Parseval's theorem can be interpreted as:

... the total energy of a signal can be calculated by summing power-per-sample across time or spectral power across frequency.

For the case of a signal $x(t)$ and its Fourier transform $X(\omega)$, the theorem says:

$$ \int{|x(t)|^2 \; dt} = \int{|X(\omega)|^2 \; d\omega} $$

For the case of discrete wavelet transform (DWT), or wavelet packet decomposition (WPD), we get a 2D array of coefficients along the time and frequency (or scale) axis:

        | c{1,f}
        | ...
freq    | c{1,2}
        | c{1,1} c{2, 1} ... c{t, 1}

Can a sum of this series somehow be understood as a signal's energy? Is there an equivalent rule to Parseval's theorem?


2 Answers 2


Yes indeed! In theory as long as the wavelet is orthogonal, the sum of the squares of all the coefficients should be equal to the energy of the signal. In practice, one should be careful that:

  • the decomposition is not "expansive", i.e. the number of samples and of coefficients is the same.
  • wavelet filter coefficients are not re-scaled, as happens in some applications (like lifting wavelets, to keep integer computations).
  • wavelets are orthogonal (this is not the case in JPEG2000 compression).

You can verify this indirectly, looking at approximation coefficients. At each level, their number of samples is halved, and their amplitudes have around a $1.4$ scale factor, which is just $\sqrt{2}$. This feature is used for instance to estimate the Gaussian noise power from wavelet coefficients:

$$ \hat{\sigma} = \textrm{median} (w_i)/0.6745$$

A little further, there is a notion that generalizes (orthonormal) bases: frames. A set of functions $(\phi_i)_{i\in \mathcal{I}}$ ($\mathcal{I}$ is a finite or infinite index set) is a frame if for all vectors $x$:

$$ C_\flat\|x\|^2 \le \sum_{i\in \mathcal{I}} |<x,\phi_i>|^2\le C_\sharp\|x\|^2$$

with $0<C_\flat,C_\sharp < \infty$. This is a more general Parseval-Plancherel-like result used for general wavelets.

In other words, it "approximately preserves energy" by projection (inner product). If the constants $ C_\flat$ and $C_\sharp $ are equal, the frame is said to be tight. Orthonormal bases are non-redundant sets of vectors with $ C_\flat=C_\sharp = 1 $.

For those using Matlab, you should care about the native border extension, which is obtained by dwtmode('status'). Some add tails to the data to help inversion with little border artifacts. With a periodic mode dwtmode('per') and a number of samples that can be divided by $2^L$ where $L$ is the wavelet level, you can get a good match in energy, with tiny differences:

Energy in the DWT domain

  • $\begingroup$ Can you elaborate on what orthogonality means in this case? $\endgroup$
    – hazrmard
    Commented Jun 26, 2019 at 15:22
  • $\begingroup$ When the scalar product vanishes for two distinct elements? $\endgroup$ Commented Jun 26, 2019 at 16:23
  • $\begingroup$ So in this case, does that mean that the sum of the inner product of a wavelet with each of its scaled and translated versions will be 0? $\endgroup$
    – hazrmard
    Commented Jun 26, 2019 at 16:44
  • 1
    $\begingroup$ This is my understanding: when a signal is projected on to orthogonal wavelets, there is no "overlap" in the representation of the signal between wavelets. Exactly like when you project a vector on an orthogonal basis, each basis vector carries exclusive information about the projection. So when you sum the squares of the projection (i.e. coefficients) there is no double-counting - hence "energy" is conserved. $\endgroup$
    – hazrmard
    Commented Jun 26, 2019 at 16:54
  • 1
    $\begingroup$ I tried to verify this in Matlab, in which I used the functions WAVEDEC - and so DWT - and DETCOEF and APPCOEF. I don't have a very good match, especially if the original signals have many samples (~70000). Is this because the wavelet decomposition is expansive using the MATLAB code DWT? $\endgroup$
    – EmThorns
    Commented Jun 11, 2020 at 13:19

Yes. The instantaneous frequency, multiplied by the square modulus of the transformed function, upon integration, yields a Parseval theorem result for the original signal - provided that it has been first reduced to an analytic signal. If it has not been, then the result is integral of the ± square modulus of the signal, in the frequency domain, with a + sign for positive frequency components and a - sign for negative frequency components.

For the wavelet transform $\tilde f(τ,σ)$, of a function $f(t)$, the derivative of $(\arg \tilde f(τ,σ))$ with respect to $τ$ is $2π$ times the instantaneous frequency at $ν_f(τ,σ)$ at $(τ,σ)$. The integral of $ν_f(τ,σ) |\tilde f(τ,σ)|^2$, up to a normalization factor, is the desired result.

For this result, it does not matter what the inverse transform is - or if there even is one. So, it applies to any transform whose forward direction is a wavelet transform. In particular, it applies to both the S and Q transforms; except that the Q transform is normally only defined discretely, not as a continuous transform, though it can be, in which case it becomes a generalization of the S transform.

In particular, if $f(t)$ is the signal and its Fourier transform is defined by $$\hat f(ν) = ∫ f(t) 1^{-νt} dt$$ where - for convenience - I am writing complex sinusoidals as $e^{2πix} = 1^x$, and if the wavelet transform is written as $$\tilde f(τ,p) = ∫ f(t) \overline{ψ(p(t - τ)) |p|^A} dt$$ where - for convenience - I'm using $p = 1/σ$ in place of $σ$ (it plays a role analogous to the frequency in the time-frequency plane), then $$∫ ∫ |\tilde f(τ,p)|^2 ν_f(τ,p) {dτ dp \over |p|^{2A}} = \left(∫ \left|\hat f(ν)\right|^2 sign(ν) dν\right) \left(∫ |\hat ψ(P)|^2 dP\right).$$ The choice $A = ½$ makes this an integral over the log scale in $p$ (or in $σ$), with the integral measure $dτ dp/|p|$ or $dτ dσ/|σ|$, while $A = 0$ makes it an integral with over $p$ as a linear scale with measure $dτ dp$, and $A = 1$ makes it an integral with measure $dτ dσ$, with $σ$ now being on a linear scale.

This requires that $$0 < ∫ |\hat ψ(P)|^2 dP < ∞$$

For the wavelet transform, since the same windowing function is used in both the forward and reverse directions (with $A = ½$), there is already another admissibility condition $$0 < ∫ |\hat ψ(P)|^2 {dP \over |P|} < ∞$$ so that it can be normalized. So, the integral admissibility condition has to hold in addition to the transform's admissibility condition.

A much weaker admissibility condition is required for the S transform that can, for instance, be satisfied by enveloped complex sinusoidals; so, it is easier to do.

The actual expression for the integrand could be written in a form evocative of what's often seen in the literature in Quantum Field Theory $$|\tilde f(τ,p)|^2 ν_f(τ,p) = \overline{\tilde f(τ,p)} \left({1 \over {4πi}} \overleftrightarrow{∂ \over ∂τ}\right) \tilde f(τ,p),$$ where $\overleftrightarrow{∂/∂τ}$ is $∂/∂τ$ applied to the right minus $∂/∂τ$ applied to the left. (Edit: I previously wrote, by mistake, that it was half right minus half left. That convention is also in use in the Physics literature.) In particular, $$∫ \overline{1^{ν'τ}} \left({1 \over {4πi}} \overleftrightarrow{∂ \over ∂τ}\right) 1^{ντ} dτ = ∫ 1^{-ν'τ} \left(ν + ν' \over 2\right) 1^{ντ} dτ = {ν + ν' \over 2} δ(ν - ν') = ν δ(ν - ν').$$

The wavelet transform can be written in the frequency domain as $$\tilde f(τ,p) = ∫ \hat f(ν) \overline{\hat ψ\left({ν \over p}\right)} 1^{ντ} |p|^{A-1} dν.$$ Throwing this all together, we get $$\begin{align} ∫ \overline{\tilde f(τ,p)} \left({1 \over {4πi}} \overleftrightarrow{∂ \over ∂τ}\right) \tilde f(τ,p) dτ & = ∫ ∫ \overline{\hat f(ν')} \hat ψ\left(ν' \over p\right) |p|^{2A-2} ν δ(ν - ν') \hat f(ν) \overline{\hat ψ\left(ν \over p\right)} dν' dν \\ & = ∫ {\left|\hat f(ν)\right|}^2 {\left|\hat ψ\left(ν \over p\right)\right|}^2 ν |p|^{2A-2} dν. \end{align}$$ Applying this to the integration with $dp/|p|^{2A}$, we have $$\begin{align} ∫ ∫ |\tilde f(τ,p)|^2 ν_f(τ,p) {dτ dp \over |p|^{2A}} & = ∫ ∫ {\left|\hat f(ν)\right|}^2 {\left|\hat ψ\left(ν \over p\right)\right|}^2 {ν dν dp \over |p|^2} \\ & = ∫ ∫ {\left|\hat f(ν)\right|}^2 sign(ν) {\left|\hat ψ\left(ν \over p\right)\right|}^2 dν {|ν| dp \over |p|^2}, \end{align}$$ and finally, with the substitution $P = ν/p$ and the separation of the $ν$ and $P$ integrals out from one another, we have result originally cited above.


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