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注册时间2011-6-3
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数值分析解jitter vs distortion
看你们和大家坛上炒得挺热,我发表点意见:
上程序说话,scilab的仿真,结果jitter只要不大于产生误码的阈值,就不会产生失真。对于一个24/192的spdif来说,频率为12.288m,检测阈值为1/(4*12.288M)s=20.35ns,即jitter小于20ns就不会对音频产生影响。对于宽带pll的接收器,如dir9001,jitter的容忍度更大,可达100多ns!
以下程序原创,转的留名!!!
用Scilab写的,不知道的看高中数学书去!!!
samplerate=44100;//unit: sps
jitter=1;//unit: ns
frequency=1000;//signal frequency
xorig=0:1/samplerate:1;
xpolluted=xorig+2.828*jitter*10^(-9)*(0.5-rand());
yorig=sin(2*%pi*frequency*xorig);
ypolluted=sin(2*%pi*frequency*xpolluted);
forig=abs(fft(yorig));
fpolluted=abs(fft(ypolluted));
dio=forig(frequency*2+1)^2+forig(frequency*3+1)^2+forig(frequency*4+1)^2+forig(frequency*5+1)^2;
dip=fpolluted(frequency*2+1)^2+fpolluted(frequency*3+1)^2+fpolluted(frequency*4+1)^2+fpolluted(frequency*5+1)^2;
dio=sqrt(dio)/forig(frequency+1);
dip=sqrt(dip)/fpolluted(frequency+1);
ddi=dip/dio;
ddi=log10(ddi)*20;
printf("The distortion increasement is %f dBn",ddi);
结果是0dB,也就是无干扰!!! |
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