Uncertainty principle - Duration bandwidth principle

Question

In one of NPTEL courses about time-frequency analysis, the professor said that the duration bandwidth principle is $\sigma_t^2 \sigma_\omega^2 \ge \frac{1}{4}$.

He added that the formula making use of time resolution and frequency resolution is a false one. The time and frequency resolutions correspond to the time "distance" between two samples of the signal and frequency resolution corresponds to the "distance" between two samples in the frequency domain (two successive samples of the Fourier transform of the time domain signal). Can anyone, please, clarify this to me as I get to see the one using time and frequency resolutions in so many papers?

NB: Here is the link to the course I'm talking about http://nptel.ac.in/courses/103106114/

Answer 1

An important theorem, known as Weyl's, 1931, is: if function $s(t)$ and related functions $ts(t)$, $s'(t)$ are in $L^2$ (square integrable) with the related $\|\cdot\|$ $L_2$ norm symbol then: $$ \| s(t) \|^2 \leq 2\| ts(t) \| \| s'(t) \|$$ Equality is attained when $s(t)$ is a modulated Gaussian/Gabor elementary function defined as: $$ s'(t) / s(t) \propto t $$ or practically as: $$s(t) = C \exp [-\alpha(t - t_m)^2 + \imath 2 \pi \nu_m (t - t_m)]$$ found by integration by part + Cauchy–Bunyakovsky–Schwarz. If one defines time or frequency location, as a center of mass related to energy as:

$$ E = \int |s(t)|^2 dt = \int |S(f)|^2 df$$

and

$$\overline{t} = 1/E \int t|s(t)|^2 dt \qquad \overline{f} = 1/E \int f|S(f)|^2 df$$ and energy dispersion as:

$$\Delta t = \sqrt{1/E \int (t - \overline{t})^2 |s(t)|^2 dt }$$ $$\Delta f = \sqrt{1/E \int (f - \overline{f})^2 |S(f)|^2 df }$$

then for finite-energy every signal $s(t)$, with $\Delta t$ and $\Delta f$ finite, one gets: $$\Delta t\Delta f \geq \frac{1}{4\pi}$$

The limit is attained for some Gaussian variations. My interpretation of the question is is more about the term 'time/frequency dispersion' than the fuzzy concept of 'time/frequency resolution'.