSamplesQ.*filterSamples
freqSignal=gSamplesI-1j*gSamplesQ
freqSignal=reshape(freqSignal,fftSize,numBlocks)
outSignal=ifft(freqSignal,fftSize)
outSignal=abs(outSignal(:))%Rayleigh distributed signal
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