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ROBINSON'S ENERGY DELAY THEOREM

We will now show that a minimum-phase wavelet has less energy delay than any other one-side wavelet with the same spectrum. More precisely, we will show that the energy summed from zero to any time t for the minimum-phase wavelet is greater than or equal to that of any other wavelet with the same spectrum. Refer to Figure 3-2.

 
3-2
Figure 2
Percent of total energy in a filter between time and time t.
3-2
view

We will compare two wavelets $P_{\rm in}$ and $P_{\rm out}$ which are identical except for one zero, which is outside the unit circled for $P_{\rm out}$ and inside for $P_{\rm in}$.We may write this as
\begin{eqnarraystar}
P_{\rm out}(Z) &= & (b + sZ)\, P(Z) \\ P_{\rm in}(Z) &= & (s + bZ)\, P(Z) \end{eqnarraystar}
where b is bigger than s and P is arbitrary but of degree n. Next we tabulate the terms in question.

t $P_{\rm out}$ $P_{\rm in}$ $P^2_{\rm out} - P^2_{\rm in}$ $\sum^t_{k = 0} (P^2_{\rm out} - P^2_{\rm in})$
  bp0 sp0 $(b^2 - s^2)\, {p_0}^2 $ $(b^2 - s^2)\, {p_0}^2 $
1 bp1 + sp0 sp1 + bp0 $(b^2 - s^2)\, ({p_1}^2 - {p_0}^2)$ $(b^2 - s^2)\, {p_1}^2$
$\vdots$ $\vdots$      
k bpk + spk -1 spk + bpk -1 $(b^2 - s^2)\, ({p_k}^2 - p_{k - 1}^2)$ $(b^2 - s^2)\, {p_k}^2$
$\vdots$ $\vdots$      
n + 1 spn bpn (b2 - s2) (-pn2)  

The difference, which is given in the right-hand column, is clearly always positive.

To prove that the minimum-phase wavelet delays energy the least, the preceding argument is repeated with each of the roots until they are all outside the unit circle.

EXERCISES:

  1. Do the foregoing minimum-energy-delay proof for complex-valued b, s, and P. [CAUTION:  Does $P_{\rm in} = (s + bZ)\, P$ or $P_{\rm in} = (\bar s + \bar b Z)P$?]

previous up next print clean
Next: THE TOEPLITZ METHOD Up: Spectral factorization Previous: ROOT METHOD
Stanford Exploration Project
10/30/1997