如何用Matlab比较两个音频信号 [英] how to compare two audio signals with Matlab

问题描述

我有两个要使用Matlab比较的音频信号，我的问题是我可以清晰听到它们之间的区别，但是当使用功能pwelch比较它们的PSD时，我看不到它们之间的区别，任何想法我该如何将它们与其他方法进行比较！提前致谢 ！ PS:1.我已经问过DSP.stackexchange中的问题，但没有答案！ 2.我不要求输入密码，这就是为什么我不放我的密码！

I have two audio signas that I want to compare using Matlab, my problem is that I can hear the difference between them loud and clear, but when use the function pwelch to compare their PSD ,I don't see much difference between them,any idea how can I compare them with a different methode ! thanks in advance ! PS: 1. I already ask the quesiton in DSP.stackexchange but had no answer! 2. I'm not asking for codes, that'S why I didn't put mine !

更新

在@Bas Swinckels的答案之后，这里是我正在使用结果的代码，而我仍然看不到描述此方法的方式?

after the answer of @Bas Swinckels here the code par that I'm using the result and I still can't see a way that the describ this ?

  clear;
clc;
[x0,Fs] = audioread('Nonoise.wav');
x0 = x0(:,1);
[Y0,G] = pwelch(x0,hamming(512));
plot(G,10*log10(Y0));
grid on 

[x50,Fs] = audioread('50% noise.wav');
x50 = x50(:,1);
[Y50,G] = pwelch(x50,hamming(512));
hold on ;
plot(G,10*log10(Y50),'r');


[x100,Fs] = audioread('100% noisy.wav');
x100 = x100(:,1);
[Y100,G] = pwelch(x100,hamming(512));
hold on ;
plot(G,10*log10(Y100),'g');

%% spectrogram  
spectrogram(x0,hann(64),32,64,Fs)
figure();
spectrogram(x50,hann(64),32,64,Fs)
figure();
spectrogram(x100,hann(64),32,64,Fs)

这是结果:

有人可以告诉我，我添加到信号源中的噪声或影响如何影响我的频谱?

can anybody tell me how does the noise or the effect that I'm adding to the source influences my spectrum ?

推荐答案

您可以尝试绘制频谱图:

spectrogram(x, hann(nfft), nfft/2, nfft, fsample)

使用pwelch计算PSD时，您只会得到平均频谱，并且会丢失所有时间信息:如果信号在开始或结束时响亮，则不会显示给您.频谱图的计算也与您的耳朵所感知的非常相似，它们根据时间来测量某种频谱.因此，如果您能听到差异，那么它应该在频谱图中显示为具有足够SNR的差异.

When you calculate a PSD with pwelch, you only get the average spectrum and you lose all temporal information: it will not show you if a signal is louder at the beginning or at the end. Calculating a spectrogram is also pretty similar to what your ears perceive, they measure a sort of spectrum as a function of time. So if you can hear the difference, it should show up as a difference in the spectrogram with enough SNR.

制作频谱图时，应尝试使用频率和时间分辨率，因为它们遵循某种海森堡定律.使FFT窗口变短(通过选择较小的nfft)将为您提供良好的时间分辨率，但会降低频率分辨率，反之亦然.

When you make a spectrogram, you should try to play with the frequency and temporal resolution, since they follow some sort of Heisenberg law. Making the FFT window short (by chosing a small nfft) will give you good temporal resolution, but worse frequency resolution, and the other way around.

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