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<math>E = \int_{0}^{1} |e^{4t}|^2\ dt \!</math> | <math>E = \int_{0}^{1} |e^{4t}|^2\ dt \!</math> | ||
| − | <math> | + | <br><br><math> = \int_{0}^{2} e^{8t}\ dt \!</math> |
<math> = \frac{1}{8}[e^{8t}]_{t=0}^{t=1} \!</math> | <math> = \frac{1}{8}[e^{8t}]_{t=0}^{t=1} \!</math> | ||
<math> = \frac{1}{8}(e^8 -1)\!</math> | <math> = \frac{1}{8}(e^8 -1)\!</math> | ||
| + | |||
| + | == Signal Power == | ||
| + | |||
| + | Average signal power between <math>[t_1,t_2]\!</math> is <math>P_{avg}=\frac{1}{t_2-t_1}\int_{t_1}^{t_2} |x(t)|^2\ dt \!</math>. | ||
| + | |||
| + | <math>P_{avg}=\frac{1}{1-0} \int_{0}^{1} |e^{4t}|^2\ dt \!</math> | ||
| + | <br><br><math> = \int_{0}^{1} e^{8t}\ dt \!</math> | ||
| + | <br><br><math> = \frac{1}{8}[e^{8t}]_{t=0}^{t=1} \!</math> | ||
| + | <br><br><math> = \frac{1}{8}(e^8 -1)\!</math> | ||
Latest revision as of 08:53, 5 September 2008
Signal Energy
$ E=\int_{t_1}^{t_2}x(t)dt $
find the signal energy of $ x(t)=e^{4t}\! $ on $ [0,1]\! $
$ E = \int_{0}^{1} |e^{4t}|^2\ dt \! $
$ = \int_{0}^{2} e^{8t}\ dt \! $
$ = \frac{1}{8}[e^{8t}]_{t=0}^{t=1} \! $ $ = \frac{1}{8}(e^8 -1)\! $
Signal Power
Average signal power between $ [t_1,t_2]\! $ is $ P_{avg}=\frac{1}{t_2-t_1}\int_{t_1}^{t_2} |x(t)|^2\ dt \! $.
$ P_{avg}=\frac{1}{1-0} \int_{0}^{1} |e^{4t}|^2\ dt \! $
$ = \int_{0}^{1} e^{8t}\ dt \! $
$ = \frac{1}{8}[e^{8t}]_{t=0}^{t=1} \! $
$ = \frac{1}{8}(e^8 -1)\! $
