在Python中使用LPC估计共振峰

sta*_*345 8 python matlab signal-processing numpy scipy

我是信号处理的新手(对此事,还有numpy,scipy和matlab)。我正在尝试通过修改以下matlab代码来估计LPC中的元音共振峰:

http://www.mathworks.com/help/signal/ug/formant-estimation-with-lpc-coefficients.html

到目前为止,这是我的代码:

#!/usr/bin/env python
import sys
import numpy
import wave
import math
from scipy.signal import lfilter, hamming
from scikits.talkbox import lpc

"""
Estimate formants using LPC.
"""

def get_formants(file_path):

    # Read from file.
    spf = wave.open(file_path, 'r') # http://www.linguistics.ucla.edu/people/hayes/103/Charts/VChart/ae.wav

    # Get file as numpy array.
    x = spf.readframes(-1)
    x = numpy.fromstring(x, 'Int16')

    # Get Hamming window.
    N = len(x)
    w = numpy.hamming(N)

    # Apply window and high pass filter.
    x1 = x * w
    x1 = lfilter([1., -0.63], 1, x1)

    # Get LPC.
    A, e, k = lpc(x1, 8)

    # Get roots.
    rts = numpy.roots(A)
    rts = [r for r in rts if numpy.imag(r) >= 0]

    # Get angles.
    angz = numpy.arctan2(numpy.imag(rts), numpy.real(rts))

    # Get frequencies.
    Fs = spf.getframerate()
    frqs = sorted(angz * (Fs / (2 * math.pi)))

    return frqs

print get_formants(sys.argv[1])
Run Code Online (Sandbox Code Playgroud)

使用此文件作为输入,我的脚本返回以下列表:

[682.18960189917243, 1886.3054773107765, 3518.8326108511073, 6524.8112723782951]
Run Code Online (Sandbox Code Playgroud)

我什至没有走到最后一步,因为它们不按照带宽过滤频率,因为列表中的频率不正确。根据Praat所说,我应该得到这样的信息(这是元音中间的共振峰列表):

Time_s     F1_Hz        F2_Hz         F3_Hz         F4_Hz
0.164969   731.914588   1737.980346   2115.510104   3191.775838 
Run Code Online (Sandbox Code Playgroud)

我究竟做错了什么?

非常感谢

更新:

我改变了这个

x1 = lfilter([1., -0.63], 1, x1)

x1 = lfilter([1], [1., 0.63], x1)

按照沃伦·韦克瑟(Warren Weckesser)的建议,现在越来越

[631.44354635609318, 1815.8629524985781, 3421.8288991389031, 6667.5030877036006]

由于F3的关闭,我觉得自己缺少了一些东西。

更新2:

我意识到,由于采样频率的差异,order被传递到的状态scikits.talkbox.lpc已关闭。更改为:

Fs = spf.getframerate()
ncoeff = 2 + Fs / 1000
A, e, k = lpc(x1, ncoeff)
Run Code Online (Sandbox Code Playgroud)

现在我得到:

[257.86573127888488, 774.59006835496086, 1769.4624576002402, 2386.7093679399809, 3282.387975973973, 4413.0428174593926, 6060.8150432549655, 6503.3090645887842, 7266.5069407315023]

更接近Praat的估计!

sta*_*345 3

该问题与传递给 lpc 函数的顺序有关。2 + fs / 1000其中fs是采样频率,根据以下经验法则:

http://www.phon.ucl.ac.uk/courses/spsci/matlab/lect10.html