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We will now show that a minimumphase wavelet
has less energy delay than any other oneside wavelet with the same spectrum.
More precisely, we will show
that the energy summed from zero to any time t for the minimumphase
wavelet is greater than or equal to that of any other wavelet with the same
spectrum.
Refer to Figure 32.
32
Figure 2
Percent of total energy in a filter between
time and time t.

 
We will compare two wavelets and which are
identical except for one zero,
which is outside the unit circled for
and inside for .We may write this as
where b is bigger than s and P is arbitrary but of degree n.
Next we tabulate the terms in question.
t 





bp_{0} 
sp_{0} 


1 
bp_{1} + sp_{0} 
sp_{1} + bp_{0} 







k 
bp_{k} + sp_{k 1} 
sp_{k} + bp_{k 1} 







n + 1 
sp_{n} 
bp_{n} 
(b^{2}  s^{2}) (p_{n}^{2}) 

The difference, which is given in the righthand column, is clearly
always positive.
To prove that the minimumphase wavelet delays energy the least,
the preceding argument is repeated with each of the roots
until they are all outside the unit circle.
EXERCISES:
 Do the foregoing minimumenergydelay proof for complexvalued b,
s, and P. [CAUTION: Does or
?]
Next: THE TOEPLITZ METHOD
Up: Spectral factorization
Previous: ROOT METHOD
Stanford Exploration Project
10/30/1997