# I Fourier transform of a sum of shifted Gaussians

1. Sep 4, 2016

### Kavorka

My first thought was simply that the Fourier transform of a sum of Gaussians functions that are displaced from the origin by different amounts would just be another sum of Gaussians:

F{G1(x) + G2(x)} = F{G1(x)} + F{G1(x)}

where a generalized shifted Gaussian is: G(x) = G0exp[-(x - x0)2 / 2σ2]

with amplitude G0, standard deviation σ and offset on the x-axis of x0.

but I've been told this is wrong. I've found that the result becomes more complicated when you add more Gaussians, and that the Fourier transform of just two shifted Gaussians is a Gaussian multiplied by a periodic function (cosine). This comes from representing a shifted Gaussian as the non-shifted gaussian multipled by a shifted delta function. My first question is why you cannot do this in the way I first considered.

My second problem was with solving the problem in the new way. I found that the the fourier transform of a Gaussian multipled by a shifted delta function is just a periodic function, not multiplied by any Gaussian function. It follows that the fourier transform of a sum of delta functions multiplied by Gaussians is just a sum of perodic functions. But I know this result isn't true, and am tripping up on where I went wrong.

I'm trying to generalize an expression in 3-d space, but in 2-d space the function I'm transforming is:

δ(x-x0)*G0exp[-x2 / 2σ2]

The result I get ends up being a complicated constant multiplied by eix0u with u being the fourier space variable. This can be expanded into a real cosine component and an imaginary sine component, and it is the real part I am interested in. Obviously this is not a Gaussian multiplied by a cosine, which is what I expected. The end behavior of the transform should go to 0. Also, I'm not finding how the transform becomes more complicated as you add more Gaussians, because you should just be able to sum the transforms.

Any idea where I am going wrong? I feel like I'm missing a fundamental understanding. I also tried to use the convolution theorem to solve this, but it seems to lead to more complicated work than just directly solving it. Can also show more work if needed, just not used to formatting

Last edited: Sep 4, 2016
2. Sep 5, 2016

### Orodruin

Staff Emeritus
The Fourier transform is linear. It holds that
$$F(f+g) = F(f) + F(g)$$
for any functions f and g (for which the Fourier transform is well defined).

3. Sep 5, 2016

### mathman

Are you adding the Gaussians or the underlying processes?

4. Sep 5, 2016

### olivermsun

Multiplied with or convolved with?

5. Sep 5, 2016

### Kavorka

I mean convolved with. I ended up working it out, my first assumption was correct I just did the work wrong. I was wondering if anyone can take a look at my result