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Assume that I manually compute the Fourier transform, $X_a(f)$, of a continuous-time aperiodic signal, $x_a(t)$, and then take equally-spaced samples of this Fourier transform, $X(k) \equiv X_a(k F_0)$. Is there a discrete-time periodic signal, $x(n)$, whose Fourier series coefficients satisfy the relation $c_k = X(k)$, where $c_k$ is defined as $c_k = \frac{1}{N}\sum_0^{N-1} x(n) e^{-j2\pi kn/N}$?

A continuous-time signal, $x_a(t)$, has no notion of sampling time, but when we take samples, $X(k)$, of its Fourier transform, $X_a(f)$, for computer analysis, we are discretizing the signal in frequency-domain. I am trying to understand if this frequency-domain discretization would give back a discrete-time signal, $x(n)$, (through inverse Fourier transform/series) that has some relation to the original continuous-time signal.

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