如何找到吉他弦音的基本频率？

Question

我想为Iphone制作一个吉他调音器应用程序.我的目标是找到吉他弦产生的声音的基本频率.我使用了Apple提供的aurioTouch样本中的一些代码来计算频谱,我找到了幅度最高的频率.它适用于纯声音(只有一个频率的声音),但对于来自吉他弦的声音,它会产生错误的结果.我读过这是因为吉他弦产生的泛音可能比基本弦更高.如何找到基频,以便它适用于吉他弦？C/C++/Obj-C中是否有用于声音分析(或信号处理)的开源库？

Answer 1

您可以使用信号的自相关,它是DFT幅度平方的逆变换.如果您以44100个样本/秒进行采样,那么82.4 Hz基波是大约535个样本,而1479.98 Hz是大约30个样本.寻找该范围内的峰值正滞后(例如从28到560).确保您的窗口至少是最长基本的两个周期,这里将是1070个样本.对于下一个2的幂,这是一个2048样本的缓冲区.为了获得更好的频率分辨率和更少的偏差估计,请使用更长的缓冲区,但不要太久以至于信号不再是近似静止的.这是Python中的一个例子:

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 

峰值滞后为538,即44100/538 = 81.97 Hz.第一次通过非循环DFT显示箱61处的基波,其为82.10 +/- 0.67Hz.第二遍使用的窗口长度为538*15 = 8070,因此DFT频率包括字符串的基本周期和谐波.这使得无偏的循环自相关能够以更少的谐波扩展来改进PSD估计(即,相关性可以周期性地环绕窗口).

编辑:更新以使用Welch的方法来估计自相关.重叠窗口可以补偿汉明窗口.我还计算了汉明窗口的锥形偏差以消除自相关.

编辑:添加了具有循环相关性的第二遍以清除功率谱密度.此过程使用3个不重叠的矩形窗口长度538*15 = 8070(足够短以至于几乎静止不动).循环相关的偏差是常数,而不是汉明窗的锥形偏差.