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<math>E_\infty = \int_{-\infty}^\infty |3\cos(4t + \frac{\pi}{3})|^2\,dt</math> | <math>E_\infty = \int_{-\infty}^\infty |3\cos(4t + \frac{\pi}{3})|^2\,dt</math> | ||
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| + | <math>E_\infty = [\frac{9sin(4t + \frac{\pi}{3})cos(4t + \frac{\pi}{3}) + 4t}{8}]_{-\infty}^{\infty}</math> | ||
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| + | |||
| + | <math>E_\infty = \infty</math> | ||
Revision as of 09:01, 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}]_{-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 $
