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Say the analog signal $x(t)$ and its spectrum $X_a(\Omega)$. After sampling with frequency $F_s$ or sampling period $T$ we get

$$x[n] \triangleq x(nT) = x\left(\frac{n}{F_s}\right)$$

and its spectrum $X_d(\omega)$.

The relationship between the two spectrums is

$$X_d(\omega) =\frac1T \sum ^{\infty}_{n=-\infty}X_a\left(\frac{\omega + 2\pi n}{T}\right)$$ $$\omega=T\Omega$$

For band-limited analog signal having spectrum in range $[-B,B]$. $X_a(\Omega)=0$ for $|\Omega| \geq B$ if $F_s$ is above Nyquist rate then $$X_d(\omega) =\frac1T X_a(\Omega)$$

For digital signal, its frequency is restrict in $[-\pi,\pi]$ but not the case of analog signal. The equation above means that $X_d(\omega)$ is not restricted in $[-\pi,\pi]$

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  • $\begingroup$ The DTFT always gives you something in $[-\pi,\pi]$. It's part of the definition -- it maps a function from $\mathbb{Z}$ to $\mathbb{C}$ to $\mathbb{C}$-valued functions on the unit circle. $\endgroup$
    – Batman
    Dec 16, 2015 at 21:21
  • $\begingroup$ Please change the Lower index of summation into "negative infinity" because when I try to do so this robocop says it is less than 6 characters and considers it as a password cracking cyber-terrorist attack on the web site possibly... $\endgroup$
    – Fat32
    Dec 18, 2015 at 0:37

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The DTFT $X_d(\omega)$ of a discrete-time signal is periodic with period $2\pi$, so it's sufficient to consider the interval $[-\pi,\pi]$. "Restricted" to this interval means that due to the inherent periodicity there can be no new frequency components outside the interval $[-\pi,\pi]$. Of course, $\pi$ corresponds to half the sampling frequency.

If $X_a(\Omega)$ is the spectrum of a band-limited continuous-time signal, and if the sampling frequency is greater than twice the band-width, then the equality

$$X_d(\omega)=\frac{1}{T}X_a\left(\frac{\omega}{T}\right)$$

holds in the interval $[-\pi,\pi]$. Outside that interval, $X_a(\Omega)$ is zero, and $X_d(\omega)$ is continued periodically.

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  • $\begingroup$ Sorry because my choice of word. I means if $X_d$ is just scale of $X_a$ then if $X_a$ is defined in $[-B,B]$ and $B>\pi$ then $X_d$ is not periodic in $[-\pi,\pi]$ $\endgroup$
    – aukxn
    Dec 16, 2015 at 16:18
  • $\begingroup$ @aukxn: $X_d(\omega)$ is always periodic. The second equation in your question holds in the interval $[-\pi,\pi]$. $\endgroup$
    – Matt L.
    Dec 16, 2015 at 16:26
  • $\begingroup$ @MattL. Otherwise said, if the DTFT co-efficients are known then $X_a\left(\dfrac{\omega}{T}\right)$ is the scaled version of DTFT co-efficients $X_d(\omega)$ by a factor $T$ How is mapping done now? From $[-\pi, \pi]$ to which frequency range? $\endgroup$
    – jomegaA
    Feb 10, 2020 at 13:58

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