# Partitioning the energy of a signal between its components

In physics, the total energy of a system is the summed energy of its component subsystems, and energy must be conserved, ie, adding the energy of the components gives the total energy.

In signal analysis (discrete time signals), one would expect that the energy of a signal would be defined in such a way that energy would again be conserved. This would make it consistent with normal usage in physics.

There are many types of component analysis. In general, there is no requirement that the components are zero-mean or orthogonal, and in data-dependent component extractions, the components are not even predictable, but in all cases it is expected that they linearly sum to the original signal values.

This is incompatible with the usual definition of signal energy as the sum of the squared component values, since the component energies do not sum to the original signal energy.

This is confusing. Is there another approach to defining the component energies that conserves energy but does not destroy the requirement that the linear sum of components must be equal to the original signal?

## 2 Answers

This question goes deep into the notion of selecting "invariant" features between representations.

From your first paragraph, I understand that your component subsystems are mutually exclusive (they don't overlap), and their union yields the whole system. This decomposition can be called covering, or segmentation for signals and images. Such nonlinear operations (intersection, union) are well-known, and used notably in mathematical morphology. They obey mathematical structure called lattices, Boolean algebras, etc. But not vector spaces that are common with linear decomposition.

If you split a signal into non-overlapping chunks, those subsystems have energy preservation, as above. If you move to other domains (frequency), orthogonal systems (or tight frames) allow to ensure that energy is preserved (eg managed overlap in the frequency domain).

But as long as one does not constrains representations, one cannot hope to preserve energy. Indeed, a linear equation is the derivative of a squared problem: the average $$\hat{m}$$ is the quantity that minimizes the sum of squared distance:

$$\arg \min \sum_1^N (x_n -m)^2$$

because the derivative vanishes when:

$$\sum_1^N (x_n -m) =0$$ which is reached when $$\sum_1^N x_n = N\hat{m} \,.$$

Of course, without constraint, $$0 = n \times s - (n-1)\times s$$

but you cannot sum the energies of components in a sound manner. Energy is energy. NOte however that for signal decompositions, one has extended the notion of orthogonality. For instance, frames are sets of vectors for which the energy is approximately preserved. That is: the ratio of the energy of the decomposition by the signal's energy is bounded above and below. If the bounds are equal, this is a tight frame.

If you want to preserve quantitative metrics, you can look at other features and measures: the whole families of entropies, other norms and divergences, norm ratios, etc. But the objective ought to be more precise.

• "as long as one does not constrain representations, one cannot hope to preserve energy" – kordes Jul 10 '20 at 21:28
• thanks for the confirmation that you can't hope to conserve energy when the component representation is unrestrained. This was the conclusion I'd reached, but as I am self-taught I wondered if there was some clever solution using other definitions. I might have also sent a fragmentary comment by accident - apologies. – kordes Jul 10 '20 at 21:46
• As for signal decompositions, one has extended the notion of orthogonality. For instance, frames are sets of vectors for which the energy is approximately preserved. That is: the ratio of the energy of the decomposition by the signal's energy is bounded above and below. If the bounds are equal, this is a tight frame – Laurent Duval Jul 12 '20 at 11:04

Say you decompose the signal $$x(t)$$ into two components: $$s(t) = x(t) + y(t).$$ As you say, there are myriad ways to do this. The energy of $$s(t)$$ is $$E_s = \int_{-\infty}^\infty s^2(t) dt.$$

If we calculate it using the decomposition, we obtain:

\begin{align}E_s &= \int_{-\infty}^\infty (x(t) + y(t))^2(t) dt \\ &= \int_{-\infty}^\infty x^2(t) dt + \int_{-\infty}^\infty y^2(t) dt + 2\int_{-\infty}^\infty x(t)y(t) dt \end{align}

Clearly, if you require that $$E_s = \int_{-\infty}^\infty x^2(t) dt + \int_{-\infty}^\infty y^2(t) dt$$, then it is necessary that $$\int_{-\infty}^\infty x(t)y(t) dt = 0$$ or, in other words, $$x(t)$$ and $$y(t)$$ must be orthogonal.

By the way, I have never seen it claimed that all existing decompositions have the energy-conserving property you are looking for, not even in the context of physics rather than engineering.