AMR 文件的 FFT 计算中的问题
我正在尝试开发一个能够识别动物声音片段的应用程序。我正在做的是，我正在接收 AMR 记录并从中读取字节数组，然后通过 FFT 发送这些数据并相应地计算幅度。
AMR文件采样频率8 KHz（标准AMR为15秒）
FFT点数4096，输入8192个值
然后我通过amplitude=2 * FFT点值/8192计算幅度
所以我现在的目的是在与频率相关的频率处获得尖峰到最高振幅，问题是最高振幅的尖峰与同一动物的其他声音片段不一致。对于另一个声音片段，与最高振幅相关的频率发生变化。这是有原因的吗？对此的任何帮助和指导将不胜感激。提前致谢。
I’m trying to develop an application that is capable of identifying a sound clip of an animal. What I’m doing is that I’m taking in an AMR recording and reading the byte array from it and sending those data through FFT and calculate amplitudes accordingly.
AMR file sample frequency 8 KHz (Standard AMR of 15 seconds)
Number of FFT points 4096 for input of 8192 values
Then I calculate amplitude by amplitude=2 * FFT point value/8192
So my intention now is to get a spike at the frequency related to the highest amplitude, The issue is that the spike at the highest aplitude is not Consistant for the same animal's some other sound clip. For another sound clip the frequency related to the highest amplitude changes. Is there a reason for this?. Any help and guidance for this will be appreciated. Thanks in advance.
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您的文件的采样频率为 8KHz，但我认为人类平均听觉频率约为 20KHz，因此您确定您尊重样本的奈奎斯特频率（.wav 文件通常具有至少 48KHz 的采样率）？
奈奎斯特频率规定，如果要对给定信号进行采样，则必须使用至少是给定信号最大频率两倍的采样频率。
此外，同一动物可以并且将会发出不同的声音，因此两个不同样本的平均频率永远不会相同。您是否有考虑不同平均频率的容忍阈值？
your file has a sample frequency of 8KHz, but I think that the average human hearing frequency is of somewhat 20KHz, so are you sure that your are respecting the nyquist frequency of your samples (.wav files usually have a sample rate of at least 48KHz)?
The nyquist frequency states that if you are to sample a given signal you must use a sample frequency that is at least twice the maximum frequency the given signal.
Also, the same animal can and will make different sounds, so your average frequency will never be the same for two different samples. Do you have a tolerance threshold that accounts for different average frequencies?