subscript indices must either be real positive integers or logicals( X_re(k_r)=XD(1,kd); can anyone help me why i am getting this error
1 次查看(过去 30 天)
显示 更早的评论
clc clear all N=1024; tic; k=50; d=28; aa=6; b=10; a=3; X=zeros(1,N); % xp_unsolved2=[]; % X_p=randperm(N,k); % X_v1=rand(1,k); % X_v=aa*X_v1+b; % X(X_p)=X_v; sss=[0,1,2,3,4,5,6]; ss=[0,1,2,3];
for snr_num=1:9; snr=snr_num*5+0; for exp_num=1:100; % X=zeros(1,N); X_p=randperm(N,k); X_v1=rand(1,k); X_v=aa*X_v1+b; X(X_p)=X_v; % X_v=[aa+b]; X_re=zeros(1,N); x0=ifft((X));
x=awgn(x0,snr,'measured') ;
enerx=sum(x.^2);
energy=abs(sum(x0.^2));
noise=sum((x-x0).^2);
db=10*log10(abs(enerx/noise));
Len=length(x); x1=0.3*(b/d)^2; x2=0.151; x3=4; X4=8; ssss=6; ss=[0,1,2,3]; for l=0:2*a-1; xd(l,:)=downsample(x(ss(l)+1:Len),d); XD(l,:)=fft(xd(l,:)); end % for l=0:2*a-1 % xd(l+1,:)=downsample(x(1+l:Len),d); % % end % D_len=size(xd,2); % d1=2*d; % XD=fft(xd,D_len,2)*d;
xp_unsolved2=[];
k_zeros=[]; k_two=[]; k_1=[]; kk=0; kk1=0; num_1=0; num_2=0; for kd=1:N/d;
if XD(1,kd).^2 < x1;
k_zeros=[k_zeros,kd+N/d];
else
if abs(abs(XD(1,kd).^2)-abs(XD(2,kd).^2)) <x2;
% if abs(XD(1,kd).^2)-(XD(2,kd).^2) <x2;
k_1 =[];
num_1=num_1+1;
for hhh=1:d
k_1=[k_1,kd+(hhh-1)*N/d];
end
kkk_1(num_1,:)=k_1;
phase= mod(angle(XD(2,kd)/XD(1,kd)),2*pi);
k_estim=phase*N/2/pi+1;
k_com=k_1-k_estim;
[value,posn]=min(abs(k_com));
k_r=k_1(posn);
X_re(k_r)=XD(1,kd);
1 个评论
Adam
2016-7-20
Simplest way to solve this is just to use the debugger when it should become obvious. Without formatting your code correctly it is very difficult to see easily where your indices are coming from for your arrays.
k_r appears to be the result of a chain of operations which is hard to verify if it is a positive integer or logical or not.
回答(1 个)
Adam
2016-7-20
At a glance N/d is 1024 / 28 which is not an integer so I imagine
k_1=[k_1,kd+(hhh-1)*N/d];
produces non-integer values and you then use results from this array to index into another array.
0 个评论
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Multirate Signal Processing 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!