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I know that a band-limited signal is a signal which have a limited set of frequencies.

What is then the relation between band limiting and $[-\pi \; \pi]$ since $-\pi$ and $\pi$ are related to angles and frequencies are in Hertz ?

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$\pi$ is much more that something related to angles. Frequencies in Hertz are reciprocal to time in seconds.

But for an abstract function $f: \mathbb{R}\to \mathbb{R}$, taken as a model of generic continuous-variable signals, ie without unit in the first ordinal variable, it does makes sense to be band-limited with a unitless constant bounding the interval in frequency.

The characterization of band-limited signals is given by the Paley-Wiener theorem.

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Now first of all concept of bandlimited signal exists in the context of analog signals not digital ones. Digital signals are always (by their definition and existance) bandlimited.

Second, frequency in analog signals is expressed either as $f$ in Hertz (cycle per second, repetation per second) or as $\Omega$ in radians per second, where they are related as $$\Omega = 2 \pi f$$ So for example a frequency of $f=1$ kHz would equivalently make a radian-frequency of $\Omega=2 \pi 1k \approx 6283$ rad/sec.

When a bandlimited analog signal $x_a(t)$ is sampled by a period of $T_s$ (and digitized afterwards in practice), the resulting discrete-time sequence $x[n]$ will have a discrete Fourier transform expression of $$ H(e^{j\omega}) = \frac{1}{T_s} \sum_k X_a( \frac{\omega + 2\pi k}{T_s})$$

such that it will be frequency normalized to the discrete-time frequency range of $-\pi \leq \omega < \pi$. Where $\omega$ is the discrete-time frequency in radians per sample. Where $X_a$ is the continuous-time Fourier transform (frequency content) of the analog signal $x_a(t)$.

for $k=0$ the base period of $H(e^{j \omega})$ becomes $$ H(e^{j \omega}) = \frac{1}{T_s} X_a( \frac{\omega}{T_s}) $$ where the frequency normalization becomes more apparent as setting $ \Omega = \frac{\omega}{T_s} $ yields the those analog radian-frequencies for $X_a(\Omega)$ when $\omega$ of $H(e^{j \omega})$ is selected from $\omega \in [-\pi,\pi)$

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  • $\begingroup$ In other words saying, that a signal is -pi ou band limited means that it is a sampled signal? $\endgroup$ – ChiPlusPlus Sep 26 '17 at 19:41
  • $\begingroup$ no it's not. if a signal is stated as bandlimited to $\pi$ radians per second.then It's still a continuous time bandlimited analog signal whose bandwidth is $W=\pi$ rad/sec. Discrete-time signals are never described as bandlimited (unless you are talking about digital sample-rate conversion systems). However a discrete-time signal's Fourier transform can be nonzeros only for $|w| < \omega_0$ and zero otherwise can be considered to be confiend to a range of (band of) frequencies w. $\endgroup$ – Fat32 Sep 26 '17 at 20:10

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