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I'm new to DSP. As I reading the textbook, I cannot understand the formula $X_{s}(f)=\frac{1}{T}\sum_{n=-\infty}^{\infty} X(f-nf_{s})$. Could you please give me some keywords so I can learn the theorem and understand it?

From spectral analysis, the original spectrum (frequency components) $X(f)$ and the sampled signal spectrum $X_s(f)$ in terms of Hz are related as $X_{s}(f)=\frac{1}{T}\sum_{n=-\infty}^{\infty} X(f-nf_{s})$ where $X(f)$ is assumed to be the original baseband spectrum, while $X_s(f)$ is its sampled signal spectrum, consisting of the original baseband spectrum $X(f)$ and its replicas $X(f-nf_s)$. Since Equation (2.2) is a well-known formula, the derivation is omitted here and can be found in well-known texts (Ahmed and Natarajan, 1983; Alkin, 1993; Ambardar, 1999; Oppenheim and Schafer, 1975; Proakis and Manolakis, 1996).

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    $\begingroup$ $X_s(f)$ is just a periodic extension of $X(f)$ scaled by $\frac1T=f_s$ with the period in $f$ as $f_s$. $\endgroup$ Commented Jan 12, 2023 at 4:16
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    $\begingroup$ Periodization $\endgroup$ Commented Jan 12, 2023 at 7:00
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    $\begingroup$ Google "sampling theorem". Did you consult the "well-known texts"? $\endgroup$
    – Matt L.
    Commented Jan 12, 2023 at 8:30

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It's not a theorem, but a result that is part of the sampling theorem, and that shows the sampling operation in the frequency domain:

The sampling operation with frequency $f_s = \dfrac{1}{T}$ can be defined as: $$x_s(t) = x(nT) = x(t)\sum_{n=-\infty}^{\infty}\delta(t-nT) = x(t) \frac{1}{T}\sum_{k=-\infty}^{\infty}e^{jk\omega_st} = \frac{1}{T}\sum_{k=-\infty}^{\infty}x(t)e^{jk\omega_st}$$ where:

  • $\omega_s = 2\pi/T$
  • $\delta(t-nT)$ is an impulse train with period $T$ (i.e sampling frequency $f_s =\tfrac{1}{T}$), and since it's periodic with period $T$, we can use its Fourier series $\frac{1}{T}\sum_{k=-\infty}^{\infty}e^{jk\omega_st}$

In the frequency domain, taking the Fourier transform of $x_s(t)$ (you can prove this using the shifting property), we get: $$X_s(\omega) = \frac{1}{T}\sum_{k=-\infty}^{\infty}X(\omega - k\omega_s)$$ Note that $\omega = 2\pi f$, so you can replace the dependency on $\omega$ by $f$: $$X_s(f) = \frac{1}{T}\sum_{k=-\infty}^{\infty}X(f- kf_s)$$

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from what I understood I believe it to be linked to the Poisson formula. https://en.wikipedia.org/wiki/Poisson_summation_formula

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1.- The key sentence you need to start with is

.. where X(f) is assumed to be the original baseband spectrum, while Xs(f) is its sampled signal spectrum ..

2.- On detecting signals you are after unknown x(t). But you may already have a fair guess, because you are already watching a certain frequency band and attempting detection of X(f) with a specific test set.

To obtain xs(t) that is close enough to x(t) you have to window X(f) you are detecting.

You cannot listen too short in time because you are certain you are going to miss something.

And you cannot listen for too long either, or you are going to pick up irrelevant things, or simply waste detection time.

Windowing X(f) too broad and there are going to be empty gaps at both sides of each cycle of Xs(f) therefore losing resolution.

Window too close and each cycle of Xs(f) is going to show overlapping, that in turn is going to distort the resulting xs(t).

3.- So, basically you assume x(t) to follow a time cycle, but you have a windowed measurement of X(f) only.

Having done such frequency windowing of X(f) correctly with a fair guess of the such time period Ts=1/fs implies that downconverting 1 frequency cycle of Xs(f) to baseband, and obtaining xs(t) is going to be quite the same as sought x(t).

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