$x[n]=\sin(\pi n)$

Since this is a discrete signal, and n can take only values like ...,1,2,3 ... the value of the power is getting zero, and since it gets zero does it mean that it's not a power signal?


There a few variations on the definitions of "energy signal" and "power signal". For example, Sklar in his textook "Digital Communications" defines "energy signal" as a signal with energy $E$ such that $0 < E < \infty$, and a "power signal" as a signal with power $P$ such that $0 < P <\infty$.

However, others (see for example this answer) define "energy signal" as a signal with $E < \infty$.

In your case, $x[n]=0$ for all $n$; then this signal has zero energy and zero power. Whether it is classified as an energy or a power signal depends on which specific definition you adopt. For instance, if you go with Sklar's, then this signal is neither an energy signal nor a power signal.

  • 1
    $\begingroup$ Well, it is both an energy signal (with zero energy) and a power signal (with zero power). Note that $\frac{1}{2N+1}\sum_{n=-N}^N |x[n]|^2 $ "approaches" a finite limit as $N \to \infty$ and so the signal is a power signal. That the average is already at the limit $0$ when $N=0$ and just stays there forever is not germane. $\endgroup$ – Dilip Sarwate Feb 25 '18 at 3:34
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    $\begingroup$ Dilip, I guess there are different possible definitions. Sklar's "Digital Communications" defines energy signals as signals with finite but non-zero energy, and does the same for power signals. I had this book nearby when I answered the question, so I went with his definition. $\endgroup$ – MBaz Feb 25 '18 at 23:03
  • $\begingroup$ @DilipSarwate I've clarified my answer to account for the different definitions of energy and power signals. $\endgroup$ – MBaz Feb 25 '18 at 23:10
  • $\begingroup$ OK, and a +1 for the edits. But I don't regard Sklar's book as definitive, and signals with zero energy are of some interest in more theoretical studies (often as counterexamples which must be carefully avoided in crafting formal definitions). $\endgroup$ – Dilip Sarwate Feb 26 '18 at 2:14
  • $\begingroup$ @DilipSarwate Agreed on all counts. $\endgroup$ – MBaz Feb 26 '18 at 3:08

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