Determining the picking position

We choose the convention that the picking position of a wavelet is the first break of the minimum phase part and the ending point of the maximum phase part. The minimum phase part is causal and the maximum phase part is anti-causal:

$$\begin{array} {lll} W_{min}(Z) & = & \displaystyle{\prod^{N_... ...isplaystyle{\prod^{N_{max}}_{k=1}(1-\beta_kZ^{-1})},\end{array}$$ (4)

where $\alpha_k$ and $\beta_k$ are generally complex with their norm less than one, and N_min+N_max=N. Clearly, this choice ensures that for a minimum phase wavelet, the first break of the wavelet is picked, and that for a zero phase wavelet, the center point of the wavelet is picked.

To estimate n₀, we need to examine the phase response of S(Z). Let $\Phi_s(\omega)$ be the phase spectrum of $S(e^{j\omega})$,and $\Phi_w(\omega)$ the phase spectrum of $W(e^{j\omega})$. Equating the the phase spectra of the two sides of equation (3) yields  

$$\Phi_s(\omega) = \Phi_s(0)+\omega n_0+\Phi_w(\omega).$$ (5)

Because w(n) is a real sequence and has causal minimum phase and anti-causal maximum phase parts, its phase spectrum is anti-symmetric with respect to $\omega=\pi$:

$$\Phi_w(2\pi-\omega) = -\Phi_w(\omega) \ \ \ \ 0 \le \omega \le \pi.$$

Using this condition, we can solve equation (5) for n₀ as follows:

$$n_0 = {1 \over 2\pi}[\Phi_s(2\pi-\omega)+\Phi_s(\omega)-2\Phi_s(0)].$$ (6)

By setting $\omega=\pi$, we see that n₀ is the phase delay of the received signal at the Nyquist frequency:

$$n_0 = {\Phi_s(\pi)-\Phi_s(0) \over \pi}.$$