Suppose that $E(k)$ be the energy distribution of a signal in spectral domain with a finite total energy. Is there anyway to concentrates all the energy of signal around a specific point?

  • $\begingroup$ Is the "specific point" that you are thinking about in the frequency domain or in the time domain? Are nonlinear methods acceptable or are you thinking about linear filtering only? $\endgroup$ Feb 23 '19 at 17:07
  • $\begingroup$ Bandpass filter with a very narrow passband plus normalization to the original level? $\endgroup$
    – bipll
    Feb 23 '19 at 18:13

Without understanding the context of your question, you might consider a Class C power amplifier used in broad cast radio.

It uses a high Q resonator with nonlinear feedback. It is not a DSP technique but does concentrate energy if you have a loose definition of input.

The best you can do with a linear process is to excite a high Q resonance.


Yes indeed. Bear with me, I am in a mood for a long explanation.

Remember physics of motion, and how one concentrates the information related to the motion of an object on its center of mass, that capture a unique point from a distribution of mass.

A similar reasoning can be applied to positive quantities $q(k)$ (like energy) whose sum is finite. Thus, you can normalize them to unitless and positive quantities $q(k)/\sum_k q(k)$ that sums to $1$, similarly to probability distributions.

There, to characterize or summarize distributions with a few values, you can use the notion of moments (probably borrowed from the same moment notion in physics) or cumulants:


The $\alpha$ moments $m_q(c,\alpha)$ of $q(\cdot)$, centered around $c$, are given by:

$$m_q(c,\alpha) = \sum_l (l-c)^\alpha \frac{q(l)}{\sum_k q(k)} = \frac{\sum_l (l-c)^\alpha q(l)}{\sum_k q(k)} \,.$$

Here, the $q$s can be interpreted as unit-sum weights, affecting the "distance" $(l-c)^\alpha$ (more a power-norm, to be honest).

So, in discrete signal processing, using orthogonal or energy-preserving transformations, it is customary to take $q(\cdot)$ as the energy of signal samples, either in the time or the frequency domain. And $l,k$ are discrete indices. For distribution location, we usually set $\alpha=1$.

So for finite-energy signal $s[n]$, you can define its "center mass" in time with:

$$ \overline{s} =\frac{\sum_n n |s[n]|^2}{\sum_n |s[n]|^2}$$ and in frequency by: $$ \overline{E} =\frac{\sum_k k E[k]}{\sum_k E[k]}$$ but beware in summing on positive frequencies only for a meaninful interpretation for real signals (as spectra are symmetric, and would yield a center of frequency mass at $0$).

To illustrate it with a picture and a Matlab code (using FFTR.m):

Center of mass, in time and frequency spectrum

clear all;
timeStart = 0;
timeStop = 1;
nPoint = 1024;
time = linspace(timeStart,timeStop,nPoint)';
dataFreqSampling = 1/median(diff(time));
dataFreq = 56;
dataSpread  = 0.2;
data = sin(2*pi*dataFreq*time).*exp(-((time-(timeStop-timeStart)/16)/(dataSpread)).^2);

dataEnergy = data.^2;
timeCenterOfMass = sum(time.*dataEnergy)/sum(dataEnergy);

[fftR,fftAxe] = FFTR(data,1/dataFreqSampling);
dataEnergyFreq = fftR.^2;
freqCenterOfMass = sum(fftAxe.*dataEnergyFreq)/sum(dataEnergyFreq);

subplot(2,1,1);hold on
plot(time,data);axis tight;
plot(timeCenterOfMass,0,'o');axis tight;grid on
xlabel('Time (s)');ylabel('Amplitude')
legend('Time Signal','Center of mass')

subplot(2,1,2);hold on
plot(fftAxe,fftR);axis tight;
plot(freqCenterOfMass,0,'o');axis tight;grid on
xlabel('Frequency (Hz)');ylabel('Amplitude')
legend('Frequency Spectrum','Center of mass')

More details about:

  • $\begingroup$ Thanks. But I don't understand well your response. In question I meant for example transition of whole the energy around a specific point. Consider a gaussian like spreading of a signal around a point which has energy equal to the original signal. $\endgroup$
    – math14
    Feb 23 '19 at 11:45
  • 1
    $\begingroup$ I do apologize, I knew I could spread confusion. Yet, I don't really get what you mean by "transition of whole the energy", "a gaussian like spreading". I think an illustration could be informative (on both sides :) $\endgroup$ Feb 23 '19 at 11:50
  • 1
    $\begingroup$ No problem @Laurent. :) $\endgroup$
    – math14
    Feb 23 '19 at 11:52
  • $\begingroup$ I have added a picture and some code, please to intereact $\endgroup$ Feb 23 '19 at 18:11

Your Answer

