Topic: Signal Energy and Power

## Question

Compute the energy $E_\infty$ and the power $P_\infty$ of the following continuous-time signal

$x(t)= \sin (2 \pi t)$


What properties of the complex magnitude can you use to check your answer?

\begin{align} E_{\infty}&=\int_{-\infty}^\infty |\sin(2 \pi t)|^2 dt \\ &=\int_{-\infty}^\infty \sin^2(2 \pi t) dt \end{align}

But $\cos(2x) = \cos^2(x)-\sin^2(x)=1-2\sin^2(x).$

and therefore $\sin^2x = \frac{1-\cos(2x)}{2}$.

\begin{align} E_{\infty}&=\int_{-\infty}^\infty \frac{1-\cos(4 \pi t)}{2} dt \\ &=\int_{-\infty}^\infty \frac{1}{2} dt - \int_{-\infty}^\infty \frac{\cos(4\pi t)}{2} dt \\ &\\ &=\infty \end{align}

So $E_{\infty} = \infty$.

\begin{align} P_{\infty}&=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T |\sin(2\pi t)|^2 dt \quad \\ \text{Similar to math above, the expression can be derived towards}\\ &= \lim_{T\rightarrow \infty} {1 \over {2T}} (\int_{-T}^T \frac{1}{2} dt - \int_{-T}^T \frac{1}{2} * \cos(4\pi t) dt) \quad \\ & = \lim_{T\rightarrow \infty} {1 \over {2T}} (\frac{1}{2} t \Big| ^T _{-T} - \frac{1}{8\pi} \int_{-T}^T \cos(4\pi t) d(4\pi t)) \quad \\ & = \lim_{T\rightarrow \infty} {1 \over {2T}} ((\frac{1}{2}T - \frac{1}{2}(-T)) - \frac{1}{8\pi} (\sin(4\pi t)) \Big| ^T _{-T}) \quad \\ & = \lim_{T\rightarrow \infty} {1 \over {2T}} (T - \frac{1}{8\pi} (\sin(4\pi T) - \sin(4\pi T)) \quad \\ &= \lim_{T\rightarrow \infty} {1 \over {2T}} (T) \quad \\ &= \lim_{T\rightarrow \infty} {1 \over {2}} \quad \\ &= \frac{1}{2} \quad \\ \end{align}

So $P_{\infty} = \frac{1}{2}$.