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Appendix 1. Proof of Lemma 1

The wavelet transform in Eq. ([*]) is a complex representation for the output of analytic filter in Eq. ([*]).

 
Xk(t) = $\displaystyle S_k(t)e^{j(\omega_k t + \phi_k(t))}$  
  := $\displaystyle \tilde{f}(a,b),\qquad a=\alpha^{k-\frac{K}{2}},b=t.$ (32)

Representing an absolute value for both terms, we obtain

\begin{displaymath}\vert X_k(t)\vert=S_k(t)=\vert\tilde{f}(\alpha^{k-\frac{K}{2}},t)\vert.
\end{displaymath} (33)

Similarly, comparing the phase terms between Eqs. ([*]) and ([*]), we obtain

 \begin{displaymath}\omega_k t +\phi_k(t)=\arg (\tilde{f}(a,b)).
\end{displaymath} (34)

Since the phase spectrum $\arg(\tilde{f}(a,b))$ is represented by

\begin{displaymath}\arg(\tilde{f}(a,b))=\tan^{-1} \frac{{\it Im}\{\tilde{f}(a,b)\}}{{\it
Re}\{\tilde{f}(a,b)\}},
\end{displaymath} (35)

it becomes a periodical ramp function within $
-\pi \leq \arg(\tilde{f}(a,b)) \leq \pi.
$Differentiating both terms in Eq. ([*]), it becomes

\begin{displaymath}\omega_k+\frac{d\phi_k(t)}{dt}=\frac{\partial}{\partial
t}\arg(\tilde{f}(\alpha^{k-\frac{K}{2}},t)).
\end{displaymath}

After clearing, we obtain

\begin{displaymath}\frac{d\phi_k(t)}{dt}=\frac{\partial}{\partial
t}\arg(\tilde{f}(\alpha^{k-\frac{K}{2}},t))-\omega_k .
\end{displaymath}

Hence, the output phase $\phi_k(t)$ is represented by

\begin{displaymath}\phi_k(t)=\int
\left(\frac{d}{dt}\arg\left(\tilde{f}(\alpha^{k-\frac{K}{2}},t)\right)-\omega_k\right)dt.
\end{displaymath}

$\Box$


next up previous
Next: Biography Up: No Title Previous: Bibliography
Masashi Unoki
2000-10-26