如何找到一个吉它弦的声音的基本频率？ [英] How to find the fundamental frequency of a guitar string sound?
问题描述
我要建立一个IPHONE的吉他调谐器应用程序。我的目标是找到由吉他弦产生的声音的基本频率。我用code位来自苹果提供aurioTouch样本来计算频谱和我找到最高幅度的频率。它工作正常的纯音（只具有一个频率的那些），但是从一个吉他弦音它会产生错误的结果。我已阅读，这是因为色彩的产生由可能比最根本的较高幅度的吉他弦。我如何才能找到基波频率，因此适用于吉他弦？有没有在C / C ++ /对象 - 声音分析（或信号处理）的开放源代码库？
I want to build a guitar tuner app for Iphone. My goal is to find the fundamental frequency of sound generated by a guitar string. I have used bits of code from aurioTouch sample provided by Apple to calculate frequency spectrum and I find the frequency with the highest amplitude . It works fine for pure sounds (the ones that have only one frequency) but for sounds from a guitar string it produces wrong results. I have read that this is because of the overtones generate by the guitar string that might have higher amplitudes than the fundamental one. How can I find the fundamental frequency so it works for guitar strings? Is there an open-source library in C/C++/Obj-C for sound analyzing (or signal processing)?
推荐答案
您可以使用该信号的自相关性，这是平方DFT的幅度的逆变换。如果你在44100样本/秒的采样，然后是82.4赫兹根本的是535的样品，而1479.98赫兹大约是30个样本。寻找在该范围内（例如，从28到560）的峰值正的滞后。确保你的窗口中最长的根本，这将是这里1070的样品至少有两个周期。到2的下一个动力，这是一个2048样本缓冲区。为了更好的频率分辨率和一个不太偏估计，使用较长的缓冲器，但并不长，所以信号不再大致静止。下面是在Python的例子：
You can use the signal's autocorrelation, which is the inverse transform of the magnitude squared of the DFT. If you're sampling at 44100 samples/s, then a 82.4 Hz fundamental is about 535 samples, whereas 1479.98 Hz is about 30 samples. Look for the peak positive lag in that range (e.g. from 28 to 560). Make sure your window is at least two periods of the longest fundamental, which would be 1070 samples here. To the next power of two that's a 2048-sample buffer. For better frequency resolution and a less biased estimate, use a longer buffer, but not so long that the signal is no longer approximately stationary. Here's an example in Python:
from pylab import *
import wave
fs = 44100.0 # sample rate
K = 3 # number of windows
L = 8192 # 1st pass window overlap, 50%
M = 16384 # 1st pass window length
N = 32768 # 1st pass DFT lenth: acyclic correlation
# load a sample of guitar playing an open string 6
# with a fundamental frequency of 82.4 Hz (in theory),
# but this sample is actually at about 81.97 Hz
g = fromstring(wave.open('dist_gtr_6.wav').readframes(-1),
dtype='int16')
g = g / float64(max(abs(g))) # normalize to +/- 1.0
mi = len(g) / 4 # start index
def welch(x, w, L, N):
# Welch's method
M = len(w)
K = (len(x) - L) / (M - L)
Xsq = zeros(N/2+1) # len(N-point rfft) = N/2+1
for k in range(K):
m = k * ( M - L)
xt = w * x[m:m+M]
# use rfft for efficiency (assumes x is real-valued)
Xsq = Xsq + abs(rfft(xt, N)) ** 2
Xsq = Xsq / K
Wsq = abs(rfft(w, N)) ** 2
bias = irfft(Wsq) # for unbiasing Rxx and Sxx
p = dot(x,x) / len(x) # avg power, used as a check
return Xsq, bias, p
# first pass: acyclic autocorrelation
x = g[mi:mi + K*M - (K-1)*L] # len(x) = 32768
w = hamming(M) # hamming[m] = 0.54 - 0.46*cos(2*pi*m/M)
# reduces the side lobes in DFT
Xsq, bias, p = welch(x, w, L, N)
Rxx = irfft(Xsq) # acyclic autocorrelation
Rxx = Rxx / bias # unbias (bias is tapered)
mp = argmax(Rxx[28:561]) + 28 # index of 1st peak in 28 to 560
# 2nd pass: cyclic autocorrelation
N = M = L - (L % mp) # window an integer number of periods
# shortened to ~8192 for stationarity
x = g[mi:mi+K*M] # data for K windows
w = ones(M); L = 0 # rectangular, non-overlaping
Xsq, bias, p = welch(x, w, L, N)
Rxx = irfft(Xsq) # cyclic autocorrelation
Rxx = Rxx / bias # unbias (bias is constant)
mp = argmax(Rxx[28:561]) + 28 # index of 1st peak in 28 to 560
Sxx = Xsq / bias[0]
Sxx[1:-1] = 2 * Sxx[1:-1] # fold the freq axis
Sxx = Sxx / N # normalize S for avg power
n0 = N / mp
np = argmax(Sxx[n0-2:n0+3]) + n0-2 # bin of the nearest peak power
# check
print "\nAverage Power"
print " p:", p
print "Rxx:", Rxx[0] # should equal dot product, p
print "Sxx:", sum(Sxx), '\n' # should equal Rxx[0]
figure().subplots_adjust(hspace=0.5)
subplot2grid((2,1), (0,0))
title('Autocorrelation, R$_{xx}$'); xlabel('Lags')
mr = r_[:3 * mp]
plot(Rxx[mr]); plot(mp, Rxx[mp], 'ro')
xticks(mp/2 * r_[1:6])
grid(); axis('tight'); ylim(1.25*min(Rxx), 1.25*max(Rxx))
subplot2grid((2,1), (1,0))
title('Power Spectral Density, S$_{xx}$'); xlabel('Frequency (Hz)')
fr = r_[:5 * np]; f = fs * fr / N;
vlines(f, 0, Sxx[fr], colors='b', linewidth=2)
xticks((fs * np/N * r_[1:5]).round(3))
grid(); axis('tight'); ylim(0,1.25*max(Sxx[fr]))
show()
输出：
Average Power
p: 0.0410611012542
Rxx: 0.0410611012542
Sxx: 0.0410611012542
峰值滞后是538，这是五百三十八分之四万四千百= 81.97赫兹。第一通无环的DFT表示在仓61，这是82.10 +/- 0.67赫兹的根本。在第2次使用的538 * 15 = 8070的窗口的长度，所以DFT的频率包括字符串的基本周期和谐波。这使得与谐波少US preading改进的PSD评估的ubiased循环自相关（即相关性可以在窗口包裹定期）。
The peak lag is 538, which is 44100/538 = 81.97 Hz. The first-pass acyclic DFT shows the fundamental at bin 61, which is 82.10 +/- 0.67 Hz. The 2nd pass uses a window length of 538*15 = 8070, so the DFT frequencies include the fundamental period and harmonics of the string. This enables an ubiased cyclic autocorrelation for an improved PSD estimate with less harmonic spreading (i.e. the correlation can wrap around the window periodically).
编辑：更新后可使用韦尔奇的方法来估计自相关。重叠窗口补偿汉明窗口。我也算Hamming窗的锥形偏置无偏自相关。
Updated to use Welch's method to estimate the autocorrelation. Overlapping the windows compensates for the Hamming window. I also calculate the tapered bias of the hamming window to unbias the autocorrelation.
编辑：增加了第2次与循环相关清理的功率谱密度。这通使用3个非重叠的，矩形窗口长度538 * 15 = 8070（短到足以几乎静止的）。对于循环相关的偏置是一个常数，代替的汉明窗口的锥形偏置
Added a 2nd pass with cyclic correlation to clean up the power spectral density. This pass uses 3 non-overlapping, rectangular windows length 538*15 = 8070 (short enough to be nearly stationary). The bias for cyclic correlation is a constant, instead of the Hamming window's tapered bias.
这篇关于如何找到一个吉它弦的声音的基本频率？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
