Variance of spectral density is related to the gradient of signal?

Question

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 $\frac{dx(t)}{dt}$?

Answer 1

It turns out that the relationship is obvious when I use Parseval's Theorem. First I am rewriting my $v$ (variance term) here:

$$ v = \int^{\infty}_{-\infty}|i \omega X(\omega))|^2 d\omega$$

$ i \omega X(\omega)$ is Fourier transform of $\frac{dx(t)}{dt}$. Using Perseval's Theorem,

$$\int^{\infty}_{-\infty}|i \omega X(\omega))|^2 d\omega = \int^{\infty}_{-\infty} | \frac{dx(t)}{dt} |^2 dt$$

Therefore,

$$ \sigma^2 = \frac{\int^{\infty}_{-\infty} | \frac{dx(t)}{dt} |^2 dt}{\int^{\infty}_{-\infty} | x(t) |^2 dt} $$

This makes sense, because the variance of the frequency, $\sigma^2$, is basically a metric that tells us how smooth a function is, which can be summarized with L2 of the derivative of the signal. I am confident with this, but please leave a comment if this can be improved or needs correction.