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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)|^2 \\ |
\cos^2(5t) = \frac{1+\cos(10t)}{2} | \cos^2(5t) = \frac{1+\cos(10t)}{2} | ||
\end{align}</math> | \end{align}</math> | ||
| Line 13: | Line 13: | ||
<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+\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 \\ | ||
| Line 24: | Line 24: | ||
<math class="inline">E_{\infty} = \infty </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}} (T - \frac{1}{8\pi} (\sin(4\pi T) - \sin(4\pi T)) \quad \\ | ||
| + | &= \lim_{T\rightarrow \infty} {1 \over {2T}} (T) \quad \\ | ||
| + | &= \lim_{T\rightarrow \infty} {1 \over {2}} \quad \\ | ||
| + | &= \frac{1}{2} \quad \\ | ||
| + | \end{align} | ||
| + | </math> | ||
<math class="inline">P_{\infty} = \frac{1}{2} </math>. | <math class="inline">P_{\infty} = \frac{1}{2} </math>. | ||
Revision as of 18:58, 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 \\ &= (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}} (T - \frac{1}{8\pi} (\sin(4\pi T) - \sin(4\pi T)) \quad \\ &= \lim_{T\rightarrow \infty} {1 \over {2T}} (T) \quad \\ &= \lim_{T\rightarrow \infty} {1 \over {2}} \quad \\ &= \frac{1}{2} \quad \\ \end{align} $
$ P_{\infty} = \frac{1}{2} $.
