Define the frequency variance as:
$$ \sigma^2 = \int^\infty_{-\infty}\omega^2 P(\omega) d\omega$$
Where $P(\omega)$ is the spectral density function, which is the same as normalized power. Therefore,
$$ \sigma^2 = \frac{\int^\infty_{-\infty}\omega^2 X(\omega)\bar{X}(\omega) d\omega}{\int^\infty_{-\infty} X(\omega)\bar{X}(\omega) d\omega}$$
$X(\omega)$ is the Fourier transform of the signal $x(t)$. We can rewrite the numerator ($v$) as:
$$ v = \int^{\infty}_{-\infty}(i \omega X(\omega)e^{i\omega t})(-i \omega \bar{X}(\omega)e^{-i\omega t}) d\omega$$
$$ = \int^{\infty}_{-\infty}|i \omega X(\omega)e^{i\omega t}|^2 d\omega$$
I am trying to relate $v$ to the following expression of the gradient of the signal $x(t)$:
$$ \frac{dx(t)}{dt}=\int^{\infty}_{-\infty} i \omega X(\omega)e^{i\omega t} d\omega$$
However, all I can come up with is the following inequality:
$$v \geq \left(\frac{dx(t)}{dt} \right)^2 $$
which doesn't make sense since $v$ is independent of time (and frequency) but $\left(\frac{dx(t)}{dt} \right)^2$ is dependent on time.
What is the best way to express $v$ in terms of $\frac{dx(t)}{dt}$?
Best Answer
My answer is: $$ \sigma^2 = \frac{\int^{\infty}_{-\infty} | \frac{dx(t)}{dt} |^2 dt}{\int^{\infty}_{-\infty} | x(t) |^2 dt} $$
I had posted the same question in Mathoverflow and just posted the answer to it.
The relationship turns out to be obvious if you identify that $i \omega X(\omega)$ is a Fourier transform of $\frac{dx(t)}{dt}$, and then we can use Perseval's theorem.