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==Problem== | ==Problem== | ||
| − | Given complex signal <math>f(t)=e^{jt} = cos(t) + j sin(t)</math>, find <math>E_\infty</math> and <math>P_\infty</math>. | + | Given complex signal <math>f(t)=e^{jt} = \cos(t) + j \sin(t)</math>, find <math>E_\infty</math> and <math>P_\infty</math>. |
==Background Knowledge== | ==Background Knowledge== | ||
| − | <math>E_\infty(x(t)) = \int_{-\infty}^\infty |x(t)|^2\,dt | + | <math>E_\infty(x(t)) = \int_{-\infty}^\infty |x(t)|^2\,dt</math> |
| − | <math>P_\infty(x(t)) = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T |x(t)|^2\,dt | + | <math>P_\infty(x(t)) = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T |x(t)|^2\,dt)</math> |
| + | |||
| + | ==Solution== | ||
| + | *<math>|x(t)| = |\cos(t) + j \sin(t)| = \sqrt{(\cos(t))^2+(\sin(t))^2} = \sqrt{1} = 1</math> | ||
| + | |||
| + | *<math>E_\infty(x(t)) = \int_{-\infty}^\infty |x(t)|^2\,dt = \int_{-\infty}^\infty 1\,dt = t|_{-\infty}^{\infty} = \infty</math> | ||
| + | |||
| + | *<math>P_\infty(x(t)) = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T |x(t)|^2\,dt) = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T 1,dt) = \lim_{T \to \infty} (\frac{1}{2T} t|_{-T}^T)</math> | ||
| + | |||
| + | <math> = \lim_{T \to \infty} (\frac{1}{2T} (2T)) = \lim_{T \to \infty} (1) = 1 </math> | ||
Revision as of 17:23, 4 September 2008
Problem
Given complex signal $ f(t)=e^{jt} = \cos(t) + j \sin(t) $, find $ E_\infty $ and $ P_\infty $.
Background Knowledge
$ E_\infty(x(t)) = \int_{-\infty}^\infty |x(t)|^2\,dt $
$ P_\infty(x(t)) = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T |x(t)|^2\,dt) $
Solution
- $ |x(t)| = |\cos(t) + j \sin(t)| = \sqrt{(\cos(t))^2+(\sin(t))^2} = \sqrt{1} = 1 $
- $ E_\infty(x(t)) = \int_{-\infty}^\infty |x(t)|^2\,dt = \int_{-\infty}^\infty 1\,dt = t|_{-\infty}^{\infty} = \infty $
- $ P_\infty(x(t)) = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T |x(t)|^2\,dt) = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T 1,dt) = \lim_{T \to \infty} (\frac{1}{2T} t|_{-T}^T) $
$ = \lim_{T \to \infty} (\frac{1}{2T} (2T)) = \lim_{T \to \infty} (1) = 1 $
