# Fourier transform of the sampled signal

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?

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. $$(2)$$.

• Thanks a lot. Is there any connection to DTFT in this case? Jul 5 '20 at 14:30
• Of course there is: the DTFT of the discrete signal $x_d[n]=x(nT)$ is equal to the expression in Eq. (1). Jul 5 '20 at 14:59