On Wed, Nov 16, 2011 at 8:21 AM, Kerry Wentworth wrote: > Think of it this way: > > Assume you have a sine wave with a period of 360 uS at 2V p-p. > 30 degrees after a 0 crossing, the voltage will be .5V. =A0At 45 degrees > it will be .707V, and at 60 degrees it will be .866V. =A0This is the > definition of a sin wave. > You can break it down further, and say that, for any 2 points (T1 and > T2), the voltage at T2 would be > V2 =3D sin(T2) * (V1 / sin(T1)) > If you have a different value of V2, then you don't have a sin wave, > even if it LOOKS like a sin wave. THANKS! That makes a lot of sense. Finally clicks. Also, I found this to be very useful: Demo 1: Time Domain vs. Frequency Domain of a Sinusoid http://library.wolfram.com/infocenter/Demos/7774/ --=20 http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .