Appendix 1. Proof of Lemma 1

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

 

$$S_k(t)e^{j(\omega_k t + \phi_k(t))}$$ X_k(t) =

$$\tilde{f}(a,b),\qquad a=\alpha^{k-\frac{K}{2}},b=t.$$ := (32)

Representing an absolute value for both terms, we obtain

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

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

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

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

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

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

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

After clearing, we obtain

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

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

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

$\Box$