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



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


