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| − | + | [[Category:energy]] | |
| + | [[Category:signal]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:practice problem]] | ||
| − | |||
| − | <math>E_\infty = \int_{-\infty}^\infty | + | ==Problem== |
| + | Calculate the energy <math>E_\infty</math> and the average power <math>P_\infty</math> for the CT signal | ||
| + | <math>f(t)=5 j \sin (t)</math> | ||
| + | ---- | ||
| + | ==Solution 1== | ||
| + | <math>E_\infty = \int_{-\infty}^\infty |5sin(t)|^2\,dt</math> | ||
| − | <math>E_\infty = \int_{-\infty}^\infty | + | <math>E_\infty = \int_{-\infty}^\infty 25sin(t)^2 \,dt</math> |
| − | <math>E_\infty | + | <math>E_\infty = \int_{-\infty}^\infty 25(.5 + .5cos(2t)),dt</math> |
| − | <math>E_\infty =\infty-0 = \infty</math> | + | <math>E_\infty =\frac{25t}{2} + \frac{25sin(2t)}{4}\bigg]_{-\infty}^\infty)</math> |
| + | |||
| + | <math>E_\infty =\infty-0 = \infty</math> <span style="color:violet"> (*) </span> | ||
| + | |||
| + | <math>P_\infty calculation</math> | ||
| + | |||
| + | <math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T|5sin(t)|^2dt</math> | ||
| + | |||
| + | <math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T \frac{25}{2} + \frac{25cos(2t)}{2}dt</math> | ||
| + | |||
| + | <math>= lim_{T \to \infty} \frac{1}{2T}*(\frac{25t}{2} + \frac{25sin(t)}{4}|_{-T}^T)</math> <span style="color:green"> (*) </span> <span style="color:orange">(*)</span> | ||
| + | |||
| + | <math>= lim_{T \to \infty} \frac{1}{2T}*(25T + \frac{25sin(T)}{4}-\frac{25sin(-T)}{4}</math> <span style="color:orange">(*)</span> | ||
| + | |||
| + | <math>= lim_{T \to \infty} \frac{1}{2T}*(25T + \frac{25sin(T)}{2})</math> <span style="color:orange">(*)</span> | ||
| + | |||
| + | <math>P\infty= \frac{25}{2} + 0</math> | ||
| + | |||
| + | *<span style="color:orange"> * I would not use the star symbol to denote multiplication here. It is usually reserved for convolution in electrical engineering.</span> | ||
| + | *<span style="color:green"> * You are missing a 2 inside the sine.</span> | ||
| + | *<span style="color:violet"> *You should not put a zero here, as the limit when t goes to infinity of of sin(t) is not defined. </span> | ||
| + | ---- | ||
| + | ==Solution 2== | ||
| + | <math>E_\infty = \int_{-\infty}^\infty |5sin(t)|^2\,dt = \int_{-\infty}^\infty 25sin(t)^2 dt = \infty,</math> since the function integrated is always non-negative and does not decrease as t approaches <math>\pm \infty </math>. | ||
| + | |||
| + | |||
| + | <math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T|5sin(t)|^2dt= lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T25 \sin^2 (t) dt = lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T \frac{25}{2} + \frac{25cos(2t)}{2}dt</math> | ||
| + | |||
| + | <math>= lim_{T \to \infty} \frac{1}{2T}\left.\left( \frac{25t}{2} + \frac{25sin(2t)}{4} \right)\right|_{-T}^T \, </math> | ||
| + | |||
| + | <math>= lim_{T \to \infty} \frac{1}{2T}\left( 25T + \frac{25sin(2T)}{4}-\frac{25sin(-2T)}{4}\right) </math> | ||
| + | |||
| + | <math>= lim_{T \to \infty} \frac{25}{2}+ \frac{25sin(2T)}{4T}</math> | ||
| + | |||
| + | <math>= \frac{25}{2} + 0</math>, since <math> \sin (2T) </math> is bounded by (-1) and 1 | ||
| + | |||
| + | <math> = \frac{25}{2} </math> | ||
| + | *<span style="color:green"> Looks good!</span> | ||
| + | ---- | ||
| + | ---- | ||
| + | [[Signal_energy_CT|Back to CT signal energy page]] | ||
Latest revision as of 17:55, 25 February 2015
Problem
Calculate the energy $ E_\infty $ and the average power $ P_\infty $ for the CT signal $ f(t)=5 j \sin (t) $
Solution 1
$ E_\infty = \int_{-\infty}^\infty |5sin(t)|^2\,dt $
$ E_\infty = \int_{-\infty}^\infty 25sin(t)^2 \,dt $
$ E_\infty = \int_{-\infty}^\infty 25(.5 + .5cos(2t)),dt $
$ E_\infty =\frac{25t}{2} + \frac{25sin(2t)}{4}\bigg]_{-\infty}^\infty) $
$ E_\infty =\infty-0 = \infty $ (*)
$ P_\infty calculation $
$ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T|5sin(t)|^2dt $
$ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T \frac{25}{2} + \frac{25cos(2t)}{2}dt $
$ = lim_{T \to \infty} \frac{1}{2T}*(\frac{25t}{2} + \frac{25sin(t)}{4}|_{-T}^T) $ (*) (*)
$ = lim_{T \to \infty} \frac{1}{2T}*(25T + \frac{25sin(T)}{4}-\frac{25sin(-T)}{4} $ (*)
$ = lim_{T \to \infty} \frac{1}{2T}*(25T + \frac{25sin(T)}{2}) $ (*)
$ P\infty= \frac{25}{2} + 0 $
- * I would not use the star symbol to denote multiplication here. It is usually reserved for convolution in electrical engineering.
- * You are missing a 2 inside the sine.
- *You should not put a zero here, as the limit when t goes to infinity of of sin(t) is not defined.
Solution 2
$ E_\infty = \int_{-\infty}^\infty |5sin(t)|^2\,dt = \int_{-\infty}^\infty 25sin(t)^2 dt = \infty, $ since the function integrated is always non-negative and does not decrease as t approaches $ \pm \infty $.
$ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T|5sin(t)|^2dt= lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T25 \sin^2 (t) dt = lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T \frac{25}{2} + \frac{25cos(2t)}{2}dt $
$ = lim_{T \to \infty} \frac{1}{2T}\left.\left( \frac{25t}{2} + \frac{25sin(2t)}{4} \right)\right|_{-T}^T \, $
$ = lim_{T \to \infty} \frac{1}{2T}\left( 25T + \frac{25sin(2T)}{4}-\frac{25sin(-2T)}{4}\right) $
$ = lim_{T \to \infty} \frac{25}{2}+ \frac{25sin(2T)}{4T} $
$ = \frac{25}{2} + 0 $, since $ \sin (2T) $ is bounded by (-1) and 1
$ = \frac{25}{2} $
- Looks good!
