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PCM編碼的matlab實現并分析其性能

代碼片段和文件信息
%show?the?pcm?encode?and?decode
clear?all
close?all
t=0:0.01:10;
vm1=-70:1:0;?%輸入的正弦信號幅度不同
vm?=?10.^(vm1/20);
figure(1)
for?k=1:length(vm)
for?m=1:2
x=vm(k)*sin(2*pi*t+2*pi*rand(1));
v=1;
xx=x/v;?%normalize
sxx?=?floor(xx*4096);
y?=?pcm_encode(sxx);
yy?=?pcm_decode(yv);
nq(m)=sum((x-yy).*(x-yy))/length(x);
sq(m)=mean(yy.^2);
snr(m)=(sq(m)/nq(m));
drawnow
subplot(211)
plot(tx);
title(‘sample?sequence‘);
subplot(212)
plot(tyy)
title(‘pcm?decode?sequence‘);
end
snrq(k)=10*log10(?mean(snr)?);
end
figure(2)
plot(vm1snrq);
axis([-60?0?0?60]);
grid;
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件????????594??2014-03-17?19:25??PCM.m
-----------?---------??----------?-----??----
??????????????????594????????????????????1
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