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| − | == | + | == For a Continuous Time Signal== |
| + | Energy from <math>t_{1} </math> to <math>t_{2}</math> | ||
| − | == | + | <math>E = \int_{t_1}^{t_2}\!|x(t)|^2\ dt</math> |
| + | |||
| + | <math>E = \int_{t_1}^{t_2}\!|\sqrt{t}|^2\ dt</math> | ||
| + | |||
| + | <math>E = \int_{t_1}^{t_2}\!t\ dt</math> | ||
| + | |||
| + | <math>E = \frac{1}{2}t^{2}|^{t_{2}}_{t_{1}}</math> | ||
| + | |||
| + | <math>E = \frac{1}{2}(t^{2}_{2}-t^{2}_{1})</math> | ||
| + | |||
| + | Average power in time interval from [<math>t_{1},t_{2} </math>]: | ||
| + | |||
| + | <math>P_{avg} = \frac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}\!|x(t)|^2\ dt</math> | ||
| + | |||
| + | <math>P_{avg} = \frac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}\!|\sqrt{t}|^2\ dt</math> | ||
| + | |||
| + | <math>P_{avg} = \frac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}\!t\ dt</math> | ||
| + | |||
| + | <math>P_{avg} = \frac{1}{{t_2}-{t_1}}(\frac{1}{2}t^{2}|^{t_{2}}_{t_{1}})</math> | ||
| + | |||
| + | <math>P_{avg} = \frac{1}{{t_2}-{t_1}}(\frac{1}{2}(t^{2}_{2}-t^{2}_{1}))</math> | ||
Latest revision as of 12:34, 5 September 2008
For a Continuous Time Signal
Energy from $ t_{1} $ to $ t_{2} $
$ E = \int_{t_1}^{t_2}\!|x(t)|^2\ dt $
$ E = \int_{t_1}^{t_2}\!|\sqrt{t}|^2\ dt $
$ E = \int_{t_1}^{t_2}\!t\ dt $
$ E = \frac{1}{2}t^{2}|^{t_{2}}_{t_{1}} $
$ E = \frac{1}{2}(t^{2}_{2}-t^{2}_{1}) $
Average power in time interval from [$ t_{1},t_{2} $]:
$ P_{avg} = \frac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}\!|x(t)|^2\ dt $
$ P_{avg} = \frac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}\!|\sqrt{t}|^2\ dt $
$ P_{avg} = \frac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}\!t\ dt $
$ P_{avg} = \frac{1}{{t_2}-{t_1}}(\frac{1}{2}t^{2}|^{t_{2}}_{t_{1}}) $
$ P_{avg} = \frac{1}{{t_2}-{t_1}}(\frac{1}{2}(t^{2}_{2}-t^{2}_{1})) $
