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<math>\begin{align} | <math>\begin{align} | ||
| − | E_{\infty}&=\sum_{n=0}^N | | + | E_{\infty}&=\sum_{n=0}^N |c|^2 \\ |
&= \sum_{n=0}^N (\left(\frac{1}{1+j}\right)^n * \left(\frac{1}{1-j}\right)^n) \\ | &= \sum_{n=0}^N (\left(\frac{1}{1+j}\right)^n * \left(\frac{1}{1-j}\right)^n) \\ | ||
&= \sum_{n=0}^N \left(\frac{1}{(1+j)(1-j)}\right)^n \\ | &= \sum_{n=0}^N \left(\frac{1}{(1+j)(1-j)}\right)^n \\ | ||
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<math>\begin{align} | <math>\begin{align} | ||
| − | P_{\infty}&=\lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n= | + | P_{\infty}&=\lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^N |\left(\frac{1}{1+j}\right)^n|^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{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 {(\sqrt{-j^2})}^2 \\ | ||
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So <math>P_{\infty} = 1</math>. | So <math>P_{\infty} = 1</math>. | ||
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=== Answer 2 === | === Answer 2 === | ||
Revision as of 10:51, 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} \left(\frac{1}{1+j}\right)^n & \text{ if } n>=0,\\ 0 & \text{otherwise}. \end{array} \right. $
Answer 1
$ \begin{align} E_{\infty}&=\sum_{n=0}^N |c|^2 \\ &= \sum_{n=0}^N (\left(\frac{1}{1+j}\right)^n * \left(\frac{1}{1-j}\right)^n) \\ &= \sum_{n=0}^N \left(\frac{1}{(1+j)(1-j)}\right)^n \\ &= \sum_{n=0}^N (\frac{1}{2})^n \\ &= \frac{1}{1-\frac{1}{2}} \\ &= 2 \\ \end{align} $
So $ E_{\infty} = 2 $
$ \begin{align} P_{\infty}&=\lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^N |\left(\frac{1}{1+j}\right)^n|^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 $.
Answer 2
write it here.
Answer 3
write it here.
