'大家好，我是一名电气工程专业的学生，刚刚接触python，我有一个信号处理课程的作业。作业是记录我自己，执行fft，将其通过BPF，并计算平均频率和频率的标准差。接下来，我想用这个结果来分离出我声音的二次和三次谐波，但我在网上找不到答案。
我录下了自己的声音，并执行了FFT，将其传递到巴特沃思带通滤波器，然后计算平均频率，得到了一个合理的答案。当我尝试计算标准差时，结果比平均频率大70倍左右，对于我尝试过的任何星座，这都是没有意义的。我尝试将文件的幅度归一化为最大值，以及频率到其最大值，我还尝试将频率（f）乘以幅度的平方，然后除以幅度平方和的列表，还尝试使用np.std（）函数。我在这里添加代码，希望有人能启发我的眼睛，感谢前面的代码。

from scipy.fftpack import fft
from scipy.io import wavfile as wav
import numpy as np
import matplotlib.pyplot as plt
import scipy.signal as sig

def mean_freq(data, sample_rate):
    x = fft(data) # List of magnitudes in frequency domain
    f = np.linspace(0.0, sample_rate / 2.0, len(x) // 2)
    abs_x = abs(x)[0:int(len(x) / 2)]
    x_squared = abs_x ** 2
    numerator = sum(f * x_squared)
    denominator = sum(x_squared)
    average_frequency = round(numerator / denominator)
    stdv = np.sqrt((np.sum(f - average_frequency )** 2 ) / len(x)) ### NEED TO CHECK
    return average_frequency, round(stdv)

# Read the soundfile
fs, signal =wav.read('RECORDEDFILE.wav')
# Design characteristics
n = 3 # Filter's order
nyquist = fs/2
t = np.arange(0, 21, 1/fs) # Creating time variable


# --------------------------------------------------------------------------------------------------
# Design BPF of first harmony, BW=70 Hz
lcf = 85/nyquist # Low cut off (Normalized)
hcf = 155/nyquist # High cut off
b, a = sig.butter(n, [lcf , hcf], btype='bandpass', analog=False, output='ba') # Obtain the filter's coefficients

# Activate the filter
filtered_signal = sig.filtfilt(b, a, signal, axis=0)

# First harmony data
mean1, stdv1 = mean_freq(filtered_signal, fs)
print(f'The average frequency of the first harmony is: {mean1} [Hz]')
print(f'The standard deviation of the first harmony is: {stdv1} [Hz]')