%%%%??這個程序是正確的事實上一旦考慮自適應，就根本不用考慮波束指向了
%%%%??波束指向只針對靜態方向圖有用
clear?all;clc;
close?all;
tic
N=24;????????????%%??the?element?number
c=pi;????????????%%??c=2*pi*d/lamda?where?d=0.5*lamda
thetaj=40;??????????????????????%%%%???干擾信號方向
theta0=10;??????????????????????%%%%???波束方向
j=sqrt（-1）;
index=0;
step=0.01
for?w=-90:step:90
????index=index+1;
????aw=exp（-j*（0:N-1）*c*（sin（w*pi/180）-sin（theta0*pi/180）））;
????PSQ（index）=20*log10（abs（ones（1N）*aw‘））;????????????%%?PSQ靜態和方向圖??P:pattern;?S:sum;????????Q:queisent;
end
w=-90:step:90;
figure;plot（wPSQ-max（PSQ））;xlabel（‘Angle?（?\circ）‘）;?ylabel（‘Power?（dB）‘）;?title（‘Queisent?sum?pattern‘）
temp1=sort（PSQ（find（diff（sign（diff（PSQ）））==-2）+1））;?????%%%??靜態和方向圖所有極大值由小到大排列
mainlobeQS=temp1（length（temp1））;????????????????????????%%%??主瓣峰值
sidelobeQS=temp1（length（temp1）-1）;??????????????????????%%%??旁瓣峰值
if?mainlobeQS==sidelobeQS
????sidelobeQS=t