| (11 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
Compute the energy and the power of the CT sinusoidal signal below: | Compute the energy and the power of the CT sinusoidal signal below: | ||
| + | |||
<math>x(t)= \cos (5t)</math> | <math>x(t)= \cos (5t)</math> | ||
| + | |||
| + | |||
| + | |||
==Solution== | ==Solution== | ||
<math>\begin{align} | <math>\begin{align} | ||
| − | \left|\cos(5t)\right|^{2} = \cos^2(5t) \\ | + | \left|\cos(5t)\right|^{2} = |\cos^2(5t)|^2 \\ |
| − | \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)|^2 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(10t) dt \\ | ||
| + | &= (\frac{1}{2}t + {\frac{1}{10}}\sin(10t)) \Big| ^T _{-T}\\ | ||
| + | &=\infty\\ | ||
| + | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
| + | |||
| + | <math class="inline">E_{\infty} = \infty </math>. | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | <math> | ||
| + | \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) \\ | ||
| + | &= 1 \\ | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | <math class="inline">P_{\infty} = 1 </math>. | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | |||
| + | Conclusion: | ||
| + | |||
| + | <math class="inline">E_{\infty} = \infty </math>, | ||
| + | |||
| + | <math class="inline">P_{\infty} = 1 </math>. | ||
Latest revision as of 19:48, 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) \\ &= 1 \\ \end{align} $
$ P_{\infty} = 1 $.
Conclusion:
$ E_{\infty} = \infty $,
$ P_{\infty} = 1 $.
