Let $I$ be a finite nonempty set, $(x_i)_{i\in I}\subseteq[0,1)^2$ and $$f(k):=\sum_{i\in I}e^{-{\rm i}2\pi\langle k,\:x_i\rangle}\;\;\;\text{for }k\in\mathbb Z^2.$$
Assume $x_i\ne x_j$ for all $i,j\in I$ with $i\ne j$. I want to choose $(x_i)_{i\in I}$ such that they are "uniformly" spread over $[0,1)^2$, but simultaneously satisfy that $|f(k)|$ is "small" whenever $|k|$ is "small". What would be a suitable condition to ensure the latter?
For example, I've read that the latter is satisfied when the $x_i$ have a suitable minimum distance to each other. But how do we see this? Clearly, $$|f(k)|^2=\sum_{i,\:j\:\in\:I}\cos\left(2\pi\langle k,x_i-x_j\rangle\right)\tag1,$$ but how do we see the desired property from this expression?
Consider the trivial case of a regualr grid: Let $m,n\in\mathbb N$ and assume $$x_{ij}=\left(\frac{i+\frac12}m,\frac{j+\frac12}n\right)\;\;\;\text{for all }(i,j)\in I:=\{0,\ldots,m-1\}\times\{0,\ldots,n-1\}.$$ For $m=n=64$ and $k\in\{-256,\ldots,255\}^2$ a splot of $|f(k)|^2$ looks as follows:
The white dots correspond to the points $x_i$. And for all other $k$, the value of $|f(k)|^2$ is $0$. But how do we see this analytically?
Set $k=(k_1,l_2)$. Then $$\langle k,x_{i,j} \rangle = k_1\frac{2i+1}{m} + k_2\frac{2j+1}{m}.$$ Applying the function $t \mapsto e^{-i2\pi t}$, one gets the product of a function of $i$ by a function of $j$.
Therefore, the double sum over $i,j$ splits into the product of a sum over $i$ by a sum over $j$. Both sums are geometric sums and can be computed explicitly (separate the cases where $k_1 = 0$ or $k_2=0$).