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!


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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett