| Line 1: | Line 1: | ||
| − | = Compute the energy <math>E_\infty</math> and the power <math>P_\infty</math> of the following discrete-time signal = | + | = [[:Category:Problem_solving|Practice Problem]]: Compute the energy <math>E_\infty</math> and the power <math>P_\infty</math> of the following discrete-time signal = |
<span class="texhtml">''x''[''n''] = ''j''</span> | <span class="texhtml">''x''[''n''] = ''j''</span> | ||
Revision as of 09:16, 11 November 2011
Contents
Practice Problem: Compute the energy $ E_\infty $ and the power $ P_\infty $ of the following discrete-time signal
x[n] = j
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}&=\lim_{N\rightarrow \infty}\sum_{n=-N}^N |j|^2 \\ &= \lim_{N\rightarrow \infty}\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.
