如果你做一个谷歌搜索C＃FFT你会发现可能的代码示例。这一个看起来相当不错，我发现here。有意思的是，如果需要的话，您可以在每秒钟将0的数据用作表格FFT的输入，这些表格需要复数。
请务必确保您拥有2个采样计数功率的数据包。
/// <summary>
/// Compute the forward or inverse Fourier Transform of data, with data
/// containing complex valued data as alternating real and imaginary
/// parts. The length must be a power of 2. This method caches values
/// and should be slightly faster on than the FFT method for repeated uses.
/// It is also slightly more accurate. Data is transformed in place.
/// </summary>
/// <param name="data">The complex data stored as alternating real
/// and imaginary parts</param>
/// <param name="forward">true for a forward transform, false for
/// inverse transform</param>
public void TableFFT(double[] data, bool forward)
{
var n = data.Length;
// checks n is a power of 2 in 2's complement format
if ((n & (n - 1)) != 0)
throw new ArgumentException(
"data length " + n + " in FFT is not a power of 2"
);
n /= 2; // n is the number of samples
Reverse(data, n); // bit index data reversal
// make table if needed
if ((cosTable == null) || (cosTable.Length != n))
Initialize(n);
// do transform: so single point transforms, then doubles, etc.
double sign = forward ? B : -B;
var mmax = 1;
var tptr = 0;
while (n > mmax)
{
var istep = 2 * mmax;
for (var m = 0; m < istep; m += 2)
{
var wr = cosTable[tptr];
var wi = sign * sinTable[tptr++];
for (var k = m; k < 2 * n; k += 2 * istep)
{
var j = k + istep;
var tempr = wr * data[j] - wi * data[j + 1];
var tempi = wi * data[j] + wr * data[j + 1];
data[j] = data[k] - tempr;
data[j + 1] = data[k + 1] - tempi;
data[k] = data[k] + tempr;
data[k + 1] = data[k + 1] + tempi;
}
}
mmax = istep;
}
// perform data scaling as needed
Scale(data, n, forward);
}
/// <summary>
/// Compute the forward or inverse Fourier Transform of data, with
/// data containing real valued data only. The output is complex
/// valued after the first two entries, stored in alternating real
/// and imaginary parts. The first two returned entries are the real
/// parts of the first and last value from the conjugate symmetric
/// output, which are necessarily real. The length must be a power
/// of 2.
/// </summary>
/// <param name="data">The complex data stored as alternating real
/// and imaginary parts</param>
/// <param name="forward">true for a forward transform, false for
/// inverse transform</param>
public void RealFFT(double[] data, bool forward)
{
var n = data.Length; // # of real inputs, 1/2 the complex length
// checks n is a power of 2 in 2's complement format
if ((n & (n - 1)) != 0)
throw new ArgumentException(
"data length " + n + " in FFT is not a power of 2"
);
var sign = -1.0; // assume inverse FFT, this controls how algebra below works
if (forward)
{ // do packed FFT. This can be changed to FFT to save memory
TableFFT(data, true);
sign = 1.0;
// scaling - divide by scaling for N/2, then mult by scaling for N
if (A != 1)
{
var scale = Math.Pow(2.0, (A - 1)/2.0);
for (var i = 0; i < data.Length; ++i)
data[i] *= scale;
}
}
var theta = B * sign * 2 * Math.PI/n;
var wpr = Math.Cos(theta);
var wpi = Math.Sin(theta);
var wjr = wpr;
var wji = wpi;
for (var j = 1; j <= n/4; ++j)
{
var k = n/2 - j;
var tkr = data[2 * k]; // real and imaginary parts of t_k = t_(n/2 - j)
var tki = data[2 * k + 1];
var tjr = data[2 * j]; // real and imaginary parts of t_j
var tji = data[2 * j + 1];
var a = (tjr - tkr) * wji;
var b = (tji + tki) * wjr;
var c = (tjr - tkr) * wjr;
var d = (tji + tki) * wji;
var e = (tjr + tkr);
var f = (tji - tki);
// compute entry y[j]
data[2 * j] = 0.5 * (e + sign * (a + b));
data[2 * j + 1] = 0.5 * (f + sign * (d - c));
// compute entry y[k]
data[2 * k] = 0.5 * (e - sign * (b + a));
data[2 * k + 1] = 0.5 * (sign * (d - c) - f);
var temp = wjr;
// todo - allow more accurate version here? make option?
wjr = wjr * wpr - wji * wpi;
wji = temp * wpi + wji * wpr;
}
if (forward)
{
// compute final y0 and y_{N/2}, store in data[0], data[1]
var temp = data[0];
data[0] += data[1];
data[1] = temp - data[1];
}
else
{
var temp = data[0]; // unpack the y0 and y_{N/2}, then invert FFT
data[0] = 0.5 * (temp + data[1]);
data[1] = 0.5 * (temp - data[1]);
// do packed inverse (table based) FFT. This can be changed to regular inverse FFT to save memory
TableFFT(data, false);
// scaling - divide by scaling for N, then mult by scaling for N/2
//if (A != -1) // todo - off by factor of 2? this works, but something seems weird
{
var scale = Math.Pow(2.0, -(A + 1)/2.0)*2;
for (var i = 0; i < data.Length; ++i)
data[i] *= scale;
}
}
}
/// <summary>
/// Determine how scaling works on the forward and inverse transforms.
/// For size N=2^n transforms, the forward transform gets divided by
/// N^((1-a)/2) and the inverse gets divided by N^((1+a)/2). Common
/// values for (A,B) are
/// (0, 1) - default
/// (-1, 1) - data processing
/// (1,-1) - signal processing
/// Usual values for A are 1, 0, or -1
/// </summary>
public int A { get; set; }
/// <summary>
/// Determine how phase works on the forward and inverse transforms.
/// For size N=2^n transforms, the forward transform uses an
/// exp(B*2*pi/N) term and the inverse uses an exp(-B*2*pi/N) term.
/// Common values for (A,B) are
/// (0, 1) - default
/// (-1, 1) - data processing
/// (1,-1) - signal processing
/// Abs(B) should be relatively prime to N.
/// Setting B=-1 effectively corresponds to conjugating both input and
/// output data.
/// Usual values for B are 1 or -1.
/// </summary>
public int B { get; set; }
/// <summary>
/// Scale data using n samples for forward and inverse transforms as needed
/// </summary>
/// <param name="data"></param>
/// <param name="n"></param>
/// <param name="forward"></param>
void Scale(double[] data, int n, bool forward)
{
// forward scaling if needed
if ((forward) && (A != 1))
{
var scale = Math.Pow(n, (A - 1)/2.0);
for (var i = 0; i < data.Length; ++i)
data[i] *= scale;
}
// inverse scaling if needed
if ((!forward) && (A != -1))
{
var scale = Math.Pow(n, -(A + 1)/2.0);
for (var i = 0; i < data.Length; ++i)
data[i] *= scale;
}
}
你知道为什么每秒钟的数字是0吗？那看起来很奇怪... – thalm
林不知道。它对我来说也有点奇怪。也许如果我把我用来处理我的pcm数据的代码，你可以为我指出 –
你增加我两个，并写入数据[我]，所以你只写入每个第二个领域。你需要增加一个。例如i ++并使用BitConverter.ToInt16（wave，i * 2）读出字节。我还想知道为什么当你读取数据时你没有考虑到PCM头的偏移？也许'我'应该从44开始？ – thalm