现代数字信号处理及应用 第五章仿真题答案.docxVIP

5.10 (1) (2) (3) 特征值分解 eig(R2)=diag{0.4704,93.6270} Eig(R3)=diag{0.3148,0.9362,139.8951} 特征值扩展： X(R2)=199.0370 X(R3)=444.4107 （4）程序 clear all clc; L=10000; sigma_v1=0.93627; A1 = zeros(L, 2); EA1 = zeros(L, 1); for i=1:100 v=sqrt(sigma_v1)*randn(L,1); a1=-0.99; u(1)=v(1); for k=2:L u(k)=-a1*u(k-1)+v(k); end % u=u(500:end); M=2; w(1,:)=zeros(1,M); e(1)=u(1); mu=0.001; uu=zeros(1,M); w(2,:)=w(1,:)+mu*e(1)*uu; uu=[u(1) uu(1:M-1)]; dd=(w(2,:)*uu); e(2)=u(2)-dd; for k=3:L w(k,:)=w(k-1,:)+mu*e(k-1)*uu; uu=[u(k-1) uu(1:M-1)]; dd=(w(k,:)*uu); e(k)=u(k)-dd; end A1 = A1 + conj(w); EA1 = EA1 +(e.^2); end figure(1); plot(EA1/100); xlabel(迭代次数);ylabel(均方误差);title(迭代500次，步长0.001); figure(2); plot(A1/100);hold on; plot(conj(w)); xlabel(迭代次数);ylabel(权向量);title(权向量收敛曲线); 5.11 clear all clear; clc; for i=1:1500 N=1000; M=5; L=2; h=[0.389 1 0.389]; sigma=1e-3; vn=sqrt(sigma)*randn(2*M+N,1); H=zeros(2*M+1,2*M+L+1); for k=1:2*M+1 H(k,k:1:k+L)=h; end s=randsrc(2*M+L+N,1); S=zeros(2*M+L+1,N); V=zeros(2*M+1,N); for k=1:N S(:,k)=s(2*M+L+k:-1:k); V(:,k)=vn(2*M+k:-1:k); end U=H*S+V; dn=S(M+L+1,:); if (i=500) mu=0.01; elseif (i500i=1000) mu=0.025; else mu=0.05; end a=size(U); M=a(1); N=a(2); err=zeros(N,1); w=zeros(M,N); w((M-1)/2+1,1)=1; err(1)=dn(1)-w(:,1)*U(:,1); for k=1:N-1 w(:,k+1)=w(:,k)+mu*U(:,k)*conj(err(k)); err(k+1)=dn(k+1)-w(:,k+1)*U(:,k+1); end if (i=500) ee1(:,i)=mean(abs(err).^2,2); elseif (i500i=1000) ee2(:,i)=mean(abs(err).^2,2); else ee3(:,i)=mean(abs(err).^2,2); end end ep1=mean(ee1); ep2=mean(ee2); ep3=mean(ee3); figure(1); plot(ep1); hold on; plot(ep2); hold on; plot(ep3) xlabe