# 366 statistical and adaptive signal processing

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366 Statistical and Adaptive Signal Processing - Solution Manual %Lattice Ef(m) = lambda*Efold(m) + alphaold(m)*conj(ef(m))*ef(m); cf(m) = lambda*Efold(m)/Ef(m); sf(m) = alphaold(m)*ef(m)/Ef(m); Eb(m) = lambda*Ebold(m) + alpha(m)*conj(eb(m))*eb(m); cb(m) = lambda*Ebold(m)/Eb(m); sb(m) = alpha(m)*eb(m)/Eb(m); ef(m+1) = ef(m) + conj(kfold(m))*ebold(m); kf(m) = cbold(m)*kfold(m) - sbold(m)*conj(ef(m)); eb(m+1) = ebold(m) + conj(kbold(m))*ef(m); kb(m) = cf(m)*kbold(m) - sf(m)*conj(ebold(m)); alpha(m+1) = alpha(m) - abs(alpha(m)*eb(m))*abs(alpha(m)*eb(m))/Eb(m); % Ladder e(m+1)=e(m)-conj(kcold(m))*eb(m); kc(m)= cb(m)*kcold(m) + sb(m)*conj(e(m)); end er(n,i)=e(M+1);erf(n,i)=ef(M);erb(n,i)=eb(M); Erf(n,i)=Eb(1); Erb(n,i)=Eb(M); al(n,i)=alpha(M); % Time Delay cbold=cb; cfold=cf; sbold=sb; sfold=sf; Ebold=Eb; Efold=Ef; ebold=eb; efold=ef; kfold=kf; kbold=kb; kcold=kc; alphaold=alpha; end i end er2=er.^2; er2m=mean(er2,2); t=(1:N)’; %plot(t,Erf,’r’,t,Erb,’b’); %text(250,300,’\leftarrow\fontsize{9pt} E^f’); %text(100,200,’E^b \rightarrow\fontsize{9pt} ’); Jmin1=1; %subplot(’position’,[0.1,0.55,0.85,0.4]); semilogy(t,er2m,’r’); axis([0,N,10^(-4), 10^0]); title([’MSE Learning Curve (\lambda=.99, W=’ num2str(W) ’ )’],’fontsize’,10); xlabel(’Number of iterations (n)’,’fontsize’,8); ylabel(’Mean squared error’,’fontsize’,8);

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Statistical and Adaptive Signal Processing - Solution Manual 367 0 100 200 300 400 500 600 700 800 900 1000 10 -4 10 -3 10 -2 10 -1 10 0 MSE Learning Curve ( λ =.99, W=2.9 ) Number of iterations (n) Mean squared error Figure 10.43: MSE learning curve in P10.43 set(gca,’xtick’,[0:100:N],’fontsize’,8); print -deps2 P1043.eps 10.44 In this problem we discuss the derivation of the normalized lattice-ladder RLS algorithm, which uses a smaller number of time and order updating recursions and has better numerical behavior due to the normalization of its variables. (a) Define the energy and angle normalized variables ¯ e f m ( n ) = ε f m ( n ) α m ( n ) E f m ( n ) ¯ e b m ( n ) = ε b m ( n ) α m ( n ) E b m ( n ) ¯ e m ( n ) = ε m ( n ) α m ( n ) E m ( n ) ¯ k m ( n ) = β m ( n ) E f m ( n ) E b m ( n 1 ) ¯ k c m ( n ) = β c m ( n ) E m ( n ) E b m ( n ) and show that the normalized errors and the partial correlation coefficients ¯ k m ( n ) and ¯ k c m ( n ) have magni- tude less than 1.
368 Statistical and Adaptive Signal Processing - Solution Manual (b) Derive the following normalized lattice-ladder RLS algorithm: E f 0 ( 1 ) = E 0 ( 1 ) = δ > 0 For n = 0 , 1 , 2 , . . . E f 0 ( n ) = λ E f 0 ( n 1 ) + | x ( n ) | 2 , E 0 ( n ) = λ E 0 ( n 1 ) + | y ( n ) | 2 ¯ e f 0 ( n ) = ¯ e b 0 ( n ) = x ( n ) E f 0 ( n ) , ¯ e 0 ( n ) = y ( n ) E 0 ( n ) For m = 0 to M 1 ¯ k m ( n ) = 1 − |¯ e f m ( n ) | 2 1 − |¯ e b m ( n 1 ) | 2 ¯ k m ( n 1 ) + ¯ e f m ( n ) ¯ e b m ( n 1 ) ¯ e f m + 1 ( n ) = 1 − | e b m ( n 1 ) | 2 1 − | ¯ k m ( n ) | 2 1 e f m ( n ) ¯ k m ( n ) ¯ e b m ( n 1 ) ] ¯ e b m + 1 ( n ) = 1 − |¯ e f m ( n ) | 2 1 − | ¯ k m ( n ) | 2 1 e b m ( n 1 ) ¯ k m ( n ) ¯ e f m ( n ) ] ¯ k c m ( n ) = 1 − |¯ e m ( n ) | 2 1 − |¯ e b m ( n ) | 2 ¯ k c m ( n 1 ) + ¯ e m ( n ) ¯ e b m ( n ) ¯ e m + 1 ( n ) = 1 − |¯ e b m ( n ) | 2 1 − | ¯ k c m ( n ) | 2 1 e m ( n ) ¯ k c m ( n ) ¯ e b m ( n ) ] (c) Write a Matlab function to implement the derived algorithm, and test its validity by using the equalization experiment in Example 10.5.2.

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• Fall '08
• Papamichalis,P

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