Signal

$ f(t) = \sin^2(t) $

Energy

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

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

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

Since $ \sin^2(t) = \frac{1-\cos(2t)}{2} $

$ E = \frac{1}{4}\int_0^{2\pi}(1-\cos(2t))^2 $

$ E = \frac{1}{4}\int_0^{2\pi}(1-2\cos(2t)+\cos^2(2t)) $

$ E = \frac{1}{4}[1]^{2\pi}_{0} - \frac{1}{4}[sin(2t)]^{2\pi}_{0} + \frac{1}{4}\int_0^{2\pi}(\cos^2(2t)) $

$ E = \frac{1}{2}\pi - 0 + \frac{1}{4}\int_0^{2\pi}(\cos^2(2t)) $

Since $ \cos^2(2t) = \frac{1+\cos(4t)}{2} $

$ E = \frac{1}{2}\pi + \frac{1}{8}\int_0^{2\pi}(1+\cos(4t)) $

$ E = \frac{1}{2}\pi + \frac{1}{8}[1]^{2\pi}_{0} + \frac{1}{32}[\sin(4t)]^{2\pi}_{0} $

$ E = \frac{1}{2}\pi + \frac{1}{8}(2\pi) + 0 $

$ E = \frac{1}{2}\pi + \frac{1}{4}\pi $

$ E = \frac{3}{4}\pi $

Average Power

$ P = \frac{1}{t_2 - t_1}\int_{t_1}^{t_2}\!|f(t)|^2\ dt $

$ P = \frac{1}{t_2 - t_1}\int_{t_1}^{t_2}\!|\sin^2(t)|^2\ dt $

$ P = \frac{1}{2\pi - 0}\int_{0}^{2\pi}\!|\sin^4(t)|\ dt $

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

$ P = \frac{1}{2\pi}(\frac{3}{4}\pi) $

$ P = \frac{3}{8} $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett