混淆测试fftw3 - 泊松方程2d测试 [英] Confusion testing fftw3 - poisson equation 2d test
问题描述
我无法解释/理解以下现象：
要测试fftw3我使用2d poisson测试用例：
laplacian（f ，y））= -g（x，y）与周期边界条件。在应用傅立叶变换到方程式后，我们得到：F（kx，ky）=如果我取g（x，y）= sin（x）+ sin（y），则G（kx，ky）/（kx 2 + ky 2）（1）
< （x，y）\in [0,2 \pi]我立即有f（x，y）= g（x，y）
我试图用fft获取：
我用g进行正向傅里叶变换计算g
最后，我用后向傅里叶变换计算f（不用忘记归一化1 /（nx *）），我们可以计算f的傅立叶变换。（b）
（例如，幅度为更糟糕的是，如果我尝试g（x，y）= sin（x）* sin（y），则N = 256是N = 512时获得的振幅的两倍）
< ，该曲线甚至没有相同的解的形式。
（注意，我必须改变方程; i在这种情况下除以2的拉普拉斯算子：（1）变为F（kx，ky）= 2 * G（kx，ky）/（kx 2 + ky 2）
这是代码：
/ *
* fftw test - double precision
* /
#include< iostream>
#include< stdio.h>
#include< stdlib.h>
#include< math.h>
#include< fftw3.h>
using namespace std;
int main（）
{
int N = 128;
int i，j;
double pi = 3.14159265359;
double * X，* Y;
X =（double *）malloc（N * sizeof（double））;
Y =（double *）malloc（N * sizeof（double））;
fftw_complex * out1，* in2，* out2，* in1;
fftw_plan p1，p2;
double L = 2. * pi;
double dx = L /（N-1）;
in1 =（fftw_complex *）fftw_malloc（sizeof（fftw_complex）*（N * N））;
out2 =（fftw_complex *）fftw_malloc（sizeof（fftw_complex）*（N * N））;
out1 =（fftw_complex *）fftw_malloc（sizeof（fftw_complex）*（N * N））;
in2 =（fftw_complex *）fftw_malloc（sizeof（fftw_complex）*（N * N））;
p1 = fftw_plan_dft_2d（N，N，in1，out1，FFTW_FORWARD，FFTW_MEASURE）;
p2 = fftw_plan_dft_2d（N，N，in2，out2，FFTW_BACKWARD，FFTW_MEASURE）;
for（i = 0; i
for（j = 0; j
in1 [i * N + j] [0] = sin（X [i]）+ sin（Y [j] // row major ordering
// in1 [i * N + j] [0] = sin（X [i]）* sin（Y [j]）; //第二个测试用例
in1 [i * N + j] [1] = 0;
}
}
fftw_execute（p1）; // FFT forward
for（i = 0; i
in2 [i * N + j] [1] = out1 [i * N + j] [1] /（i * i + j * j + 1e-16）。
// in2 [i * N + j] [0] = 2 * out1 [i * N + j] [0] /（i * i + j * j + 1e-16） //第二测试用例
// in2 [i * N + j] [1] = 2 * out1 [i * N + j] [1] /（i * i + j * j + 1e- ;
}
}
fftw_execute（p2）; // FFT backward
//检查计算结果
double erl1 = 0 .;
for（i = 0; i
cout<< i<<<< j<< sin（X [i]）+ sin（Y [j]）<< out2 [i * N + j] [0] / N / N < endl; //>输出
}
}
cout<< erl1<< endl; // L1 error
fftw_destroy_plan（p1）;
fftw_destroy_plan（p2）;
fftw_free（out1）;
fftw_free（out2）;
fftw_free（in1）;
fftw_free（in2）;
return 0;
}
我在我的代码中找不到任何错误上周的fftw3库）和我没有看到一个问题的数学或者，但我不认为这是fft的错。因此我的困境。
任何帮助解决这个难题都会非常感激。
注意：
执行：./ a.out> output
并且我使用gnuplot绘制曲线：
（在gnuplot中）splotoutputu 1：2：4对于计算解）
这里有一些小修改：
-
您需要考虑所有小频率，包括负频率！索引
i 对应于频率2PI i / N，但也对应于频率2PI ）/ N。在傅里叶空间中，数组的结束与开始一样重要！在我们的例子中，我们保持最小的频率：对于数组的前半部分，它是2PI i / N，在后半部分是2PI（iN）/ N。 p> -
当然，如Paul所说，
N-1应该是N在double dx = L /（N-1）;=>c $ c>N-1不对应于连续的周期性信号。 -
缩放...我是凭经验做的
我获得的结果更接近预期结果，对于这两种情况。这是代码：
/ *
* fftw test - double precision
* /
#include< iostream>
#include< stdio.h>
#include< stdlib.h>
#include< math.h>
#include< fftw3.h>
using namespace std;
int main（）
{
int N = 128;
int i，j;
double pi = 3.14159265359;
double * X，* Y;
X =（double *）malloc（N * sizeof（double））;
Y =（double *）malloc（N * sizeof（double））;
fftw_complex * output1，* in2，* out2，* in1;
fftw_plan p1，p2;
double L = 2. * pi;
double dx = L /（N）;
in1 =（fftw_complex *）fftw_malloc（sizeof（fftw_complex）*（N * N））;
out2 =（fftw_complex *）fftw_malloc（sizeof（fftw_complex）*（N * N））;
out1 =（fftw_complex *）fftw_malloc（sizeof（fftw_complex）*（N * N））;
in2 =（fftw_complex *）fftw_malloc（sizeof（fftw_complex）*（N * N））;
p1 = fftw_plan_dft_2d（N，N，in1，out1，FFTW_FORWARD，FFTW_MEASURE）;
p2 = fftw_plan_dft_2d（N，N，in2，out2，FFTW_BACKWARD，FFTW_MEASURE）;
for（i = 0; i
for（j = 0; j
in1 [i * N + j] [0] = sin（X [i]）+ sin（Y [j] // row major ordering
// in1 [i * N + j] [0] = sin（X [i]）* sin（Y [j]）; //第二个测试用例
in1 [i * N + j] [1] = 0;
}
}
fftw_execute（p1）; // FFT forward
for（i = 0; i
double fact = 0;
in2 [i * N + j] [0] = 0;
in2 [i * N + j] [1] = 0;
if（2 * i
} else {
fact =（（double）（N-i）*（N-i））;
}
if（2 * j
} else {
fact + =（（double）（N-j）*（N-j））;
}
if（fact！= 0）{
in2 [i * N + j] [0] = out1 [i * N + j] [0]
in2 [i * N + j] [1] = out1 [i * N + j] [1]
} else {
in2 [i * N + j] [0] = 0;
in2 [i * N + j] [1] = 0;
}
// in2 [i * N + j] [0] = out1 [i * N + j] [0]
// in2 [i * N + j] [1] = out1 [i * N + j] [1]
// in2 [i * N + j] [0] = out1 [i * N + j] [0] *（1.0 /（i * i + 1e-16）+ 1.0 / 1e-16）+1.0 /（（Ni）*（Ni）+ 1e-16）+1.0 /（（Nj）*（Nj）+ 1e-16）
// in2 [i * N + j] [1] = out1 [i * N + j] [1] *（1.0 /（i * i + 1e-16）+ 1.0 / 1e-16）+1.0 /（（Ni）*（Ni）+ 1e-16）+1.0 /（（Nj）*（Nj）+ 1e-16）
// in2 [i * N + j] [0] = 2 * out1 [i * N + j] [0] /（i * i + j * j + 1e-16） //第二测试用例
// in2 [i * N + j] [1] = 2 * out1 [i * N + j] [1] /（i * i + j * j + 1e- ;
}
}
fftw_execute（p2）; // FFT backward
//检查计算结果
double erl1 = 0 .;
for（i = 0; i
cout<< i<<<< j<< sin（X [i]）+ sin（Y [j]）<< out2 [i * N + j] [0] /（N * N）< endl; //>输出
// cout<< i<<<< j<< sin（X [i]）* sin（Y [j]）<< out2 [i * N + j] [0] /（N * N）< endl; //>输出
}
}
cout< erl1<< endl; // L1 error
fftw_destroy_plan（p1）;
fftw_destroy_plan（p2）;
fftw_free（out1）;
fftw_free（out2）;
fftw_free（in1）;
fftw_free（in2）;
return 0;
}
这段代码远非完美，它既不优化也不美观。但它几乎给出了预期的结果。
再见，
I am having trouble explaining/understanding the following phenomenon: To test fftw3 i am using the 2d poisson test case:
laplacian(f(x,y)) = - g(x,y) with periodic boundary conditions.
After applying the fourier transform to the equation we obtain : F(kx,ky) = G(kx,ky) /(kx² + ky²) (1)
if i take g(x,y) = sin (x) + sin(y) , (x,y) \in [0,2 \pi] i have immediately f(x,y) = g(x,y)
which is what i am trying to obtain with the fft :
i compute G from g with a forward Fourier transform
From this i can compute the Fourier transform of f with (1).
Finally, i compute f with the backward Fourier transform (without forgetting to normalize by 1/(nx*ny)).
In practice, the results are pretty bad?
(For instance, the amplitude for N = 256 is twice the amplitude obtained with N = 512)
Even worse, if i try g(x,y) = sin(x)*sin(y) , the curve has not even the same form of the solution.
(note that i must change the equation; i divide by two the laplacian in this case : (1) becomes F(kx,ky) = 2*G(kx,ky)/(kx²+ky²)
Here is the code:
/*
* fftw test -- double precision
*/
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <fftw3.h>
using namespace std;
int main()
{
int N = 128;
int i, j ;
double pi = 3.14159265359;
double *X, *Y ;
X = (double*) malloc(N*sizeof(double));
Y = (double*) malloc(N*sizeof(double));
fftw_complex *out1, *in2, *out2, *in1;
fftw_plan p1, p2;
double L = 2.*pi;
double dx = L/(N - 1);
in1 = (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*(N*N) );
out2 = (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*(N*N) );
out1 = (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*(N*N) );
in2 = (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*(N*N) );
p1 = fftw_plan_dft_2d(N, N, in1, out1, FFTW_FORWARD,FFTW_MEASURE );
p2 = fftw_plan_dft_2d(N, N, in2, out2, FFTW_BACKWARD,FFTW_MEASURE);
for(i = 0; i < N; i++){
X[i] = -pi + i*dx ;
for(j = 0; j < N; j++){
Y[j] = -pi + j*dx ;
in1[i*N + j][0] = sin(X[i]) + sin(Y[j]) ; // row major ordering
//in1[i*N + j][0] = sin(X[i]) * sin(Y[j]) ; // 2nd test case
in1[i*N + j][1] = 0 ;
}
}
fftw_execute(p1); // FFT forward
for ( i = 0; i < N; i++){ // f = g / ( kx² + ky² )
for( j = 0; j < N; j++){
in2[i*N + j][0] = out1[i*N + j][0]/ (i*i+j*j+1e-16);
in2[i*N + j][1] = out1[i*N + j][1]/ (i*i+j*j+1e-16);
//in2[i*N + j][0] = 2*out1[i*N + j][0]/ (i*i+j*j+1e-16); // 2nd test case
//in2[i*N + j][1] = 2*out1[i*N + j][1]/ (i*i+j*j+1e-16);
}
}
fftw_execute(p2); //FFT backward
// checking the results computed
double erl1 = 0.;
for ( i = 0; i < N; i++) {
for( j = 0; j < N; j++){
erl1 += fabs( in1[i*N + j][0] - out2[i*N + j][0]/N/N )*dx*dx;
cout<< i <<" "<< j<<" "<< sin(X[i])+sin(Y[j])<<" "<< out2[i*N+j][0]/N/N <<" "<< endl; // > output
}
}
cout<< erl1 << endl ; // L1 error
fftw_destroy_plan(p1);
fftw_destroy_plan(p2);
fftw_free(out1);
fftw_free(out2);
fftw_free(in1);
fftw_free(in2);
return 0;
}
I can't find any (more) mistakes in my code (i installed the fftw3 library last week) and i don't see a problem with the maths either but i don't think it's the fft's fault. Hence my predicament. I am all out of ideas and all out of google as well.
Any help solving this puzzle would be greatly appreciated.
note :
compiling : g++ test.cpp -lfftw3 -lm
executing : ./a.out > output
and i use gnuplot in order to plot the curves : (in gnuplot ) splot "output" u 1:2:4 ( for the computed solution )
Here are some little points to be modified :
You need to account for all small frequencies, including the negative ones ! Index
icorresponds to the frequency2PI i/Nbut also to the frequency2PI (i-N)/N. In the Fourier space, the end of the array matters as much as the beginning ! In our case, we keep the smallest frequency : it's2PI i/Nfor the first half of the array, and 2PI(i-N)/N on the second half.Of course, as Paul said,
N-1should beNindouble dx = L/(N - 1);=>double dx = L/(N );N-1does not correspond to a continious periodic signal. It woud be hard to use it as a test case...Scaling...I did it empirically
The result i obtain is closer to the expected one, for both cases. Here is the code :
/*
* fftw test -- double precision
*/
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <fftw3.h>
using namespace std;
int main()
{
int N = 128;
int i, j ;
double pi = 3.14159265359;
double *X, *Y ;
X = (double*) malloc(N*sizeof(double));
Y = (double*) malloc(N*sizeof(double));
fftw_complex *out1, *in2, *out2, *in1;
fftw_plan p1, p2;
double L = 2.*pi;
double dx = L/(N );
in1 = (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*(N*N) );
out2 = (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*(N*N) );
out1 = (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*(N*N) );
in2 = (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*(N*N) );
p1 = fftw_plan_dft_2d(N, N, in1, out1, FFTW_FORWARD,FFTW_MEASURE );
p2 = fftw_plan_dft_2d(N, N, in2, out2, FFTW_BACKWARD,FFTW_MEASURE);
for(i = 0; i < N; i++){
X[i] = -pi + i*dx ;
for(j = 0; j < N; j++){
Y[j] = -pi + j*dx ;
in1[i*N + j][0] = sin(X[i]) + sin(Y[j]) ; // row major ordering
// in1[i*N + j][0] = sin(X[i]) * sin(Y[j]) ; // 2nd test case
in1[i*N + j][1] = 0 ;
}
}
fftw_execute(p1); // FFT forward
for ( i = 0; i < N; i++){ // f = g / ( kx² + ky² )
for( j = 0; j < N; j++){
double fact=0;
in2[i*N + j][0]=0;
in2[i*N + j][1]=0;
if(2*i<N){
fact=((double)i*i);
}else{
fact=((double)(N-i)*(N-i));
}
if(2*j<N){
fact+=((double)j*j);
}else{
fact+=((double)(N-j)*(N-j));
}
if(fact!=0){
in2[i*N + j][0] = out1[i*N + j][0]/fact;
in2[i*N + j][1] = out1[i*N + j][1]/fact;
}else{
in2[i*N + j][0] = 0;
in2[i*N + j][1] = 0;
}
//in2[i*N + j][0] = out1[i*N + j][0];
//in2[i*N + j][1] = out1[i*N + j][1];
// in2[i*N + j][0] = out1[i*N + j][0]*(1.0/(i*i+1e-16)+1.0/(j*j+1e-16)+1.0/((N-i)*(N-i)+1e-16)+1.0/((N-j)*(N-j)+1e-16))*N*N;
// in2[i*N + j][1] = out1[i*N + j][1]*(1.0/(i*i+1e-16)+1.0/(j*j+1e-16)+1.0/((N-i)*(N-i)+1e-16)+1.0/((N-j)*(N-j)+1e-16))*N*N;
//in2[i*N + j][0] = 2*out1[i*N + j][0]/ (i*i+j*j+1e-16); // 2nd test case
//in2[i*N + j][1] = 2*out1[i*N + j][1]/ (i*i+j*j+1e-16);
}
}
fftw_execute(p2); //FFT backward
// checking the results computed
double erl1 = 0.;
for ( i = 0; i < N; i++) {
for( j = 0; j < N; j++){
erl1 += fabs( in1[i*N + j][0] - out2[i*N + j][0]/(N*N))*dx*dx;
cout<< i <<" "<< j<<" "<< sin(X[i])+sin(Y[j])<<" "<< out2[i*N+j][0]/(N*N) <<" "<< endl; // > output
// cout<< i <<" "<< j<<" "<< sin(X[i])*sin(Y[j])<<" "<< out2[i*N+j][0]/(N*N) <<" "<< endl; // > output
}
}
cout<< erl1 << endl ; // L1 error
fftw_destroy_plan(p1);
fftw_destroy_plan(p2);
fftw_free(out1);
fftw_free(out2);
fftw_free(in1);
fftw_free(in2);
return 0;
}
This code is far from being perfect, it is neither optimized nor beautiful. But it gives almost what is expected.
Bye,
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