Gaussian Blur with FFT Questions

Question

I have a current implementation of Gaussian Blur using regular convolution. It is efficient enough for small kernels, but once the kernels size gets a little bigger, the perform

Answer 1

1. The 2-D FFT is seperable and you are correct in how to perform it except that you must multiply by the 2-D FFT of the 2D kernel. If you are using kissfft, an easier way to perform the 2-D FFT is to just use kiss_fftnd in the tools directory of the kissfft package. This will do multi-dimensional FFTs.

2. The kernel size does not have to be any particular size. If the kernel is smaller than the image, you just need to zero-pad up to the image size before performing the 2-D FFT. You should also zero pad the image edges since the convoulution being performed by multiplication in the frequency domain is actually circular convolution and results wrap around at the edges.

So to summarize (given that the image size is M x N):

1. come up with a 2-D kernel of any size (U x V)
2. zero-pad the kernel up to (M+U-1) x (N+V-1)
3. take the 2-D fft of the kernel
4. zero-pad the image up to (M+U-1) x (N+V-1)
5. take the 2-D FFT of the image
6. multiply FFT of kernel by FFT of image
7. take inverse 2-D FFT of result
8. trim off garbage at edges

If you are performing the same filter multiple times on different images, you don't have to perform 1-3 every time.

Note: The kernel size will have to be rather large for this to be faster than direct computation of convolution. Also, did you implement your direct convolution taking advantage of the fact that a 2-D gaussian filter is separable (see this a few paragraphs into the "Mechanics" section)? That is, you can perform the 2-D convolution as 1-D convolutions on the rows and then the columns. I have found this to be faster than most FFT-based approaches unless the kernels are quite large.

Response to Edit

1. If the input is real, the output will still be complex except for rare circumstances. The FFT of your gaussian kernel will also be complex, so the multiply must be a complex multiplication. When you perform the inverse FFT, the output should be real since your input image and kernel are real. The output will be returned in a complex array, but the imaginary components should be zero or very small (floating point error) and can be discarded.

2. If you are using the same image, you can reuse the image FFT, but you will need to zero pad based on your biggest kernel size. You will have to compute the FFTs of all of the different kernels.

3. For visualization, the magnitude of the complex output should be used. The log scale just helps to visualize smaller components of the output when larger components would drown them out in a linear scale. The Decibel scale is often used and is given by either 20*log10(abs(x)) or 10*log10(x*x') which are equivalent. (x is the complex fft output and x' is the complex conjugate of x).

4. The input and output of the FFT will be the same size. Also the real and imaginary parts will be the same size since one real and one imaginary value form a single sample.

Answer 2

Remember that convolution in space is equivalent to multiplication in frequency domain. This means that once you perform FFT of both image and mask (kernel), you only have to do point-by-point multiplication, and then IFFT of the result. Having said that, here are a few words of caution.

You probably know that in digital signal processing, we often use circular convolution, not linear convolution. This happens because of curious periodicity. What this means in simple terms is that DFT (and FFT which is its computationally efficient variant) assumes that you signal is periodic, and when you filter your signal in such manner -- suppose your image is N x M pixels -- that it takes pixel at (1,m) to the the neighbor or pixel at (N, m) for some m<M. You signal virtually wraps around onto itself. This means that your Gaussian mask will be averaging pixels on the far right with pixels on the far left, and same goes for top and bottom. This might or might not be desired, but in general one has to deal with edging artifacts anyway. It is however much easier to forget about this issue when dealing with FFT multiplication because the problem stops being apparent. There are many ways to take care of this problem. The best way is to simply pad your image with zeros and remove the extra pixels later.

A very neat thing about using a Gaussian filter in frequency domain is that you never really have to take its FFT. It si a well-know fact that Fourier transform of a Gaussian is a Gaussian (some technical details here). All you would have to do then is pad you image with zeros (both top and bottom), generate a Gaussian in the frequency domain, multiply them together and take IFFT. Then you're done.

Hope this helps.