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In signal processing, there is a time-frequency duality. Band-limited signals have infinite time duration and time-limited signals have infinite bandwidth.

Where can I find a proof of this and is there some intuitive explanation for this?

I am not sure if it has to do something with fouriers uncertainty principle?

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  • $\begingroup$ I think you should check the literature for a proof... $\endgroup$ – Fat32 May 1 at 0:16
  • $\begingroup$ Possible duplicate of Uncertainty principle - Duration bandwidth principle $\endgroup$ – MBaz May 1 at 1:30
  • $\begingroup$ @MBaz I don't think the uncertainty principle says anything about the question being asked here. $\endgroup$ – Dilip Sarwate May 1 at 4:01
  • $\begingroup$ @DilipSarwate I misunderstood, then. I have answered with what I believe is the traditional argument. $\endgroup$ – MBaz May 1 at 16:25
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There are several proofs of this important property, e.g., on this wikipedia page on band-limiting, and on this dsp blog page. I haven't checked them in detail, so it's up to you to decide whether these proofs are sufficiently rigorous.

I want to show another proof which is very simple, if you accept the truth of the Paley-Wiener condition:

A square-integrable function $A(\omega)\ge 0$ is the magnitude of the Fourier transform of a causal function if and only if $$\int_{-\infty}^{\infty}\frac{|\log A(\omega)|}{1+\omega^2}d\omega<\infty\tag{1}$$

Assume that $f(t)$ is a time-limited function, i.e., $f(t)=0$ for $|t|>T$, and $F(\omega)$ is its Fourier transform. Then $f(t-T)$ is causal and its Fourier transform is $F(\omega)e^{-j\omega T}$. According to the Paley-Wiener condition for causal functions, $A(\omega)=|F(\omega)e^{-j\omega T}|=|F(\omega)|$ cannot be zero for $|\omega|>\Omega$, regardless of the choice of $\Omega$, because otherwise the integral in $(1)$ is not finite. Consequently, $f(t)$ cannot be band-limited.

Note that this proof shows that a time-limited function cannot be band-limited, but it also shows that a band-limited function cannot be time-limited, because if the inverse Fourier transform of a band-limited function were time-limited, it could be made causal by an appropriate shift in the time domain, which would not change its magnitude spectrum. So it would need to satisfy the Paley-Wiener condition, which we've shown to be impossible for a band-limited function.

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Here is an argument that doesn't require the uncertainty principle.

Assume $y(t)$ is a signal that is not identical to zero, with Fourier transform $Y(f)$. Let $s(t) = \text{rect}(t/T)$ a rectangular pulse of duration $T$ and Fourier transform $S(f) = |T| \text{sinc}(T f)$.

Then, $x(t) = y(t)s(t)$ is a time-limited signal with Fourier transform $X(f) = Y(f) \ast S(f)$. Since $S(f)$ has infinite support, so does $X(f)$.

Since any time-limited signal $x(t)$ can be defined in this way for a suitable choice of $y(t)$ and $s(t)$ (and a possible time-shift that doesn't affect the result), then we can say that all such signals have infinite bandwidth.

The same argument can be applied for the inverse Fourier transform: band-limiting a signal involves multiplication by a rectangular pulse in the frequency domain, or a convolution with a sinc in the time domain, resulting in an infinite-duration signal.

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  • $\begingroup$ What is $t_1$ and how does it relate to $T$? $\endgroup$ – Dilip Sarwate May 1 at 19:35
  • $\begingroup$ @DilipSarwate I wrote this with $t_1$ first and then changed my mind and decided to use $T$; thanks for catching it! $\endgroup$ – MBaz May 1 at 21:01
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    $\begingroup$ "Since $S(f)$ has infinite support, so does $X(f)$." I don't think that this qualifies as a proof. In general, if one of two functions has infinite support, it is not guaranteed that the convolution of the two has infinite support as well. Of course, in this case it is true, but that's exactly what we'd like to prove! $\endgroup$ – Matt L. May 1 at 21:02
  • $\begingroup$ @MattL. That's the reason I qualified this as an "argument" and not a proof... Perhaps I should say "this should convince you if you accept that the convolution in this case is infinite." :) $\endgroup$ – MBaz May 1 at 21:08
  • $\begingroup$ @MBaz: Well, fair enough, I've just never been really happy with that "intuitive" argument. Anyway, it's probably worth leaving it as it is. $\endgroup$ – Matt L. May 1 at 21:12

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