Difference between revisions of "E infinity and P infinity for CT signal - Tyler Mattmuller (smattmul)" - Rhea

Revision as of 19:35, 20 June 2009

Compute $ E\infty $

$ x(t)=\cos(t)+j*\sin(t) $

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

$ E\infty=\int_{-\infty}^\infty |\cos(t)+j\sin(t)|^2dt $

$ E\infty=\int_{-\infty}^\infty |e^{j*t}|^2dt $

$ E\infty=\int_{-\infty}^\infty |e^2*e^{j*t}|dt $

$ E\infty=e^2/j*(\infty-0) $

$ E\infty=\infty $

Compute $ P\infty $

$ x(t)=\cos(t)+j*\sin(t) $

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

$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T |\cos(t)+j\sin(t)|^2dt $

$ P\infty=\lim_{T \to \infty}\int_{-T}^T|e^{j*t}|^2dt $

$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T|e^2*e^{j*t}|dt $

$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T|e^2*e^{j*t}|dt $

$ P\infty=\frac{\int_{-\infty}^\infty |e^2*e^{j*t}|dt}{\lim_{T \to \infty}2*T*\int_{-T}^T|e^2*e^{j*t}|dt)} $

$ P\infty=\frac{1}{\lim_{T \to \infty}2*T} $

$ P\infty=0 $