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What is the procedure to check the signal type?

example:

$ x(t) = A \sin (\omega t) $

$ y(t) = A e^ {-\lambda |t|} $

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4 Answers 4

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Let's assume the signal, $x(t)$ is not identically zero for all $t$.

An "energy signal" (what I would prefer to call a "finite energy signal") is such a signal, $x(t)$ with a finite energy: $$ 0 \ < \ \int\limits_{-\infty}^{\infty} \big|x(t)\big|^2 \ \mathrm{d}t \ < \ \infty $$

BTW, sometimes for mathematical ease, we require a stricter sense of finite "energy": $$ 0 \ < \ \int\limits_{-\infty}^{\infty} \big|x(t)\big| \ \mathrm{d}t \ < \ \infty $$

And a "power signal" (what I would prefer to call a "finite power signal") is such a signal, $x(t)$ with finite power: $$ 0 \ < \ \lim_{T \to \infty} \frac{1}{T}\int\limits_{-\frac{T}2}^{\frac{T}2} \big|x(t)\big|^2 \ \mathrm{d}t \ < \ \infty $$

I think that is the most fundamental definitions of the two classes of continuous-time signals.


You can do a very similar definitions for discrete-time signals, $x[n]$. Let's assume the signal, $x[n]$ is not identically zero for all $n$.

An "energy signal" (what I would prefer to call a "finite energy signal") is such a signal, $x[n]$ with a finite energy:

$$ 0 \ < \ \sum\limits_{n=-\infty}^{\infty} \big|x[n]\big|^2 \ < \ \infty $$

And for mathematical ease, we sometimes require a stricter sense of finite "energy": $$ 0 \ < \ \sum\limits_{n=-\infty}^{\infty} \big|x[n]\big| \ < \ \infty $$

And a "power signal" (what I would prefer to call a "finite power signal") is such a signal, $x[n]$ with finite power:

$$ 0 \ < \ \lim_{N \to \infty} \frac{1}{2N+1}\sum\limits_{n=-N}^N \big|x[n]\big|^2 \ < \ \infty $$

I think that is the most fundamental definitions of the two classes of discrete-time signals.

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  • $\begingroup$ in other words to check the signal type i need to calculate the improper integral? $\endgroup$ Commented Jun 13, 2014 at 12:24
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    $\begingroup$ "calculate" as in calculus. plug $A \sin (\omega t)$ into the finite energy integral, and see that it blows up and does not converge to a finite value. so it's not finite energy. but if you plug it into the integral and limit that defines mean power, you will see that there is a finite limit. likewise, if you plug $A e^ {-\lambda |t|}$ into the energy definition, you will see it come out as a finite value, but if you plug it into the power definition, in the limit it will go to 0. all those two expressions show are the total energy over all time and the mean power over all time. $\endgroup$ Commented Jun 13, 2014 at 12:48
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    $\begingroup$ Tu sum up Asin(ωt) is a power signal and Ae−λ|t| is an energy signal. In general infinite non bounded signals are analyzed as power signals because they have infinite energy. Bounded signals are analyzed as energy signals and have 0 power $\endgroup$
    – VMMF
    Commented Dec 16, 2021 at 13:11
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All bounded periodic signals are power signals, because they do not converge to a finite value so their energy is infinite and their power is finite. So we say that a signal is a power signal if its power is finite and its energy is infinite. And the signal is an energy signal if its energy is finite and power is zero.

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Of course, all real world signals are bounded in power and energy, as infinite time is unobservable, and infinite power is unrealizable, at least by mere mortals.

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The assumption is that the signals under consideration are deterministic.

Power signals

A periodic signal is always a power signal if its Fourier transform is a set of discrete components comprising of a fundamental and its harmonics. In case of sinusoid it has only the fundamental component.

Energy Signal

An aperiodic signal is an energy signal if its Fourier transform is continuous.

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  • $\begingroup$ In fact, bounded periodic signals are power signals. But not all aperiodic signals are energy signal, not even the bounded ones. Simple counter examples include constant power noise or a simple sweep. So even if you exclude indeterministic signals your statement is not true. $\endgroup$
    – Jazzmaniac
    Commented Jun 15, 2014 at 14:04
  • $\begingroup$ It's also not true that aperiodic signals necessarily have a 'continuous' Fourier transform. Again, a simple counter example is the sum of three sines with non-rationally related frequencies. $\endgroup$
    – Jazzmaniac
    Commented Jun 15, 2014 at 14:07
  • $\begingroup$ @Jazzmaniac thankx. Ill modify my answer accordingly. $\endgroup$
    – learner
    Commented Jun 15, 2014 at 16:39

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