我有java代码,它给出了来自实际输入的FFT输出。我需要执行MCLT。目前我有以下格式的FFT输出。我已经看到一些快速的MCLT算法(https://www.microsoft.com/en-us/research/wp-content/uploads/2016/02/tr-2005-02.pdf),用Matlab编码,但无法理解它。有人可以帮我写相应的java代码。
Java代码起点:
int dtLength = data.length/2;
double[] realPart = new double[dtLength];
double[] imagPart = new double[dtLength];
Matlab代码:
function X = fmclt(x)
% FMCLT - Compute MCLT of a vector via double-length FFT
%
% H. Malvar, September 2001 -- (c) 1998-2001 Microsoft Corp.
%
% Syntax: X = fmclt(x)
%
% Input: x : real-valued input vector of length 2*M
%
% Output: X : complex-valued MCLT coefficients, M subbands
% in Matlab, by default j = sqrt(-1)
% determine # of subbands, M
L = length(x);
M = L/2;
% normalized FFT of input
U = sqrt(1/(2*M)) * fft(x);
% compute modulation function
k = [0:M]';
c = W(8,2*k+1) .* W(4*M,k);
% modulate U into V
V = c .* U(1:M+1);
% compute MCLT coefficients
X = j * V(1:M) + V(2:M+1);
return;
% Local function: complex exponential
function w = W(M,r)
w = exp(-j*2*pi*r/M);
return;
答案 0 :(得分:1)
尽管这个问题对于SO来说有点边缘,但论文非常有趣,所以我决定投入一些时间来阅读它并尝试将Matlab
代码转换为Java
。结果如下:
import org.apache.commons.math3.complex.Complex;
public class MCLT
{
public static void main(String args[])
{
Complex[] x = new Complex[16];
for (int i = 1; i <= 16; ++i)
x[(i - 1)] = new Complex((double)i, 0.0d);
Complex[] result = fmclt(x);
for (int i = 0; i < result.length; ++i)
System.out.println(result[i]);
}
public static Complex[] fmclt(Complex[] x)
{
int L = x.length;
int M = L / 2;
double z = Math.sqrt(1.0d / (2.0d * M));
Complex[] F = fft(x);
Complex[] U = new Complex[F.length];
for (int i = 0; i < F.length; ++i)
U[i] = F[i].multiply(z);
double[] k = new double[(M + 1)];
for (int i = 0; i <= M; ++i)
k[i] = (double)i;
Complex[] c = new Complex[(M + 1)];
for (int i = 0; i <= M; ++i)
c[i] = W(8.0d, ((2.0d * k[i]) + 1.0d)).multiply(W((4.0d * M), k[i]));
Complex[][] V = new Complex[(M + 1)][];
for (int i = 0; i <= M; ++i)
{
V[i] = new Complex[(M + 1)];
for (int j = 0; j <= M; ++j)
V[i][j] = c[i].multiply(U[j]);
}
Complex[] V1 = new Complex[M];
for (int i = 0; i < M; ++i)
V1[i] = V[i][0];
Complex[] V2 = new Complex[M];
for (int i = 1; i <= M; ++i)
V2[(i - 1)] = V[i][0];
Complex b = new Complex(0.0d, 1.0d);
Complex[] result = new Complex[M];
for (int i = 0; i < M; ++i)
result[i] = b.multiply(V1[i]).add(V2[i]);
return result;
}
public static Complex[] fft(Complex[] x)
{
int n = x.length;
if (n == 1)
return new Complex[] { x[0] };
if ((n % 2) != 0)
throw new IllegalArgumentException("Invalid length.");
int nh = n / 2;
Complex[] even = new Complex[nh];
for (int i = 0; i < nh; ++i)
even[i] = x[(2 * i)];
Complex[] q = fft(even);
Complex[] odd = even;
for (int i = 0; i < nh; ++i)
odd[i] = x[((2 * i) + 1)];
Complex[] r = fft(odd);
Complex[] y = new Complex[n];
for (int i = 0; i < nh; ++i)
{
double kth = -2.0d * i * (Math.PI / n);
Complex wk = new Complex(Math.cos(kth), Math.sin(kth));
y[i] = q[i].add(wk.multiply(r[i]));
y[(i + nh)] = q[i].subtract(wk.multiply(r[i]));
}
return y;
}
public static Complex W(double M, double r)
{
Complex j = (new Complex(0.0d, 1.0d)).multiply(-1.0d);
double z = 2.0d * Math.PI * (r / M);
return j.multiply(z).exp();
}
}
在我看来,对于实部和虚部使用单独的双数组并不是一个好的设计选择,所以我决定将我的代码基于Complex类的Apache Commons类。
为了计算快速傅里叶变换,我决定使用一些现成的代码。我的fft
函数基于this implementation,它似乎非常可靠并且使用了前面提到的Complex
类。
使用相同的值向量,Matlab
和Java
代码都返回相同的输出。您可以通过在this website上复制粘贴代码来在线测试代码,但您还需要先安装Apache Commons
库才能成功运行它。单击位于底部的Add External Library (from Maven Repo)
按钮,然后在输入表单中插入以下参数:
<!-- https://mvnrepository.com/artifact/org.apache.commons/commons-math3 -->
<dependency>
<groupId>org.apache.commons</groupId>
<artifactId>commons-math3</artifactId>
<version>3.6.1</version>
</dependency>