一、最小二乘一次完成算法 MATLAB 仿真 1、考虑仿真对象： z(k)-1.6 *z(k-1)+0.8*z(k-2)=0.8*u(k-1)+0.3*u(k-2)+v(k); 式中，v(k)是服从正态分布的白噪声 N（0,1）。输入信号采用 4 阶 M 序列，幅度为 1。选择如下形式的辨识模型： z(k)+a1*z(k-1)+a2*z(k-2)=b1*u(k-1)+b2*u(k-2)+v(k); （1）程序如下： u=[-1,1,-1,1,1,1,1,-1,-1,-1,1,-1,-1,1,1]; z=zeros(1,16); for k=3:16 z(k)=1.6*z(k-1)-0.8*z(k-2)+0.8*u(k-1)+0.3*u(k-2); end subplot(3,1,1) stem(u) subplot(3,1,2) i=1:1:16; plot(i,z) subplot(3,1,3) stem(z),grid on u,z L=14; HL=[-z(2) -z(1) u(2) u(1);-z(3) -z(2) u(3) u(2);-z(4) -z(3) u(4) u(3);-z(5) -z(4) u(5) u(4);-z(6) -z(5) u(6) u(5);-z(7) -z(6) u(7) u(6);-z(8) -z(7) u(8) u(7);-z(9) -z(8) u(9) u(8); -z(10) -z(9) u(10) u(9);-z(11) -z(10) u(11) u(10);-z(12) -z(11) u(12) u(11);-z(13) -z(12) u(13) u(12);-z(14) -z(13) u(14) u(13);-z(15) -z(14) u(15) u(14)] ZL=[z(3);z(4);z(5);z(6);z(7);z(8);z(9);z(10);z(11);z(12);z(13);z(14);z(15);z(16)] c1=HL'*HL;c2=inv(c1);c3=HL'*ZL;c=c2*c3 a1=c(1),a2=c(2),b1=c(3),b2=c(4) （2）程序运行结果： u = -1 1 -1 1 1 1 1 -1 -1 -1 1 -1 -1 1 1 z = 0 0 0.5000 0.3000 0.5800 1.7880 3.4968 5.2645 5.1257 2.8896 -0.5773 -2.7353 -4.4146 -5.9752 -5.5286 -2.9656 HL = 0 0 1.0000 -1.0000 -0.5000 0 -1.0000 1.0000 -0.3000 -0.5000 1.0000 -1.0000 -0.5800 -0.3000 1.0000 1.0000 -1.7880 -0.5800 1.0000 1.0000 -3.4968 -1.7880 1.0000 1.0000 -5.2645 -3.4968 -1.0000 1.0000 -5.1257 -5.2645 -1.0000 -1.0000 -2.8896 -5.1257 -1.0000 -1.0000 0.5773 -2.8896 1.0000 -1.0000 2.7353 0.5773 -1.0000 1.0000 4.4146 2.7353 -1.0000 -1.0000 5.9752 4.4146 1.0000 -1.0000 5.5286 5.9752 1.0000 1.0000 ZL = 0.5000 0.3000 0.5800 1.7880 3.4968 5.2645 5.1257 2.8896 -0.5773 -2.7353 -4.4146 -5.9752 -5.5286 -2.9656 c =-1.6000 0.8000 0.8000 0.3000 a1 = -1.6000 a2 =0.8000 b1 = 0.8000 b2 =0.3000 （3）辨识结果： 参数 a1 a2...
