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I'm self-learning signal processing now, and I've run into this question about band-limited signals:

Consider the signal $x(t) = 1$ for $0 \leq t \leq T$ and $0$ otherwise. I've found that its Fourier transform is: $$ x_\phi(\Omega) = \frac{1-e^{-j\Omega T}}{j\Omega} $$

How can I determine if this signal is band-limited signal, or not?

Thanks in advance

Sina
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DanielY
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1 Answers1

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Check if there is some $R>0$ such that $x_\phi (\Omega)=0$ holds for all $|\Omega |\geq R$. That would be the definition.

Another way is the following: Every unlimited signal is smooth (infinitely differentiable), even analytic by the Palye Wiener Theorem. Is this the case for your signal? (Is your signal continuous?)

PhoemueX
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  • I don't think, intuitivelly, that my signal holds the definition. But how can I prove this? – DanielY Dec 23 '14 at 09:29
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    Try to find a sequence $(\Omega_n)n$ with $\Omega_n \to \infty$ and $x\phi (\Omega_n)\neq 0$ for all $n$. This will.imply that the definition is not satisfied. – PhoemueX Dec 23 '14 at 09:31
  • What about $f(x) = 1$ if $x = 0$ and $f(x) = 0$ is $x\neq 0$? It's not smooth but has unlimited bandwidth. – AnonSubmitter85 Jan 04 '15 at 18:49
  • @AnonSubmitter85: First of all, you probably mean that the function has limited bandwidth. The (apparent) contradiction is resolved by noting that the actual statement is that there is a smooth function $g$ with $f=g$ almost everywhere (i.e. on the complement of a null set). I will edit the post this evening to reflect this. – PhoemueX Jan 04 '15 at 19:48
  • @PhoemueX Perhaps I am misinterpreting "unlimited signal" to mean a bandlimited signal? – AnonSubmitter85 Jan 04 '15 at 20:39