Question

我正在尝试从8分钟长的海豚声学波中形成频谱图，并将其另存为图像文件。目标是利用以下论文中提到的降噪和提取算法：https://pdfs.semanticscholar.org/9cbf/d5b23600d4976f27ea454329f183b1fe6166.pdf

基本思想是，海豚咔嗒声几乎总是表示为频谱图上的垂直宽带尖峰，而口哨声则沿对角线弯曲。该算法查找峰值强度的像素，然后查看相邻像素。如果像素邻域是垂直的，则它们将被衰减，如果不是，则它们将被维护。

此代码源自http://www.frank-zalkow.de/en/code-snippets/create-audio-spectrograms-with-python.html?ckattempt=1，因为我喜欢在plotstft def中易于调整binsize因子。

代码创建了一个频谱图，可以放大并找到足够清晰的信号，但是当我查看形成的.png时，缩放时的分辨率不足。口哨的长度通常为0.4-1.3秒，而每次点击的时间约为0.1秒。因此，我需要能够放大到足以看到适合这些长度的时间的分辨率。

此外，我正在尝试找出如何过滤所有> = 5000hz和<= 16000 hz的频率，但是我不太确定要在代码中进行哪些调整。

我已经将fig1.savefig（'Spectrogram.png'，dpi = 500）调整为增加dpi。我在1000 dpi时遇到内存错误。

我也将def plotstft（audiopath，binsize = 2xx12，plotpath = None，colormap =“ Greys”）中的binsize增加到2xx12，当我使用图中的缩放功能时，它提供了很好的分辨率。

用于过滤频率的AS。我一直在尝试找出哪个数组包含频率本身，以便可以从数组中删除超出限制的频率，但是我运气不太好。

import numpy as np
from matplotlib import pyplot as plt
import scipy.io.wavfile as wav
from numpy.lib import stride_tricks


def stft(sig, frameSize, overlapFac=0.5, window=np.hanning):
    win = window(frameSize)
    hopSize = int(frameSize - np.floor(overlapFac * frameSize))

    # zeros at beginning (thus center of 1st window should be for sample nr. 0)   
    samples = np.append(np.zeros(int(np.floor(frameSize/2.0))), sig)
    # cols for windowing
    cols = np.ceil( (len(samples) - frameSize) / float(hopSize)) + 1
    # zeros at end (thus samples can be fully covered by frames)
    samples = np.append(samples, np.zeros(frameSize))

    frames = stride_tricks.as_strided(samples, shape=(int(cols), frameSize), strides=(samples.strides[0]*hopSize, samples.strides[0])).copy()
    frames *= win

    return np.fft.rfft(frames)    


def logscale_spec(spec, sr=44100, factor=20.):
    timebins, freqbins = np.shape(spec)
    scale = np.linspace(0, 1, freqbins) ** factor
    scale *= (freqbins-1)/max(scale)
    scale = np.unique(np.round(scale))

    # create spectrogram with new freq bins
    newspec = np.complex128(np.zeros([timebins, len(scale)]))
    for i in range(0, len(scale)):        
        if i == len(scale)-1:
            newspec[:,i] = np.sum(spec[:,int(scale[i]):], axis=1)
        else:        
            newspec[:,i] = np.sum(spec[:,int(scale[i]):int(scale[i+1])], axis=1)

    # list center freq of bins
    allfreqs = np.abs(np.fft.fftfreq(freqbins*2, 1./sr)[:freqbins+1])
    freqs = []
    fmin = 5000
    fmax = 16000
    for i in range(0, len(scale)):
        if i == len(scale)-1:
            freqs += [np.mean(allfreqs[int(scale[i]):])]
        else:
            freqs += [np.mean(allfreqs[int(scale[i]):int(scale[i+1])])]


    return newspec, freqs


def plotstft(audiopath, binsize=2**12, plotpath=None, colormap="Greys"):
    samplerate, samples = wav.read(audiopath)
    s = stft(samples, binsize)

    sshow, freq = logscale_spec(s, factor=1.0, sr=samplerate)

    ims = 20.*np.log10(np.abs(sshow)/10e-6) # amplitude to decibel

    timebins, freqbins = np.shape(ims)

    print("timebins: ", timebins)
    print("freqbins: ", freqbins)

    plt.figure(figsize=(15, 7.5))
    plt.imshow(np.transpose(ims), origin="lower", aspect="auto", cmap=colormap, interpolation="none")
    plt.colorbar()

    plt.xlabel("time (s)")
    plt.ylabel("frequency (hz)")
    plt.xlim([0, timebins-1])
    plt.ylim([0, freqbins])

    xlocs = np.float32(np.linspace(0, timebins-1, 5))
    plt.xticks(xlocs, ["%.02f" % l for l in ((xlocs*len(samples)/timebins)+(0.5*binsize))/samplerate])
    ylocs = np.int16(np.round(np.linspace(0, freqbins-1, 10)))
    plt.yticks(ylocs, ["%.02f" % freq[i] for i in ylocs])

    if plotpath:
        plt.savefig(plotpath, bbox_inches="tight")
    else:
        fig1 = plt.gcf()
        plt.show()
        fig1.savefig('Spectrogram.png', dpi=500)

    plt.clf()

    return ims

ims = plotstft('8min.wav')

输出高分辨率的光谱图图像，其像素分辨率与放大后的像素分辨率相同

如何将高分辨率频谱图以.png格式输出，以进行基于频谱图的降噪和信号提取

0 个答案: