clear?all;%清理工作間變量
?L=1024;
?y1=1;y2=1;y3=1;y4=0;
?for?i=1:1024
?????x1=xor（y3y4）;
?????x2=y1;
?????x3=y2;
?????x4=y3;
?????y（i）=y4;
?????if?y（i）==0u（i）=-1;%如果M序列的值為“0“時辨識的輸入信號取“-1”
??????else?u（i）=1;%當M序列的值為“1“時辨識的輸入信號取“1”
??????end%小循環結束
??????y1=x1;y2=x2;y3=x3;y4=x4;%為下一次的輸入信號做準備
??????end%大循環結束，產生輸入信號u?
??????
??????A=179;?x0=11;?M=2^35;??KM=1024;a=0;
??????N=12;?miu=0;sigma=1;
??????b=sqrt（N/12）;
??????for?k=1:KM
??????????for?j=1:N
?????????x2=A*x0;??
?????????x1=mod?（x2M）;???
?????????v（j）=x1/M;
?????????a=a+v（j）;
?????????x0=x1;
??????????end
??????????eita（:k）=miu+[sigma*（a-N/2）]/b;
??????????a=0;
??????end
??????v=eita;%%白噪聲
??????
z=zeros（11024）;?%定義輸出觀測值的長度??
z（2）=0;z（1）=0;%取z的前兩個初始值為零??
for?k=3:1024?
z（k）=1.5*z（k-1）-0.7*z（k-2）+u（k-1）+0.5*u（k-2）+v（k）;?%用輸出值作為觀測值

kk=k-2;
zz（kk:）=z（k）‘;
end
zz

for?k=4:1024?
z（k）=1.5*z（k-1）-0.7*z（k-2）+u（k-1）+0.5*u（k-2）+v（k）;?%用輸出值作為觀測值

kkk=k-3;
zzz（kkk:）=z（k）‘;
end
zzz;
?%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%?構造數據矩陣
??????for?k1=3:L;?
??????????h1=[z（k1-1）u（k1-1）]‘;
??????????k=k1-2;
??????????H11（k:）=h1‘;%?na=1nb=1
??????end
?????
??????
??????for?k2=3:L;?
??????????h2=[z（k2-1）u（k2-1）u（k2-2）]‘?;
??????????k=k2-2;
??????????H12（k:）=h2‘;%?na=1nb=2
??????end

??????
??????for?k3=3:L;?
??????????h3=[z（k3-1）z（k3-2）u（k3-1）]‘?;
??????????k=k3-2;
??????????H21（k:）=h3‘;%?na=2nb=1
??????end
???

??????for?k4=3:L;?
??????????h4=[z（k4-1）z（k4-2）u（k4-1）u（k4-2）]‘;
??????????k=k4-2;
??????????H22（k:）=h4‘;%?na=2nb=2
??????end
?????
??????
??????for?k5=4:L;?
??????????h5=[z（k5-1）z（k5-2）z（k5-3）u（k5-1）]‘;
??????????k=k5-3;
??????????H31（k:）=h5‘;%?na=3nb=1
??????end
???
??????for?k6=4:L;?
??????????h6=[z（k6-1）z（k6-2）z（k6-3）u（k6-1）u（k6-2）]‘;
??????????k=k6-3;
??????????H32（k:）=h6‘;%?na=3nb=2
??????end??
??????
??????for?k7=4:L;?
??????????h7=[z（k7-1）z（k7-2）z（k7-3）u（k7-1）u（k7-2）u（k7-3）]‘;
??????????k=k7-3;
??????????H33（k:）=h7‘;%?na=3nb=3
??????end???
??????
??????for?k8=4:L;?
??????????h8=[z（k8-1）u（k8-1）u（k8-2）u（k8-3）]‘;
??????????k=k8-3;
??????????H13（k:）=h8‘;%?na=1nb=3
??????end???????
??????
??????for?k9=4:L;?
??????????h9=[z（k9-1）z（k9-2）u（k9-1）u（k9-2）u（k9-3）]‘;
??????????k=k9-3;
??????????H23（k:）=h9‘;%?na=2nb=3
??????end??????
?%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%計算AIC定階準則
??????c11=inv（H11‘*H11）*H11‘*zz;
??????d11=1/L*（zz-H11*c11）‘*（zz-H11*c11）;
??????AIC（11）=L*log（d11）+2*（1+1）;

??????c12=inv（H12‘*H12）*H12‘*zz;
??????d12=1/L*（zz-H12*c12）‘*（zz-H12*c12）;
??????AIC（12）=L*log（d12）+2*（1+2）;
??????
??????c21=inv（H21‘*H21）*H21‘*zz;
??????d21=1/L*（zz-H21*c21）‘*（zz-H21*c21）;
??????AIC（21）=L*log（d21）+2*（2+1）;
??????
??????c22=inv（H22‘*H22）*H22‘*zz;
??????d22=1/L*（zz-H22*c22）‘*（zz-H22*c22）;
??????AIC（22）=L*log（d22）+2*（2+2）;
??????
??????c31=inv（H31‘*H31）*H31‘*zzz;
??????d31=1/L*（zzz-H31*c31）‘*（zzz-H31*c31）;
??????AIC（31）=L*log（d31）+2*（3+1）;
?????

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

?????文件???????3880??2009-05-24?22:52??辨識\c（z）=1-1.5（z-1）+0.7（z-2）AIC定階法\AIC_jieci.m

?????文件???????1451??2009-05-24?22:56??辨識\c（z）=1-1.5（z-1）+0.7（z-2）一次性完成\onetimeone.m

?????文件???????1962??2009-05-24?22:53??辨識\c（z）=1-1.5（z-1）+0.7（z-2）行列式定階法\hanglieshi.m

?????文件???????2321??2009-05-24?21:21??辨識\c（z）=1-1.5（z-1）+0.7（z-2）遞推完成\rwlsone.m

?????文件???????3060??2009-05-24?22:55??辨識\c（z）=1bayes法\bayes.m

?????文件???????3033??2009-05-24?22:55??辨識\c（z）=1一次性完成\onetimetwo.m

?????文件???????2934??2009-05-24?21:26??辨識\c（z）=1極大似然法\rmlone.m

?????文件???????2705??2009-05-24?21:20??辨識\c（z）=1遞推完成\xinzengguang.m

?????文件????????904??2009-05-24?22:49??辨識\M序列和逆M序列\M_xulie.m

?????文件??????11528??2009-05-25?21:49??辨識\白噪聲\noise.mat

?????文件??????11528??2009-05-25?21:34??辨識\白噪聲\noise.mat.mat

?????文件????????603??2009-05-25?21:49??辨識\白噪聲\white.m

?????文件?????520192??2010-11-15?18:33??辨識\系統辨識作業.doc

?????文件????????690??2009-05-24?23:01??辨識\辨識m文件說明.txt

?????目錄??????????0??2010-08-12?14:42??辨識\c（z）=1-1.5（z-1）+0.7（z-2）AIC定階法

?????目錄??????????0??2010-08-12?14:42??辨識\c（z）=1-1.5（z-1）+0.7（z-2）一次性完成

?????目錄??????????0??2010-08-12?14:42??辨識\c（z）=1-1.5（z-1）+0.7（z-2）行列式定階法

?????目錄??????????0??2010-08-12?14:42??辨識\c（z）=1-1.5（z-1）+0.7（z-2）遞推完成

?????目錄??????????0??2010-08-12?14:42??辨識\c（z）=1bayes法

?????目錄??????????0??2010-08-12?14:42??辨識\c（z）=1一次性完成

?????目錄??????????0??2010-08-12?14:42??辨識\c（z）=1極大似然法

?????目錄??????????0??2010-08-12?14:42??辨識\c（z）=1遞推完成

?????目錄??????????0??2010-08-12?14:42??辨識\M序列和逆M序列

?????目錄??????????0??2010-08-12?14:42??辨識\白噪聲

?????目錄??????????0??2010-11-15?18:33??辨識

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

???????????????566791????????????????????25