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我有以下代码使用 Swift 和 Accelerate 在坡道上进行 DFT

import Foundation
import Accelerate
let N = 16
var xdtar = UnsafeMutablePointer<Double>.allocate(capacity: N)
var xdtai = UnsafeMutablePointer<Double>.allocate(capacity: N)
xdtar.initialize(to: 0.0, count: N )
xdtai.initialize(to: 0.0, count: N )
var x = DSPDoubleSplitComplex(realp: xdtar, imagp: xdtai)
var ydtar = UnsafeMutablePointer<Double>.allocate(capacity: N)
var ydtai = UnsafeMutablePointer<Double>.allocate(capacity: N)
ydtar.initialize(to: 0.0, count: N )
ydtai.initialize(to: 0.0, count: N )
var y = DSPDoubleSplitComplex(realp: ydtar, imagp: ydtai)
for i in 0..<N {
    xdtar[i] = Double(i)
}
let setup = vDSP_DFT_zop_CreateSetupD(nil, vDSP_Length(N),
       vDSP_DFT_Direction.FORWARD)
vDSP_DFT_ExecuteD(setup!, x.realp, x.imagp, y.realp, y.imagp)
vDSP_DFT_DestroySetupD(setup)
for i in 0..<N {
    print(String(format: "%2d \t-> in \t(%5.4f, %5.4fi)\t out \t(%5.4f, %5.4fi)",
                i, xdtar[i], xdtai[i], ydtar[i], ydtai[i]))
}
xdtar.deinitialize(count: N); xdtar.deallocate(capacity: N)
xdtai.deinitialize(count: N); xdtai.deallocate(capacity: N)
ydtar.deinitialize(count: N); ydtar.deallocate(capacity: N)
ydtai.deinitialize(count: N); ydtai.deallocate(capacity: N)

输出是

  0     -> in   (0.0000, 0.0000i)    out    (120.0000, 0.0000i)
  1     -> in   (1.0000, 0.0000i)    out    (-8.0000, 40.2187i)
  2     -> in   (2.0000, 0.0000i)    out    (-8.0000, 19.3137i)
  3     -> in   (3.0000, 0.0000i)    out    (-8.0000, 11.9728i)
  4     -> in   (4.0000, 0.0000i)    out    (-8.0000, 8.0000i)
  5     -> in   (5.0000, 0.0000i)    out    (-8.0000, 5.3454i)
  6     -> in   (6.0000, 0.0000i)    out    (-8.0000, 3.3137i)
  7     -> in   (7.0000, 0.0000i)    out    (-8.0000, 1.5913i)
  8     -> in   (8.0000, 0.0000i)    out    (-8.0000, 0.0000i)
  9     -> in   (9.0000, 0.0000i)    out    (-8.0000, -1.5913i)
 10     -> in   (10.0000, 0.0000i)   out    (-8.0000, -3.3137i)
 11     -> in   (11.0000, 0.0000i)   out    (-8.0000, -5.3454i)
 12     -> in   (12.0000, 0.0000i)   out    (-8.0000, -8.0000i)
 13     -> in   (13.0000, 0.0000i)   out    (-8.0000, -11.9728i)
 14     -> in   (14.0000, 0.0000i)   out    (-8.0000, -19.3137i)
 15     -> in   (15.0000, 0.0000i)   out    (-8.0000, -40.2187i)

我认为这是正确的。以上是复杂到复杂的公式。基本上检查下面的代码，这是一个真正到复杂的 FFT。

import Foundation
import Accelerate
let N = 16
let log2N = vDSP_Length(4)
var xdtar = UnsafeMutablePointer<Double>.allocate(capacity: N)
xdtar.initialize(to: 0.0, count: N )
var x = DSPDoubleSplitComplex(realp: xdtar, imagp: xdtar + N/2)
var ydtar = UnsafeMutablePointer<Double>.allocate(capacity: N/2)
var ydtai = UnsafeMutablePointer<Double>.allocate(capacity: N/2)
ydtar.initialize(to: 0.0, count: N )
ydtai.initialize(to: 0.0, count: N )
var y = DSPDoubleSplitComplex(realp: ydtar, imagp: ydtai)
for i in 0..<N {
    xdtar[i] = Double(i)
}
let setup = vDSP_create_fftsetupD(log2N, Int32(2))
vDSP_fft_zropD(setup!, &x, vDSP_Stride(1), &y, vDSP_Stride(1), log2N, Int32(1))
vDSP_destroy_fftsetupD(setup)
for i in 0..<N/2 {
    print(String(format: "%2d \t-> in \t(%5.4f, %5.4fi)\t out \t(%5.4f, %5.4fi)",
                 i, x.realp[i], x.imagp[i], ydtar[i], ydtai[i]))
}
xdtar.deinitialize(count: N); xdtar.deallocate(capacity: N)
ydtar.deinitialize(count: N/2); ydtar.deallocate(capacity: N/2)
ydtai.deinitialize(count: N/2); ydtai.deallocate(capacity: N/2)

它的输出是

 0  -> in   (0.0000, 8.0000i)    out    (240.0000, -128.0000i)
 1  -> in   (1.0000, 9.0000i)    out    (-8.0000, 40.2187i)
 2  -> in   (2.0000, 10.0000i)   out    (-8.0000, 19.3137i)
 3  -> in   (3.0000, 11.0000i)   out    (-8.0000, 11.9728i)
 4  -> in   (4.0000, 12.0000i)   out    (-8.0000, 8.0000i)
 5  -> in   (5.0000, 13.0000i)   out    (-8.0000, 5.3454i)
 6  -> in   (6.0000, 14.0000i)   out    (-8.0000, 3.3137i)
 7  -> in   (7.0000, 15.0000i)   out    (-8.0000, 1.5913i)

所以在我看来，条目 0 中的打包输出应该是 120，-8，但事实并非如此。有人对这里发生的事情有任何建议吗？我正在探索这些功能，因此很容易出错，但足够多的输出是正确的，以至于我不理解打包的条目。

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1 回答 1

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基于非常仓促的阅读，我看到了两个可能的错误，这两个错误都在文档中进行了讨论。

具体来说，您的输入数据布局不正确（请参阅“Real FFT 的数据打包”）；所有偶数项都应在 x 的“实部”部分中，奇数项应在“虚部”部分中。修复后，我得到：

0 -> in (0.0000, 1.0000i)    out (240.0000, -16.0000i)
1 -> in (2.0000, 3.0000i)    out (-16.0000, 80.4374i)
2 -> in (4.0000, 5.0000i)    out (-16.0000, 38.6274i)
3 -> in (6.0000, 7.0000i)    out (-16.0000, 23.9457i)
4 -> in (8.0000, 9.0000i)    out (-16.0000, 16.0000i)
5 -> in (10.0000, 11.0000i)  out (-16.0000, 10.6909i)
6 -> in (12.0000, 13.0000i)  out (-16.0000, 6.6274i)
7 -> in (14.0000, 15.0000i)  out (-16.0000, 3.1826i)

更好，但差了两倍；如果我们查看“Scaling for Fourier Transforms”，我们会看到以下注释：

为了提供最佳的执行速度，vDSP 库的函数并不总是严格遵守傅里叶变换的教科书公式，必须相应地进行缩放。以下部分指定了 vDSP 库实现的每种类型的傅里叶变换的缩放比例。参考章节中函数定义附带的公式中也明确说明了比例因子。

...

真正的前向变换：RF_imp = RF_math * 2

这就是额外的二分之一的来源。

于 2017-07-09T02:39:35.830 回答