I read that a infinite extension continuous time signal with decreasing amplitude will always be an energy signal. Though if you take the integral of, for example, f(t)=1/(t+1) we can see that is diverges to infinity. Doesn't the saying that if a signal has finite area then it is an energy signal? But that doesn't go the other way, meaning that if a signal is power signal then it has a finite area? I suppect something happens when you square the signal to calculate it's energy but I don't want to dive into caclulus and series convergence and such

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    $\begingroup$ @havakok "Energy signal" is a common term, and the basis for determining the existence of certain transforms; it's a signal whose integrated square is finite. And no, not all signals have energy! Even fewer have finite energy, and these we call Energy Signals. $\endgroup$ Commented Jan 26, 2021 at 10:05
  • $\begingroup$ @havakok sure! $s(t)=1$ has no finite energy. $\endgroup$ Commented Jan 26, 2021 at 10:21
  • $\begingroup$ @havakok what's its energy then? $\endgroup$ Commented Jan 26, 2021 at 10:24
  • $\begingroup$ @havakok $$s(x) = \frac{\sqrt x }{\sqrt{\left[1 + x^2\right]}} $$ doesn't have enegry, either. $\endgroup$ Commented Jan 26, 2021 at 10:25
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    $\begingroup$ you can do that, but it's an interesting discussion, and I think it has helped OP's understanding! $\endgroup$ Commented Jan 26, 2021 at 10:29

3 Answers 3


What you read is clearly wrong. Also, don't forget that you need to integrate the square of the function to obtain its energy. Hence, for a signal to be an energy signal, the area of its squared magnitude must be finite:


The condition that $x(t)$ must decay for $|t|\to\infty$ is only necessary, not sufficient, as shown by the example in Marcus Müller's answer.

Note that due to Parseval's theorem you can also compute a signal's energy in the frequency domain:


where $X(\omega)$ is the Fourier transform of $x(t)$.

For power signals, the integral $(1)$ doesn't converge, and the Fourier transform doesn't exist, at least not in the conventional sense. A signal's power is computed as


Clearly, for energy signals, the expression $(3)$ equals zero.


I read that a infinite extension continuous time signal with decreasing amplitude will always be an energy signal.

That is wrong. It suffices to show a single counter example to show it's wrong: Let

$$s(t) = \frac1{\sqrt{\left\lvert t\right\rvert}},$$

which clearly is infinitely extending and always getting smaller the further you go from $t=0$. Then its energy is

\begin{align} E_s &= \int\limits_{-\infty}^{\infty} \left(\frac1{\sqrt{\left\lvert t\right\rvert}}\right)^2\,\mathrm dt\\ &=2\int\limits_{0}^{\infty} \frac1{\left(\sqrt{t}\right)^2}\,\mathrm dt \end{align}

and as "$\sqrt\infty$" doesn't ever exist, doesn't converge, and hence, this is clearly not an energy signal.

Your example works, too, and you get an "$\log\infty$" as a result. So, don't bother too much basing questions on a wrong claim.

Doesn't the saying that if a signal has finite area then it is an energy signal?

Well, "area" doesn't work great for signals that aren't real, or that aren't piece-wise continuous, or…

But yeah, if you can assign a finite area to it, then it's an energy signal, as that requires bounded absolute values. But thinking in "areas between curve and axis" really isn't the best level of abstraction for signal processing.


I suspect that the continous assumption is misleading here.

For either continuous or discrete signals, being positive and decreasing does not suffice, as shown by the colleagues. Indeed, such signals converge (a common property for bounded monotonous functions or sequences) to a finite limit. And if the limit is not zero, say $\eta\neq 0$, the sum or squared sum can easily be lower-bounded by $K\eta/2$ or $K(\eta/2) ^2$ for any $K$, thus infinite. Even if you suppose that the limit is zero, then you have counterexamples.

However, something differs in general between continuous and discrete signals, when they are action signals, eg with finite absolute area (or are summable or integrable). While a continuous function may reach arbitrary high-values and be integrable, this is not the case for a discrete sequence.

In other words, if $x_k$ is a summable sequence, ie $\sum |x_k|<\infty$, then necessarily $x_k\to 0$. Consequently, for some $K$, $\forall k>K$ we have $ |x_k|<1$, thus $ |x_k|^2<|x_k|<1$. Finally, if $x_k$ is summable, it is also square summable, and of finite energy. In the space of sequences, one says that $\ell_2(\mathbb{C})\subset \ell_1(\mathbb{C})$. The decreasing assumption is not needed.

Such an inclusion does not exist for continuous functions.


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