PDA

查看完整版本 : 如何从MATLAB中的自动相关数据中提取特定值?


poster
2019-12-10, 20:48
我一直在努力从看起来像这样的图形中提取峰值(请参阅上一个问题) (https://stackoverflow.com/questions/3495892/how-can-i-extract-peak-values-from-auto-correlated-data-in-matlab) :

https://i.stack.imgur.com/Ch1PE.png

但我注意到,对于某些我一直在处理的值的xcorr图,结果并没有达到预期效果,通常看起来更像这样:

https://i.stack.imgur.com/o4cou.jpg

和这个:

https://i.stack.imgur.com/UnkJA.jpg (https://i.stack.imgur.com/UnkJA.jpg)

与其尝试像第一个图中的代码那样尝试选择峰值,不如尝试选择向下倾斜暂时使自身平整的值(如图3所示)?

当我尝试对图2和3所示的数据以当前状态运行代码时,我没有得到任何有用的数据作为回报。

我认为我需要在“查找极值点”部分中使用if语句或类似语句,但是我不确定这是否正确。到目前为止,我对该函数的.M代码如下所示:

[inputname, pathname] = uigetfile('*.wav', 'Select WAV-file'); thumb1 = inputname; %# Get filename information fprintf('\n%s is being turned into a 30s thumbnail...\n', thumb1); fprintf('Please wait..!\n\n'); %# load the signal [y, fs, nb] = wavread(thumb1); y = mean(y,2); %# stereo, take avrg of 2 channels %# Calculate frame energy fWidth = round(fs*1); %# 10ms numFrames = floor(length(y)/fWidth); energy = zeros(1,numFrames); for f=1:numFrames energy(f) = sum( y((f-1)*fWidth+1:f*fWidth).^2 ); end %# smooth the signal (moving average with window size = 1% * length of data) WINDOW_SIZE = round(length(energy) * 0.01); %# 200 XX = filtfilt(ones(1,WINDOW_SIZE)/WINDOW_SIZE, 1, energy); %# auto-correlation [r,lags] = xcorr(XX, 'biased'); %# find extrema points dr = diff(r); eIdx = find(dr(1:end-1) .* dr(2:end)