I was asked one day why this full-scale sine wave was being measured by our signal analysis tools as -3 dB FS, even though it hits the maximum and the minimum sample values:

The answer is "because it's a sine wave, not a square wave." The intensity of a signal can be calculated from the following formula:

The inner integral does not depend on *t* - it's just the average sample value - so it's usually precalculated:

The importance of taking dc into account during analysis can be appreciated if you try to calculate the intensity of a signal with a high dc.

*Exercise:* calculate the intensity of *x*(*t*) ≡ -0.5 using the formulas above; calculate the "naïve intensity" by using the last formula above and omitting dc. Note the difference.

Now that we have the necessary formulas, let's analyze our full-scale sine wave signal. Plugging in a full-scale sine wave and analyzing over a full period we get:

As expected, the average value of the signal over a single period is 0.

Evaluating the intensity requires finding the antiderivative of (sin(t))^{2}. This can be ascertained most easily by plotting a few values and realizing that (sin(t))^{2} = (1 - cos(2t))/2:

This is a dimensionless number that ranges from 0 (signal is a flat line) to 1 (... we'll get to that later.)

We can convert this into a dB FS measurement using the formula I_{dB FS} = 20 log_{10} I_{RMS}:

Et voilà - that's where the -3 dB comes from.

In contrast to a sine wave, a full-scale square wave centered at 0 has an intensity of 0 dB FS:

To finish up, a couple of advanced exercises:

Advanced Exercise: prove that, provided t_{1} and t_{2} are sufficiently far apart, the intensity of a general sine wave x(t) = a sin(ωt + φ) + c depends only on a and not on ω, φ, or c.

Advanced Exercise: completely characterize the set of digital signals that achieve 0 dB FS intensity. If you have, say, 1000 samples of mono 16-bit integer audio to play with, how many distinct signals achieve 0 dB FS intensity?