Swift 对于vDSP_fft_zropD，压缩结果似乎是错误的。这是一个错误还是我做错了什么

我有下面的代码，它使用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)

我相信这是正确的。以上是一个复杂到复杂的公式。基本上检查下面的代码，这是一个实到复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)

---

所以在我看来，条目0中的压缩输出应该是120，-8，但事实并非如此。有人对这里发生的事情有什么建议吗？我正在研究这些函数，因此很容易出现错误，但足够多的输出是正确的，因此我不理解压缩的条目。

基于非常仓促的阅读，我看到两个可能的错误，这两个错误都在中讨论

具体而言，您的输入数据布局不正确（请参阅“真实FFT的数据打包”）；所有偶数项应在x的“实”部分，奇数项应在“虚”部分。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)

比较好，但相差两倍；如果我们看一下“傅里叶变换的缩放”，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循傅立叶变换的教科书公式，必须相应地进行缩放。以下各节指定了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随附的公式中也明确说明了比例因子

实正向变换：RF_imp=RF_math*2

这就是二的额外因子的来源

 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)

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)