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| − | E_{\infty}&= | + | E_{\infty}&=\sum_{n=0}^N |(\frac{1}{1+j})^n|^2 \\ |
| − | &= | + | &= \sum_{n=-N}^N {(\sqrt{jj^*})}^2 \\ |
&= \lim_{N\rightarrow \infty}\sum_{n=-N}^N {(\sqrt{-j^2})}^2 \\ | &= \lim_{N\rightarrow \infty}\sum_{n=-N}^N {(\sqrt{-j^2})}^2 \\ | ||
&= \lim_{N\rightarrow \infty}\sum_{n=-N}^N 1 \\ | &= \lim_{N\rightarrow \infty}\sum_{n=-N}^N 1 \\ | ||
Revision as of 10:40, 22 January 2018
Practice Question on "Signals and Systems"
Topic: Signal Energy and Power
Contents
Question
Compute the energy $ E_\infty $ and the power $ P_\infty $ of the following discrete-time signal
$ x[n] = \left\{ \begin{array}{ll} (\frac{1}{1+j})^n & \text{ if } n>=0,\\ 0 & \text{otherwise}. \end{array} \right. $
Answer 1
$ \begin{align} E_{\infty}&=\sum_{n=0}^N |(\frac{1}{1+j})^n|^2 \\ &= \sum_{n=-N}^N {(\sqrt{jj^*})}^2 \\ &= \lim_{N\rightarrow \infty}\sum_{n=-N}^N {(\sqrt{-j^2})}^2 \\ &= \lim_{N\rightarrow \infty}\sum_{n=-N}^N 1 \\ &=\infty. \\ \end{align} $
So $ E_{\infty} = \infty $.
- Instructors comment: Good job! The answer is correct and the justification is very clear. Now can someone compute the power? --Mboutin 19:31, 13 January 2011 (UTC)
$ \begin{align} P_{\infty}&=\lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N |j|^2 \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N {(\sqrt{jj^*})}^2 \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N {(\sqrt{-j^2})}^2 \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N 1 \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^{2N} 1 \\ &= \lim_{N\rightarrow \infty}{2N+1 \over {2N+1}} \\ &= \lim_{N\rightarrow \infty}{1}\\ &= 1 \\ \end{align} $
So $ P_{\infty} = 1 $.
--Rgieseck 21:35, 12 January 2011
Answer 2
write it here.
Answer 3
write it here.
