Problem
Compute the energy and the power of the CT sinusoidal signal below:
$ x(t)= \cos (5t) $
Solution
$ \begin{align} \left|\cos(5t)\right|^{2} = |\cos^2(5t)|^2 \\ \cos^2(5t) = \frac{1+\cos(10t)}{2} \end{align} $
$ \begin{align} E_{\infty} &=\int_{-\infty}^\infty |\cos^2(5t)|^2 dt \\ &=\int_{-\infty}^\infty \frac{1+\cos(10t)}{2} dt \\ &=\int_{-\infty}^\infty \frac{1}{2} + {\frac{1}{2}}\cos(10t) dt \\ &= (t + {\frac{1}{10}}\sin(10t)) \Big| ^T _{-T}\\ &=\infty\\ \end{align} $
$ E_{\infty} = \infty $.
$ \begin{align} P_{\infty}&=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T |\cos^2(5t)|^2 dt \\ &= \lim_{T\rightarrow \infty} {1 \over {2T}} (\int_{-T}^T \frac{1+\cos(10t)}{2} dt \quad \\ & = \lim_{T\rightarrow \infty} {1 \over {2T}} ((t + {\frac{1}{10}}\sin(10t)) \Big| ^T _{-T} \quad \\ & = \lim_{T\rightarrow \infty} {1 \over {2T}} (\frac{1}{2}T - \frac{1}{10}(-T) )+ \frac{1}{10} (\sin(10T) - \sin(-10T)) \quad \\ &= \lim_{T\rightarrow \infty} {1 \over {2T}} (T) \quad \\ &= \lim_{T\rightarrow \infty} {1 \over {2}} \quad \\ &= \frac{1}{2} \quad \\ \end{align} $
$ P_{\infty} = \frac{1}{2} $.
