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Swift vDSP rFFT与Python np.fft.rfft()结果不同

pythonswiftnumpyfftvdsp
4

我正在尝试在iOS中实现实时FFT，为此我正在使用Accelerate框架。这里是我的Swift代码。

class FFT {

private var fftSetup: FFTSetup?
private var log2n: Float?
private var length: Int?

func initialize(count: Int){
    length = count
    log2n = log2(Float(length!))
    self.fftSetup = vDSP_create_fftsetup(vDSP_Length(log2n!), FFTRadix(kFFTRadix2))!
}

func computeFFT(input: [Float]) -> ([Float], [Float]) {
    
    var real = input
    var imag = [Float](repeating: 0.0, count: input.count)
    var splitComplexBuffer = DSPSplitComplex(realp: &real, imagp: &imag)

    let halfLength = (input.count/2) + 1
    real = [Float](repeating: 0.0, count: halfLength)
    imag = [Float](repeating: 0.0, count: halfLength)

    // input is alternated across the real and imaginary arrays of the DSPSplitComplex structure
    splitComplexBuffer = DSPSplitComplex(fromInputArray: input, realParts: &real, imaginaryParts: &imag)

    // even though there are 2 real and 2 imaginary output elements, we still need to ask the fft to process 4 input samples
    vDSP_fft_zrip(fftSetup!, &splitComplexBuffer, 1, vDSP_Length(log2n!), FFTDirection(FFT_FORWARD))

    // zrip results are 2x the standard FFT and need to be scaled
    var scaleFactor = Float(1.0/2.0)
    vDSP_vsmul(splitComplexBuffer.realp, 1, &scaleFactor, splitComplexBuffer.realp, 1, vDSP_Length(halfLength))
    vDSP_vsmul(splitComplexBuffer.imagp, 1, &scaleFactor, splitComplexBuffer.imagp, 1, vDSP_Length(halfLength))
    
    return (real, imag)
    
}

func computeIFFT(real: [Float], imag: [Float]) -> [Float]{
    
    var real = [Float](real)
    var imag = [Float](imag)
    
    var result : [Float] = [Float](repeating: 0.0, count: length!)
    var resultAsComplex : UnsafeMutablePointer<DSPComplex>? = nil

    result.withUnsafeMutableBytes {
        resultAsComplex = $0.baseAddress?.bindMemory(to: DSPComplex.self, capacity: 512)
    }
    
    var splitComplexBuffer = DSPSplitComplex(realp: &real, imagp: &imag)
    
    vDSP_fft_zrip(fftSetup!, &splitComplexBuffer, 1, vDSP_Length(log2n!), FFTDirection(FFT_INVERSE));
    
    vDSP_ztoc(&splitComplexBuffer, 1, resultAsComplex!, 2, vDSP_Length(length! / 2));
    //
    //// Neither the forward nor inverse FFT does any scaling. Here we compensate for that.
    var scale : Float = 1.0/Float(length!);
    var copyOfResult = result;
    vDSP_vsmul(&result, 1, &scale, &copyOfResult, 1, vDSP_Length(length!));
    result = copyOfResult
    
    return result
}

func deinitialize(){
    vDSP_destroy_fftsetup(fftSetup)
}

}

这是Python代码，用于计算rFFT和irFFT

# calculate fft of input block
    in_block_fft = np.fft.rfft(np.squeeze(in_buffer)).astype("complex64")

# apply mask and calculate the ifft
    estimated_block = np.fft.irfft(in_block_fft * out_mask)

问题:

Swift 如果我对512帧计算rFFT，并对rFFT的结果应用irFFT，我会得到相同的原始数组。

Python 同样适用于Python，如果我进行rFFT和irFFT，我会得到相同的原始数组。

问题 问题出现在我比较Swift rFFT和Python rFFT的结果时。它们的结果在小数值上有所不同。有时实部相同，但虚部完全不同。

我尝试了Python中的不同框架，如Numpy、SciPy和TensorFlow，它们的结果完全相同（小数部分略有不同）。但是，当我在iOS上使用上面的Swift代码对相同的输入计算rfft时，结果不同。

如果有经验的加速框架和FFT知识方面的人能帮助我解决这个问题，那将非常有帮助。我对FFT的了解有限。

2个回答
1

有两点需要注意：首先，你对DSPSplitComplex(realp: &real, imagp: &imag)的调用会创建临时指针，而fromInputArray初始化程序已经被弃用。请查看Data Packing for Fourier Transforms以了解最佳实践。

其次，bindMemory容量是元素数量。以下代码行创建了512个复杂元素，但应该（如果我理解正确的话）创建256个复杂元素来表示512个实数元素。

        result.withUnsafeMutableBytes {
            resultAsComplex = $0.baseAddress?.bindMemory(to: DSPComplex.self, capacity: 512)
        }

西蒙

0
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