Question

我有对5秒钟的wav文件执行FFT的代码。我不太擅长Python，因此我写了非常基本的代码，该文件拆分了wav文件并每秒计算FFT。还有其他更方便的方法吗？

由于范围部分，我也不确定它们是否显示每个频率及其相关振幅。我将信号分成5个部分，但也可能将频率分为5个部分。

以数字结尾的变量名是我的加法运算，通常我每个变量名只有一个来计算wav上的整个FFT。任何建议都会很棒。（由于偏见，我先抹掉了第二个字，您应该检查图）

#!/usr/bin/env python
# -*- coding: utf-8 -*-

from __future__ import print_function
import scipy.io.wavfile as wavfile
import scipy
import scipy.fftpack
import numpy as np
from matplotlib import pyplot as plt

fs_rate, signal = wavfile.read("db1.wav")
#print ("Frequency sampling", fs_rate)
l_audio = len(signal.shape)
#print ("Channels", l_audio)
if l_audio == 2:
    signal = signal.sum(axis=1) / 2

signal2 = signal + 480000000    
N = signal2.shape[0]
#print ("Complete Samplings N", N)
secs = N / float(fs_rate)
#print ("secs", secs)
Ts = 1.0/fs_rate # sampling interval in time
#print ("Timestep between samples Ts", Ts)
t = scipy.arange(0, secs, Ts) # time vector as scipy arange field / numpy.ndarray

#FFT1 = abs(scipy.fft(signal2[0:44100]))
FFT2 = abs(scipy.fft(signal2[44100:88200]))
FFT3 = abs(scipy.fft(signal2[88200:132300]))
FFT4 = abs(scipy.fft(signal2[132300:176400]))
FFT5 = abs(scipy.fft(signal2[176400:220500]))

#FFT_side1 = FFT1[range(N//20)] # one side FFT range
FFT_side2 = FFT2[range(N//20)] # one side FFT range
FFT_side3 = FFT3[range(N//20)] # one side FFT range
FFT_side4 = FFT4[range(N//20)] # one side FFT range
FFT_side5 = FFT5[range(N//20)] # one side FFT range

#freqs1 = scipy.fftpack.fftfreq(signal2[0:44100].size, t[1]-t[0])
freqs2 = scipy.fftpack.fftfreq(signal2[44100:88200].size, t[1]-t[0])
freqs3 = scipy.fftpack.fftfreq(signal2[88200:132300].size, t[1]-t[0])
freqs4 = scipy.fftpack.fftfreq(signal2[132300:176400].size, t[1]-t[0])
freqs5 = scipy.fftpack.fftfreq(signal2[176400:220500].size, t[1]-t[0])

#fft_freqs = np.array(freqs)

#freqs_side1 = freqs1[range(N//20)] # one side frequency range
freqs_side2 = freqs2[range(N//20)] # one side frequency range
freqs_side3 = freqs3[range(N//20)] # one side frequency range
freqs_side4 = freqs4[range(N//20)] # one side frequency range
freqs_side5 = freqs5[range(N//20)] # one side frequency range


#fft_freqs_side = np.array(freqs_side)

#abs(FFT_side1)
abs(FFT_side2)
abs(FFT_side3)
abs(FFT_side4)
abs(FFT_side5)

for a in range(60):
    #FFT_side1[a] = 0
    FFT_side2[a] = 0
    FFT_side3[a] = 0
    FFT_side4[a] = 0
    FFT_side5[a] = 0

plt.subplot(611)
p1 = plt.plot(t, signal2, "r") # plotting the signal
plt.xlabel('Time')
plt.ylabel('Amplitude')

# plt.subplot(612)
# p3 = plt.plot(freqs_side1, FFT_side1, "b") # plotting the positive fft spectrum
# plt.xlabel('Frequency (Hz)')
# plt.ylabel('Amplitude')

plt.subplot(613)
p3 = plt.plot(freqs_side2, FFT_side2, "g") # plotting the positive fft spectrum
plt.xlabel('Frequency (Hz)')
plt.ylabel('Amplitude')

plt.subplot(614)
p3 = plt.plot(freqs_side3, FFT_side3, "g") # plotting the positive fft spectrum
plt.xlabel('Frequency (Hz)')
plt.ylabel('Amplitude')

plt.subplot(615)
p3 = plt.plot(freqs_side4, FFT_side4, "g") # plotting the positive fft spectrum
plt.xlabel('Frequency (Hz)')
plt.ylabel('Amplitude')

plt.subplot(616)
p3 = plt.plot(freqs_side5, FFT_side5, "g") # plotting the positive fft spectrum
plt.xlabel('Frequency (Hz)')
plt.ylabel('Amplitude')

plt.show()

Answer 1

通过使用STFT的标准命令，您可以在单个通道中完成所需的操作

Answer 2

以前没有在Python中使用FFT玩过，借此机会玩了一下。我录制了一些音频-在Audacity中，因为它还具有不错的FFT功能，因此我有很好的参考来检查我是否得到了“正确”的答案

import numpy as np
from numpy import fft
from scipy.io import wavfile
import matplotlib.pyplot as plt

# recording of me whistling at ~1.2 and 1khz
fs_rate, signal = wavfile.read("whistle-1170_970hz.wav")

# convert stereo to mono
signal = signal.mean(axis=1)

# generate time in seconds
t = np.arange(signal.shape[0]) / fs_rate

# plot everything
plt.plot(t, signal);

# create some plots
fig, axs = plt.subplots(
    2, sharex=True, sharey=True,
    figsize=(8, 6))

for ax, i in zip(axs, range(0, signal.shape[0], fs_rate)):
    # pull out sample for this second
    ss = signal[i:i + fs_rate]

    # generate FFT and frequencies
    sp = fft.fft(ss)
    freq = fft.fftfreq(len(ss), 1 / fs_rate)

    # plot the first few components
    ax.plot(freq[:2000], np.abs(sp.real[:2000]));

这对我来说在正确的位置给出了峰的峰值-底部的曲线。在重复代码和重复索引操作之后，我遇到了麻烦，但认为您可能做对了！

Answer 3

这是一个Python示例，该示例接受任何WAV并将其按样本转换为FFT。可以将样本（time_period）配置为介于0.05秒和10秒之间。

输出显示原始声音（用于最终样本），FFT输出（在存储桶中），输出的一维图像和二维图像表示。

Python 3代码：

#!/usr/bin/env python
# -*- coding: utf-8 -*-

from __future__ import print_function
import scipy.io.wavfile as wavfile
import scipy
import scipy.fftpack
from scipy.signal import argrelextrema
import numpy as np
from matplotlib import pyplot as plt

filename = "audio/pysynth_anthem.wav"
filename = "audio/pysynth_chopin.wav"
filename = "audio/menuet.wav"
filename = "audio/bach_violin.wav"

# ==============================================

time_period = 0.1 # FFT time period (in seconds). Can comfortably process time frames from 0.05 seconds - 10 seconds

# ==============================================

fs_rate, signal_original = wavfile.read(filename)
total_time = int(np.floor(len(signal_original)/fs_rate))
sample_range = np.arange(0,total_time,time_period)
total_samples = len(sample_range)

print ("Frequency sampling", fs_rate)
print ("total time: ", total_time)
print ("sample time period: ", time_period)
print ("total samples: ", total_samples)

output_array = []
for i in sample_range:

    print ("Processing: %d / %d (%d%%)" % (i/time_period + 1, total_samples, (i/time_period + 1)*100/total_samples))

    sample_start = int(i*fs_rate)
    sample_end = int((i+time_period)*fs_rate)
    signal = signal_original[sample_start:sample_end]

    l_audio = len(signal.shape)
    #print ("Channels", l_audio)

    if l_audio == 2:
        signal = signal.sum(axis=1) / 2
    N = signal.shape[0]
    #print ("Complete Samplings N", N)

    secs = N / float(fs_rate)
    # print ("secs", secs)
    Ts = 1.0/fs_rate # sampling interval in time
    #print ("Timestep between samples Ts", Ts)

    t = scipy.arange(0, secs, Ts) # time vector as scipy arange field / numpy.ndarray

    FFT = abs(scipy.fft(signal))
    FFT_side = FFT[range(int(N/2))] # one side FFT range
    freqs = scipy.fftpack.fftfreq(signal.size, t[1]-t[0])
    fft_freqs = np.array(freqs)
    freqs_side = freqs[range(int(N/2))] # one side frequency range
    fft_freqs_side = np.array(freqs_side)

    # Reduce to 0-5000 Hz
    bucket_size = 5
    buckets = 16

    FFT_side = FFT_side[0:bucket_size*buckets]
    fft_freqs_side = fft_freqs_side[0:bucket_size*buckets]

    # Combine frequencies into buckets
    FFT_side = np.array([int(sum(FFT_side[current: current+bucket_size])) for current in range(0, len(FFT_side), bucket_size)])
    fft_freqs_side = np.array([int(sum(fft_freqs_side[current: current+bucket_size])) for current in range(0, len(fft_freqs_side), bucket_size)])

    # FFT_side: Normalize (0-1)
    max_value = max(FFT_side)
    if (max_value != 0):
        FFT_side_norm = FFT_side / max_value

    # Append to output array
    output_array.append(FFT_side_norm)

# ============================================

# Plotting

plt.figure(figsize=(8,10))

plt.subplot(411)
plt.plot(t, signal, "g") # plotting the signal
plt.xlabel('Time')
plt.ylabel('Amplitude')

plt.subplot(412)
diff = np.diff(fft_freqs_side)
widths = np.hstack([diff, diff[-1]])
plt.bar(fft_freqs_side, abs(FFT_side_norm), width=widths) # plotting the positive fft spectrum
plt.xticks(fft_freqs_side, fft_freqs_side, rotation='vertical')
plt.xlabel('Frequency (Hz)')
plt.ylabel('Count single-sided')

FFT_side_norm_line = FFT_side_norm.copy()
FFT_side_norm_line.resize( (1,buckets) )

plt.subplot(413)
plt.imshow(FFT_side_norm_line)
plt.axis('off')
plt.title('Image Representation (1D)')

width_img = int(np.sqrt(buckets))
height_img = int(np.ceil(buckets / int(np.sqrt(buckets))))
FFT_side_norm_rect = FFT_side_norm.copy()
FFT_side_norm_rect.resize( (width_img,height_img) )

plt.subplot(414)
plt.imshow(FFT_side_norm_rect)
plt.axis('off')
plt.title('Image Representation (2D): %d x %d' % (width_img,height_img))

plt.show()

# =======================================================

使用Python在wav文件上每秒执行FFT

3 个答案: