# Infinite extent of spectrum, but also in time in Oppenheim's Discrete Time Signal Processing?

In Oppenheim's Discrete Time Signal Processing there's on p. 323 no limited band in both time and frequency - wouldn't that violate the Heisenberg Principle?

• what is the $grd$ operator ? – AlexTP Apr 29 '17 at 9:21
• @AlexTP: "group delay" – Matt L. Apr 29 '17 at 10:32

Not at all. The Uncertainty Principle says that a function cannot be both limited in time and limited in frequency. More specifically, the product of the signal's widths in time and in frequency (i.e., its time extension $\Delta_t$ and its bandwidth $\Delta_f$) is bounded from below:

$$\Delta_t\cdot\Delta_f\ge C\tag{1}$$

where the constant $C$ depends on the definition of bandwidth and time extension.

Note that $(1)$ is a lower bound, not an upper bound, so both widths can be infinite without contradicting $(1)$.

If a function is sharply localized in time then, by the Uncertainty Principle, it cannot be sharply localized in frequency, and vice versa. However, if - as in your example - a function is NOT localized in one of the two domains then this does not mean that it must be localized in the other domain. In may very well be non-localized in both domains, as is the case in the given example for non-integer $\alpha$.

Also take a look at this question and its answers for more details on the the Uncertainty Principle.

• @Starhowl: The author clearly understands the principle. And my answer is very much related to your question. Your question comes from a misunderstanding of the uncertainty principle. In my answer I tried to point out that a signal with infinite time extension and infinite bandwidth clearly does NOT violate the uncertainty principle. – Matt L. Apr 29 '17 at 10:31
• @Starhowl, your question wrongly implies that, if a signal's frequency spectrum has infinite support, than its time domain support must be finite according to uncertainty principle. As Matt L described, the principle does not say that. For signal processing applications it's interpreted in the way that a signal cannot be both time and frequency limited... So just like a Gaussian signal $x(t)=e^{-a|t|}$, it can have infinite support in both time and frequency domains. – Fat32 Apr 29 '17 at 11:15