I want to convolve discrete signals A and B. I can compute their energies beforehand by squaring the samples and summing the squares, but I'm curious if I can compute the energy of the signal I will get if I convolve A and B using their energies (or anything about them really) without even doing convolution? Is there any mathematical relationship there?

  • 1
    $\begingroup$ I think the best you can do is to find an upper bound on the energy of $A * B$ -- think in terms of B being a unity-gain filter with a certain bandpass. Then the most energy that $A * B$ can have is if $A$'s energy is all concentrated in $B$'s bandpass, and the energy of $A * B$ is equal to the energy of $A$. But if $A$ has energy outside of $B$'s bandpass, $A * B$ will have less energy than $A$. Trying to turn that intuition into math makes my head explode, however. $\endgroup$
    – TimWescott
    May 7, 2022 at 14:25
  • $\begingroup$ I've added some new bounds. $\endgroup$ May 11, 2022 at 17:03
  • $\begingroup$ Are A and B knowns? Is it theoretical question or in practice you're trying to preserve some computational efforts? $\endgroup$
    – Royi
    May 16, 2022 at 20:52

3 Answers 3


Probably not an equality directly, but upper bounds. Let us look at the continuous case first, which is easier to derive. There is a Young's convolution inequality: with proper integrability conditions ($A$ is $L_p$ integrable, $B$ is $L_q$ integrable), $1\le p,q\le\infty$ and conjugation: $$\frac{1}{p}+\frac{1}{q}=\frac{1}{r}+1\,.$$

On the one hand then: $$\|A\ast B \|_r \le c_{p,q} \|A \|_p \|B\|_q$$

for some constant $c_{p,q}$ that can be computed. On the other hand, there are non-zero vectors whose convolution vanishes, see When does the convolution of 2 signals equal zero?. Therefore, knowing something about $A$ and $B$ separately won't get you a precise idea about their convolution: you can hope for lower and upper bounds, but no equality. If you are only interested in the energy of the convolution, set $r=2$.

In the discrete case, with standard or circular convolution, I remember that results were more complicated to derive, but you can hope for inequalities as well (see works of Beckner and related, like Optimal Young's inequality and its converse: a simple proof.

  • 1
    $\begingroup$ ha! Beautiful! That's exactly the motivation for matched filtering! $\endgroup$ May 7, 2022 at 12:03
a = ones(64,1);
b = [1 -1];
c = conv(a,b);

In this case, a and b have some non-zero energy while their convolution is exactly zero besides head and tail conditions.

  • $\begingroup$ With circular convolution, I guess $\endgroup$ May 7, 2022 at 10:47
  • $\begingroup$ Not to be picky here, but your variable c is NOT all zeros. $\endgroup$
    – Hilmar
    May 7, 2022 at 17:01
  • $\begingroup$ I did say except head and tail conditions, @Hilmar? $\endgroup$
    – Knut Inge
    May 7, 2022 at 17:08

We can get exact bounds from cross-domain energy relations. Given a signal $x(t)$ and a kernel $h(t)$ to convolve with, the frequency-domain result is $X(f)H(f)$, and we know from Parseval-Plancherel's theorem that

$$ ||x * h|| = ||XH||. $$


$$ A \leq |H(f)|^2 \leq B $$ that is, the energy of any frequency of $h$ is bound between $A$ and $B$. Then, multiplying by $|X(f)|^2$, we get

$$ A |X(f)|^2 \leq |H(f)X(f)|^2 \leq B |X(f)|^2 $$

applying integration,

$$ A \int |X(f)|^2 \leq \int |H(f)X(f)|^2 \leq B \int |X(f)|^2 $$

then the theorem

$$ A \int |x(t)|^2 \leq \int |x * h|^2 \leq B \int |x(t)|^2 \\ \Leftrightarrow \\ A||x||^2 \leq ||x * h||^2 \leq B ||x||^2 \\ $$

which is

$$ \boxed{\text{min}(|H(f)|^2) ||x||^2 \leq ||x * h||^2 \leq \text{max}(|H(f)|^2) ||x||^2} $$

Hence, the energy of convolving with $h$ is bound by the minimum and maximum of energy of any frequency of $h$. Since $x * h = h * x$, one can repeat this argument starting with $X(f)$:

$$ \boxed{\text{min}(|X(f)|^2) ||h||^2 \leq ||x * h||^2 \leq \text{max}(|X(f)|^2) ||h||^2} $$

This holds in both continuous and discrete domains, where the discrete energy relation is

$$ \sum_{n=0}^{N-1} |x[n]|^2 = \frac{1}{N}\sum_{k=0}^{N-1} |X[k]|^2 $$

Without going to frequency domain, working with time-domain energies alone, we can't tell much, as the energy can exceed $||x||^2 \cdot ||h||^2$ or even be zero while neither's energy is.


We can have $||a|| > 0$ and $||b|| > 0$ while $||a * b|| = 0$:

import numpy as np
from numpy.fft import fft, ifft

def E(x):
    return np.sum(np.abs(x)**2)

x = np.random.randn(128)
x -= x.mean()
h = np.ones(128)
conv = ifft(fft(x) * fft(h))

assert np.allclose(conv, 0)
print(E(x), E(h), E(conv))
117.73576304545941 128.0 1.6155871338926322e-27

because $\text{min}(||X[k]||^2) = 0$ (the DC bin, due to x -= x.mean()). Note, this is for circular convolution, but since we can reformulate it as an exact equivalent of linear, one simply needs to account for spectral effects of zero padding to derive a similar relation (and since zeros don't add energy, the upper bound won't change).

Empirical validation of bounds

We trial many examples of a constant convolved with complex white noise, compute distances with upper and lower bounds, and report minimum distances for each:

def E(x):
    return np.sum(np.abs(x)**2)

dists_min, dists_max = [], []
N = 1024
for _ in range(1000):
    x = np.ones(N)
    h = np.random.randn(N) + np.random.randn(N)*1j
    xf, hf = fft(x), fft(h)
    mn, mx = np.min(np.abs(hf)**2), np.max(np.abs(hf)**2)
    Ex = E(x)
    cconv = ifft(xf * hf)
    dists_min.append(E(cconv) - mn*Ex)
    dists_max.append(mx*Ex - E(cconv))

print(np.min(dists_min), np.min(dists_max))
-9.094947017729282e-13 3.725290298461914e-09

The upper bound is always strictly met, while the lower bound fails only within float precision. If both signals are noise, we're much more within bounds.

  • $\begingroup$ I see where you're going with $E(a * b) = \frac{1}{N}E(AB)$, but if you're using $E(\cdot)$ to denote the energy of something, then I'd go with $E(x) = E\left (\mathcal F \{x\}\right)$ by definition, and, hence, $E\left( \mathcal F \{x\} \right) = \frac{1}{N} \left \| \mathcal F \{ x \} \right \|$ -- even if it's a bit more confusing on the surface. $\endgroup$
    – TimWescott
    May 7, 2022 at 13:59
  • $\begingroup$ @TimWescott Is it by definition for DFT? For CFT, we get $1/2\pi$ with $\omega$, though DFT does $2\pi k$ analogous to CFT's $2\pi f$. Question is, is energy defined as $\sum |\text{stuff}|^2$, or it only happens to work out in time domain (if so, what's the actual definition)? I could open a question $\endgroup$ May 7, 2022 at 14:35
  • 1
    $\begingroup$ I think it's worth a question. It does get a bit philosophical, but I think that for signal processing you want the time domain energy to be $\sum | \text{stuff} |^2$ or $\int |\text{stuff}|^2 dt$, and you want the frequency domain energy for a given signal to calculate out to the same number as the time-domain signal. $\endgroup$
    – TimWescott
    May 7, 2022 at 16:14

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