通过自相关法计算音高误差

Question

• 目标：音高计算

• 问题：计算的音高与预期的不匹配。例如，输出约为。 'D3'，但预期的输出是 'C5'。

• 来源声音：https://freewavesamples.com/1980s-casio-celesta-c5

源代码

library("tuneR")
library("seewave")

#0: Acquisition of sample sound
snd_smpl = readWave(paste("~/Music/sample/1980s-Casio-Celesta-C5.wav"), 
                                    from = 0, to = 1, units = "seconds")
dur_smpl = duration(snd_smpl)
len_smpl = length(snd_smpl)

#1 : Pre-Processing Stage

#1.1 : Application of Hanning Window
n = 1:len_smpl
han_win = 0.5-0.5*cos(2*pi*n/(len_smpl-1))
wind_sig = han_win*snd_smpl@left

#2.1 : Auto-Correlation Calculation
rev_wind_sig = rev(wind_sig)    #Reversing the windowed signal

acorr_1 = convolve(wind_sig, rev_wind_sig, type = "open")
# Obtaining the 2nd half of the correlation, to simplify calculation
n = 2*len_smpl-1 
acorr_2 = (1/len_smpl)*acorr_1[len_smpl:n]

#2.2 : Note Calculation
min_index = which.min(acorr_2)
print(min_index)
fs = 44100              
fo = fs/min_index #To obtain fundamental frequency

print(fo)
print(notenames(noteFromFF(fo)))

输出

> print(min_index)
[1] 37
> fs = 44100                
> fo = fs/min_index 
> print(fo)
[1] 1191.892
> print(notenames(noteFromFF(fo)))
[1] "d'''"

整个计算在时域中进行。 我目前正在使用自相关作为基础来了解更多关于音高检测和分析的信息。我试图用“Audacity”分析样本，结果是“C5”。因此，我想知道问题到底出在哪里。 大家能帮我找到吗？

此外，还有一些重要的疑问：

1. 我的分析窗口实际上应该有多小（20 毫秒、1 秒、..）？
2. 使用 AMDF 和其他类似算法强化自相关算法是否会使这个音高检测模块更加稳健？

Answer 1

整个分析似乎不正确。您不应该在时域分析中使用窗口化。

附上python语言的简短解决方案；您可以将其用作伪代码

from soundfile import read
from glob import glob
from scipy.signal import correlate, find_peaks
from matplotlib.pyplot import plot, show, xlim, title, xlabel
import numpy as np
%matplotlib inline

name = glob('*wav')[0]
samples, fs = read(name)

corr = correlate(samples, samples)
corr = corr[corr.size / 2:]
time = np.arange(corr.size) / float(fs)
ind = find_peaks(corr[time < 0.002])[0]

plot(time, corr)
plot(time[ind], corr[ind], '*')
xlim([0, 0.005])
title('Frequency = {} Hz'.format(1 / time[ind][0]))
xlabel('Time [Sec]')
show()