我正在尝试绘制一个 wav 文件的频谱,但似乎频谱总是与时域信号匹配,代码如下。
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)))
我做错了什么吗?
最佳答案
如果你想要两个独立的立体声轨道到左右声道,然后分别为每个声道制作一个单独的图表,除非你像 Frank Zalkow 说的那样将轨道设置为单声道,否则读数会准确得多。这是将立体声轨道分成左右声道的方法:
"""
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()
希望对您有所帮助。
关于python - 使用python从wav文件中绘制fft,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27306474/