(→Energy) |
(→Power) |
||
| Line 18: | Line 18: | ||
===Power=== | ===Power=== | ||
| + | :<math> y = 2e^{t} </math> from (0,10) | ||
| + | |||
| + | :<math>Average Power = \frac{1}{t2 - t1}\int_{t1}^{t2}x(t)^2 </math> | ||
| + | |||
| + | |||
| + | :<math>Average Power = \frac{1}{10}\int_{0}^{5}4e^{2t}dt </math> | ||
| + | |||
| + | :<math>Average Power = \frac{1}{10}({2}e^{20} - 2) </math> | ||
Latest revision as of 18:57, 5 September 2008
Problem
Compute the energy and power of a CT signal $ y=2e^t $ from (0,10)
Energy
$ E = \int_{t_1}^{t_2} \! |f(t)|^2\ dt $
$ =\int_0^{10}{|2e^t|^2dt} $
$ =\int_0^{10}{(4e^{2t})dt} $
$ =(2e^{2t})|_{t=0}^{t=10} $
$ =2e^{20}-2 $
$ =9.703x10^{8} $
Power
- $ y = 2e^{t} $ from (0,10)
- $ Average Power = \frac{1}{t2 - t1}\int_{t1}^{t2}x(t)^2 $
- $ Average Power = \frac{1}{10}\int_{0}^{5}4e^{2t}dt $
- $ Average Power = \frac{1}{10}({2}e^{20} - 2) $
