Signal

The signal used was $ cos(3t) $.

Energy Equation

$ E = \int_{0}^{3 \pi}(|cos(3t)|^2 dt) $

$ E = \frac{x}{2} + \frac{1}{12} sin(6x) |_{0}^{3 \pi} $

$ sin(18 \pi)/12 $ is -3.333e^-14, so it is rounded to 0.

$ E = \frac{3 \pi}{2} $

Power Equation

$ P = \frac{1}{3 \pi - 0} \int_{0}^{3 \pi}(|cos(3t)|^2 dt) $

$ P = \frac{x}{6 \pi} + \frac{1}{36 \pi} sin(6x) |_{0}^{3 \pi} $

$ sin(18 \pi)/36 \pi $ is -1.111e^-14, so it is rounded to 0.

$ P = \frac{1}{2} $

Alumni Liaison

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

Buyue Zhang