Consider the signal $ x(t)=cos(t) $ over the interval 0 to $ 4\pi $

Average Power

$ Avg. Power = {1\over(t2-t1)}\int_{t_1}^{t_2}\!|x(t)|^2 dt $


$ Avg. Power = {1\over(4\pi-0)}\int_{0}^{4\pi}\!|cos(t)|^2 dt $


$ Avg. Power = {1\over(4\pi)}{1\over2}\int_{0}^{4\pi}\!(1+cos(2t)) dt $


$ Avg. Power = {1\over8\pi}(4\pi+{1\over2}sin(8\pi)) $


$ Avg. Power = {1\over2} $

Energy

$ E = \int_{t_1}^{t_2}\!|x(t)|^2\ dt $


$ E = \int_{0}^{4\pi}\!|cos(t)|^2\ dt $


$ E = {1\over2}\int_{0}^{4\pi}\!(1+cos(2t)) dt $


$ E = {1\over2}(4\pi+{1\over2}sin(8\pi)) $


$ E = 2\pi $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang