swift - Swift 中的 DFT 结果与 MATLAB 中的不同

Question

import Cocoa
import Accelerate

let filePath = Bundle.main.path(forResource: "sinusoid", ofType: "txt")
let contentData = FileManager.default.contents(atPath: filePath!)
var content = NSString(data: contentData!, encoding: String.Encoding.utf8.rawValue) as? String

var idx = content?.characters.index(of: "\n")
idx = content?.index(after: idx!)

repeat {
    //let fromIndex = index(from: )
    content = content?.substring(from: idx!)
    idx = content?.characters.index(of: "\n")
    idx = content?.index(after: idx!)
} while content!.characters.contains("%")

let regex = try? NSRegularExpression(pattern: "[ ]+", options:[])

let delimiter = ","
var modifiedString = regex?.stringByReplacingMatches(in: content!, options: [], range: NSRange(location: 0, length: (content! as NSString).length), withTemplate: delimiter)

let lines = modifiedString?.components(separatedBy: "\n")

var s = [Double]()

for var line in lines! {
    if !line.isEmpty {
        let data = line.components(separatedBy: ",")
        s.append(Double(data[1])!)
    }
}

let length = vDSP_Length(pow(2, floor(log2(Float(s.count)))))
let L = Int(length)

// zrop or zop? 
// zrop covers real to complex, and zop covers complex
// length must be a power of 2 or specific multiples of powers of 2 if size is at least 4
let setup = vDSP_DFT_zrop_CreateSetupD(nil, length, vDSP_DFT_Direction.FORWARD)

var inputReal = UnsafeMutablePointer<Double>.allocate(capacity: L)
var inputImaginary = UnsafeMutablePointer<Double>.allocate(capacity: L)
var outputReal = UnsafeMutablePointer<Double>.allocate(capacity: L)
var outputImaginary = UnsafeMutablePointer<Double>.allocate(capacity: L)

for i in 0..<L {
    inputReal[i] = s[i]
    inputImaginary[i] = 0.0
}

vDSP_DFT_ExecuteD(setup!, inputReal, inputImaginary, outputReal, outputImaginary)

for i in 0..<L {
    print("\(outputReal[i]) + \(outputImaginary[i])i")
}

输入文件“sinusoid.txt”位于以下链接中 https://dpaste.de/M1VD

输入文件数据由频率分别为 50 和 120 的两个正弦波组成。Matlab 代码生成以下链接中给出的正确输出:

https://dpaste.de/2mdK

当对Matlab的结果进行缩放并获取幅度时，它正确地显示频率为50时的幅度为0.7，频率为120时的幅度为1。

clear all; close all; clc;
data = load('sinusoid.txt');
S = data(:,2);
Fs = 1000;
Y = fft(S);
L = length(S);
P2 = abs(Y/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
f = Fs*(0:(L/2))/L;
plot(f,P1)
title('Single-Sided Amplitude Spectrum of X(t)')
xlabel('f (Hz)')
ylabel('|P1(f)|')

与 Matlab 输出相比，Swift 代码输出完全不同且无法识别，无论应用什么缩放因子以及是否应用实数到复数或复数到复数转换:

https://dpaste.de/MUwB

有什么想法吗？

Answer 1

您的 2 FFT 的长度不同，因此结果当然不会匹配。您还将不同数量的数据传递给 2 个 FFT。

打印出 FFT 长度和输入数据向量以调试代码。在比较结果之前确保输入匹配。

此外，Apple 的 Accelerate/vDSP FFT 可以使用 2 的幂以外的长度(也允许使用 3 或 5 因子的长度)。

另外，请注意，Matlab 索引数组从 1 开始，而不是从 0 开始，这在 C 和 Swift 函数中更为常见。