Averaging group delay

Another way to overcome the problem of phase-wrap-around is to compute the phase spectrum by integrating its derivative. The phase spectrum of a signal is related to its Fourier transform as follows:

$$\Phi_s(\omega) = \Im[\ln S(\omega)].$$ (8)

Taking the derivative with respect to $\omega$ on both sides gives

$${d\Phi(\omega) \over d\omega} = \Im[{S^\prime(\omega) \over S(\omega)}].$$ (9)

If we assume $\Phi_s(0)=0$, then

$$\Phi_s(\omega)=\Im\int^{\omega}_0 {S^\prime(\omega) \over S(\omega)} d\omega.$$ (10)

The phase spectrum computed in this way is continuous. Furthermore, it can be shown (Oppenheim and Schafer, 1975) that  

$$n_0={1 \over 2\pi j}\int^\pi_{-\pi} {S^\prime(\omega) \over S(\omega)}d\omega.$$ (11)

If we define the group delay as

$$n_g = {d\Phi_s(\omega) \over d\omega},$$ (12)

then equation (11) tells us that n₀ is an average group delay over frequency range $[-\pi,\pi]$.Considering the limited band width of the seismic signals and the presence of noises, we should include uneven weights in the averaging step of equation (11) as follows:  

$$n_0={1 \over j}\int^\pi_{-\pi} M(\omega) {S^\prime(\omega) \over S(\omega)}d\omega,$$ (13)

where $M(\omega)$ is a weighting function. One choice of the weighting function is

$$M(\omega) = \displaystyle{\vert S(e^{j\omega})\vert^\gamma \... ...yle{\int^\pi_{-\pi} \vert S(e^{j\omega})\vert^\gamma d\omega}}.$$

If $\gamma=0$, the result is the evenly weighted average of the group delay. If $\gamma=\infty$, the result is equal to the group delay of the frequency corresponding to the peak amplitude spectrum.