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== Signal Energy == | == Signal Energy == | ||
| − | The signal energy expanded from <math>t_1\!</math> to <math>t_2\!</math> is defined as <math>E = \int_{t_1}^{t_2} \! | | + | The signal energy expanded from <math>t_1\!</math> to <math>t_2\!</math> is defined as <math>E = \int_{t_1}^{t_2} \! |x(t)|^2\ dt</math>. |
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<math>E = \int_{0}^{2} |e^{2t}|^2\ dt \!</math> | <math>E = \int_{0}^{2} |e^{2t}|^2\ dt \!</math> | ||
<br><br><math> = \int_{0}^{2} e^{4t}\ dt \!</math> | <br><br><math> = \int_{0}^{2} e^{4t}\ dt \!</math> | ||
| − | <br><br><math> = | + | <br><br><math> = \frac{1}{4}[e^{4t}]_{t=0}^{t=2} \!</math> |
| − | <br><br><math> = | + | <br><br><math> = \frac{1}{4}(e^8 -1)\!</math> |
== Average Signal Power== | == Average Signal Power== | ||
| − | The average signal power over an interval <math>[t_1,t_2]\!</math> is defined as <math> | + | The average signal power over an interval <math>[t_1,t_2]\!</math> is defined as <math>P_{avg}=\frac{1}{t_2-t_1} \int_{t_1}^{t_2} |x(t)|^2\ dt \!</math>. |
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| + | The following is an example problem to find the average signal power for <math>x(t)=e^{2t}\!</math> over <math>[0,2]\!</math>: | ||
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| + | <math>P_{avg}=\frac{1}{2-0} \int_{0}^{2} |e^{2t}|^2\ dt \!</math> | ||
| + | <br><br><math> = \frac{1}{2}\int_{0}^{2} e^{4t}\ dt \!</math> | ||
| + | <br><br><math> = \frac{1}{2}*\frac{1}{4}[e^{4t}]_{t=0}^{t=2} \!</math> | ||
| + | <br><br><math> = \frac{1}{8}(e^8 -1)\!</math> | ||
Latest revision as of 13:29, 4 September 2008
Signal Energy
The signal energy expanded from $ t_1\! $ to $ t_2\! $ is defined as $ E = \int_{t_1}^{t_2} \! |x(t)|^2\ dt $.
The following is an example problem to find the signal energy for $ x(t)=e^{2t}\! $ on $ [0,2]\! $:
$ E = \int_{0}^{2} |e^{2t}|^2\ dt \! $
$ = \int_{0}^{2} e^{4t}\ dt \! $
$ = \frac{1}{4}[e^{4t}]_{t=0}^{t=2} \! $
$ = \frac{1}{4}(e^8 -1)\! $
Average Signal Power
The average signal power over an interval $ [t_1,t_2]\! $ is defined as $ P_{avg}=\frac{1}{t_2-t_1} \int_{t_1}^{t_2} |x(t)|^2\ dt \! $.
The following is an example problem to find the average signal power for $ x(t)=e^{2t}\! $ over $ [0,2]\! $:
$ P_{avg}=\frac{1}{2-0} \int_{0}^{2} |e^{2t}|^2\ dt \! $
$ = \frac{1}{2}\int_{0}^{2} e^{4t}\ dt \! $
$ = \frac{1}{2}*\frac{1}{4}[e^{4t}]_{t=0}^{t=2} \! $
$ = \frac{1}{8}(e^8 -1)\! $
