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資源簡介

PCM編碼和解碼的Matlab程序,我試過的,正確的

資源截圖

代碼片段和文件信息

%show?the?pcm?encode?and?decode?
clear?all
close?all
t=0:0.01:10;%1001個數
vm1=-70:1:10;??%輸入正弦信號幅度不同???dB
vm=10.^(vm1/20);%dB---10進制
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);%燥聲功率
????????sq(m)=mean(yy.^2);%信號功率
????????snr(m)=(sq(m)/nq(m));%信燥比
????????
????????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????????973??2011-03-30?22:15??tongxin\pcm_decode.m

?????文件???????1322??2011-03-30?22:15??tongxin\pcm_encode.m

?????文件????????800??2011-03-30?22:15??tongxin\pcm1.m

?????目錄??????????0??2011-03-31?08:49??tongxin

-----------?---------??----------?-----??----

?????????????????3095????????????????????4


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