Let $N \in \mathbb N$ and $c\_n \in \mathbb C$, $t\_n \in \mathbb R$ for $n=1, \dots, N$. Suppose that $f$ is a linear combination of dirac-deltas with locations $t\_n$ and coefficients $c\_n$, i.e.


$$
f=\sum\_{n=1}^Nc\_n\delta\_{t\_n}.
$$


The $k$-th Fourier coefficient of $f$ is defined by


$$
\hat f (k) = \sum\_{n=1}^Nc\_ne^{-2\pi i k t\_n}
$$


How many Fourier coefficients $\hat f(k)$ do we need such that $f$ is uniquely determined by them? In other words: For which (minimal) $K \subset \mathbb Z$ such that
$$
\hat f(k) = \hat h (k) \ \forall k \in K
$$
is $f=h$ where $h=\sum\_{n=1}^Nd\_n\delta\_{s\_n}$ is a second linear combination of deltas.



How many Fourier coefficients of a sparse signal $f=\sum_{n=1}^Nc_n\delta_{t_n}$ are needed to determine $f$ uniquely?
Assuming you only care about $t\_n, s\_n$ mod $1$, then $2N$ values of $k$ suffice. (A dimension argument shows this is optimal). In fact, we can take $K= \{0,\dots, 2N-1\}$.


Form a $2N \times N$ matrix whose entry in the $j$th column and the $k+1$st rown is $e^{- 2 \pi i k t\_j}$. Add at most $N$ additional column to the matrix of the form $e^{ -2 \pi i k s\_j}$, but don't add an additional if $s\_j$ already equals $t\_i$ for some $i$.


Because the Vandermonde matrix is invertible, this matrix is injective. We can see that $\hat{f}(k)$ for $k$ from $0$ to $2N-1$ is the transform of the vector $( c\_1, c\_2,\dots, c\_N, 0,\dots, 0)$ under this matrix, and similarly $\hat{h}(k)$ for $k$ from $0$ to $2N-1$ is the transform of a vector with entries $d\_1,\dots, d\_N$.


Because the matrix is injective, if $\hat{f}(k) = \hat{h}(k)$ for all such $k$, then these two vectors are equal, so the coefficients of each delta function in $f$ and $h$ are equal, and thus $f=h$.


For a robust version of this, you will need to look at the reference Shil B. posted.


