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& = \lim_{T\rightarrow \infty} {1 \over {2T}} ((\frac{1}{2}T - \frac{1}{2}(-T)) - \frac{1}{8\pi} (\sin(4\pi t)) \Big| ^T _{-T}) \quad \\ | & = \lim_{T\rightarrow \infty} {1 \over {2T}} ((\frac{1}{2}T - \frac{1}{2}(-T)) - \frac{1}{8\pi} (\sin(4\pi t)) \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 - \frac{1}{8\pi} (\sin(4\pi T) - \sin(4\pi T)) \quad \\ | ||
| − | &= \lim_{T\rightarrow \infty} {1 \over {2T}} (T) | + | &= \lim_{T\rightarrow \infty} {1 \over {2T}} (T) \quad \\ |
| − | &= \lim_{T\rightarrow \infty} {1 \over {2}} | + | &= \lim_{T\rightarrow \infty} {1 \over {2}} \quad \\ |
| − | &= \frac{1}{2} | + | &= \frac{1}{2} \quad \\ |
\end{align} | \end{align} | ||
</math> | </math> | ||
| − | So <math class="inline">P_{\infty} = 1 </math>. | + | So <math class="inline">P_{\infty} = \frac{1}{2} </math>. |
Latest revision as of 09:09, 22 January 2018
Practice Question on "Signals and Systems"
Topic: Signal Energy and Power
Question
Compute the energy $ E_\infty $ and the power $ P_\infty $ of the following continuous-time signal
$ x(t)= \sin (2 \pi t) $
What properties of the complex magnitude can you use to check your answer?
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1=
$ \begin{align} E_{\infty}&=\int_{-\infty}^\infty |\sin(2 \pi t)|^2 dt \\ &=\int_{-\infty}^\infty \sin^2(2 \pi t) dt \end{align} $
But $ \cos(2x) = \cos^2(x)-\sin^2(x)=1-2\sin^2(x). $
and therefore $ \sin^2x = \frac{1-\cos(2x)}{2} $.
$ \begin{align} E_{\infty}&=\int_{-\infty}^\infty \frac{1-\cos(4 \pi t)}{2} dt \\ &=\int_{-\infty}^\infty \frac{1}{2} dt - \int_{-\infty}^\infty \frac{\cos(4\pi t)}{2} dt \\ &\\ &=\infty \end{align} $
So $ E_{\infty} = \infty $.
$ \begin{align} P_{\infty}&=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T |\sin(2\pi t)|^2 dt \quad \\ \text{Similar to math above, the expression can be derived towards}\\ &= \lim_{T\rightarrow \infty} {1 \over {2T}} (\int_{-T}^T \frac{1}{2} dt - \int_{-T}^T \frac{1}{2} * \cos(4\pi t) dt) \quad \\ & = \lim_{T\rightarrow \infty} {1 \over {2T}} (\frac{1}{2} t \Big| ^T _{-T} - \frac{1}{8\pi} \int_{-T}^T \cos(4\pi t) d(4\pi t)) \quad \\ & = \lim_{T\rightarrow \infty} {1 \over {2T}} ((\frac{1}{2}T - \frac{1}{2}(-T)) - \frac{1}{8\pi} (\sin(4\pi t)) \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} $
So $ P_{\infty} = \frac{1}{2} $.
