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.

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 where b is bigger than s and P is arbitrary but of degree n. Next we tabulate the terms in question.

$$P_{\rm out} P_{\rm in} P^2_{\rm out} - P^2_{\rm in} \sum^t_{k = 0} (P^2_{\rm out} - P^2_{\rm in})$$ t

$$(b^2 - s^2)\, {p_0}^2 (b^2 - s^2)\, {p_0}^2$$ bp₀ sp₀

$$(b^2 - s^2)\, ({p_1}^2 - {p_0}^2) (b^2 - s^2)\, {p_1}^2$$ 1 bp₁ + sp₀ sp₁ + bp₀

$$\vdots \vdots$$

$$(b^2 - s^2)\, ({p_k}^2 - p_{k - 1}^2) (b^2 - s^2)\, {p_k}^2$$ k bp_k + sp_{k -1} sp_k + bp_{k -1}

$$\vdots \vdots$$

n + 1 sp_n bp_n (b² - s²) (-p_n²)

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 $P_{\rm in} = (\bar s + \bar b Z)P$?]