# Questions about discrete signal energy calculation

I'm looking at the definition of signal energy (e.g. Wikipedia, cnx.org). For discrete signals, it's defined like the following, where $x(n)$ holds the signal:

$Energy = \sum_{n=-\infty}^{\infty} |x(n)|^2$

So my questions:

1. For a windowed, finite signal like double signal[256], the sum is from 1 to 256 (or 0 to 255 in a program) rather than $-\infty$ to $\infty$, right? (I don't even know how I would sum over infinity.)

2. Why does the energy formula have the absolute-value operator $|...|$? The result of taking the absolute value is squared anyways to produce a positive value, so taking the absolute value seems to be pointless. Is it because the $x(n)$ can be complex, so the absolute value of a complex number would be the scalar from Pythagoras' theorem?

• What do you mean by "the actual value is always obtained by mathematical manipulations rather than by explicit computation by adding terms"? How can you get a final actual value unless you do perform the summation? – stackoverflowuser2010 Mar 27 '12 at 18:01
• Yes, I know the convergence of a geometric series sum. However, I don't see why that is useful here. Why would the values of $x(n)$ take the form of 1, $x$, $x^2$, $x^3$, ...? The signals that I'm reading (microphone, accelerometer, etc.) certainly don't have values like that. – stackoverflowuser2010 Mar 27 '12 at 22:57
• I appreciate the help, but as I said, the data that I'm reading (e.g. from microphone, accelerometer) doesn't look like an exponentially decaying signal, or at least it doesn't look like it. – stackoverflowuser2010 Mar 27 '12 at 23:35
• You're a mathematician and not an engineer, right? – stackoverflowuser2010 Mar 29 '12 at 14:35

2. Yes, it is because x(n) could be complex. If we didn't take the absolute values of the complex numbers (Euclidean norm), a signal containing $[\dots,0,0,a+ia, a-ia,0,0,\dots]$ would have an energy of zero (instead of $4a^2$) although it contains non-zero samples. For real numbers however $|a|^2 = a^2$ and the absolute value doesn't matter.