- Thread Starter
- #31
Yes. The easiest way to think of it is how I described it initially. The clipped signal is no longer a sine wave; it has become a square wave. Imagine it as literally square, pi on the base, 1 tall on the sides. That has more area than a sine curve from 0 to pi, as the area when the function is increasing to pi/2 isn't filled, as well as the area from pi/2 to pi. The square wave does have that area filled. Obviously a true clipped signal doesn't have 90 degree angles, but the concept is still the same.I'm having a hard time understanding this. You are saying, comparing 2 sine waves, one that resembles a clipped signal and one that does not, that the integral of the clipped signal will have more area (or power, however you want to think of it) underneath the curve than the other? That doesn't make sense to me.
