频率选择性无线信道MATLAB仿真

[y592406439]

请问大佬们 怎样根据下面的代码得到最后面的仿真图
v=300;
fc=9e+8;
NoneZeroPaths=[1,4,9,16,20];
PDF=[0,-3,-5,-9,-11];
%TimeSignal未知
%fs未知
%t0未知
time_length=1000;
fft_length=100;
jingshu=5;
%Initialization:
%dopplerfrequency
wm=(2*pi*v*fc/3.e+8)/3.6;%v km/h
%Someparameters.
M=8;
N=4*M+2;
%Resettingofthein-phaseandQ-phasechannelresponse.
pNum=length(NoneZeroPaths);
Len=length(TimeSignal);
Xi=zeros(pNum,Len);
Xq=zeros(pNum,Len);
for ii=1:pNum
%Setbetaandsita.
beta=zeros(1,M);
sita=zeros(1,M);
for i1=1:M
beta(1,i1)=pi*i1/(M+1);
sita(1,i1)=beta(1,i1)+2*pi*(ii-1)/(M+1);
end
for i1=0:Len-1
%Calculatecosinefordifferenti2.
csin=zeros(1,M);
for i2=1:M
csin(1,i2)=cos(wm*cos(2*pi*i2/N)*(i1/fs+t0)+sita(1,i2));
end
%Channelresponse.
Xi(ii,i1+1)=2*cos(beta)*csin'+1.414*cos(wm*(i1/fs+t0));
Xq(ii,i1+1)=2*sin(beta)*csin';
end
end
CH=1.414/N^0.5*(Xi+j*Xq);
for ii=1:pNum
CH(ii,:) = sqrt(size(CH,2))*CH(ii,:)/norm(CH(ii,:));
end
MeanAmp=10.^(PDF/10);
MeanAmp=sqrt(MeanAmp/sum(MeanAmp));
for ii=1:pNum
CHpdf(ii,:)=MeanAmp(ii).*CH(ii,:);
end
TempSig=zeros(pNum,size(CH,2));
for i=1:pNum
if NoneZeroPaths(1,i)>1
TempSig(i,:)=[zeros(1,NoneZeroPaths(1,i)-1),TimeSignal(1,[1:end-NoneZeroPaths(1,i)+1])];
else
TempSig(i,:)=TimeSignal;
end
end
OutSignal=MeanAmp*(TempSig.*CH);
h=zeros(tao_max,time_length);
H1=zeros(fft_length,time_length);
for i=1:time_length
for ii=1:jingshu
h(NoneZeroPaths(ii),i)=CHpdf(ii,i);
end
H1(:,i)=fft(h(:,i),fft_length);
end
x1=1:1:fft_length;
y=1:time_length;
plot(x1,abs(H1(:,3)));%3是1~time_length取的随机值