FFT 哪些频率在哪些 bin 中？

Question

我想看看某些频率，特别是 20 - 60hz 的低低音是如何出现在一段音频中的。我将音频作为字节数组，我将其转换为短裤数组，然后通过 (short[i]/(double)short.MaxValue, 0) 转换为复数。然后我将它从 Aforge 传递给 FFT。

音频为单声道，采样率为 44100。我知道我只能在 ^2 处通过 FFT。以 4096 为例。我不明白输出箱中的频率。

如果我从 44100 采样率的音频中获取 4096 个样本。这是否意味着我需要几毫秒的音频？还是只获取一些将出现的频率？

我将 FFT 的输出添加到数组中，我的理解是，当我采用 4096 时，bin 0 将包含 0*44100/4096 = 0hz，bin 1 将包含 1*44100/4096 = 10.7666015625hz 等等在。它是否正确？还是我在这里做一些根本错误的事情？

我的目标是将频率平均在 20 到 60 赫兹之间，因此对于低音非常低、重低音的歌曲，这个数字将高于低音非常少的柔和钢琴曲。

这是我的代码。

OpenFileDialog file = new OpenFileDialog();
file.ShowDialog();
WaveFileReader reader = new WaveFileReader(file.FileName);

byte[] data = new byte[reader.Length];
reader.Read(data, 0, data.Length);

samepleRate = reader.WaveFormat.SampleRate;
bitDepth = reader.WaveFormat.BitsPerSample;
channels = reader.WaveFormat.Channels;

Console.WriteLine("audio has " + channels + " channels, a sample rate of " + samepleRate + " and bitdepth of " + bitDepth + ".");


short[] shorts = data.Select(b => (short)b).ToArray();

int size = 4096;
int window = 44100 * 10;
int y = 0;
Complex[] complexData = new Complex[size];
for (int i = window; i < window + size; i++) 
{
    Complex tmp = new Complex(shorts[i]/(double)short.MaxValue, 0);

    complexData[y] = tmp;
    y++;

}




FourierTransform.FFT(complexData, FourierTransform.Direction.Forward);


double[] arr = new double[complexData.Length];
//print out sample of conversion
for (int i = 0; i < complexData.Length; i++)
{
    arr[i] = complexData[i].Magnitude;

}

Console.Write("complete, ");


return arr; 

编辑：从 DFT 改为 FFT

Answer 1

这是您的代码的修改版本。注意以“***”开头的 cmets。

OpenFileDialog file = new OpenFileDialog();
file.ShowDialog();
WaveFileReader reader = new WaveFileReader(file.FileName);

byte[] data = new byte[reader.Length];
reader.Read(data, 0, data.Length);

samepleRate = reader.WaveFormat.SampleRate;
bitDepth = reader.WaveFormat.BitsPerSample;
channels = reader.WaveFormat.Channels;

Console.WriteLine("audio has " + channels + " channels, a sample rate of " + samepleRate + " and bitdepth of " + bitDepth + ".");

// *** NAudio "thinks" in floats
float[] floats = new float[data.Length / sizeof(float)]
Buffer.BlockCopy(data, 0, floats, 0, data.Length);

int size = 4096;
// *** You don't have to fill the FFT buffer to get valid results.  More noisy & smaller "magnitudes", but better freq. res.
int inputSamples = samepleRate / 100; // 10ms... adjust as needed
int offset = samepleRate * 10 * channels;
int y = 0;
Complex[] complexData = new Complex[size];
// *** get a "scaling" curve to make both ends of sample region 0 but still allow full amplitude in the middle of the region.
float[] window = CalcWindowFunction(inputSamples);
for (int i = 0; i < inputSamples; i++)
{
    // *** "floats" is stored as LRLRLR interleaved data for stereo audio
    complexData[y] = new Complex(floats[i * channels + offset] * window[i], 0);
    y++;
}
// make sure the back portion of the buffer is set to all 0's
while (y < size)
{
    complexData[y] = new Complex(0, 0);
    y++;
}


// *** Consider using a DCT here instead...  It returns less "noisy" results
FourierTransform.FFT(complexData, FourierTransform.Direction.Forward);


double[] arr = new double[complexData.Length];
//print out sample of conversion
for (int i = 0; i < complexData.Length; i++)
{
    // *** I assume we don't care about phase???
    arr[i] = complexData[i].Magnitude;
}

Console.Write("complete, ");


return arr;

获得结果后，假设采样率为 44100 Hz 且大小 = 4096，则元素 2 - 4 应该是您要查找的值。有一种方法可以将它们转换为 dB，但我不记得了。

祝你好运！