The aperiodic signal could be processed by DFT. DFT can process both periodic and aperiodic signal?
The Discrete Fourier Transform (DFT) defines a relationship between an N-point time-domain sequence x[n], n=0..N-1, and an N-point frequency domain sequence (samples of the Fourier transform uniformly spaced between \omega = 0 and 2\pi) X[k], k=0..N-1. The forward transformation is given by:
X[k] = 1/N \sum_{n=0}^{N-1} x[n] exp{-j 2\pi k.n/N}
which can be expressed as a matrix multiply
X = D x
where x and X are N-element column vectors corresponding to the time and frequency domains, and D is the N-by-N DFT matrix,
D_{kn} = 1/N exp{-j 2\pi k.n/N}
(and thus the inverse transform is trivially derived from the matrix inverse of D).
As such, you can calculate X[k] for any N-point input sequence x[n], and it doesn't even make that much sense to define periodicity for a finite-length sequence. If x[n] can be broken into several pieces that exactly repeat (e.g., a repeated N/2 point sequence), then we will see corresponding structure in X[k] (all the odd-numbered spectral samples will be zero for this example).
Now, you can interpret the DFT as the Fourier transform of an infinite-duration, periodic sequence consisting of infinitely many repeats of the N-point time-domain sequence you start with. In this case, the DFT values X[k] correspond to the weights of the Dirac deltas that constitute the spectrum of this infinite-energy (but finite-power) sequence.
But you can also interpret it as sampling the Fourier transform of finite-length sequence, equivalently an infinite-duration sequence that happens to be nonzero only over a finite, N-point range. In this case, the X[k] values are the finite-valued samples of the the full, continuous-frequency Fourier transform.