如何应用低通滤波器,其截止频率从例如线性变化(或比线性曲线更一般)。沿时间从10000hz到200hz,具有numpy/scipy,可能没有其他库?

例子:

  • 在00:00,000,低通截止= 10000hz
  • at 00:05,000,低通截止= 5000hz
  • at 00:09,000,低通截止= 1000hz
  • 然后截止在10秒内保持在1000hz,然后截止降低到200hz

  • 这是一个简单的100hz低通的方法:
    from scipy.io import wavfile
    import numpy as np
    from scipy.signal import butter, lfilter
    
    sr, x = wavfile.read('test.wav')
    b, a = butter(2, 100.0 / sr, btype='low')  # Butterworth
    y = lfilter(b, a, x)
    wavfile.write('out.wav', sr, np.asarray(y, dtype=np.int16))
    

    但是如何使临界值变化呢?

    注意:我已经读过Applying time-variant filter in Python,但是答案很复杂(它通常适用于多种过滤器)。

    最佳答案

    一种比较简单的方法是使滤波器保持固定并改为调制信号时间。例如,如果信号时间快10倍,则10KHz低通将像标准时间中的1KHz低通一样起作用。
    为此,我们需要解决一个简单的ODE

    dy       1
    --  =  ----
    dt     f(y)
    
    此处t是实时调制时间y,而f是时间y所需的截止时间。
    原型(prototype)实现:
    from __future__ import division
    import numpy as np
    from scipy import integrate, interpolate
    from scipy.signal import butter, lfilter, spectrogram
    
    slack_l, slack = 0.1, 1
    cutoff = 50
    L = 25
    
    from scipy.io import wavfile
    sr, x = wavfile.read('capriccio.wav')
    x = x[:(L + slack) * sr, 0]
    x = x
    
    # sr = 44100
    # x = np.random.normal(size=((L + slack) * sr,))
    
    b, a = butter(2, 2 * cutoff / sr, btype='low')  # Butterworth
    
    # cutoff function
    def f(t):
        return (10000 - 1000 * np.clip(t, 0, 9) - 1000 * np.clip(t-19, 0, 0.8)) \
            / cutoff
    
    # and its reciprocal
    def fr(_, t):
        return cutoff / (10000 - 1000 * t.clip(0, 9) - 1000 * (t-19).clip(0, 0.8))
    
    # modulate time
    # calculate upper end of td first
    tdmax = integrate.quad(f, 0, L + slack_l, points=[9, 19, 19.8])[0]
    span = (0, tdmax)
    t = np.arange(x.size) / sr
    tdinfo = integrate.solve_ivp(fr, span, np.zeros((1,)),
                                 t_eval=np.arange(0, span[-1], 1 / sr),
                                 vectorized=True)
    td = tdinfo.y.ravel()
    # modulate signal
    xd = interpolate.interp1d(t, x)(td)
    # and linearly filter
    yd = lfilter(b, a, xd)
    # modulate signal back to linear time
    y = interpolate.interp1d(td, yd)(t[:-sr*slack])
    
    # check
    import pylab
    xa, ya, z = spectrogram(y, sr)
    pylab.pcolor(ya, xa, z, vmax=2**8, cmap='nipy_spectral')
    pylab.savefig('tst.png')
    
    wavfile.write('capriccio_vandalized.wav', sr, y.astype(np.int16))
    
    样本输出:
    python - 具有时变截止频率的低通滤波器,使用Python-LMLPHP
    BWV 826 Capriccio的前25秒频谱图,通过时弯实现时变低通。

    关于python - 具有时变截止频率的低通滤波器,使用Python,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/52403589/

    10-12 21:43