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!
