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===Answer 1=== | ===Answer 1=== | ||
| − | + | <math> | |
| + | \begin{align} | ||
| + | E_{\infty}&=\lim_{T\rightarrow \infty}\int_{-T}^T |j|^2 dx | ||
| + | &= \lim_{T\rightarrow \infty}\int_{-T}^T {(\sqrt{jj*})}^2 dx | ||
| + | &= \lim_{T\rightarrow \infty}\int_{-T}^T {(\sqrt{-j^2})}^2 dx | ||
| + | & = \lim_{T\rightarrow \infty}\int_{-T}^T 1 dx | ||
| + | &= \lim_{T\rightarrow \infty} t \Big| ^T _{-T} | ||
| + | &=\infty. | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | So <math class="inline">E_{\infty} = \infty</math>. | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | P_{\infty}&=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T |j|^2 dx | ||
| + | &= \lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T 1 dx | ||
| + | & = \lim_{T\rightarrow \infty} {1 \over {2T}} t \Big| ^T _{-T} | ||
| + | & = \lim_{T\rightarrow \infty} {1 \over {2T}} T - {1 \over {2T}} | ||
| + | & = \lim_{T\rightarrow \infty} {1 \over {2}} + {1 \over {2}} | ||
| + | &= 1 | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | So <math class="inline">P_{\infty} = 1 </math>. | ||
| + | |||
| + | --[[User:Rgieseck|Rgieseck]] 21:35, 12 January 2011 [[Category:ECE301Spring2011Boutin]] | ||
===Answer 2=== | ===Answer 2=== | ||
write it here. | write it here. | ||
Revision as of 17:34, 12 January 2011
Contents
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_{T\rightarrow \infty}\int_{-T}^T |j|^2 dx &= \lim_{T\rightarrow \infty}\int_{-T}^T {(\sqrt{jj*})}^2 dx &= \lim_{T\rightarrow \infty}\int_{-T}^T {(\sqrt{-j^2})}^2 dx & = \lim_{T\rightarrow \infty}\int_{-T}^T 1 dx &= \lim_{T\rightarrow \infty} t \Big| ^T _{-T} &=\infty. \end{align} $
So $ E_{\infty} = \infty $.
$ \begin{align} P_{\infty}&=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T |j|^2 dx &= \lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T 1 dx & = \lim_{T\rightarrow \infty} {1 \over {2T}} t \Big| ^T _{-T} & = \lim_{T\rightarrow \infty} {1 \over {2T}} T - {1 \over {2T}} & = \lim_{T\rightarrow \infty} {1 \over {2}} + {1 \over {2}} &= 1 \end{align} $
So $ P_{\infty} = 1 $.
--Rgieseck 21:35, 12 January 2011
Answer 2
write it here.
Answer 3
write it here.
