%距離和角度二維估計(jì)by?MUSIC;
clear?all
close?all
rad?=?pi/180;???????????%??deg?->?rad
c=300;f=100;????%??光速;頻率600MHz
lamba=c/f;
sensor?=?16;?????????????%??陣列數(shù)量
k=1:sensor;
dd?=lamba/2;???????????????%?間隔?
d=0:dd:（sensor-1）*dd;???%?
source?=2;??????????????%?number?of?DOA
theta=[34?36];
range=[14?16];
%?theta1?=25;theta2?=65;??????????????%??角度
%?range1?=15;range2?=15;????????????????%??距離
snr?=?20;???????????????%??input?SNR?（dB）
n?=?2;????????????????%??snapshots
%??????????ai=-2*pi*d.*sin（theta*derad）;%%%第i個(gè)信源的DOA和距離的非線性函數(shù)
%??????????bi=pi*d.^2*[cos（theta*derad）].^2/range;%%%第i個(gè)信源的DOA和距離的非線性函數(shù)?
for?i=1:source
????theta1=theta（:i）;
????range1=range（:i）;?
A（i:）=exp（j*（k.*（-2*pi*d.*sin（theta1*rad）./lamba）+k.^2.*（pi*d.^2*[cos（theta1*rad）].^2/（range1./lamba））））;%%%%?direction?matrix
%A（2:）=exp（j*（k.*（-2*pi*d.*sin（theta2*rad）./lamba）+k.^2.*（pi*d.^2*[cos（theta2*rad）].^2/（range2./lamba