clc;
clear?all;
%?---------------------------參數(shù)初始化------------------------------------
M=32;?????????????????????????%?陣元數(shù)目
L=1024;??????????????????????%?信號(hào)長(zhǎng)度/快拍數(shù)?
lam=0.15;????????????????????%?信號(hào)波長(zhǎng)
d=1/2*lam;???????????????????%?陣元間距
angle=[-pi/3?pi/6?35*pi/180];??????%?信號(hào)入射角度
snr=-10;?????????????????????????????%信噪比
N=3;?????????????????????????%信源數(shù)
%?---------------------------信號(hào)模型--------------------------------------
A=zeros（NM）;
for?k=1:N
A（k:）=exp（-j*2*pi/lam*（[0:M-1]*d）*sin（angle（k）））;???
end
A=A‘;?????????????????????????????????????????????%?導(dǎo)向向量
SS=2*（randn（1L）+j*randn（1L））;
Wn=[0.36?0.38];
[ba]?=?fir1（512Wn‘bandpass‘）;??????%?設(shè)計(jì)帶通濾波器
SS1=filter（baSS）;?
Wn=[0.38?0.4];
[ba]?=?fir1（512Wn‘bandpass‘）;??????%?設(shè)計(jì)帶通濾波器
SS2=filter（baSS）;?
Wn=[0.4?0.42];
[ba]?=?fir1（512Wn‘bandpass‘）;??????%?設(shè)計(jì)帶通濾波器
SS3=filter（baSS）;?
SS=zeros（NL）;???SS=[SS1;SS2;SS3];
RS=SS*SS‘/L;?
gls=trace（RS）/N;
R=A*RS*A‘+（gls*10^（-snr/10））*eye（M）;?????????????%白噪聲協(xié)方差矩陣為對(duì)角陣
%?---------------------------角度估計(jì)--------------------------------------
theta=-pi/2:pi/3600:pi/2;???????%?ULA估計(jì)角度變化的范圍和頻率選擇?
for?k2=1:length（theta）??????%?角度估計(jì)
????AA=exp（-j*2*pi/lam*（[0:M-1]*d）*si