绘制 .wav 文件的体积时间图

Question

我正在尝试获取 .wav 文件的体积时间图。首先，我通过 android 将声音（患者呼气）录制为 .wav 文件，但是当我在 MATLAB 中读取此 .wav 文件时，它具有负值。负值是什么意思？其次，MATLAB 专家，请您检查一下下面的代码是否与我的 cmets 中编写的代码相同？还有另一个问题。 Y = fft(WindowArray); p = abs(Y).^2; 我利用了从 fft 返回的值的力量...正确吗？这一步的目标是什么？？

[data, fs] =  wavread('newF2');
% read exhalation audio wav  file (1 channel, mono)
% frequency is 44100 HZ
% windows of 0.1 s and overlap of 0.05 seconds
WINDOW_SIZE = fs*0.1; %4410 = fs*0.1
array_size = length(data); % array size of data
numOfPeaks = (array_size/(WINDOW_SIZE/2)) - 1;
step = floor(WINDOW_SIZE/2); %step size used in loop
transformed = data;
start =1;
k = 1;
t = 1;
g = 1;
o = 1;
% performing fft on each window and finding the peak of windows 
while(((start+WINDOW_SIZE)-1)<=array_size) 
    j=1;
    i =start;
    while(j<=WINDOW_SIZE)
        WindowArray(j) = transformed(i);
        j = j+1;
        i = i +1;
    end
    Y = fft(WindowArray);
    p = abs(Y).^2; %power
      [a, b] = max(abs(Y)); % find max a and its indices b
      [m, i] = max(p); %the maximum of the power m and its indices i
      maximum(g) = m;
      index(t) = i;
      power(o) = a;
      indexP(g) = b;
      start = start + step;
      k = k+1;
      t = t+1;
      g = g+1;
      o=o+1;  
  end
% low pass filter 
% filtering noise: ignor frequencies that are less than 5% of maximum frequency
for u=1:length(maximum)
    M = max(maximum); %highest value in the array
    Accept = 0.05* M;
    if(maximum(u) > Accept)
        maximum = maximum(u:length(maximum));
        break;
    end
end
% preparing the time of the graph, 
% Location of the Peak flow rates are estimated
TotalTime = (numOfPeaks * 0.1);
time1 = [0:0.1:TotalTime];
if(length(maximum) > ceil(numOfPeaks));
maximum = maximum(1:ceil(numOfPeaks)); 
end
time = time1(1:length(maximum));
% plotting frequency-time graph
figure(1);
plot(time, maximum);
ylabel('Frequency');
xlabel('Time (in seconds)');
% plotting volume-time graph
figure(2);
plot(time, cumsum(maximum)); % integration over time to get volume 
ylabel('Volume');
xlabel('Time (in seconds)');

Answer 1

（我只回答我理解的部分）

默认情况下，Matlab 将音频波标准化为 - 1...1 范围。如果需要整数数据，请使用 native 选项。

Answer 2

首先，在您的代码中它应该是p = abs(Y)**2，这是对从 FFT 返回的值求平方的正确方法。您采用 FFT 返回值的绝对值的原因是因为这些数字是具有实部和虚部的复数，因此虚数的绝对值（或模数）是该数字的大小。获得权力的目标可能是潜在地获得整体幅度值的 RMS 值（均方根），但您也可能有其他想法。当您说音量时间时，我假设您想要分贝，因此请尝试以下操作：

def plot_signal(file_name):

sampFreq, snd = wavfile.read(file_name)

snd = snd / (2.**15) #Convert sound array to floating point values 
                     #Floating point values range from -1 to 1

s1 = snd[:,0] #left channel

s2 = snd[:,1] #right channel

timeArray = arange(0, len(snd), 1)
timeArray = timeArray / sampFreq
timeArray = timeArray * 1000  #scale to milliseconds

timeArray2 = arange(0, len(snd), 1)
timeArray2 = timeArray2 / sampFreq
timeArray2 = timeArray2 * 1000  #scale to milliseconds

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

m = len(s2) 
p2 = fft(s2)

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

plt.plot(timeArray, 10*log10(p), color='k')
plt.xlabel('Time (ms)')
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


plt.plot(timeArray2, 10*log10(p2), color='k')
plt.xlabel('Time (ms)')
plt.ylabel('RightChannel_Power (dB)')
plt.show()

我希望这会有所帮助。