(New page: <math> x(t)=e^t </math> [1,0] == Signal Energy == <math>E=\int_{t_1}^{t_2}x(t)dt</math> find the signal energy of <math>x(t)=e^{4t}\!</math> on <math>[0,1]\!</math> <math>E = \int_{0}^...) |
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| − | + | == Signal == | |
| − | == | + | <math> x(t)=e^t </math> [0,1] |
| + | |||
| + | == Energy == | ||
<math>E=\int_{t_1}^{t_2}x(t)dt</math> | <math>E=\int_{t_1}^{t_2}x(t)dt</math> | ||
| − | find the signal energy of <math>x(t)=e^{ | + | find the signal energy of <math>x(t)=e^{t}\!</math> on <math>[0,1]\!</math> |
<math>E = \int_{0}^{1} |e^{4t}|^2\ dt \!</math> | <math>E = \int_{0}^{1} |e^{4t}|^2\ dt \!</math> | ||
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<math> = \frac{1}{8}(e^8 -1)\!</math> | <math> = \frac{1}{8}(e^8 -1)\!</math> | ||
| − | == | + | == 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>. | 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>. | ||
Revision as of 14:42, 5 September 2008
Signal
$ x(t)=e^t $ [0,1]
Energy
$ E=\int_{t_1}^{t_2}x(t)dt $
find the signal energy of $ x(t)=e^{t}\! $ 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)\! $
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)\! $
