Problems with snr function
显示 更早的评论
I want to find the snr for signal xn and noise ns but I keep getting an error.
clear;
n = [0:1023];
omega = 0.25*pi;
xn = sin(omega*n);
count = 1024;
ns = sqrt(0.2)*randn(1,count);
r = snr(xn,ns);
plot(r);
??? Undefined function or method 'snr' for input arguments of type 'double'.
采纳的回答
Image Analyst
2014-9-20
How about:
theRatio = xn ./ ns;
theSNR = mean(theRatio);
2 个评论
Dick Rusell
2014-9-20
编辑：Dick Rusell
2014-9-20
Image Analyst
2014-9-20
xn is your signal. ns is your noise. The SNR is theRatio xn/ns, but this gives the SNR element by element. So you have a bunch of SNR's - one for each element. So to get it down to just one SNR I took the mean of all the individual SNRs. If you want something different, then say what you want.
更多回答（2 个）
Guillaume
2014-9-20
Sounds like you don't have the signal processing toolbox.
ver
will tell you which toolboxes you have installed.
2 个评论
Image Analyst
2014-9-20
编辑：Image Analyst
2014-9-20
Or it's an antique version. I think snr() has not always been part of the Signal Processing Toolbox, but it's simple enough to calculate manually.
Dick Rusell
2014-9-20
编辑：Dick Rusell
2014-9-20
Youssef Khmou
2014-9-20
编辑：Youssef Khmou
2014-9-20
Generally the formula is SNR=20log10(std(signal)/std(noise)) , in your case you have :
snr=20*log10(std(xn)/std(ns)) % 3.8dB
类别
在 帮助中心 和 File Exchange 中查找有关 Signal Operations 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
