(New page: =Signal Power= The average power over an interval of time <math>[t_1,t_2]\!</math> is <math>P_{avg}=\frac{1}{t_2-t_1}\int_{t_1}^{t_2} |x(t)|^2\ dt \!</math>. Given that <math>y(t) = cos(2...) |
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The average power over an interval of time <math>[t_1,t_2]\!</math> is <math>P_{avg}=\frac{1}{t_2-t_1}\int_{t_1}^{t_2} |x(t)|^2\ dt \!</math>. | The average power over an interval of time <math>[t_1,t_2]\!</math> is <math>P_{avg}=\frac{1}{t_2-t_1}\int_{t_1}^{t_2} |x(t)|^2\ dt \!</math>. | ||
| − | Given that <math>y(t) = | + | Given that <math>y(t) = t \!</math> and time interval <math>[0,2]\!</math>, |
| − | 1/ | + | |
| + | <math>P_{avg} = \frac{1}{2-0} \int_{0}^{2} |t|^2\ dt \!</math> | ||
| + | <math>=\frac{1}{2} [\frac{1}{3} t^3]_{0}^{2} \! </math> | ||
| + | <math>=\frac{1}{6}[8-0] = \frac{4}{3} \!</math> | ||
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<math>E=\int_{t_1}^{t_2}x(t)dt</math> | <math>E=\int_{t_1}^{t_2}x(t)dt</math> | ||
| − | using the same equation <math>y(t) = | + | using the same equation <math>y(t) = t \!</math> over a time interval <math>[0,2]\!</math> as the previous section, |
| + | |||
| + | <math>E = \int_{0}^{2} t\ dt = \frac{1}{2}[t^2]_{0}^{2} = \frac{1}{2}[4 - 0] = 2 \!</math> | ||
Latest revision as of 10:47, 5 September 2008
Signal Power
The average power over an interval of time $ [t_1,t_2]\! $ is $ P_{avg}=\frac{1}{t_2-t_1}\int_{t_1}^{t_2} |x(t)|^2\ dt \! $.
Given that $ y(t) = t \! $ and time interval $ [0,2]\! $,
$ P_{avg} = \frac{1}{2-0} \int_{0}^{2} |t|^2\ dt \! $ $ =\frac{1}{2} [\frac{1}{3} t^3]_{0}^{2} \! $ $ =\frac{1}{6}[8-0] = \frac{4}{3} \! $
Signal Energy
The equation to calculate signal energy is as follows: $ E=\int_{t_1}^{t_2}x(t)dt $
using the same equation $ y(t) = t \! $ over a time interval $ [0,2]\! $ as the previous section,
$ E = \int_{0}^{2} t\ dt = \frac{1}{2}[t^2]_{0}^{2} = \frac{1}{2}[4 - 0] = 2 \! $
