Wiki writes iCWT as

$$ f(t) = C_{\psi}^{-1} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} W_f(a,b) \frac{1}{|a|^{1/2}} \tilde\psi \left(\frac{t - b}{a}\right) db \frac{da}{a^2}, \tag{1} $$

where $\tilde\psi$ is the dual wavelet of $\psi$ in CWT, $W_f$ is the CWT of $f$, and $C_\psi$ is the admissible constant

$$ C_\psi = \int_{-\infty}^{\infty} \frac{\bar{\hat{\psi}}(\omega) \hat{\tilde{\psi}}(\omega)}{|\omega|} d\omega, \tag{2} $$

Hat = $\mathcal{F}$, overbar = complex conjugate. Some wavelets are own duals, $\psi = \tilde\psi$; then the integrand is $|\hat\psi(\omega)|^2 / |\omega|$.

How does $(1)$ invert the CWT? And why is there a $a^{-2}$ normalizing factor? Above wavelets are L2-normed - do we normalize iCWT differently if they are L1-normed (i.e. $|a|^{-1}$)?

The normalization is particularly troubling; in forward-transform, we do $|a|^{-1/2}$, so in inversion one might expect $|a|^{1/2}$ for cancellation, but it only goes the other way: another $|a|^{-1/2}$, along $a^{-2}$. The role of $C_\psi ^{-1}$ is also unclear.

  • 1
    $\begingroup$ So you ask a question and answer it within a minute. $\endgroup$ – Laurent Duval Oct 30 '20 at 17:03
  • $\begingroup$ @LaurentDuval If that's meant to be a complaint, self Q&A's are neither prohibited nor discouraged in SE; my complaint would be this network's complaints on someone putting effort into providing value to others. $\endgroup$ – OverLordGoldDragon Oct 30 '20 at 17:39
  • $\begingroup$ I am only concerned about made-up questions $\endgroup$ – Laurent Duval Oct 30 '20 at 18:31
  • $\begingroup$ @LaurentDuval You're welcome. $\endgroup$ – OverLordGoldDragon Oct 30 '20 at 18:34

Summary: the dual wavelet's role is analogous to that of $e^{j\omega t}$; it undoes the wavelet's convolving with the signal (integrated inner product). The main intricacy's indeed in normalization; there's no exact answer without digging in math (done below) - but intuitively, the seemingly matter-worsening $a^{-2}$ is due partly to the dilating nature of the wavelet, i.e. $\psi(t/a)$, which upon inversion requires downscaling not only to properly preserve energy, but reconstruct correctly by undoing the inner product.

Below are more "proofs" rather than "derivations", as they work backwards to show equality with $f(t)$, but I term them latter as full-fledged proofs are beyond the scope of this answer.

Derivation 1: Convolution Theorem

Pasted for convenience the CWT and iCWT

$$ W_f(a, b) = \int_{-\infty}^{\infty} f(t) \frac{1}{|a|^{1/2}} \psi \left(\frac{t - b}{a}\right) dt \tag{0} $$ $$ f(t) = \frac{1}{C_{\psi}} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} W_f(a,b) \frac{1}{|a|^{1/2}} \psi \left(\frac{t - b}{a}\right) db \frac{da}{a^2}, \tag{1} $$

Begin by rewriting the CWT as a convolution: $W_f(a, b) = f \star {\bar\psi_a}(b)$, where $\psi_a(t)=a^{-1/2}\psi(t/a)$. Inserting into $(1)$, and letting the right side equal $b(t)$,

$$ \begin{align} b(t) &= \frac{1}{C_\psi} \int_0^{\infty} W_f(.,a) \star \psi_a(t) \frac{da}{a^2} \\ &= \frac{1}{C_\psi} \int_0^{\infty} f \star {\bar\psi_a}(b) \star \psi_a(t) \frac{da}{a^2} \tag{2} \end{align} $$

where we also expressed the $db$ integral as a convolution, and $"."$ indicates the variable over which convolution is computed. We prove $b=f$ by showing their Fourier Transforms are equal. With $\hat{} = \mathcal{F}$,

$$ \begin{align} {\hat b}(\omega) &= \frac{1}{C_\psi} \int_0^{\infty} {\hat f}(\omega) \sqrt{a} {\bar{\hat \psi}}(a\omega) \sqrt{a} {\hat\psi}(a\omega) \frac{da}{a^2} \tag{3a} \\ &= \frac{{\hat f}(\omega)}{C_\psi} \int_0^{\infty} \tag{3b} |{\hat \psi}(a\omega)|^2 \frac{da}{a} \end{align} $$

With change of variables $\xi = a\omega$, we thus 'prove'

$$ {\hat b}(\omega) = {\hat f}(\omega) \left(\frac{1}{C_\psi} \int_0^\infty \frac{|{\hat \psi}(\xi)|^2}{\xi} d\xi \right) = {\hat f}(\omega). \tag{4} $$


It's tempting to attribute the $a^{-2}$ to the wavelet dilations, $\psi(t/a)$, as they manifest as rescalings of Frequency-domain components ($\cdot \sqrt{a}$), and while that's true, it's not the crux.

Recall when I said it's not just about energy? Consider what happens if the normalization is anything other than $a^{-2}$. To arrive at $(4)$, we relied on $a^1$ in denominator in $(3b)$, else there is no cancellation with $C_\psi$, so no reconstruction. Thus, one can conclude that $a^{-2}$ is set in retrospect - i.e. "what do we make it to cancel $C_\psi$".

Note that we can't go the other way, i.e. redefine $C_\psi$ such that $a^{-2}$ is no longer needed - at least not entirely; per above derivation, it's clear that $\psi$ must somehow drop, thus $C_\psi$ must include it. The usage of exactly $C_\psi$ might be convention-based, since $C_\psi$'s applicability spans much outside the iCWT as the fundamentally defining criterion of wavelets (Farge, 1992).

So how to invert L1-normed CWT? Obvious from above.


  • Proof followed Mallat, Theorem 4.4
  • Split integrals based on identity
  • Assumed $f \in \mathbf{L^2}(\mathbb{R})$ (square-integrable)
  • $\psi$ was also assumed real, and note the tilde is missing. I'm unsure why such a restriction was made, or whether Mallat implies that all real $\psi$ are self-dual, but the rescaling identities (e.g. $a \Rightarrow a^{-1}$ via $\mathcal{F}$) should hold regardless. Derivation 2 does not restrict $\psi$ to real.

Derivation 2: Resolution of Identity

For all $f, g \in \mathbf{L^2}(\mathbb{R})$ (real or complex), below holds:

$$ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{da db}{a^2} W_f(a, b) \overline{ W_g (a, b)} = C_\psi \left< f, g \right> \tag{5} $$

Begin proof by transforming to frequency domain:

$$ \begin{align} (5) = \int \int \frac{da db}{a^2} & \left[ \int d\xi {\hat f}(\xi) |a|^{1/2} e^{-jb\xi} \overline{ {\hat \psi}(a\xi)} \right] \\ \cdot & \left[ \int d\xi' \overline{ {\hat g}(\xi')} |a|^{1/2} e^{j b\xi'} {\hat \psi}(a\xi') \right] \tag{6} \end{align} $$

This is similar to derivation 1, except $\psi$ replaced with another CWT of some different function $g$. In the brackets we have the Fourier transform of $F_a(\xi) = |a|^{1/2} {\hat f(\xi)} \overline{\hat \psi (a\xi)}$ times the Fourier transform of complex conjugate of $G_a(\xi) = |a|^{1/2} {\hat g(\xi)} \overline{\hat \psi (a\xi)}$. By the unitarity of the Fourier transform, we have

$$ \begin{align} (6) &= 2\pi \int \frac{da}{a^2} \int d\xi F_a(\xi) \overline{G_a(\xi)} \\ &= 2\pi \int \frac{da}{|a|} \int d\xi {\hat f(\xi)} \overline{ {\hat g}(\xi)} |{\hat \psi}(a\xi)|^2 \\ &= 2\pi \int d\xi {\hat f(\xi)} \overline{ {\hat g}(\xi)} \int \frac{da}{|a|} |{\hat \psi}(a\xi)|^2 \ \ \ \ \text{(Fubini's theorem)} \\ &= C_\psi \left< f,g \right>, \end{align} $$

where in last step a similar change of variables was used, $\zeta = a\xi$. $(6)$ can be read as

$$ f = C_\psi^{-1} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{da db}{a^2} W_f(a, b) \psi^{a, b}, \tag{7} $$

quoting the source,

with convergence of the integral "in the weak sense", i.e., taking the inner product of both sides of $(7)$ with any $g \in \mathbf{L^2}(\mathbb{R})$, and commuting the inner product with the integral over $a, b$ in the right-hand side, leads to a true formula.

This seems to say that the original derivation was for the inner product $\left< f, g \right>$, whereas $(7)$ reads only for $(6)$, which will have somewhat different properties. The "original" is recovered by convolving with a $g$. -- I'm also unsure what exactly "in the weak sense" means; this is the most I found (clarifications welcome).

Once again, it is evident that $a^{-2}$ was set in retrospect, depending on how the forward transform $W$ was normalized.

Source: Daubechies, Ten Lectures on Wavelets, Chapter 2


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