資源簡介
基于當前模型的matlab 程序由兩部分組成 運行check_current 可以出圖

代碼片段和文件信息
clc;
clear;
TS=0.01;???????????????????????????????%?simulation?step?????
alphax?=?1/3;
alphay?=?1/3;
F?=?[0?0?1?0?0?0;??????????????????????%?current?model?
????0?0?0?1?0?0;?
????0?0?0?0?1?0;?
????0?0?0?0?0?1;?
????0?0?0?0?-alphax?0;?
????0?0?0?0?0?-alphay];
F1?=[?0?0?1?0?0?0;???????????????????????%?real?target?model
??????0?0?0?1?0?0;?
??????0?0?0?0?1?0;?
??????0?0?0?0?0?1;?
??????0?0?0?0?0?0;?
??????0?0?0?0?0?0];
G?=?[0?0;?0?0;0?0;?0?0;?alphax?0;?0?alphay];
H?=?[1?0?0?0?0?0?;?
????0?1?0?0?0?0?];
[order_zorder]=?size(H);
syms?s?t;
PHAI?=?ilaplace(inv(eye(orderorder)*s-F));
PHAI1?=?ilaplace(inv(eye(orderorder)*s-F1));
GK?=?int(PHAI*Gt0t);
PHAI?=?subs(PHAItTS);
PHAI1?=?subs(PHAI1tTS);
GK?=?subs(GKtTS);
qx11=2*alphax*(1-exp(-2*alphax*TS)+2*alphax*TS+2*((alphax*TS)^3)/3-2*(alphax^2)*(TS^2)-4*alphax*TS*exp(-alphax*TS))/(2*alphax^5);
qx12=2*alphax*(1+exp(-2*alphax*TS)-2*exp(-alphax*TS)+2*alphax*TS*exp(-alphax*TS)-2*alphax*TS+alphax^2*TS^2)/(2*alphax^4);
qx13=2*alphax*(1-exp(-2*alphax*TS)-2*alphax*TS*exp(-alphax*TS))/(2*alphax^3);
qx22=2*alphax*(4*exp(-alphax*TS)-3-exp(-2*alphax*TS)+2*alphax*TS)/(2*alphax^3);
qx23=2*alphax*(1+exp(-2*alphax*TS)-2*exp(-alphax*TS))/(2*alphax^2);
qx33=2*alphax*(1-exp(-2*alphax*TS))/(2*alphax);
qy11=2*alphay*(1-exp(-2*alphay*TS)+2*alphay*TS+2*((alphay*TS)^3)/3-2*(alphay^2)*(TS^2)-4*alphay*TS*exp(-alphay*TS))/(2*alphay^5);
qy12=2*alphay*(1+exp(-2*alphay*TS)-2*exp(-alphay*TS)+2*alphay*TS*exp(-alphay*TS)-2*alphay*TS+alphay^2*TS^2)/(2*alphay^4);
qy13=2*alphay*(1-exp(-2*alphay*TS)-2*alphay*TS*exp(-alphay*TS))/(2*alphay^3);
qy22=2*alphay*(4*exp(-alphay*TS)-3-exp(-2*alphay*TS)+2*alphay*TS)/(2*alphay^3);
qy23=2*alphay*(1+exp(-2*alphay*TS)-2*exp(-alphay*TS))/(2*alphay^2);
qy33=2*alphay*(1-exp(-2*alphay*TS))/(2*alphay);
Q=[?qx11?0?qx12?0?qx13?0;
????0?qy11?0?qy12?0?qy13;
????qx12?0?qx22?0?qx23?0;
????0?qy12?0?qy22?0?qy23;
????qx13?0?qx23?0?qx33?0;
????0?qy13?0?qy23?0?qy33];
q0?=?5;
r?=?2.5;
R?=?diag([r^2r^2]);
n?=?order;
m?=?order_z;
f=@(x)[PHAI*x(1:n1)+GK*x(n-n/3+1:n1)];????????%?nonlinear?state?equations
f1=@(x)[PHAI1*x(1:n1)];?????????????????????????%?target?state?equations
h=@(x)[H*x(1:n1)];?????????????????????????????%?measurement?equation
s=[15000;35000;0;-1300;2;0];????????????????????%?initial?state
x=[15050;35050;1;-1280;1.5;0];??????????????????%?initial?state?with?noise
P?=?diag([100100404011]);????????????????????%?initial?state?covraiance
N=?20000;??????????????????????????????????????%?total?dynamic?steps
xV?=?zeros(nN);????????????????????????????????%?estmate
sV?=?zeros(nN);????????????????????????????????%?allocate?memory
zV?=?zeros(mN);????????????????????????????????%?actual
for?k=1:N
????z?=?h(s)?+?R*randn(m1);????????????????????%?measurments
????sV(:k)=?s;?????????????????????????????????%?save?actual?state
????zV(:k)??=?z;???????????????????????????????%?save?measurment
????[x?P]?=?ukf_Current(fxPhzQRq
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件????????3778??2014-11-20?13:47??ukf_Current.m
?????文件????????3627??2014-11-20?21:19??check_dangqian.m
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