1
$\begingroup$

The energy of a continuous signal $x(t)$ recorded between $t=0$ and $t=T$ (namely $x(t \notin [0, T]) = 0$) is defined as

$$E = \int_0^T |x(t)|^2\, dt.$$

In most signal processing texts I met, the formula for the energy of the same signal sampled at $N$ regular steps separated by $\Delta t$ is given as

$$E_N = \sum_{n=0}^{N-1} |x_n|^2,$$

which implies an implicit normalisation of the signal. As I would like to keep track of the physical units, I am inclined to use

$$E_N = \sum_{n=0}^{N-1} |x_n|^2 \Delta t.$$

Now, Parseval's identity gives, in the continuous case,

$$E = \int_0^T |x(t)|^2\, dt = \int_{-\infty}^\infty |\hat{x}(f)|^2\, df,$$

where $\hat{x}$ is the Fourier transform of $x$. This leads to define the /energy density/ as

$$e(f) = |\hat{x}(t)|^2.$$

I would translate this in the discrete case as

$$E_N = \sum_{n=0}^{N-1} |x_n|^2 \Delta t = \Delta t^2 \sum_{k=0}^{N-1} a_k^2 \Delta f,$$

where $\Delta f = 1/T$, and the $(a_k)$ are the Fourier amplitudes and such that

$$\forall \; k \in \{0, N-1\}, \, \sum_{n=0}^{N-1} x_n e^{-2i\pi k \frac{n t}{T}} = a_k e^{i \phi_k},$$

Finally, I would define the discrete energy density as

$$e_N(f) = \Delta t^2 |a_k|^2 \text{ for } k \Delta f \leq f < (k+1) \Delta f.$$

Does this sounds sane to practitioners of the field?

$\endgroup$
  • $\begingroup$ Your equation for the continuous-time Parseval theorem is incorrect: 1) the limits need to be $(-\infty, +\infty)$ in both time and frequency domains, and 2) the frequency variable is $f$ not $t$. Have a look at this question over on the Math.SE. That might also provide insight. $\endgroup$ – Peter K. Jan 15 '16 at 14:49
  • $\begingroup$ @PeterK. Thank you for your corrections. I corrected the question accordingly. The bounds of the time integral do not need to be infinite because the signals is zero outside of $[0,T]$. $\endgroup$ – Alfred M. Jan 15 '16 at 15:59
  • $\begingroup$ I don't understand why you mix $t$ and $f$ in your definition of energy density. Can you clarify? $\endgroup$ – MBaz Jan 15 '16 at 16:31
  • $\begingroup$ @MBaz $\Delta f$ is for integration over the frequency domain. The $\Delta t^2$ comes from time integration leading to Parseval's identity as per en.wikipedia.org/wiki/Spectral_density#Energy_spectral_density. $\endgroup$ – Alfred M. Jan 18 '16 at 7:21
  • $\begingroup$ please check if this note answers your question: "Discrete and Continuous Time Sampling and Convolution Review and Relationships" by Chien-Hsin Lee, web.stanford.edu/group/cioffi/ee379a/extra/… $\endgroup$ – Ali Jan 21 '16 at 18:45
1
+50
$\begingroup$

First it should be noted that the nomenclature using power and energy as properties of signals is somewhat misleading. For physical energy to come out of a time integral over a squared signal, the signal would need to be of dimension $\sqrt{[\mathrm{energy}]/[\mathrm{time}]}$, which is not the case for practically all signals that you encounter.

In fact, if you have signals that fully describe the state of a physical system over time, and the energy of the system can there be expressed using just that signal, the result is usually very different and most importantly, it is a property of each time instant and not the whole signal. If you have a signal that describes a non-temporal aspect of a physical signal, like the momentary state of a string, the energy of the system will in general require the time derivative of that state.

Therefore, the terms energy and power are a bit of stretch and should be used carefully. I personally prefer speaking of a squared magnitude and a squared magnitude density.

With this preamble, the use of physical units like time and frequency in the definition of power and energy of a signal become somewhat relative. The choice of scaling and therefore of units is mostly arbitrary and depends on what you want to express and which transformations you want to keep your measures invariant.

For example, if you have a choice of sampling your physical system at different rates and you want the energy measure to be invariant under such rate changes, then you need to first understand how the amplitudes behave under a rate change. The normal behaviour of an amplitude is to be be unaffected by a change of sampling rate. Some acquisition methods may however rely on integration to record the samples and your sample values may therefore scale with the sampling rate. So your energy sum may contain different powers of the time step in order to keep the energy invariant under sampling rate changes.

Keeping track of all this for different kinds of signal behaviours quickly become tricky. That is why I made a habit of defining a discrete signal not just by giving a sequence of numbers, but also by providing a homomorphism to an infinite dimensional function space that allows me to work out all the meaningful measures and properties in a well defined framework with properly defined integral transforms and continuous variables.

So, a discrete signal is a tuple $\{ s\in \mathbb{S},\, D: \mathbb{C}\to\mathbb{S},\, \bar{D}:\mathbb{S}\to\mathbb{C} \}$ where $\mathbb{S}$ is the space the discrete signal lives in, $\mathbb{C}$ the space of the continuous time signals, and $s$ a representation of the discrete time signal in $\mathbb{S}$. Note that $D\circ\bar{D}=\mathrm{id}_\mathbb{S}$ but not necessarily $\bar{D}\circ D=\mathrm{id}_\mathbb{C}$.

You give these two maps $D$ and $\bar{D}$ all the invariance properties you would like to see. Afterwards you can pull back all definitions from $\mathbb{C}$ to $\mathbb{S}$ by applying the two domain changing maps.

This method of working with discrete signals will give you a well defined way of talking about properties define for continuous signals and allow you to get your scaling correct under sample rate changes or changes of other parameters relevant for your discrete signal representation.

I will not try to replicate your derivation using this approach, because it is not clear to me which invariance and signal properties you want to preserve and what exactly your goal is. Like I said, it depends on the context what is right. I still hope this answer helps your understanding and answers the slightly more fundamental question of how to deal with units, dimensions and scaling of discrete signals compared to continuous signals.

$\endgroup$
  • $\begingroup$ you need to first understand how the amplitudes behave under a rate change” ←This is the right question indeed. $\endgroup$ – Alfred M. Jan 22 '16 at 14:08

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.