Discrete Fourier Transform on Impulses?
For digital signals, you represent the $s(n) = \delta(n)$ dirac impulse by 

$$s[n] = \delta_\text{discrete}[n] = \begin{cases}
0&amp; \forall n \ne 0\text,\\
1 &amp; n = 0
\end{cases}\tag{*}$$

which is obvious if you think about the dirac impulse of having the function (in continuous math) to give you 

$$\int\limits_{-\infty}^\infty f(x)\delta_\text{continuous}(x) dx = f(x) $$

and in the discrete case, your integral collapses to the sum of 

$$\sum\limits_{-\infty}^\infty f[x]\delta_\text{discrete}[x] dx \overset!= f[x] \tag{$\dagger$}$$

The only function that fulfills $(\dagger)$ is $\delta$ from $(\text *)$.