Difference between revisions of "Calculating E infinity and P infinity - Jonathan Chu (Chu7)" - Rhea

Latest revision as of 17:06, 25 February 2015

Problem

Calculate the energy $ E_\infty $ and the average power $ P_\infty $ for the CT signal

$ x(t)=tu(t) $

Solution 1

$ E_\infty = \int_{-\infty}^\infty |tu(t)|^2\,dt = \int_{0}^\infty t^2\,dt=\infty $

$ P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \int_{-T}^T |tu(t)|^2\,dt = lim_{T \to \infty} \ \frac{1}{2T} \int_{0}^T t^2\,dt =\frac{\infty}{\infty}=1 $

Your energy is correct, but you distributed the limit too early when you computed the average power, so your answer came out wrong.

Solution 2

$ E_\infty = \int_{-\infty}^\infty |tu(t)|^2\,dt) $

$ E_\infty = \int_{0}^\infty t^2\,dt) $

$ E_\infty =\frac{t^3}{3}\bigg]_0^\infty) $

$ E_\infty =\infty-0 = \infty $

Calculating $ P_\infty $

$ P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \int_{-T}^T |tu(t)|^2\,dt $

$ P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \int_{0}^T t^2\,dt $

$ P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \frac{t^3}{3}\bigg]_0^T $

$ P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \frac{T^3}{3} $

$ P_\infty = lim_{T \to \infty} \ \frac{T^2}{6} $

$ P_\infty = \infty $

Looks pretty good!

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