(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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| − | <math> x(t)=e^t </math> [ | + | == Signal == |
| + | '''<math> x(t)=e^t </math>''' [0,1] | ||
| − | == | + | == Energy == |
| − | <math>E=\int_{t_1}^{t_2}x(t) | + | <math>E=\int_{t_1}^{t_2}|x(t)|^2dt</math> |
| − | + | <math>E = \int_{0}^{1} |e^{t}|^2\ dt \!</math> | |
| − | <math> | + | <br><br><math> = \int_{0}^{2} e^{2t}\ dt \!</math> |
| − | < | + | <math> = \frac{1}{2}[e^{2t}]_{t=0}^{t=1} \!</math> |
| + | <math> = \frac{1}{2}(e^2 -1)\!</math> | ||
| − | + | == Power == | |
| − | + | ||
| − | = | + | <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^{t}|^2\ dt \!</math> | |
| − | + | <br><br><math> = \int_{0}^{1} e^{2t}\ dt \!</math> | |
| − | <math>P_{avg}=\frac{1}{1-0} \int_{0}^{1} |e^{ | + | <br><br><math> = \frac{1}{2}[e^{2t}]_{t=0}^{t=1} \!</math> |
| − | <br><br><math> = \int_{0}^{1} e^{ | + | <br><br><math> = \frac{1}{2}(e^2 -1)\!</math> |
| − | <br><br><math> = \frac{1}{ | + | |
| − | <br><br><math> = \frac{1}{ | + | |
Latest revision as of 14:51, 5 September 2008
Signal
$ x(t)=e^t $ [0,1]
Energy
$ E=\int_{t_1}^{t_2}|x(t)|^2dt $
$ E = \int_{0}^{1} |e^{t}|^2\ dt \! $
$ = \int_{0}^{2} e^{2t}\ dt \! $
$ = \frac{1}{2}[e^{2t}]_{t=0}^{t=1} \! $ $ = \frac{1}{2}(e^2 -1)\! $
Power
$ 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^{t}|^2\ dt \! $
$ = \int_{0}^{1} e^{2t}\ dt \! $
$ = \frac{1}{2}[e^{2t}]_{t=0}^{t=1} \! $
$ = \frac{1}{2}(e^2 -1)\! $
