我用这个代码(基于苹果的audioRouch样品):如何获得iOS在FFT频率上加速FFT结果?
void FFTHelper::ComputeFFT(Float32* inAudioData, Float32* outFFTData)
{
if (inAudioData == NULL || outFFTData == NULL) return;
// Generate a split complex vector from the real data
vDSP_ctoz((COMPLEX *)inAudioData, 2, &mDspSplitComplex, 1, mFFTLength);
// Take the fft and scale appropriately
vDSP_fft_zrip(mSpectrumAnalysis, &mDspSplitComplex, 1, mLog2N, kFFTDirection_Forward);
vDSP_vsmul(mDspSplitComplex.realp, 1, &mFFTNormFactor, mDspSplitComplex.realp, 1, mFFTLength);
vDSP_vsmul(mDspSplitComplex.imagp, 1, &mFFTNormFactor, mDspSplitComplex.imagp, 1, mFFTLength);
// Zero out the nyquist value
mDspSplitComplex.imagp[0] = 0.0;
// Complex vector magnitudes squared; single precision.
// Calculates the squared magnitudes of complex vector A.
vDSP_zvmags(&mDspSplitComplex, 1, outFFTData, 1, mFFTLength);
}
为了计算上尽可能简单的FFT - 1Hz的正弦波(1个单位移):
Float32 waveFreq = 1.0;
int samplesCount = 1024;
Float32 samplesPerSecond = 1000; //sample rate
Float32 dt = 1/samplesPerSecond;
Float32 sd = M_PI * 2.0 * waveFreq;
FFTHelper *mFFTHelper = new FFTHelper(samplesCount);
Float32 NyquistMaxFreq = samplesPerSecond/2.0;
Float32 fftDataSize = samplesCount/2.0;
Float32 *sinusoidOriginal = (Float32 *)malloc(sizeof(Float32) * samplesCount);
Float32 *outFFTData = (Float32 *)malloc(sizeof(Float32) * fftDataSize);
// 2. Generate sin samples:
for (int i = 0; i < samplesCount; i++) {
Float32 x = dt * i;
sinusoidOriginal[i] = sin(sd * x) + 1;
[originalPlot addVector2D:GLVector2DMake(x, sinusoidOriginal[i])];
}
mFFTHelper->ComputeFFT(sinusoidOriginal, outFFTData);
for (int i = 0; i < fftDataSize; i++) {
Float32 hz = ((Float32)i/(Float32)fftDataSize) * NyquistMaxFreq;
GLfloat mag = outFFTData[i];
[fftPlot addVector2D:GLVector2DMake(hz, 0)];
[fftPlot addVector2D:GLVector2DMake(hz, mag)];
}
结果我得到的是:
黑线是来自FTT的绘图仪结果,水平定位在它们的频率上。 DC值(左起第一条黑线)看起来正常,正确表示y = sin(x)+1垂直偏移。
但为什么第二条黑线表示窦性方程中存在的唯一频率,没有大小= 1,并且不完全保持在1Hz?
任何人都可以指向我的vDSP函数将FFT结果转换为输入信号的幅度单位吗?