Opt1

Find the associated power spectrum given the covariance function $B_x(\tau) = \exp\left(-\left|\frac{\tau}{\tau_0}\right|\right), \tau_0 > 0\,$ .

The power spectrum is defined as $S_x(\omega) = \frac{1}{\pi}\int_0^\infty B_x(\tau)\cos(\omega \tau) \,d\tau\,$ .

$= \frac{1}{\pi} \int_0^\infty \exp\left(-\left|\frac{\tau}{\tau_0}\right|\right) \cos(\omega \tau)\,d\tau\,$

$= \frac{1}{\pi} \int_0^\infty \exp\left(-\frac{\tau}{\tau_0}\right) \cos(\omega \tau)\,d\tau\,$ since $\tau_0>0\,$ and $\tau>0\,$ .

A useful integral relation is:

$\int_0^\infty x^{\mu-1}e^{-ax}\cos(bx)\,dx = \frac{\Gamma(\mu)}{(a^2+b^2)^{\mu/2}}\cos(\mu \arctan(b/a)), \,\mu>0, \,a>0\,$

In the present case, $\mu=1,\,a=\tau_0^{-1},\,b=\omega\,$ , so the problem continues:

$=\frac{1}{\pi}\frac{\Gamma(1)}{\left(\tau_0^{-2}+\omega^2\right)^{1/2}}\cos(\arctan(\omega \tau_0))\,$

It is true that $\Gamma(1)=1\,$ and $\cos(\arctan(x)) = \frac{1}{\sqrt{x^2+1}}\,$ , so the problem continues:

$=\frac{1}{\pi}\left[(\tau_0^{-2}+\omega^2)(\omega^2\tau_0^2+1)\right]^{-1/2}\,$

$=\frac{1}{\pi}(\omega^2+\tau_0^{-2}+\omega^4\tau_0^2+\omega^2)^{-1/2}\,$

$=\frac{1}{\pi}\left(\left[ 1+2\omega^2\tau_0^{2}+\omega^4\tau_0^4 \right]\frac{1}{\tau_0^2}\right)^{-1/2}\,$

$=\frac{1}{\pi}\left(\left[ 1+\omega^2\tau_0^{2}\right]^2\frac{1}{\tau_0^2}\right)^{-1/2}\,$

$=\frac{1}{\pi} \frac{\tau_0}{1+\omega^2\tau_0^2} \,$