This is really a comment that could be added on to existing answers, but I can't add comments. As a particularly simple example that follows Ben's answer, as well as an example of SandeepJ's observation about changing the 'size' of the sum, consider counting solutions to some linear equation L = 0 in a subset A of a finite abelian group G. Let's say $L = x + y - + z$. To count solutions to $L(x,y,z) = 0$ with $x,y,z \in A$ one could look at the sum $\sum_{(x, y, z) \in H} F(x, y, z)$, where $F(x, y, z) = 1_A(x)1_A(y)1_A(z)$ and $H \leq G^3$ is the subgroup of solutions $(x,y,z) \in G^3$ such that $L(x,y,z) = 0$. Up to some normalization factors, Poisson summation then says something like $$\sum_{\mathbf{x} \in H} F(\mathbf{x}) = \sum_{\chi \in H^\perp} \widehat{F}(\chi) = \sum_{\gamma \in \widehat{G}} \widehat{1_A}(\gamma)^2 \overline{\widehat{1_A}(\gamma)},$$ widehat{1_A}(\gamma)^3,$$ since $H^\perp \cong \widehat{G}$. So the 'two-dimensional' sum on the left has become a 'one-dimensional' sum on the right.
This is of course the basis of a lot of arguments in additive combinatorics, and as Ben said one might fruitfully consider the term $\widehat{F}(0)$ in the sum on the right-hand side above.
