Difference between revisions of "Calculating E infinity and P infinity - Stuart Pulliam (spulliam)" - Rhea

Latest revision as of 17:05, 25 February 2015

Problem

Calculate the energy $ E_\infty $ and the average power $ P_\infty $ for the CT signal $ x(t)=2t^2 $

Solution 1

$ E_{\infty} $

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

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

$ E_{\infty}=2\int_{-\infty}^\infty t^4\,dt $

$ E_{\infty}=\frac{2}{5}t^5|_{-\infty}^\infty $

$ E_{\infty}=\infty $

$ P_{\infty} $

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

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

$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{T}\int_{-T}^{T}t^4\,dt $

$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{5T}t^5|_{-T}^{T} $

$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{5T}[T^5 - (-T^5)] $

$ P_{\infty}=\lim_{T \to \infty} \ \frac{2T^5}{5T} $

$ P_{\infty}=\lim_{T \to \infty} \ \frac{2T^4}{5} $

$ P_{\infty}=\infty $

Both of your final answer are correct, but there are small mistakes (constant multipliers) in your computation.

Solution 2

$ E_{\infty}=\int_{-\infty}^\infty |x(t)|^2\,dt =\int_{-\infty}^\infty |2t^2|^2\,dt= 4\int_{-\infty}^\infty t^4\,dt =\infty $

$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|x(t)|^2\,dt= \lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|2t^2|^2\,dt =\lim_{T \to \infty} \ \frac{2}{T}\int_{-T}^{T}t^4\,dt =\lim_{T \to \infty} \ \frac{2}{5T}t^5|_{-T}^{T}=\lim_{T \to \infty} \ \frac{2}{5T}[T^5 - (-T^5)] =\lim_{T \to \infty} \ \frac{4T^4}{5}=\infty $

Looks pretty good!

Solution 3

$ E_{\infty}=\int_{-\infty}^\infty |x(t)|^2\,dt =\int_{-\infty}^\infty |2t^2|^2\,dt= 4\int_{-\infty}^\infty t^4\,dt = 4 \frac{t^5}{5}=\infty. $

$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|x(t)|^2\,dt= \lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|2t^2|^2\,dt =\lim_{T \to \infty} \ \frac{2}{T}\int_{-T}^{T}t^4\,dt =\frac{\infty}{\infty}=1. $

The energy computation looks good. But in the power computation you distributed the limit too early and so your final answer is wrong.

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