Work By Ryne Rayburn (rrayburn)
$ x(t)=\sqrt{t} $
$ E\infty=\int_{-\infty}^\infty |x(t)|^2\,dt $
$ E\infty=\int_{-\infty}^\infty |\sqrt{t}|^2\,dt=\int_0^\infty t\,dt $ (due to sqrt limiting to positive Real numbers) $ E\infty=.5*t^2| $(from $ 0 $ to $ \infty $) $ E\infty=.5(\infty^2-0^2)=\infty $
$ P\infty=lim_{T \to \infty} \ 1/(2T)\int_{-T}^{T} |x(t)|^2\,dt $
$ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T |\sqrt{t}|^2\,dt=\int_0^T t\,dt $ $ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*.5t^2| $(from $ 0 $ to $ T $) $ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(.5T^2) $ $ P\infty=lim_{T \to \infty} \ (.25T)=\infty $
$ x(t)=\cos(t)+\jmath\sin(t) $
$ |x(t)|=|\cos(t)+\jmath\sin(t)|=\sqrt{\cos^2(t)+\sin^2(t)}=1 $
$ E\infty=\int_{-\infty}^\infty |x(t)|^2\,dt $
$ E\infty=\int_{-\infty}^\infty |1|^2\,dt=t| $(from $ -\infty $ to $ \infty $) $ E\infty=\infty $
$ P\infty=lim_{T \to \infty} \ 1/(2T)\int_{-T}^{T} |x(t)|^2\,dt $
$ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int|1|^2dt $ $ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}t| $(from $ -T $ to $ T $) $ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}(T-(-T)) $ $ P\infty=lim_{T \to \infty} \ 1 $ $ P\infty=1 $