從使用python

Question

我試圖繪製wav文件的頻譜，但它似乎是頻譜始終是時域信號相匹配，用下面的代碼wav文件繪製FFT。

import matplotlib.pyplot as plt 
import numpy as np 


def plot(data): 
    plt.plot(data, color='steelblue') 
    plt.figure() 
    plt.show() 

rate, wav_data = wavfile.read("audio_self/on/on.wav") 
plot(wav_data) 
plot(np.abs(np.fft.fft(wav_data))) 

我做錯了什麼？

Answer 1

如果你想兩個獨立的立體聲音軌，以左，右聲道，然後把每一個單獨的圖形，它會除非你把賽道單像弗蘭克Zalkow說是很多更準確的讀數的。這是如何將立體聲音軌分成左，右聲道：

""" 
Plot 
""" 
#Plots a stereo .wav file 
#Decibels on the y-axis 
#Frequency Hz on the x-axis 

import matplotlib.pyplot as plt 
import numpy as np 

from pylab import* 
from scipy.io import wavfile 


def plot(file_name): 

    sampFreq, snd = wavfile.read(file_name) 

    snd = snd/(2.**15) #convert sound array to float pt. values 

    s1 = snd[:,0] #left channel 

    s2 = snd[:,1] #right channel 

    n = len(s1) 
    p = fft(s1) # take the fourier transform of left channel 

    m = len(s2) 
    p2 = fft(s2) # take the fourier transform of right channel 

    nUniquePts = ceil((n+1)/2.0) 
    p = p[0:nUniquePts] 
    p = abs(p) 

    mUniquePts = ceil((m+1)/2.0) 
    p2 = p2[0:mUniquePts] 
    p2 = abs(p2) 

''' 
Left Channel 
''' 
    p = p/float(n) # scale by the number of points so that 
      # the magnitude does not depend on the length 
      # of the signal or on its sampling frequency 
    p = p**2 # square it to get the power 




# multiply by two (see technical document for details) 
# odd nfft excludes Nyquist point 
    if n % 2 > 0: # we've got odd number of points fft 
     p[1:len(p)] = p[1:len(p)] * 2 
    else: 
     p[1:len(p) -1] = p[1:len(p) - 1] * 2 # we've got even number of points fft 

    freqArray = arange(0, nUniquePts, 1.0) * (sampFreq/n); 
    plt.plot(freqArray/1000, 10*log10(p), color='k') 
    plt.xlabel('LeftChannel_Frequency (kHz)') 
    plt.ylabel('LeftChannel_Power (dB)') 
    plt.show() 

''' 
Right Channel 
''' 
    p2 = p2/float(m) # scale by the number of points so that 
      # the magnitude does not depend on the length 
      # of the signal or on its sampling frequency 
    p2 = p2**2 # square it to get the power 




# multiply by two (see technical document for details) 
# odd nfft excludes Nyquist point 
    if m % 2 > 0: # we've got odd number of points fft 
     p2[1:len(p2)] = p2[1:len(p2)] * 2 
    else: 
     p2[1:len(p2) -1] = p2[1:len(p2) - 1] * 2 # we've got even number of points fft 

    freqArray2 = arange(0, mUniquePts, 1.0) * (sampFreq/m); 
    plt.plot(freqArray2/1000, 10*log10(p2), color='k') 
    plt.xlabel('RightChannel_Frequency (kHz)') 
    plt.ylabel('RightChannel_Power (dB)') 
    plt.show() 

我希望這有助於。