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E_{\infty} &=\int_{-\infty}^\infty \cos^2(5t) dt \\ | E_{\infty} &=\int_{-\infty}^\infty \cos^2(5t) dt \\ | ||
&=\int_{-\infty}^\infty \frac{1+\cos(10t)}{2} dt \\ | &=\int_{-\infty}^\infty \frac{1+\cos(10t)}{2} dt \\ | ||
− | &=\int_{-\infty}^\infty \frac{1}{2} + {\frac{1}{2}}\cos( | + | &=\int_{-\infty}^\infty \frac{1}{2} + {\frac{1}{2}}\cos(10t) dt \\ |
+ | &= (t + {\frac{1}{10}}\sin(10t) \Big| ^T _{-T} | ||
\end{align} | \end{align} | ||
</math> | </math> |
Revision as of 18:42, 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(10t) dt \\ &= (t + {\frac{1}{10}}\sin(10t) \Big| ^T _{-T} \end{align} $