我是 DSP 的新手，試圖為f(0)音頻文件的每個分段幀計算基頻（ ）。F0估計的方法可以分為三類：基于信號時域的時間動態；基于頻率結構頻域，以及混合方法。大多數示例都是基于頻率結構頻域估計基頻，我正在尋找基于信號時域的時間動態。本文提供了一些信息，但我仍然不清楚如何在時域中計算它？https://gist.github.com/endolith/255291這是我發現的到目前為止使用的代碼：def freq_from_autocorr(sig, fs): """ Estimate frequency using autocorrelation """ # Calculate autocorrelation and throw away the negative lags corr = correlate(sig, sig, mode='full') corr = corr[len(corr)//2:] # Find the first low point d = diff(corr) start = nonzero(d > 0)[0][0] # Find the next peak after the low point (other than 0 lag). This bit is # not reliable for long signals, due to the desired peak occurring between # samples, and other peaks appearing higher. # Should use a weighting function to de-emphasize the peaks at longer lags. peak = argmax(corr[start:]) + start px, py = parabolic(corr, peak) return fs / px如何在時域進行估計？提前致謝！
1 回答
慕容3067478
TA貢獻1773條經驗 獲得超3個贊
這是一個正確的實現。不是很健壯，但肯定有效。為了驗證這一點，我們可以生成一個已知頻率的信號，看看我們會得到什么結果：
import numpy as np
from scipy.io import wavfile
from scipy.signal import correlate, fftconvolve
from scipy.interpolate import interp1d
fs = 44100
frequency = 440
length = 0.01 # in seconds
t = np.linspace(0, length, int(fs * length))
y = np.sin(frequency * 2 * np.pi * t)
def parabolic(f, x):
xv = 1/2. * (f[x-1] - f[x+1]) / (f[x-1] - 2 * f[x] + f[x+1]) + x
yv = f[x] - 1/4. * (f[x-1] - f[x+1]) * (xv - x)
return (xv, yv)
def freq_from_autocorr(sig, fs):
"""
Estimate frequency using autocorrelation
"""
corr = correlate(sig, sig, mode='full')
corr = corr[len(corr)//2:]
d = np.diff(corr)
start = np.nonzero(d > 0)[0][0]
peak = np.argmax(corr[start:]) + start
px, py = parabolic(corr, peak)
return fs / px
結果
運行freq_from_autocorr(y, fs)得到我們~442.014 Hz，大約 0.45% 的錯誤。
獎金 - 我們可以改進
我們可以通過稍微多一點的編碼使它更精確和健壯：
def indexes(y, thres=0.3, min_dist=1, thres_abs=False):
"""Peak detection routine borrowed from
https://bitbucket.org/lucashnegri/peakutils/src/master/peakutils/peak.py
"""
if isinstance(y, np.ndarray) and np.issubdtype(y.dtype, np.unsignedinteger):
raise ValueError("y must be signed")
if not thres_abs:
thres = thres * (np.max(y) - np.min(y)) + np.min(y)
min_dist = int(min_dist)
# compute first order difference
dy = np.diff(y)
# propagate left and right values successively to fill all plateau pixels (0-value)
zeros, = np.where(dy == 0)
# check if the signal is totally flat
if len(zeros) == len(y) - 1:
return np.array([])
if len(zeros):
# compute first order difference of zero indexes
zeros_diff = np.diff(zeros)
# check when zeros are not chained together
zeros_diff_not_one, = np.add(np.where(zeros_diff != 1), 1)
# make an array of the chained zero indexes
zero_plateaus = np.split(zeros, zeros_diff_not_one)
# fix if leftmost value in dy is zero
if zero_plateaus[0][0] == 0:
dy[zero_plateaus[0]] = dy[zero_plateaus[0][-1] + 1]
zero_plateaus.pop(0)
# fix if rightmost value of dy is zero
if len(zero_plateaus) and zero_plateaus[-1][-1] == len(dy) - 1:
dy[zero_plateaus[-1]] = dy[zero_plateaus[-1][0] - 1]
zero_plateaus.pop(-1)
# for each chain of zero indexes
for plateau in zero_plateaus:
median = np.median(plateau)
# set leftmost values to leftmost non zero values
dy[plateau[plateau < median]] = dy[plateau[0] - 1]
# set rightmost and middle values to rightmost non zero values
dy[plateau[plateau >= median]] = dy[plateau[-1] + 1]
# find the peaks by using the first order difference
peaks = np.where(
(np.hstack([dy, 0.0]) < 0.0)
& (np.hstack([0.0, dy]) > 0.0)
& (np.greater(y, thres))
)[0]
# handle multiple peaks, respecting the minimum distance
if peaks.size > 1 and min_dist > 1:
highest = peaks[np.argsort(y[peaks])][::-1]
rem = np.ones(y.size, dtype=bool)
rem[peaks] = False
for peak in highest:
if not rem[peak]:
sl = slice(max(0, peak - min_dist), peak + min_dist + 1)
rem[sl] = True
rem[peak] = False
peaks = np.arange(y.size)[~rem]
return peaks
def freq_from_autocorr_improved(signal, fs):
signal -= np.mean(signal) # Remove DC offset
corr = fftconvolve(signal, signal[::-1], mode='full')
corr = corr[len(corr)//2:]
# Find the first peak on the left
i_peak = indexes(corr, thres=0.8, min_dist=5)[0]
i_interp = parabolic(corr, i_peak)[0]
return fs / i_interp, corr, i_interp
運行freq_from_autocorr_improved(y, fs)產量~441.825 Hz，大約 0.41% 的誤差。這種方法在更復雜的情況下會表現得更好，并且需要兩倍的時間來計算。
通過更長的采樣時間（即設置length為例如 0.1 秒），我們將獲得更準確的結果。
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