Line 15: Line 15:
 
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(5t)  dt \\
+
&=\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} $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva