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SAR面目標的回波仿真 利用matlab有效仿真出平頂樓 三角錐等立體目標的回波仿真

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代碼片段和文件信息

tic
clear?all;
c=3e8;????????????????????????????????
fc=75e9;???
lamda=c/fc;?

%?X0=200;????????????????????????????????
Rc=865000;
theta=23/180*pi;
Rr=Rc*sin(theta);
Z=Rc*cos(theta);
?

%?Lsar=lamda*Rc/6;
beta=66*pi/180;
%?vr=0;
%?vr=0;
%?va=0;
%?vh=0;
%?R0=150;???????????????????????????????
?

B=500;
alpha=-6*pi/180;
Z1=Z;
Y1=0;
Z2=Z+B*sin(alpha);
Y2=B*cos(alpha);

%?R0=Rc*cos(theta);
%?B1=B*cos(theta-alpha);
%?d=B1/R0/sin(theta);
%?
%?fen=lamda/2/d;
?

Tr=4e-6;??????????????????????????
Br=50e6;????????????????????????????
Kr=Br/Tr;????????????????????????????
Nr=512;
fs=100e6;
dt=1/fs;????????????????????????????
?

v=7098.2;???????????????????????????????????
Lsar=lamda*Rc/6;????????????????????????????
Na=512;
fp=2*v/6*2;????????????????????????????????????????????%4.5e3;
?
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%?????載入地形數據???????%%%%%%%%%%%%%%%%%%%%%%%%
load?zhui.mat;
%?load?duo.mat;
%?load?ping.mat;

%?mesh(plane);
%?figure
%?mesh(plane);

tmp=size(find(plane));
Axis=zeros(3tmp(1));?
flagrow=1;
for?flagp=1:size(plane(:1))???????????????????????????????????????????????????%%%?flagp?是plane的行數
????tmp=max(abs(plane(flagp:)));??????????????????????????????????????????????%%%?tmp是取plane中每一行的最大值?為了尋找不等于0的
????if?tmp
????????tmpp=find(plane(flagp:));?????????????????????????????????????????????%%%?tmpp是記錄?plane中?在flagp行的不等于0的數的位置
????????tmpa=size(tmpp);???????????????????????????????????????????????????????%%%?tmpa(2)是記錄?plane中?在flagp行的不等于0的數的個數
????????tmpa=tmpa(2);
????????flagrow=flagrow+tmpa;
????????Axis(1flagrow-tmpa:flagrow-1)=(flagp-30)*3;
????????Axis(2flagrow-tmpa:flagrow-1)=(tmpp-20)*3;
????????Axis(3flagrow-tmpa:flagrow-1)=plane(flagptmpp)*1.015;%*1.00487443452357*4/5;
????end
end
?
y=Axis(2:)*cos(beta);
z=-Axis(2:)*sin(beta);
y=y+Axis(3:)*sin(beta);
z=z+Axis(3:)*cos(beta);
Axis(2:)=y;
Axis(3:)=z;???????????????????????
?

echo=zeros(512512);
ttmp=(-256:255)*dt;
tr=ttmp+2*Rc/c;??????????????
%?p=2*Rc/c;
x=(-256:255)/fp*v;
for?flaga=1:Na
????for?flagdot=1:flagrow-1
????????rd=sqrt((Rr-Y1+Axis(2flagdot))^2+(x(flaga)-Axis(1flagdot))^2+(Z1-Axis(3flagdot))^2);
????????td=2*rd/c;
????????wr=rectpuls(tr-tdTr);????????
????????wa=rectpuls(x(flaga)-Axis(1flagdot)Lsar);
????????echo(flaga:)=echo(flaga:)+exp(i*2*pi*(-fc*td+1/2*Kr*(tr-td).^2)).*wr*wa;
????end
end
figure?
imagesc(abs(echo));
save?planechoh1.mat?echo;
disp?1
%%%%%%%%%%%%%%%%?????第二組回波???%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
echo=zeros(512512);
ttmp=(-256:255)*dt;
tr=ttmp+(sqrt((Rr-Y2)^2+Z2^2)+Rc)/c;?????????????????
p=(sqrt((Rr-Y2)^2+Z2^2)+Rc)/c;
x=(-256:255)/fp*v;
for?flaga=1:Na
????for?flagdot=1:flagrow-1
????????rd2=sqrt((Rr-Y2+Axis(2flagdot))^2+(x(flaga)-Axis(1flagdot))^2+(Z2-Axis(3flagdot))^2);
%?????????rd1=sqrt((Rr-Y1+Axis(2flagdot))^2+(x(flaga)-Axis(1flagdot))^2+(Z1-Axis(3flagdot))^2);
????????td=2*

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----

?????文件????????334??2012-06-19?01:23??huibo\duo.mat

?????文件???????3333??2012-06-21?03:00??huibo\huibo.m

?????文件????????207??2012-06-19?01:24??huibo\ping.mat

?????文件????2313592??2012-06-19?01:22??huibo\zhui.mat

?????目錄??????????0??2013-03-11?10:58??huibo

-----------?---------??----------?-----??----

??????????????2317466????????????????????5


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