Revision as of 09:01, 5 September 2008 by Longja (Talk)

Power and Energy Problem

$ x(t) = 3\cos(4t + \frac{\pi}{3}) $

$ P_\infty = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T |3\cos(4t + \frac{\pi}{3})|^2\,dt) $


$ P_\infty = \lim_{T \to \infty} (\frac{1}{2T} [\frac{9sin(4t + \frac{\pi}{3})cos(4t + \frac{\pi}{3}) + 4t}{8}]_{-T}^{T}) $


$ E_\infty = \int_{-\infty}^\infty |3\cos(4t + \frac{\pi}{3})|^2\,dt $


$ E_\infty = [\frac{9sin(4t + \frac{\pi}{3})cos(4t + \frac{\pi}{3}) + 4t}{8}]_{-\infty}^{\infty} $


$ E_\infty = \infty $


  • Bonus Problem!


$ x(t) = e^{j(\pi t-1)} $


$ P_\infty = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T |e^{j(\pi t-1)}|^2\,dt) $


$ P_\infty = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T 1\,dt) $


$ P_\infty = \lim_{T \to \infty} (\frac{1}{2T} 2T) $


$ P_\infty = 1 $



$ E_\infty = \int_{-\infty}^\infty |e^{j(\pi t-1)}|^2\,dt $


$ E_\infty = \int_{-\infty}^\infty 1,dt $


$ E_\infty = \infty $

Alumni Liaison

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

Dr. Paul Garrett