(Power and Energy Problem)
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<math>P_\infty = \lim_{T \to \infty} (\frac{1}{2T}  \int_{-T}^T |3\cos(4t + \frac{\pi}{3})|^2\,dt)</math>  
 
<math>P_\infty = \lim_{T \to \infty} (\frac{1}{2T}  \int_{-T}^T |3\cos(4t + \frac{\pi}{3})|^2\,dt)</math>  
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<math>P_\infty = \lim_{T \to \infty} (\frac{1}{2T} \frac{9sin(4t + \frac{pi}{3})cos(4t + \frac{pi}{3}) + 4t}{8})</math>
  
  

Revision as of 08:54, 5 September 2008

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}) $


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


  • 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 $

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

Buyue Zhang