Problem
Calculate the energy $ E_\infty $ and the average power $ P_\infty $ for the CT signal $ x(t)=\sqrt{2t}u(t) $
Solution 1
Compute $ E\infty $
$ E\infty=\int_{-\infty}^\infty |x(t)|^2dt $
$ E\infty=\int_{-\infty}^\infty |\sqrt{2t}u(t)|^2dt $ (*)
$ E\infty=\int_{-\infty}^\infty |2tu(t)|dt $ (*)
$ E\infty=\int_{0}^\infty 2tdt $ (*)
$ E\infty=t^2|_{0}^{\infty} $
$ E\infty= \infty-0 $
$ E\infty=\infty $
Compute $ P\infty $
$ 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 |\sqrt(2t)u(t)|^2dt $
$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T|2tu(t)|dt $
$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{0}^T2tdt $
$ P\infty=\lim_{T \to \infty}\frac{|t^2|_{0}^{T}}{2*T} $ (*)
$ P\infty=\lim_{T \to \infty}\frac{T^2-0}{2*T} $
$ P\infty=\lim_{T \to \infty}\frac{T}{2} $
$ P\infty=\infty $
- It would be better not to use the start symbol to denote multiplication in this context. The star symbol in electrical engineering is usually denoting convolution.
- *This is probably not the best way to order these steps, but it worked out at the end.
- *You should not have the absolute values at this point.
Solution 2
$ E_\infty=\int_{-\infty}^\infty |x(t)|^2d=\int_{-\infty}^\infty |\sqrt{2t}u(t)|^2dt =\int_{0}^\infty | \sqrt{2t}|^2dt =\int_{0}^\infty 2tdt= \infty $
$ P\infty=\lim_{T \to \infty}\frac{1}{2T}\int_{-T}^T |x(t)|^2dt =\lim_{T \to \infty}\frac{1}{2T}\int_{-T}^T \left| \sqrt{2t} u(t) \right|^2dt =\lim_{T \to \infty}\frac{1}{2T}\int_{0}^T 2t dt= \lim_{T \to \infty}\frac{ t^2|_{0}^{T}}{2 T}=\lim_{T \to \infty}\frac{T^2-0}{2T}= \infty $
Short and sweet!
