Multiplication on frequency domain after DFT
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I have a 100 x 100 image and 5x5 Kernel and I want to apply this kernel to this image
1- I add borders (5-1)/2 to the image and take DFT of image
2- I create another matrix which is of the same size as the image's and fill it with 0s. Then I put my original kernel on the upper left hand side of this new matrix.I will call it as filter now
3- Then do the same for the filter
Now I have imaginary and real parts of the image and filter. I multiply them as follows
for i to 102
for j to 102
newReal = realImage * realKernel - imaginaryImage * imaginaryKernel
newImaginary = realImage * imaginaryKernel + imaginaryImage * realKernel
and do InverseFourierTransform(newReal, newImaginary)
4- and then it produces an almost black result which, I think, is because of this padding operation with 0s.
If i do not apply a filter and just take dft and inverseDftmy code works. What am i missing here?
image-processing filter dft
asked Nov 11 at 2:37
Thunfische
1151313
add a comment |
up vote
0
down vote
favorite
I have a 100 x 100 image and 5x5 Kernel and I want to apply this kernel to this image
1- I add borders (5-1)/2 to the image and take DFT of image
2- I create another matrix which is of the same size as the image's and fill it with 0s. Then I put my original kernel on the upper left hand side of this new matrix.I will call it as filter now
3- Then do the same for the filter
Now I have imaginary and real parts of the image and filter. I multiply them as follows
for i to 102
for j to 102
newReal = realImage * realKernel - imaginaryImage * imaginaryKernel
newImaginary = realImage * imaginaryKernel + imaginaryImage * realKernel
and do InverseFourierTransform(newReal, newImaginary)
4- and then it produces an almost black result which, I think, is because of this padding operation with 0s.
If i do not apply a filter and just take dft and inverseDftmy code works. What am i missing here?
image-processing filter dft
asked Nov 11 at 2:37
Thunfische
1151313
I think it's better to use Euler's formula to represent the DFT image here. This way we have a frequency image and a image of phase shift (angle). Then just multiply the frequency image with the frequency image of the filter, and add the phase shift image with the phase shift image of the filter. It's just a multiply of complex numbers. That's not the problem of padding with zeros.
– Kaiwen Chang
Nov 15 at 16:40
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a 100 x 100 image and 5x5 Kernel and I want to apply this kernel to this image
1- I add borders (5-1)/2 to the image and take DFT of image
2- I create another matrix which is of the same size as the image's and fill it with 0s. Then I put my original kernel on the upper left hand side of this new matrix.I will call it as filter now
3- Then do the same for the filter
Now I have imaginary and real parts of the image and filter. I multiply them as follows
for i to 102
for j to 102
newReal = realImage * realKernel - imaginaryImage * imaginaryKernel
newImaginary = realImage * imaginaryKernel + imaginaryImage * realKernel
and do InverseFourierTransform(newReal, newImaginary)
4- and then it produces an almost black result which, I think, is because of this padding operation with 0s.
If i do not apply a filter and just take dft and inverseDftmy code works. What am i missing here?
image-processing filter dft
asked Nov 11 at 2:37
Thunfische
1151313
I have a 100 x 100 image and 5x5 Kernel and I want to apply this kernel to this image
1- I add borders (5-1)/2 to the image and take DFT of image
2- I create another matrix which is of the same size as the image's and fill it with 0s. Then I put my original kernel on the upper left hand side of this new matrix.I will call it as filter now
3- Then do the same for the filter
Now I have imaginary and real parts of the image and filter. I multiply them as follows
for i to 102
for j to 102
newReal = realImage * realKernel - imaginaryImage * imaginaryKernel
newImaginary = realImage * imaginaryKernel + imaginaryImage * realKernel
and do InverseFourierTransform(newReal, newImaginary)
4- and then it produces an almost black result which, I think, is because of this padding operation with 0s.
If i do not apply a filter and just take dft and inverseDftmy code works. What am i missing here?
image-processing filter dft
image-processing filter dft
asked Nov 11 at 2:37
Thunfische
1151313
asked Nov 11 at 2:37
Thunfische
1151313
asked Nov 11 at 2:37
Thunfische
1151313
asked Nov 11 at 2:37
Thunfische
1151313
asked Nov 11 at 2:37
Thunfische
1151313
1151313
I think it's better to use Euler's formula to represent the DFT image here. This way we have a frequency image and a image of phase shift (angle). Then just multiply the frequency image with the frequency image of the filter, and add the phase shift image with the phase shift image of the filter. It's just a multiply of complex numbers. That's not the problem of padding with zeros.
– Kaiwen Chang
Nov 15 at 16:40
add a comment |
I think it's better to use Euler's formula to represent the DFT image here. This way we have a frequency image and a image of phase shift (angle). Then just multiply the frequency image with the frequency image of the filter, and add the phase shift image with the phase shift image of the filter. It's just a multiply of complex numbers. That's not the problem of padding with zeros.
– Kaiwen Chang
Nov 15 at 16:40
I think it's better to use Euler's formula to represent the DFT image here. This way we have a frequency image and a image of phase shift (angle). Then just multiply the frequency image with the frequency image of the filter, and add the phase shift image with the phase shift image of the filter. It's just a multiply of complex numbers. That's not the problem of padding with zeros.
– Kaiwen Chang
Nov 15 at 16:40
I think it's better to use Euler's formula to represent the DFT image here. This way we have a frequency image and a image of phase shift (angle). Then just multiply the frequency image with the frequency image of the filter, and add the phase shift image with the phase shift image of the filter. It's just a multiply of complex numbers. That's not the problem of padding with zeros.
– Kaiwen Chang
Nov 15 at 16:40
add a comment |
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I think it's better to use Euler's formula to represent the DFT image here. This way we have a frequency image and a image of phase shift (angle). Then just multiply the frequency image with the frequency image of the filter, and add the phase shift image with the phase shift image of the filter. It's just a multiply of complex numbers. That's not the problem of padding with zeros.
– Kaiwen Chang
Nov 15 at 16:40