通过 Chirp Z 变换 (CZT) 的快速逆 FFT (IFFT)

Question

对于任意样本大小(样本不等于 2^N)，我已经能够使用 iOS Accelerate 的 FFT 函数(仅适用于等于 2^N 的样本)通过线性调频 Z 变换 (CZT) 实现 FFT ).

结果很好，与任意长度序列(信号)的 Matlab FFT 输出相匹配。我粘贴下面的代码。

我的下一个挑战是使用 iOS Accelerate 的 FFT 函数(仅适用于等于 2^N 的样本)来完成任意样本大小(样本不等于 2^N)的逆 FFT。

由于我的 CZT 现在可以完成任意长度的 FFT(见下文)，我希望逆 CZT (ICZT) 可以使用 iOS Accelerate 的 FFT 函数(仅适用于等于 2^N 的样本)完成任意长度的 IFFT。

有什么建议/指导吗？

// FFT IOS ACCELERATE FRAMEWORK (works only for 2^N samples)
import Accelerate

public func fft(x: [Double], y: [Double], type: String) -> ([Double], [Double]) {

    var real = [Double](x)

    var imaginary = [Double](y)

    var splitComplex = DSPDoubleSplitComplex(realp: &real, imagp: &imaginary)

    let length = vDSP_Length(floor(log2(Float(real.count))))

    let radix = FFTRadix(kFFTRadix2)

    let weights = vDSP_create_fftsetupD(length, radix)

    switch type.lowercased() {

    case ("fft"): // CASE FFT
        vDSP_fft_zipD(weights!, &splitComplex, 1, length, FFTDirection(FFT_FORWARD))
        vDSP_destroy_fftsetup(weights)

    case ("ifft"): // CASE INVERSE FFT
        vDSP_fft_zipD(weights!, &splitComplex, 1, length, FFTDirection(FFT_INVERSE))
        vDSP_destroy_fftsetup(weights)
        real = real.map({ $0 / Double(x.count) }) // Normalize IFFT by sample count
        imaginary = imaginary.map({ $0 / Double(x.count) }) // Normalize IFFT by sample count

    default: // DEFAULT CASE (FFT)
        vDSP_fft_zipD(weights!, &splitComplex, 1, length, FFTDirection(FFT_FORWARD))
        vDSP_destroy_fftsetup(weights)
    }

    return (real, imaginary)
}

// END FFT IOS ACCELERATE FRAMEWORK (works only for 2^N samples)

// DEFINE COMPLEX NUMBERS
struct Complex<T: FloatingPoint> {
    let real: T
    let imaginary: T
    static func +(lhs: Complex<T>, rhs: Complex<T>) -> Complex<T> {
        return Complex(real: lhs.real + rhs.real, imaginary: lhs.imaginary + rhs.imaginary)
    }

    static func -(lhs: Complex<T>, rhs: Complex<T>) -> Complex<T> {
        return Complex(real: lhs.real - rhs.real, imaginary: lhs.imaginary - rhs.imaginary)
    }

    static func *(lhs: Complex<T>, rhs: Complex<T>) -> Complex<T> {
        return Complex(real: lhs.real * rhs.real - lhs.imaginary * rhs.imaginary,
                       imaginary: lhs.imaginary * rhs.real + lhs.real * rhs.imaginary)
    }
}

extension Complex: CustomStringConvertible {
    var description: String {
        switch (real, imaginary) {
        case (_, 0):
            return "\(real)"
        case (0, _):
            return "\(imaginary)i"
        case (_, let b) where b < 0:
            return "\(real) - \(abs(imaginary))i"
        default:
            return "\(real) + \(imaginary)i"
        }
    }
}

// DEFINE COMPLEX NUMBERS

// DFT BASED ON CHIRP Z TRANSFORM (CZT)
public func dft(x: [Double]) -> ([Double], [Double]) {

    let m = x.count // number of samples

    var N: [Double] = Array(stride(from: Double(0), through: Double(m - 1), by: 1.0))

    N = N.map({ $0 + Double(m) })

    var NM: [Double] = Array(stride(from: Double(-(m - 1)), through: Double(m - 1), by: 1.0))

    NM = NM.map({ $0 + Double(m) })

    var M: [Double] = Array(stride(from: Double(0), through: Double(m - 1), by: 1.0))

    M = M.map({ $0 + Double(m) })

    let nfft = Int(pow(2, ceil(log2(Double(m + m - 1))))) // fft pad

    var p1: [Double] = Array(stride(from: Double(-(m - 1)), through: Double(m - 1), by: 1.0))

    p1 = (zip(p1, p1).map(*)).map({ $0 / Double(2) }) // W = WR + j*WI has to be raised to power p1

    var WR = [Double]()
    var WI = [Double]()

    for i in 0 ..< p1.count { // Use De Moivre's formula to raise to power p1
        WR.append(cos(p1[i] * 2.0 * M_PI / Double(m)))
        WI.append(sin(-p1[i] * 2.0 * M_PI / Double(m)))
    }

    var aaR = [Double]()
    var aaI = [Double]()

    for j in 0 ..< N.count {
        aaR.append(WR[Int(N[j] - 1)] * x[j])
        aaI.append(WI[Int(N[j] - 1)] * x[j])
    }

    let la = nfft - aaR.count

    let pad: [Double] = Array(repeating: 0, count: la) // 1st zero padding

    aaR += pad

    aaI += pad

    let (fgr, fgi) = fft(x: aaR, y: aaI, type: "fft") // 1st FFT

    var bbR = [Double]()
    var bbI = [Double]()

    for k in 0 ..< NM.count {
        bbR.append((WR[Int(NM[k] - 1)]) / (((WR[Int(NM[k] - 1)])) * ((WR[Int(NM[k] - 1)])) + ((WI[Int(NM[k] - 1)])) * ((WI[Int(NM[k] - 1)])))) // take reciprocal
        bbI.append(-(WI[Int(NM[k] - 1)]) / (((WR[Int(NM[k] - 1)])) * ((WR[Int(NM[k] - 1)])) + ((WI[Int(NM[k] - 1)])) * ((WI[Int(NM[k] - 1)])))) // take reciprocal
    }

    let lb = nfft - bbR.count

    let pad2: [Double] = Array(repeating: 0, count: lb) // 2nd zero padding

    bbR += pad2

    bbI += pad2

    let (fwr, fwi) = fft(x: bbR, y: bbI, type: "fft") // 2nd FFT

    let fg = zip(fgr, fgi).map { Complex<Double>(real: $0, imaginary: $1) } // complexN 1

    let fw = zip(fwr, fwi).map { Complex<Double>(real: $0, imaginary: $1) } // complexN 2

    let cc = zip(fg, fw).map { $0 * $1 } // multiply above 2 complex numbers fg * fw

    var ccR = cc.map { $0.real } // real part (vector) of complex multiply

    var ccI = cc.map { $0.imaginary } // imag part (vector) of complex multiply

    let lc = nfft - ccR.count

    let pad3: [Double] = Array(repeating: 0, count: lc) // 3rd zero padding

    ccR += pad3

    ccI += pad3

    let (ggr, ggi) = fft(x: ccR, y: ccI, type: "ifft") // 3rd FFT (IFFT)

    var GGr = [Double]()
    var GGi = [Double]()
    var W2r = [Double]()
    var W2i = [Double]()

    for v in 0 ..< M.count {
        GGr.append(ggr[Int(M[v] - 1)])
        GGi.append(ggi[Int(M[v] - 1)])
        W2r.append(WR[Int(M[v] - 1)])
        W2i.append(WI[Int(M[v] - 1)])
    }

    let ggg = zip(GGr, GGi).map { Complex<Double>(real: $0, imaginary: $1) }

    let www = zip(W2r, W2i).map { Complex<Double>(real: $0, imaginary: $1) }

    let y = zip(ggg, www).map { $0 * $1 }

    let yR = y.map { $0.real } // FFT real part (output vector)

    let yI = y.map { $0.imaginary } // FFT imag part (output vector)

    return (yR, yI)
}

// END DFT BASED ON CHIRP Z TRANSFORM (CZT)

// CHIRP DFT (CZT) TEST
let x: [Double] = [1, 2, 3, 4, 5] // arbitrary sample size
let (fftR, fftI) = dft(x: x)
print("DFT Real Part:", fftR)
print(" ")
print("DFT Imag Part:", fftI)

// Matches Matlab FFT Output
// DFT Real Part: [15.0, -2.5000000000000018, -2.5000000000000013, -2.4999999999999991, -2.499999999999996]
// DFT Imag Part: [-1.1102230246251565e-16, 3.4409548011779334, 0.81229924058226477, -0.81229924058226599, -3.4409548011779356]

// END CHIRP DFT (CZT) TEST

Answer 1

发表我的评论作为结束这个问题的答案—

如果您确定要使用 ICZT 作为 IFFT 的等价物，那么让您的 dft 函数接受一个 type: String 参数，就像您的 fft。当type为ifft时，你只需要翻转这里的符号:

WI.append(sin(-p1[i] * 2.0 * M_PI/Double(m)))

正向 FFT 为负，反向 FFT (IFFT) 为正。

---

这是我为演示 CZT 编写的一些 Octave/Matlab 代码:gist.github.com/fasiha/42a21405de92ea46f59e .该演示展示了如何使用 czt2 执行 fft。 czt2 的第三个参数(在代码中称为 w)是用于 FFT 的 exp(-2j * pi/Nup)。只需将它与 exp(+2j * pi/Nup) 结合即可获得 IFFT。

这就是翻转 WI 中的 sin 符号的作用。