I want to calculate Fourier transform of the sampled signal in two ways. Let $$s(t) = \sum_{k = -\infty}^{\infty}\delta(t - kT)$$And $z(t) = x(t)s(t)$. So we have $$z(t) = \sum_{k = -\infty}^{\infty}x(kT)\delta(t - kT)$$Directly computing Fourier transform leads to $$\mathcal{F}(z(t)) = \sum_{k = -\infty}^{\infty}x(kT)e^{-j\omega kT} $$ On the other hand, if we use $\mathcal{F}(x(t)s(t)) = \frac{1}{2\pi}X(j\omega)\star S(j\omega)$ we have $$\mathcal{F}(z(t)) = \frac{1}{T}\sum_{k = -\infty}^{\infty}X(j(\omega - k\frac{2\pi}{T}))$$ How these are equivalent to each other mathematically? I mean how we can prove $$\frac{1}{T}\sum_{k = -\infty}^{\infty}X(j(\omega - k\frac{2\pi}{T})) = \sum_{k = -\infty}^{\infty}x(kT)e^{-j\omega kT}$$ And is there any intuition for this connection in terms of other subjects like sampling theorem or DTFT?


1 Answer 1


The equality

$$\frac{1}{T}\sum_{k = -\infty}^{\infty}X(j(\omega - k\frac{2\pi}{T})) = \sum_{k = -\infty}^{\infty}x(kT)e^{-j\omega kT}\tag{1}$$

is an instance of Poisson's sum formula. The term on the right-hand side of $(1)$ is just the Fourier series representation of the periodic function on the left-hand side of $(1)$.

The samples $x(kT)$ of the time domain signal are basically the Fourier series coefficients of the periodic spectrum of the sampled signal.

The significance of $(1)$ is that it shows that the discrete-time Fourier transform (DTFT) $X_d(e^{j\Omega})$ of the sequence $x_d[k]=x(kT)$ equals a periodized version of the Fourier transform of the corresponding continuous-time signal $x(t)$:

$$X_d(e^{j\Omega})=\sum_{k=-\infty}^{\infty}x_d[k]e^{-jk\Omega}=\frac{1}{T}\sum_{k=-\infty}^{\infty}X\left(\frac{j(\Omega-2\pi k)}{T}\right),\qquad\Omega=\omega T\tag{2}$$

Clearly, if $X(j\omega)$ is band-limited with a maximum frequency $\omega_c<\pi/T$, then the shifted spectra in the sum on the right-hand side of $(2)$ won't overlap, i.e., there is no aliasing and the signal $x(t)$ can be reconstructed without error from its samples $x(kT)$. That means that the basic form of the sampling theorem is implicit in Eq. $(2)$.

  • $\begingroup$ Thanks a lot. Is there any connection to DTFT in this case? $\endgroup$
    – S.H.W
    Jul 5, 2020 at 14:30
  • $\begingroup$ Of course there is: the DTFT of the discrete signal $x_d[n]=x(nT)$ is equal to the expression in Eq. (1). $\endgroup$
    – Matt L.
    Jul 5, 2020 at 14:59

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