| Line 16: | Line 16: | ||
&=\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(10t) dt \\ | &=\int_{-\infty}^\infty \frac{1}{2} + {\frac{1}{2}}\cos(10t) dt \\ | ||
| − | &= (t + {\frac{1}{10}}\sin(10t)) \Big| ^T _{-T}\\ | + | &= (\frac{1}{2}t + {\frac{1}{10}}\sin(10t)) \Big| ^T _{-T}\\ |
&=\infty\\ | &=\infty\\ | ||
| Line 29: | Line 29: | ||
&= \lim_{T\rightarrow \infty} {1 \over {2T}} (\int_{-T}^T \frac{1+\cos(10t)}{2} dt \quad \\ | &= \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}} ((t + {\frac{1}{10}}\sin(10t)) \Big| ^T _{-T} \quad \\ | ||
| − | & = \lim_{T\rightarrow \infty} {1 \over {2T}} (\frac{1}{2}T | + | & = \lim_{T\rightarrow \infty} {1 \over {2T}} (\frac{1}{2}T + \frac{1}{10} (\sin(10T) - (\frac{1}{2}(-T) + \sin(-10T)) \quad \\ |
| − | &= \lim_{T\rightarrow \infty} {1 \over {2T}} ( | + | &= \lim_{T\rightarrow \infty} {1 \over {2T}} (2T) \quad \\ |
| − | + | &= \frac 1 \quad \\ | |
| − | &= \frac | + | |
\end{align} | \end{align} | ||
</math> | </math> | ||
<math class="inline">P_{\infty} = \frac{1}{2} </math>. | <math class="inline">P_{\infty} = \frac{1}{2} </math>. | ||
Revision as of 19:34, 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)|^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 \\ &= (\frac{1}{2}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} (\sin(10T) - (\frac{1}{2}(-T) + \sin(-10T)) \quad \\ &= \lim_{T\rightarrow \infty} {1 \over {2T}} (2T) \quad \\ &= \frac 1 \quad \\ \end{align} $
$ P_{\infty} = \frac{1}{2} $.
