DFT for 2D image using C++ - bad result
1
I try write DFT in C++ Qt. I use version with exp function. I tried many versions and always result is the same. First DFT for rows, and next for columns. Image 400 x 400 RGB. Is it correct? If not, what is wrong?
using namespace std;
complex<double> *tab;
tab = new complex<double>[640000];
complex<double> i(0.0, 1.0);
for (int j = 0; j < 400; ++j) {
for (int k = 0; k < 400; ++k) {
complex<double> sum(0.0, 0.0);
for (int n = 0; n < 400; ++n) {
int t = (j * 400 + n) << 2;
complex<double> arg = -i * ((2.0 * M_PI * k * n) / 400);
complex<double> val(tmp[t] / 255.0, 0.0); //tmp - unsigned char, values R G B
sum += val * exp(arg);
}
int t = (j * 400 + k) << 2;
tab[t] = sum;
tab[t + 1] = sum;
tab[t + 2] = sum;
}
}
for (int j = 0; j < 400; ++j) {
for (int k = 0; k < 400; ++k) {
complex<double> sum(0.0, 0.0);
for (int n = 0; n < 400; ++n) {
int t = (n * 400 + j) << 2;
complex<double> arg = -i * ((2.0 * M_PI * k * n) / 400);
sum += tab[t] * exp(arg);
}
int t = (k * 400 + j) << 2;
unsigned char p;
p = static_cast<unsigned char>(static_cast<int>(round(abs(sum))) % 256);
buf[t] = p;
buf[t + 1] = p;
buf[t + 2] = p;
}
}
delete tab;
I use image:
Result:
c++ image-processing fft dft
edited Nov 10 at 17:58
eyllanesc
72.5k93054
asked Nov 10 at 17:13
Przemysław Lewandowski
61
add a comment |
1
I try write DFT in C++ Qt. I use version with exp function. I tried many versions and always result is the same. First DFT for rows, and next for columns. Image 400 x 400 RGB. Is it correct? If not, what is wrong?
using namespace std;
complex<double> *tab;
tab = new complex<double>[640000];
complex<double> i(0.0, 1.0);
for (int j = 0; j < 400; ++j) {
for (int k = 0; k < 400; ++k) {
complex<double> sum(0.0, 0.0);
for (int n = 0; n < 400; ++n) {
int t = (j * 400 + n) << 2;
complex<double> arg = -i * ((2.0 * M_PI * k * n) / 400);
complex<double> val(tmp[t] / 255.0, 0.0); //tmp - unsigned char, values R G B
sum += val * exp(arg);
}
int t = (j * 400 + k) << 2;
tab[t] = sum;
tab[t + 1] = sum;
tab[t + 2] = sum;
}
}
for (int j = 0; j < 400; ++j) {
for (int k = 0; k < 400; ++k) {
complex<double> sum(0.0, 0.0);
for (int n = 0; n < 400; ++n) {
int t = (n * 400 + j) << 2;
complex<double> arg = -i * ((2.0 * M_PI * k * n) / 400);
sum += tab[t] * exp(arg);
}
int t = (k * 400 + j) << 2;
unsigned char p;
p = static_cast<unsigned char>(static_cast<int>(round(abs(sum))) % 256);
buf[t] = p;
buf[t + 1] = p;
buf[t + 2] = p;
}
}
delete tab;
I use image:
Result:
c++ image-processing fft dft
edited Nov 10 at 17:58
eyllanesc
72.5k93054
asked Nov 10 at 17:13
Przemysław Lewandowski
61
Did you scale the FFT properly?
– Matthieu Brucher
Nov 10 at 17:28
2
You may already know that to acheive "optical" image FFT (produced by lens) you'll need to swap quadrants (1st with 3rd, 2nd with 4th), so the 0th (yours brightest) term lays in the middle. Also I'd advise to use logarithmic scale for viewing as generally your result looks alright however since FFT may produce a wide range of values with high ones being orders of magnitude larger than low ones.
– michelson
Nov 10 at 18:14
1
To check your program, did you try to perform DFT then IDFT and compare the result with the input ?
– Damien
Nov 10 at 18:38
add a comment |
1
1
1
I try write DFT in C++ Qt. I use version with exp function. I tried many versions and always result is the same. First DFT for rows, and next for columns. Image 400 x 400 RGB. Is it correct? If not, what is wrong?
using namespace std;
complex<double> *tab;
tab = new complex<double>[640000];
complex<double> i(0.0, 1.0);
for (int j = 0; j < 400; ++j) {
for (int k = 0; k < 400; ++k) {
complex<double> sum(0.0, 0.0);
for (int n = 0; n < 400; ++n) {
int t = (j * 400 + n) << 2;
complex<double> arg = -i * ((2.0 * M_PI * k * n) / 400);
complex<double> val(tmp[t] / 255.0, 0.0); //tmp - unsigned char, values R G B
sum += val * exp(arg);
}
int t = (j * 400 + k) << 2;
tab[t] = sum;
tab[t + 1] = sum;
tab[t + 2] = sum;
}
}
for (int j = 0; j < 400; ++j) {
for (int k = 0; k < 400; ++k) {
complex<double> sum(0.0, 0.0);
for (int n = 0; n < 400; ++n) {
int t = (n * 400 + j) << 2;
complex<double> arg = -i * ((2.0 * M_PI * k * n) / 400);
sum += tab[t] * exp(arg);
}
int t = (k * 400 + j) << 2;
unsigned char p;
p = static_cast<unsigned char>(static_cast<int>(round(abs(sum))) % 256);
buf[t] = p;
buf[t + 1] = p;
buf[t + 2] = p;
}
}
delete tab;
I use image:
Result:
c++ image-processing fft dft
edited Nov 10 at 17:58
eyllanesc
72.5k93054
asked Nov 10 at 17:13
Przemysław Lewandowski
61
I try write DFT in C++ Qt. I use version with exp function. I tried many versions and always result is the same. First DFT for rows, and next for columns. Image 400 x 400 RGB. Is it correct? If not, what is wrong?
using namespace std;
complex<double> *tab;
tab = new complex<double>[640000];
complex<double> i(0.0, 1.0);
for (int j = 0; j < 400; ++j) {
for (int k = 0; k < 400; ++k) {
complex<double> sum(0.0, 0.0);
for (int n = 0; n < 400; ++n) {
int t = (j * 400 + n) << 2;
complex<double> arg = -i * ((2.0 * M_PI * k * n) / 400);
complex<double> val(tmp[t] / 255.0, 0.0); //tmp - unsigned char, values R G B
sum += val * exp(arg);
}
int t = (j * 400 + k) << 2;
tab[t] = sum;
tab[t + 1] = sum;
tab[t + 2] = sum;
}
}
for (int j = 0; j < 400; ++j) {
for (int k = 0; k < 400; ++k) {
complex<double> sum(0.0, 0.0);
for (int n = 0; n < 400; ++n) {
int t = (n * 400 + j) << 2;
complex<double> arg = -i * ((2.0 * M_PI * k * n) / 400);
sum += tab[t] * exp(arg);
}
int t = (k * 400 + j) << 2;
unsigned char p;
p = static_cast<unsigned char>(static_cast<int>(round(abs(sum))) % 256);
buf[t] = p;
buf[t + 1] = p;
buf[t + 2] = p;
}
}
delete tab;
I use image:
Result:
c++ image-processing fft dft
c++ image-processing fft dft
edited Nov 10 at 17:58
eyllanesc
72.5k93054
asked Nov 10 at 17:13
Przemysław Lewandowski
61
edited Nov 10 at 17:58
eyllanesc
72.5k93054
asked Nov 10 at 17:13
Przemysław Lewandowski
61
edited Nov 10 at 17:58
eyllanesc
72.5k93054
edited Nov 10 at 17:58
eyllanesc
72.5k93054
edited Nov 10 at 17:58
eyllanesc
72.5k93054
72.5k93054
asked Nov 10 at 17:13
Przemysław Lewandowski
61
asked Nov 10 at 17:13
Przemysław Lewandowski
61
asked Nov 10 at 17:13
Przemysław Lewandowski
61
61
Did you scale the FFT properly?
– Matthieu Brucher
Nov 10 at 17:28
2
You may already know that to acheive "optical" image FFT (produced by lens) you'll need to swap quadrants (1st with 3rd, 2nd with 4th), so the 0th (yours brightest) term lays in the middle. Also I'd advise to use logarithmic scale for viewing as generally your result looks alright however since FFT may produce a wide range of values with high ones being orders of magnitude larger than low ones.
– michelson
Nov 10 at 18:14
1
To check your program, did you try to perform DFT then IDFT and compare the result with the input ?
– Damien
Nov 10 at 18:38
add a comment |
Did you scale the FFT properly?
– Matthieu Brucher
Nov 10 at 17:28
2
You may already know that to acheive "optical" image FFT (produced by lens) you'll need to swap quadrants (1st with 3rd, 2nd with 4th), so the 0th (yours brightest) term lays in the middle. Also I'd advise to use logarithmic scale for viewing as generally your result looks alright however since FFT may produce a wide range of values with high ones being orders of magnitude larger than low ones.
– michelson
Nov 10 at 18:14
1
To check your program, did you try to perform DFT then IDFT and compare the result with the input ?
– Damien
Nov 10 at 18:38
Did you scale the FFT properly?
– Matthieu Brucher
Nov 10 at 17:28
Did you scale the FFT properly?
– Matthieu Brucher
Nov 10 at 17:28
2
2
You may already know that to acheive "optical" image FFT (produced by lens) you'll need to swap quadrants (1st with 3rd, 2nd with 4th), so the 0th (yours brightest) term lays in the middle. Also I'd advise to use logarithmic scale for viewing as generally your result looks alright however since FFT may produce a wide range of values with high ones being orders of magnitude larger than low ones.
– michelson
Nov 10 at 18:14
You may already know that to acheive "optical" image FFT (produced by lens) you'll need to swap quadrants (1st with 3rd, 2nd with 4th), so the 0th (yours brightest) term lays in the middle. Also I'd advise to use logarithmic scale for viewing as generally your result looks alright however since FFT may produce a wide range of values with high ones being orders of magnitude larger than low ones.
– michelson
Nov 10 at 18:14
1
1
To check your program, did you try to perform DFT then IDFT and compare the result with the input ?
– Damien
Nov 10 at 18:38
To check your program, did you try to perform DFT then IDFT and compare the result with the input ?
– Damien
Nov 10 at 18:38
add a comment |
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Did you scale the FFT properly?
– Matthieu Brucher
Nov 10 at 17:28
2
You may already know that to acheive "optical" image FFT (produced by lens) you'll need to swap quadrants (1st with 3rd, 2nd with 4th), so the 0th (yours brightest) term lays in the middle. Also I'd advise to use logarithmic scale for viewing as generally your result looks alright however since FFT may produce a wide range of values with high ones being orders of magnitude larger than low ones.
– michelson
Nov 10 at 18:14
1
To check your program, did you try to perform DFT then IDFT and compare the result with the input ?
– Damien
Nov 10 at 18:38