(New page: Let us find the energy and average power of a signal x(t) = ==Energy== ==Average Power==) |
|||
| Line 1: | Line 1: | ||
| − | Let us find the energy and average power of a signal x(t) = | + | Let us find the energy and average power of a signal <math>x(t) = 5e^{5t}</math> for the time interval [0,5] |
==Energy== | ==Energy== | ||
| + | |||
| + | Energy expanded from [0,5] | ||
| + | |||
| + | <math>E = \int_0^{5}{(5e^{5t})^2dt}</math> | ||
| + | |||
| + | <math>E =\int_0^{5}{25e^{10t}dt}</math> | ||
| + | |||
| + | <math>E =25\int_0^{5}{e^{10t}dt}</math> | ||
| + | |||
| + | <math>E =25|\frac{e^{10t}}{10}|_{t=0}^{t=5}</math> | ||
| + | |||
| + | <math>E =\frac{25}{10}|e^{50}-e^0|</math> | ||
| + | |||
| + | <math>E =2.5(e^{50}-1)</math> | ||
| + | |||
==Average Power== | ==Average Power== | ||
| + | |||
| + | <math>P =\frac{1}{5-0}\int_0^{5}{(5e^{5t})^2dt}</math> | ||
| + | |||
| + | <math>P =\frac{1}{5}\int_0^{5}{25e^{10t}dt}</math> | ||
| + | |||
| + | <math>P =5\int_0^{5}{e^{10t}dt}</math> | ||
| + | |||
| + | <math>P =5\int_0^{5}{e^{10t}dt}</math> | ||
| + | |||
| + | <math>P =5|\frac{e^{10t}}{10}|_{t=0}^{t=5}</math> | ||
| + | |||
| + | <math>P =0.5[e^{50} - e^0]</math> | ||
| + | |||
| + | <math>P = 0.5(e^{50} - 1)</math> | ||
Latest revision as of 10:12, 4 September 2008
Let us find the energy and average power of a signal $ x(t) = 5e^{5t} $ for the time interval [0,5]
Energy
Energy expanded from [0,5]
$ E = \int_0^{5}{(5e^{5t})^2dt} $
$ E =\int_0^{5}{25e^{10t}dt} $
$ E =25\int_0^{5}{e^{10t}dt} $
$ E =25|\frac{e^{10t}}{10}|_{t=0}^{t=5} $
$ E =\frac{25}{10}|e^{50}-e^0| $
$ E =2.5(e^{50}-1) $
Average Power
$ P =\frac{1}{5-0}\int_0^{5}{(5e^{5t})^2dt} $
$ P =\frac{1}{5}\int_0^{5}{25e^{10t}dt} $
$ P =5\int_0^{5}{e^{10t}dt} $
$ P =5\int_0^{5}{e^{10t}dt} $
$ P =5|\frac{e^{10t}}{10}|_{t=0}^{t=5} $
$ P =0.5[e^{50} - e^0] $
$ P = 0.5(e^{50} - 1) $
