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Compute the energy and power of a CT signal <math>y=2e^t</math> from (0,10) | Compute the energy and power of a CT signal <math>y=2e^t</math> from (0,10) | ||
===Energy === | ===Energy === | ||
| + | <font size="4"> | ||
| + | |||
| + | <math>E = \int_{t_1}^{t_2} \! |f(t)|^2\ dt</math> | ||
| + | |||
| + | <math>=\int_0^{10}{|2e^t|^2dt}</math> | ||
| + | |||
| + | <math>=\int_0^{10}{(4e^{2t})dt}</math> | ||
| + | |||
| + | <math>=(2e^{2t})|_{t=0}^{t=10}</math> | ||
| + | |||
| + | <math>=2e^{20}-2</math> | ||
| + | |||
| + | <math>=9.703x10^{8}</math> | ||
| + | </font> | ||
===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) $
