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<math>\begin{align} | <math>\begin{align} | ||
\left|\cos(5t)\right|^{2} = \cos^2(5t) \\ | \left|\cos(5t)\right|^{2} = \cos^2(5t) \\ | ||
| − | \cos^2(5t) = \frac{1 | + | \cos^2(5t) = \frac{1+\cos(10t)}{2} |
\end{align}</math> | \end{align}</math> | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
| − | E_{\infty} &=\int_{-\infty}^\infty \cos^2(5t) dt | + | E_{\infty} &=\int_{-\infty}^\infty \cos^2(5t) dt \\ |
| − | &=\int_{-\infty}^\infty \frac{1 | + | &=\int_{-\infty}^\infty \frac{1+\cos(10t)}{2} dt \\ |
| + | &=\int_{-\infty}^\infty \frac{1}{2} + {\frac{1}{2}}\cos(5t) dt \\ | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 18:27, 1 December 2018
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) \\ \cos^2(5t) = \frac{1+\cos(10t)}{2} \end{align} $
$ \begin{align} E_{\infty} &=\int_{-\infty}^\infty \cos^2(5t) dt \\ &=\int_{-\infty}^\infty \frac{1+\cos(10t)}{2} dt \\ &=\int_{-\infty}^\infty \frac{1}{2} + {\frac{1}{2}}\cos(5t) dt \\ \end{align} $
