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大小: 224KB文件類型: .rar金幣: 2下載: 1 次發(fā)布日期: 2021-10-20
- 語言: Matlab
- 標(biāo)簽:
資源簡介
分析研究了用CV模型,CA模型,Singer模型對單機動目標(biāo)進行跟蹤算法的matlab仿真。比較了各模型的濾波性能,得出了一些有意義的結(jié)論。

代碼片段和文件信息
clear;
clc;
T=2;
num=50;?
N0=400/T;
N1=600/T;
%?N3=610/T;N4=660/T;N5=900/T;
x=zeros(N11);
y=zeros(N11);
vx=zeros(N11);
vy=zeros(N11);
x(1)=-10000;y(1)=2000;
vx(1)=15;vy(1)=0;
ax=0;ay=0;
var=100;
for?i=1:N1-1
????if(i>N0-1&i<=N1-1)
????????ax=-0.075;ay=0.075;
????????vx(i+1)=vx(i)+ax*T;
????????vy(i+1)=vy(i)+ax*T;
???????
????else
????????ax=0;ay=0;
????????vx(i+1)=vx(i);
????????vy(i+1)=vy(i);
????end
????x(i+1)=x(i)+vx(1)*T;
????y(i+1)=y(i)+vy(i)*T;
?end
rex(numN1)=0;
rey(numN1)=0;
%
for?m=1:num
????nx=100*randn(N11);
????ny=100*randn(N11);
????zx=x+nx;
????zy=y+ny;
????rex(m1)=-10000;
????rey(m1)=2000;
????rex(m2)=-9960;
????rey(m2)=2000;
????ki=0;
????low=1;high=0;
????u=0;ua=0;
????e=0.8;
????xks(1)=zx(1);
????yks(1)=zy(1);
????xks(2)=zx(2);
????yks(2)=zy(2);
???o=[1T000.5*T^20;0100T0;001T00.5*T^2;00010T;000010;000001];
h=[100000;001000];
g=[T^2/40;T/20;0T^2/4;0T/2;10;01];
q=[100000;010000];
%?????o=[1T00;0100;001T;0001];
%?????h=[1000;0010];
%?????g=[T^2/20;T0;0T^2/2;0T];
%?????q=[100000;010000];
%?????perr=[var^2var^2/T00;var*var/T2*var^2/(T^2)00;00var^2var^2/T;00var^2/T2*var^2/(T^2)];
perr=diag([100001000010000100001000010000]);
????vx=(zx(2)-zx(1))/2;vy=(zy(2)-zy(1))/2;
????xk=[zx(1);vx;zy(1);vy;ax;ay];
????%
????for?r=3:N1
????????????z=[zx(r);zy(r)];
????????????xk1=o*xk;
????????????perr1=o*perr*o‘;
????????????k1=perr1*h‘*inv(h*perr1*h‘+q);
????????????xk=xk1+k1*(z-h*xk1);
????????????perr=(eye(6)-k1*h)*perr1;
????????????xks(r)=xk(11);
????????????yks(r)=xk(31);
????????????vxks(r)=xk(21);
????????????vyks(r)=xk(41);
????????????xk1s(r)=xk1(11);
????????????yk1s(r)=xk1(31);
????????????vxk1s(r)=xk1(21);
????????????vyks1(r)=xk1(41);
????????????perr11(r)=perr(11);
????????????perr12(r)=perr(12);
????????????perr22(r)=perr(22);
????????????rex(mr)=xks(r);
???????????rey(mr)=yks(r);
?????end
????
end
ex=0;ey=0;
eqx=0;eqy=0;
ex1(N11)=0;ey1(N11)=0;
qx(N11)=0;qy(N11)=0;
for?i=1:N1
????for?j=1:num
????????ex=ex+x(i)-rex(ji);
????????ey=ey+y(i)-rey(ji);
?????????%eqx=eqx+(x(i)-rex(ji))^2;
????????eqx=eqx+(x(i)-rex(ji))^2;
????????eqy=eqy+(y(i)-rey(ji))^2;
????end
%?????eqx/num-(ex1(i)^2)
%sqrt((double(eqx)/num-(ex1(i)^2)));
????ex1(i)=ex/num;
????ey1(i)=ey/num;
???qx(i)=sqrt((double(eqx)/num-(ex1(i)^2)));
??qy(i)=(eqy/num-(ey1(i)^2))^0.5;
????ex=0;eqx=0;ey=0;eqy=0;
????end
figure(1);
plot(xy‘k-‘zxzy‘g:‘xksyks‘r-.‘);
legend(‘true?trace‘‘observation?samples‘‘estimated?samples‘);
figure(2);
plot(zxzy);
legend(‘observation?samples‘);
figure(3)
plot(xksyks);legend(‘estimated?trace‘);
figure(4);plot(xex1);legend(‘the?error?at?x?anix‘);
figure(5);plot(xqx);
legend(‘the?square?error?at?x?anix‘);
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件???????2898??2010-06-18?22:09??仿真大作業(yè)\ca.m
?????文件???????3378??2010-06-18?19:39??仿真大作業(yè)\cv2.m
?????文件???????4441??2010-06-18?21:59??仿真大作業(yè)\singer3.m
?????文件?????446305??2010-06-18?22:25??仿真大作業(yè)\單模型機動目標(biāo)跟蹤算法的仿真研究.doc
?????目錄??????????0??2010-06-18?22:25??仿真大作業(yè)
-----------?---------??----------?-----??----
???????????????457022????????????????????5
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