# Constructing a Gaussian kernel in the frequency domain

I'm currently learning about Fourier transform, but find the differences between spatial domain and frequency domain a bit confusing at times.

Let's say I would like to perform convolution of an image with a Gaussian kernel. As far as i understand the Fourier transform of a Gaussian is also a Gaussian i.e the Fourier transform of

$$g(x,y;\sigma) = \frac{1}{2\pi\sigma^2} \cdot e^{- \frac{x^2+y^2}{2\sigma^2}}$$ is $$G(x,y;\sigma) = e^{- 2\pi\sigma^2(x^2+y^2)}$$ However if I directly construct a kernel using $$G(x,y;\sigma)$$ the output kernel looks completely different from first computing $$g(x,y;\sigma)$$ and then doing the FFT. What is the correct approach here? I hope the question is clear, but it is essentially how I can construct a gaussian kernel directly in frequency space so i do not have to FFT the kernel.

• Indeed, $\mathscr{F}\{g(x,y,σ)\} = G(k_x,k_y,σ)$. What do you mean "the output kernel looks completely different from first computing $g(x,y;σ)$ and then doing the FFT"? Are you doing it right? I mean, when you compute in the coordinate domain, you do a convolution with the kernel, and when you compute in the frequency domain, you simply multiply the image Fourier transform and $G(k_x,k_y,σ)$. Mar 8, 2021 at 11:32
• I mean that if I first compute the kernel with a gaussian and then the FFT of that gaussian kernel(and shift to center the frequency domain), then the result is different from directly computing what is believe is the FFT of the gaussian. Maybe $x$ and $y$ are different from $k_x$ and $k_y$ and this is where i am mistaken? Mar 8, 2021 at 11:51
• I do not understand what you mean "Maybe $x$ and $y$ are different from $k_x$ and $k_y$ ". It is a convenience to use different characters to denote variables in different domains, and, yes, $x$ is not $k_x$ and $y$ is not $k_y$. $x$, $y$ are measured in meters and $k_x$, $k_y$, in inverse meters, 1/m units. Compare the units for time and frequency domain variables, $t$ [second] and $ω$ [Hertz]. "Hertz" is an inverse "second". Mar 8, 2021 at 12:08
• Different how? Is the resulting filter big in the center and drops off at the sides? Does it not conform to a Gaussian shape? Does it have a lot of imaginary values running around? There's a lot of that it could be partially right that one can only recognize with experience, but to help you out with that we need to know differently how. Sep 7, 2021 at 14:07

• Note that any straight-line section through a 2D Gaussian in Cartesian coordinates should be Gaussian. So if you take the Fourier transform of a 2D Gaussian that's a good size, then test the result for $x = 0$ and $y = 0$, it should be Gaussian, too. And that's something you should be able to easily plot. Sep 7, 2021 at 14:09