In Example 8.24 the differential pulse code modulation (DPCM) predictor was obta
ID: 3210286 • Letter: I
Question
In Example 8.24 the differential pulse code modulation (DPCM) predictor was obtained for a 2 nd order system. See Figure1 second-order esti mator is Delay Delay and the error is The following correlation values have been found for speech in,me-0.562m mana-m, m,7- m,"me-0.825, Figure 1: DPCM Predictor, 2nd order. k-2 mpr-=0.004 m., 0.308 , 0.243 m,m, where R11 = R22 = m R12 = R 21 = R01 = 0.825m2 R02 = 0.562m The optimum coefficients values were found to be al-1.1314 and a2-_0.37 14. These coefficients were obtained from the solution of the Yule-Walker equations R,i=a,R,1 + a2 Ri,t +a,Rin where Rij-x,x, and 01 al=R21 R22 R2n 02 with a mean square error [1-(0.825 a, + 0.562 aJI = 0.275 m2 and a SNR improvement SNR, = 10 logo(m2/(0.2753m)) =-10 log,o(0.2753) = 5.6 dB Use MATLAB and publish your results as a pdf. Upload to Blackboard. 1. Verify these 2d order coefficients, and the resultant mean square error. Find the 3rd and 4th order coefficients and their respective mean square errors and plot the SNR improvement vs predictor order for all the 2d through the 4th order predictors. 2.Explanation / Answer
ANSWER:
CODE
clear all;clc;
%for part (1) 2nd order coefficients
R1=[1 0.825;0.825 1];
R01=[0.825 0.562];
a1=(inv(R1))*R01';
E1=1-R01*a1;
SNR1=-10*log10(E1);
disp('2nd order coefficients');
fprintf('a1=%f. a2=%f. ',a1(1),a1(2));
fprintf('2nd order mean square error=%f(m^2). ',E1);
fprintf('2nd order SNR improvement=%fdB. ',SNR1);
%for part (2) 3rd order coefficients
R2=[1 0.825 0.562;0.825 1 0.825;0.562 0.825 1];
R02=[0.825 0.562 0.308];
a2=(inv(R2))*R02';
E2=1-R02*a2;
SNR2=-10*log10(E2);
disp('3rd order coefficients');
fprintf('a1=%f. a2=%f. a3=%f. ',a2(1),a2(2),a2(3));
fprintf('3rd order mean square error=%f(m^2). ',E2);
fprintf('3rd order SNR improvement=%fdB. ',SNR2);
%for part (2) 4th order coefficients
R3=[1 0.825 0.562 0.308;
0.825 1 0.825 0.562;
0.562 0.825 1 0.825;
0.308 0.562 0.825 1];
R03=[0.825 0.562 0.308 0.004];
a3=(inv(R3))*R03';
E3=1-R03*a3;
SNR3=-10*log10(E3);
disp('4th order coefficients');
fprintf('a1=%f. a2=%f. a3=%f. a4=%f. ',a3(1),a3(2),a3(3),a3(4));
fprintf('4th order mean square error=%f(m^2). ',E3);
fprintf('4th order SNR improvement=%f.dB ',SNR3);
or=[2 3 4];
SNR=[SNR1 SNR2 SNR3];
plot(or,SNR);
title('SNR improvement plot with respect to order of predictor');
xlabel('order of predictor');ylabel('SNR Improvemnt(dB)');
SAMPLE OUTPUT:
2nd order coefficients
a1=1.131429.
a2=-0.371429.
2nd order mean square error=0.275314(m^2).
2nd order SNR improvement=5.601713dB.
3rd order coefficients
a1=1.102511.
a2=-0.283342.
a3=-0.077854.
3rd order mean square error=0.273646(m^2).
3rd order SNR improvement=5.628116dB.
4th order coefficients
a1=1.070617.
a2=-0.399421.
a3=0.373817.
a4=-0.409674.
4th order mean square error=0.227719(m^2).
4th order SNR improvement=6.426012.dB