Appendix C: Proof of Lemma 2

The wavelet transform in Eq. (21) is a complex representation of the output of the analyzing filter in Eq. (2):

 

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

$$\tilde{f}(a,b),\qquad a=\alpha^{k-\frac{K}{2}},b=t$$ := (48)

Taking the absolute value for both terms, we obtain

$$\vert X_k(t)\vert=S_k(t)=\vert\tilde{f}(\alpha^{k-\frac{K}{2}},t)\vert.$$ (49)

Similarly, comparing phase terms between Eqs. (48) and (21), we obtain

$$\omega_k t +\phi_k(t)=\arg (\tilde{f}(a,b)).$$ (50)

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

$$\arg(\tilde{f}(a,b))=\arctan \frac{{\it Im}\{\tilde{f}(a,b)\}}{{\it Re}\{\tilde{f}(a,b)\}},$$ (51)

it becomes a periodical ramp function within

$$-\pi \leq \arg(\tilde{f}(a,b)) \leq \pi.$$ (52)

Differentiating both terms in Eq. (50), we get

$$\omega_k+\frac{d\phi_k(t)}{dt}=\frac{\partial}{\partial t}\arg(\tilde{f}(\alpha^{k-\frac{K}{2}},t)).$$ (53)

After clearing, we obtain

$$\frac{d\phi_k(t)}{dt}=\frac{\partial}{\partial t}\arg(\tilde{f}(\alpha^{k-\frac{K}{2}},t))-\omega_k.$$ (54)

Hence, the instantaneous output phase $\phi _k(t)$ is represented by

$$\phi_k(t)=\int \left(\frac{d}{dt}\arg\left(\tilde{f}(\alpha^{k-\frac{K}{2}},t)\right)-\omega_k\right)dt.$$ (55)