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This is about the Fourier uncertainty principle, which is closely related to, but not the same as the Heisenberg uncertainty principle in physics. So no $\hbar$.

This may be expressed with scaling and definitions more preferred by electrical engineers than might be the habit of mathematicians.

The definition of the continuous Fourier Transform preferred by most electrical engineers:

$$ X(f) = \mathscr{F} \Big\{ x(t) \Big\} = \int\limits_{-\infty}^{+\infty} x(t) \, e^{-i 2 \pi f t} \ \mathrm{d}t $$

and inverse:

$$ x(t) = \mathscr{F}^{-1} \Big\{ X(f) \Big\} = \int\limits_{-\infty}^{+\infty} X(f) \, e^{+i 2 \pi f t} \ \mathrm{d}f $$

Even with different signs on $i$, the elegant symmetry between the forward transform and inverse should be clear. And it makes remembering the Duality property:

If $X(f) = \mathscr{F} \Big\{ x(t) \Big\}$, then $$x(-f) = \mathscr{F} \Big\{ X(t) \Big\}$$

the "DC values":

$$ X(0) = \int\limits_{-\infty}^{+\infty} x(t) \ \mathrm{d}t $$

$$ x(0) = \int\limits_{-\infty}^{+\infty} X(f) \ \mathrm{d}f $$

and Parseval's theorem:

$$ E = \int\limits_{-\infty}^{+\infty} \Big| x(t) \Big|^2 \ \mathrm{d}t = \int\limits_{-\infty}^{+\infty} \Big| X(f) \Big|^2 \ \mathrm{d}f$$

...easy. $E$ is the total energy of the signal in both the time or frequency domains.

No nasty asymmetrical scaling factors to worry about! (Just remember the $2\pi$ in the exponent.) This is why EE's like this definition of the Fourier Transform.

If you define the the square of the time bandwidth of $x(t)$ as

$$ w^2 = \frac{1}{E} \int\limits_{-\infty}^{+\infty} t^2 \Big| x(t) \Big|^2 \ \mathrm{d}t $$

and, using the above definition, the square of the frequency bandwidth of $X(f)$ as

$$ W^2 = \frac{1}{E} \int\limits_{-\infty}^{+\infty} f^2 \Big| X(f) \Big|^2 \ \mathrm{d}f $$

and toss in this condition in the limit,

$$ \lim_{|t|\to\infty} \sqrt{|t|} \ x(t) = 0 $$

Then,

$$ w \cdot W \ge \frac{1}{4 \pi} $$

In consistent units (like time in seconds and frequency in hertz), then the time bandwidth times the frequency bandwidth must equal or exceed $\frac{1}{4 \pi}$.

I could derive this here if someone requires (it requires Cauchy-Schwarz inequality and some messing around), but my question is about something that puzzles me. Suppose I delay $x(t)$ by some large, but finite delay, $\tau$

$$ x_\tau(t) = x(t-\tau) $$

That certainly that increases the time bandwidth measure

$$\begin{align} w_\tau^2 &= \frac{1}{E} \int\limits_{-\infty}^{+\infty} t^2 \Big| x_\tau(t) \Big|^2 \ \mathrm{d}t \\ &= \frac{1}{E} \int\limits_{-\infty}^{+\infty} t^2 \Big| x(t-\tau) \Big|^2 \ \mathrm{d}t \\ &= \frac{1}{E} \int\limits_{-\infty}^{+\infty} (t+\tau)^2 \Big| x(t) \Big|^2 \ \mathrm{d}t \\ \end{align} $$

There is no assumption that the mean $t$ is zero. Perhaps $ \int_{-\infty}^{+\infty} t \Big| x(t) \Big|^2 \ \mathrm{d}t \ne 0 $.

This is like measuring the mean-square $t$, not the variance of $t$.

But delaying by $\tau$ does nothing to the magnitude of the Fourier Transform:

$$ X_\tau(f) = \mathscr{F} \Big\{ x_\tau(t) \Big\} = \mathscr{F} \Big\{ x(t-\tau) \Big\} $$ ... $$ \big| X_\tau(f) \big| = \big| X(f) \big| $$

So the frequency bandwidth $W_\tau$ is not affected:

$$\begin{align} W_\tau^2 &= \frac{1}{E} \int\limits_{-\infty}^{+\infty} f^2 \Big| X_\tau(f) \Big|^2 \ \mathrm{d}f \\ &= \frac{1}{E} \int\limits_{-\infty}^{+\infty} f^2 \Big| X(f) \Big|^2 \ \mathrm{d}f \\ &= W^2 \\ \end{align} $$

It doesn't break the inequality, but wouldn't I expect $W_\tau<W$ if $w_\tau>w$? Do we also need to require

$$ \int\limits_{-\infty}^{+\infty} t \Big| x(t) \Big|^2 \ \mathrm{d}t = 0 $$ and $$ \int\limits_{-\infty}^{+\infty} f \Big| X(f) \Big|^2 \ \mathrm{d}f = 0 $$

so that both $x(t)$ and $X(f)$ are centered about $t=0$ and $f=0$ and the time width and frequency width cannot be artificially increased just by translating in time or frequency?

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  • $\begingroup$ See Theorem 1.1 in math.stonybrook.edu/~bishop/classes/math533.S21/Notes/…, basically the offset does not matter. $\endgroup$
    – copper.hat
    Dec 4, 2023 at 7:35
  • $\begingroup$ @copper.hat , it does matter in a sense that it increases the inequality. But it does not change the sense of the inequality. I'm just bothered that increasing $w$ by translation does not decrease $W$ at all. $\endgroup$ Dec 5, 2023 at 18:30
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    $\begingroup$ The inequality still holds, that is the point. $\endgroup$
    – copper.hat
    Dec 5, 2023 at 19:04
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    $\begingroup$ Yup. We're in agreement. It's just that I expected one quantity to shrink when the other got larger. But that's not the case. I did take a look at that math533 reference. But it's about soooo much more than I was thinking about. And I know how to derive the Fourier Uncertainty Principle. It's just that there was a subtle property that I did not expect, in the outset, to be true. (But it is.) $\endgroup$ Dec 5, 2023 at 19:56
  • $\begingroup$ The reference just introduces arbitrary time/frequency offsets explicitly. $\endgroup$
    – copper.hat
    Dec 5, 2023 at 20:58

1 Answer 1

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Let’s say that $w$ is as small as possible when $\tau$ is 0. Then $w \cdot W$ satisfies the inequality. Delaying it, as you have by letting $\tau\neq 0$, increases $w$ and therefore increases $w \cdot W$, so the inequality is still valid.

An insight from your thoughts is that $w \cdot W$ satisfies the inequality even if you’re allowed to shift the function in time so that $w$ is as small as possible.

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    $\begingroup$ I'm close to check-marking this, but I really am hoping for other answers and perspective. $\endgroup$ Dec 4, 2023 at 19:07
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    $\begingroup$ @robertbristow-johnson Totally fine! Feel free to leave it unchecked forever if you think that’s better. If you found this helpful, that’s great! $\endgroup$
    – NicNic8
    Dec 5, 2023 at 15:27
  • $\begingroup$ I guess I'm not getting much other insight. This really is about mean-square vs. variance. With $\tau$ adjusted so that $w$ is minimum, we got variance. $\endgroup$ Dec 5, 2023 at 18:46

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