0
$\begingroup$

Is there a function, in continuous or (primarily) discrete time, such that convolving with it preserves the input's energy? For $x$ that is finite-valued, finitely supported in time, and:

  1. Real or complex?
  2. Real?
  3. Real and non-negative?

where "energy" refers to L2^2:

$$ \text{E}(f(x)) = \int_{a}^{b} |f(x)|^2 dx \ \ \text{or}\ \ \sum_{n=a}^{b} |f[n]|^2 $$

and following must hold

$$ \text{E}(h \star x) = \text{E}(x) $$

where $\star$ is linear or circular convolution (either works for me). If there's no such kernel, can this be proven?


I'm aware of the trivial case of the delta; $h$ should be such that $x$ is transformed non-trivially (not just a copy, conjugate, rescaling, sign flip, etc).

Idea: we can satisfy Plancherel's formulation per Envidia's answer with a distribution, example being the Dirac delta, but unsure there'd be something one could implement finitely.

$\endgroup$
1
2
$\begingroup$

Assuming that the conditions are met to invoke Plancherel's (aka Parseval's) theorem, we can reformulate this in the frequency domain where the energy equality $E(x \star h) = E(x)$ can be expressed as

$$\int_{-\infty}^{\infty}|X(f)H(f)|^2df = \int_{-\infty}^{\infty}|X(f)|^2df$$

I see no condition where energy is preserved unless $|H(f)| = 1$ (all-pass), the main example being the delta function in the time domain as you already said.

$\endgroup$
2
  • $\begingroup$ Think it should be $|H(f)|=1$, i.e. all-pass. Makes sense but I wonder if finite support changes anything - for one we can easily design an L1-preserving $h$ for non-negative $x$. $\endgroup$ May 20 at 17:27
  • 1
    $\begingroup$ Finite time support would have infinite Fourier support so the theorem stays relevant... Fair enough. $\endgroup$ May 20 at 19:38
1
$\begingroup$

I think any allpass filter should do that. Since they only change phase and not amplitude in the frequency domain, the time domain energy remains unchanged per Parseval's theorem.

$$\sum x[n]^2 = \sum |X(k)|^2$$

An all pass only changes the phase of $X(k)$ so total energy is maintained.

You could argue that an allpass is an IIR filter, so it's not suitable to be a convolution kernel (which implies FIR). But some allpass filters are FIR (delays) and you can always truncate an allpass IIR impulse response to the "desired precision".

$\endgroup$
3
  • $\begingroup$ I could've been clearer but this qualifies as "trivial" as it copies the input's spectrum magnitude. $\endgroup$ May 20 at 17:20
  • $\begingroup$ A general allpass filter might preserve «local» energy (from a...b) for periodic input but not for general input? $\endgroup$
    – Knut Inge
    May 20 at 17:51
  • $\begingroup$ If you define "trivial" as synonymous to "preserving input spectrum magnitude" then you can prove to yourself that any solution that preserves energy in all cases is "trivial", no matter how much it may smear the signal out in real life. $\endgroup$
    – TimWescott
    May 20 at 19:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.