问题描述
我有一个音频样本，采样率为8khz，约为14秒.我正在使用librosa从该音频文件中提取一些功能.
I have an audio sample of about 14 seconds in 8khz Sample Rate.Im using librosa to extract some features from this audio file.
y, sr = librosa.load(file_name)
stft = np.abs(librosa.stft(y, n_fft=n_fft))
# file_length = 14.650022675736961 #sec
# defaults
# n_fft =2048
# hop_length = 512 # win_length/4 = n_fft/4 = 512 (win_length = n_fft default)
#windowsTime = n_fft * Ts # (1/sr)
stft.shape
# (1025, 631)
Specshow:
librosa.display.specshow(stft, x_axis='time', y_axis='log')
[！[stft sr = 22050] [1]] [1]
[![stft sr = 22050][1]][1]
现在，我可以理解STFT的形状
Now,i can understand the shape of the STFT
631 time bins = are 4 * ( file_length / Ts * windowsTime) #overlapping
1025 frequency bins = Frames frequency gap sr/n_fft.
so there are 1025 frequencies in 0 to sr/2(Nyquest)
我不明白的是两种不同采样率的不同情节具有相同的比率.1-22050作为默认的librosa2-8khz作为采样率文件
what i cant understand is the different plot of two different sample rateswith same ratios.1 - 22050 as librosa default2 - 8khz as sampling rate file
y2, sr = librosa.load(file_name, sr=None)
n_fft2 =743 # (same ratio to get same visuals for comparsion)
hop_length = 186 # (1/4 n_fft by default)
stft2 = np.abs(librosa.stft(y2, n_fft=n_fft2))
因此，stft的威信会有所不同
so ofc the shappe of stft will be different
stft2.shape
# (372, 634)
[！[stft sr = 743] [2]] [2]
[![stft sr = 743][2]][2]
1.但是为什么绝对频率不一样呢?其相同的信号只是不被过采样，因此每个采样都是唯一的.我想念什么?是静态的y轴吗?
2.我无法理解时间段的值.我希望从前一点到文件结尾的第一个为跳数长度，第二个为windowTime时，帧数为bin.但是单位很奇怪?
我希望能够在特定的时间(帧)中提取特定Fbin的大小，或者希望能够对其中一些进行求和以得出时间范围的磁化强度.
i want to be able to extract the magnitude of a specific Fbin in a specific time (frame) or additionally be able to sum some of those to get the magnitue for time RANGE.
因此，如果我将stft [fBin的数量]取为1025 fBins的1行(stft [1025])并查看其内容，则stft [0]包含630点，对于每个频率而言，这恰好是630个时间点因此第1-1025帧中的每个帧都有相同的时间点.
Therefore, if i take stft[number of fBin] which is 1 row of 1025 fBins (stft[1025]) and look at it contents so stft[0] contains 630 points, which are exactly 630 time points for each frequency so each of the frames 1-1025 will have the same time points.
因此，如果我也采样了一个适合所有其他fbin的样本(相同时间点)，即stft [0]我将能够选择时间范围和fBin并获得特定的幅度:
so if i take one sample which suits all the other fbins as well ( same time points) which is stft[0]i would be able to choose time frame and fBin and get the spcific magnitude:
times = librosa.core.frames_to_time(stft2[0], sr=sr2, n_fft=n_fft2, hop_length=hop_length)
fft_bin = 6
time_idx = 10
print('freq (Hz)', freqs[fft_bin])
print('time (s)', times[time_idx])
print('amplitude', stft[fft_bin, time_idx])
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[1]: https://i.imgur.com/OeKzvrb.png
[2]: https://i.imgur.com/ALtba5F.png
推荐答案
问题1:
使用 specshow 时，您需要指定采样率:
You need to specify the sampling rate when using specshow:
librosa.display.specshow(stft, x_axis='time', y_axis='log', sr=sr)
否则将使用默认值(22,050 Hz)(请参阅文档).
Otherwise the default value (22,050 Hz) will be used (see docs).
问题2:
librosa.core.frames_to_time 不使用 stft [0] 作为参数，它是第一帧的频点.相反，它以帧数作为第一个参数.
librosa.core.frames_to_time does not take stft[0] as argument, which would be the frequency bins of the first frame. Instead, it takes number of frames as first argument.
想象一下，您有一个 sr = 10000 Hz的音频信号.然后，使用 n_fft = 2000 和 hop_length = 1000 在其上运行STFT.然后，您每跳获得一个 frame ，并且该跳的长度为0.1s，因为10000个样本对应于1s，而1000个样本(1个跃点)因此对应于0.1s.
Imagine you have an audio signal with sr=10000 Hz. Then you run an STFT over it using n_fft=2000 and hop_length=1000. Then you get one frame per hop and the hop is 0.1s long, because 10000 samples correspond to 1s and 1000 samples (1 hop) therefore correspond to 0.1s.
stft [0] 不是帧编号.相反，第一个 stft 的形状为(1 + n_fft/2，t)(请参阅此处).这意味着第一维是频点，第二维是帧号( t ).
stft[0] is not a frame number. Instead the first stft is of shape (1 + n_fft/2, t) (see here). This means the first dimension is the frequency bin and the second dimension is the frame number (t).
因此， stft 中的帧总数为 stft.shape [1] .要获取源音频的长度，可以执行以下操作:
The total number of frames in stft is therefore stft.shape[1].To get the length of the source audio, you could do:
time = librosa.core.frames_to_time(stft.shape[1], sr=sr, hop_length=hop_length, n_fft=n_fft)
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