The CWT (continuous wavelet transform) of a signal $x(t)$ is
$$X_w(a,b)=\frac{1}{\sqrt{|a|}} \int_{-\infty}^{\infty} x(t)\psi^{\ast}\left(\frac{t-b}{a}\right)\, dt$$
In order to reconstruct the signal, we can use this inverse formula :
$$x(t)=\frac{1}{B_\psi}\int_{0}^{\infty}\int_{-\infty}^{\infty} \frac{1}{a^2}X_w(a,b)\frac{1}{\sqrt{|a|}} \tilde\psi\left(\frac{t-b}{a}\right)\, db\ da$$
where
$$ B_\psi := \int_0^\infty \frac{\left|\hat{\psi}(u)\right|^2}{u} du.$$
But I read in some places (http://wavelets.ens.fr/ENSEIGNEMENT/COURS/CNRS/3_Marie_CWT_1D.pdf) that it is possible to do reconstruction with a different function than the original analysis wavelet (and it should give the same result ! I tried with numerical signals in Python, and it is true, it works!).
For example using $\psi(x)=\delta(x)$ leads to the Morlet reconstruction formula :
$$x(t)= \dfrac{1}{C_\psi}\int_{0}^{\infty} \frac{1}{a^2}X_w(a,t)\frac{1}{\sqrt{|a|}} \,da$$
where
$$ C_\psi := \int_0^\infty \frac{\overline{\psi(u)}}{u} du $$
Do you know where we can find a proof for this ?
Best Answer
Here's a proof of Morlet's inversion formula. We define \begin{align*} C_\psi &:= \int_0^\infty \dfrac{\overline{\hat{\psi}(u)}}{u}\, \mathrm{d}u \end{align*} and assume that this quantity is finite. Assume that $x,\psi \in L^2(\mathbb{R})$. I make this assumption so that Parseval's identity holds. Then, using Parseval's identity and basic properties of Fourier transforms (i.e., dilation becomes contraction on the Fourier side, and translation becomes phase modulation on the Fourier side), we can write
\begin{align*} X_w(a,b) &= \int_{-\infty}^\infty x(t) \frac{1}{\sqrt{a}} \overline{\psi\left(\frac{t-b}{a}\right)} \, \mathrm{d}t \\&= \int_{-\infty}^\infty \hat{x}(\xi) \frac{1}{\sqrt{a}} \overline{a\hat{\psi}\left(a\xi\right) \mathrm{e}^{-2\pi \mathrm{i} b \xi}}\, \mathrm{d}\xi \\&= \sqrt{a} \int_{-\infty}^\infty \hat{x}(\xi) \mathrm{e}^{2\pi \mathrm{i}b \xi}\overline{\hat{\psi}(a\xi)} \, \mathrm{d}\xi. \end{align*} Now, assume that $\hat{x} \in L^1(\mathbb{R})$ and that \begin{align*} A_\psi &:= \int_{0}^\infty \dfrac{\left|\hat{\psi}(u)\right|}{u} \, \mathrm{d}u \end{align*} is finite. I make these assumptions so that the precondition of the Fubini-Tonelli Theorem (i.e., absolute integrability of the double integral over $a$ and $\xi$) holds. Then, we can plug in our expression above for $X_w(a,t)$ and use the Fubini-Tonelli Theorem to switch the order of the integrals: \begin{align*} \frac{1}{C_\psi} \int_0^\infty X_w(a,t) \frac{\mathrm{d}a}{a^{3/2}} &= \frac{1}{C_\psi} \int_0^\infty \frac{\sqrt{a}}{a^{3/2}} \int_{-\infty}^\infty \hat{x}(\xi) \mathrm{e}^{2\pi \mathrm{i}b\xi} \overline{\hat{\psi}(a\xi)} \,\mathrm{d}\xi \mathrm{d}a \\&= \frac{1}{C_\psi} \int_{-\infty}^\infty \hat{x}(\xi) \mathrm{e}^{2\pi \mathrm{i}b\xi} \int_0^\infty \dfrac{\overline{\hat{\psi}(a\xi)}}{a} \, \mathrm{d}a \mathrm{d}\xi \\&= \frac{1}{C_\psi} \int_{-\infty}^\infty \hat{x}(\xi) \mathrm{e}^{2\pi \mathrm{i}b\xi} \int_0^\infty \dfrac{\overline{\hat{\psi}(u)}}{u} \, \mathrm{d}u \mathrm{d}\xi \\&= \frac{C_\psi}{C_\psi} \int_{-\infty}^\infty \hat{x}(\xi) \mathrm{e}^{2\pi \mathrm{i}b\xi} \\&= x(b), \end{align*} where in the third-to-last line we used the change-of-variable $u = \xi a$ in the inner integral, and in the last line we used Fourier inversion.
Note: The finiteness of $A_\psi$ is implied by the finiteness of $C_\psi$ provided that we're talking about Lebesgue integration. However, if we understand $C_\psi$ in the sense of an indefinite Riemann integral, then this is not necessarily guaranteed. (I can't come up with a real counterexample off the top of my head, but put for example $\hat{\psi}(\xi) = \sin(\xi)$; then $C_\psi$ is a finite indefinite Riemann integral but $A_\psi$ is infinite. Of course, this $\psi$ violates the assumption $\psi \in L^2$, so this isn't really a counterexample.)
Note 2: Regarding the idea of using different functions for reconstruction, the answer lies in the paper of Holschneider and Tchamitchian from 1991, Pointwise analysis of Riemann's "nondifferentiable" function (in Electricman's answer, the paper from Farge cites it as Holschneider & Tchamitchian 1989). Here, they prove the following theorem:
Let $\psi$ be as above and assume there exists a function $g \in L^1(\mathbb{R})$ where $g$ and $\psi$ together satisfy
Suppose then that $x$ is a function which is "weakly oscillating about 0" in the sense that
$$ \lim_{u \rightarrow \infty} \dfrac{1}{2u} \int_{t-u}^{t+u} x(y) \, \mathrm{d}y = 0$$
where the limit holds uniformly in $t$.
Then at any point $t$ where $x$ is continuous, we have the inversion formula
$$ x(t) = \int_0^\infty \int_{-\infty}^\infty X_w(a,b) g\left(\dfrac{t-b}{a}\right)a^{-2} \, \mathrm{d}b \mathrm{d}a, $$
where the outer integral is understood to be a Riemann indefinite integral (i.e., you take limits as you approach the endpoints $0$ and $\infty$), and where $X_w(a,b)$ is the wavelet transform with respect to the mother wavelet $\psi$.
Of course, taking $g$ to be the delta function here isn't technically allowed (because $g$ must be in $L^1$) but it's still interesting to see this theorem.
Thank you for asking this really interesting question! I didn't know that such a different reconstruction could be done until I saw this post.