# Reconstruction of Contiuous - Time Signals

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 $$F_s$$.
• 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 $\infty$ – Anthony Lauly May 6 '20 at 7:54
• @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 $k=\infty$. – DSP Rookie May 6 '20 at 9:43