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Why are random signals considered as power signals (i.e. signals with infinite energy and finite average power)?

Does this make any sense? What does it mean for random signals to have infinite energy even though we know that real-life signals (usually with inherent randomness in them) have finite energy!

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    $\begingroup$ You're making multiple statements that aren't true or only are half true. First of all, you define a model for your random signal. If that model has infinite energy, that's your fault. Then, yeah, the universe is finite and the sun will die one day, but for all practical purposes, all naturally occurring sources of noise tend to be an infinite source of energy. $\endgroup$
    – Marcus Müller
    Jun 7, 2017 at 5:27
  • $\begingroup$ @MarcusMüller Ok. So basically, what you are saying is that this is only true for random signals where the noise comes from a naturally occurring source (like Brownian motion for eg). Is that correct? $\endgroup$
    – Likely
    Jun 7, 2017 at 7:43
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    $\begingroup$ No, that is not correct. $\endgroup$
    – Marcus Müller
    Jun 7, 2017 at 12:31
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    $\begingroup$ Just as $\sin(wt)$ has infinite extent, is a mathematical construct and not a pysical (practical) reality, the mathematical definition of a random process should have infinite energy: The energy integral cannot converge because you cannot show that $X(t)$ goes to zero as $t$ goes to infinity (as you must show this for the integral to be convergent). Because if you were able to show it, then $X(t)$ would become a deterministic signal as t goes to infinity...( as its value is predicted with certainty, which is 0, in the limit). $\endgroup$
    – Fat32
    Jun 8, 2017 at 1:49
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    $\begingroup$ Yes its about the mathematical definition of a Random Process. On the other hand any practical aplication will observe only a finite extent of such a process and therefore will have large but finite energy. This subject is as similar as observing a DC signal. The true DC signal should have an infinite extent hence infinite energy. But the practical one will not. As a consequence of this fact, The Fourier transform of the true DC is an impulse (amplitude infinite) while the FT of a windowed (practical) DC is a sinc-pulse, finite valued, finite energy. $\endgroup$
    – Fat32
    Jun 8, 2017 at 11:35

3 Answers 3

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Note that the condition

$$\int_{-\infty}^{\infty}|f(t)|^2dt<\infty\tag{1}$$

(i.e., that the signal $f(t)$ has finite energy) is very restrictive when we try to model signals, even though obviously any actually occurring signal must have finite energy. Modeling signals as random processes means that we ignore condition $(1)$. Models are always unrealistic to a certain degree, but many signals can be described very well by random processes even though the signals have finite energy and their models do not. This aspect of the model is often irrelevant.

One example which may serve to understand this fact a bit better is the frequently used model of a (wide-sense) stationary process. Certain statistical properties of such a process do not change over time, and, consequently, realizations of such a process will generally not decay as $t\rightarrow\pm\infty$, and $(1)$ will generally not be satisfied, even though we are only interested in the properties of that process during a certain finite time window. However, power and the power spectrum can be defined for such processes, and most practically useful processes have finite power (or can easily be made to have finite power).

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  • $\begingroup$ "Modeling signals as random processes means that we ignore condition (1)", does this mean that the statement that random signals cannot have finite energy but only finite power is true? Sorry if this seems like asking the same thing twice but I'm really confused now and I think I need a clear answer to that question. $\endgroup$
    – Likely
    Jun 7, 2017 at 16:32
  • $\begingroup$ @Likely: Yes, that's true, even though specific realizations of a random process could theoretically have finite energy. $\endgroup$
    – Matt L.
    Jun 7, 2017 at 21:32
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I think simple.

We want to model a random physical phenomenon for analysis purpose. One way is to model it by a stochastic process $X(t)$, i.e. a time series of random variables $\left\lbrace X(t_k) = X(t=t_k), t_k \in \mathbb{R} \right\rbrace$.

The random variable $X(t_k)$ is associated with a probability distribution function (PDF) with some finite moments (in typical cases, the 1st and 2nd moments equivalent to mean and variance), again for analysis purpose.

The fact that the outcome of the random variable $X(t_k)$ can be infinite, even with very low probability, (in general) makes energy of realizations of the stochastic process $X(t)$ infinite in any time-windowed version of $X(t)$.

What about the power ? $$P=\lim_{T \to \infty} \frac{1}{T} \int_{-T}^{+T} |x(t)|^2 \mathrm{d}t$$

The power $P$ can be defined finite by, for example, assuming ergodicity of $X(t)$ and finite moments.

People thought this kind of model was reasonable, tried using it and have found it fit many useful processes. Thus the model is kept.

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  • $\begingroup$ -1 for the nonsense in the fourth paragraph. By definition, random variables take on values in $\mathbb R$, not $\mathbb R+$ and so realizations of $X(t)$ cannot have value $\pm \infty$ over an interval of nonzero duration, i.e. contribute nothing to the energy of the realization, whether time-windowed or not. $\endgroup$
    – Dilip Sarwate
    Jun 7, 2017 at 13:25
  • $\begingroup$ @DilipSarwate thanks, could I understand that if variables take on values in $\mathbb{R}+$ (is $(0,+\infty) \subset \mathbb{R}$ ?), realizations of $X(t)$ can have value $\pm \infty$ over an interval of nonzero duration? And how can you explain that a time-windowed random signal has infinite energy? $\endgroup$
    – AlexTP
    Jun 7, 2017 at 13:57
  • $\begingroup$ In elementary expositions of the theory of random variables, a random variable is a mapping from a sample space $\Omega$ to $\mathbb R$ and one denies the existence of any outcomes in $\omega$ that are mapped to $\pm\infty$: those values are not in $\mathbb R$. A more formal exposition allows the range $\mathbb R+$ but insists that the set of all outcomes mapped to $\pm \infty$ is an event of probability $0$. Now, energy is the integral of $|x(t)|^2$ and a momentary value of $\pm \infty$ conveys no energy; you need $|x(t)|$ to be $\infty$ over an interval of nonzero duration... $\endgroup$
    – Dilip Sarwate
    Jun 7, 2017 at 15:00
  • $\begingroup$ ... and if your random process model allows uncountably infinitely many $X(t_k)$ to have value $\pm \infty$ as it must if a realization of $X(t)$ is to hit $\pm\infty$ and stay there for a nonzero interval, you have more serious problems to worry about than trivialities about energy and power. $\endgroup$
    – Dilip Sarwate
    Jun 7, 2017 at 15:06
  • $\begingroup$ @DilipSarwate I do agree with you that my explaination about "random signal have infinite energy" is incorrect. And if I understand correctly, you have just explained that a random process model cannot have uncountably $X(t_k)$ having value $\pm \infty$ to prove that I was wrong. Can you give me some intuition why random signal is energy-infinite? The answer of Matt L "Modeling signals as random processes means that we ignore the finite energy condition" seems vague. Thanks $\endgroup$
    – AlexTP
    Jun 14, 2017 at 8:13
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In addition to Marcus Müller comment, If a signal has finite energy then the signal value must reach zero after long enough time, but for random signals your signals generally don't have such restriction.

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  • $\begingroup$ But what you are saying means that random signals do not have finite energy!! Hence it's correct to say they may have instead finite power! $\endgroup$
    – Likely
    Jun 7, 2017 at 7:45
  • $\begingroup$ Consider thermal noise of a resistor, if you could keep the temperature of that resistor non-zero eternally, then your thermal noise has infinite energy, in other word if you managed to drain energy from that resistor's thermal noise, then its temperature would drop unless you add energy to that resistor and keep its temperature over zero. $\endgroup$
    – Mohammad M
    Jun 7, 2017 at 8:47
  • $\begingroup$ @Likely The problem of infinite energy rises when signal's amplitude diverges or it has infinite duration. $\endgroup$
    – Mohammad M
    Jun 7, 2017 at 8:56
  • $\begingroup$ Okay I got that. Still you're confusing me. In your answer you gave a condition for signals to have finite energy and said random signals don't follow it, in other words random signals can't have finite energy. Now, you are giving conditions for a random signal to have infinite energy which are practically impossible, which means random signals can't have infinite energy. In the end I can't tell which is true! $\endgroup$
    – Likely
    Jun 7, 2017 at 9:16
  • $\begingroup$ I didn't say all random signals don't follow that condition, you could define random processes (random signals) with finite energy. In science we always use simplifying assumptions e.g. In celestial mechanics we assume earth is just a point without any dimension and surprisingly get satisfying results. $\endgroup$
    – Mohammad M
    Jun 7, 2017 at 19:42

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