In terms of analog signals, we can represent digital signal as :

$$ x[n] \triangleq x_{a}(nT) = \int_{-\infty}^{\infty}X_{a}(f) \, e^{j2\pi f nT} \ \mathrm{d}f $$

While if we focused on the integral on the right side and according to the Digital Signal Processing by John Proakis, chapter 6.1, we can rewrite it into :

$$ \int_{-\infty}^{\infty}X_{a}(f) \, e^{j2\pi f nT} \ \mathrm{d}f = \sum_{k=-\infty}^{\infty} \int_{(k-1/2)F_{s}}^{(k+1/2)F_{s}}X_{a}(f) \, e^{j2\pi nf/F_{s}} \ \mathrm{d}f $$

where $ F_{s} \triangleq \frac1T $. My question is how the second equation comes up ? what does the interval of $(k-1/2)F_{s}$ to $(k+1/2)F_{s}$ means ?

Furthermore, it is stated in the book that "observing the $X_{a}(f)$ in the interval of $(k-1/2)F_{s}$ to $(k+1/2)F_{s}$ is identical to $X_{a}(f-kF_{s})$ in the interval of $-F_{s}/2$ to $F_{s}/2$". Is there any explanation or derivation how both things are identical ?

Thank you so much, hope that my question is clear enough


Because sampled signals Spectrum will have copies of original spectrum at multiples of $f=F_s$. Even if aliasing happened due to choice of $F_s$, the spectrum of sampled signal $x(n.T_s)$ will be periodic in frequency always with period $F_s$. And you can take any one such $k^{th}$ copy or period and integrate from $\frac{k-1}{2}F_s$ to $\frac{k+1}{2}F_s$. Now if you vary $k$ from $-\infty$ to $\infty$, then it is equivalent to integration on LHS.

Key: Sampling at $F_s$ will create a spectrum which is periodic in frequency with period $F_s$.

  • $\begingroup$ Thanks sir, I think now I understand, just to clarify, instead of using integral for continuous summing in the interval of $ -\infty $ until $ \infty $, I can just sum a continuously in the interval of $ -\frac{1}{2} F_{s} $ until $ \frac{1}{2} F_{s} $. And because the spectrum outside that interval is just a copy, than I can just repeatly sum it for $ -\infty $ until $ \infty $ $\endgroup$ – Anthony Lauly May 6 '20 at 7:54
  • $\begingroup$ @AnthonyLauly No, you have to integrate the spectrum from $-F_s/2$ to $F_s/2$ because it is continuous in frequency. Also, $k$ indexing is necesaary because the integration has to be summed for $k=-\infty$ to $k=\infty$. $\endgroup$ – DSP Rookie May 6 '20 at 9:43
  • $\begingroup$ yup that is what I mean, I think I chose the wrong sentences to delivered. Btw thank you, I get the point. $\endgroup$ – Anthony Lauly May 6 '20 at 11:35
  • $\begingroup$ @AnthonyLauly If DSP Rookie answered your question, best practice would be to check it off as correct (or if you are looking for better / alternate answers, leave as is). Thanks! $\endgroup$ – Dan Boschen May 6 '20 at 23:16

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