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<math>\begin{align} | <math>\begin{align} | ||
E_{\infty}&=\sum_{n=0}^N |\left(\frac{1}{1+j}\right)^n|^2 \\ | E_{\infty}&=\sum_{n=0}^N |\left(\frac{1}{1+j}\right)^n|^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 (\frac{1}{2})^n \\ |
| − | &= | + | &= \frac{1}{1-\frac{1}{2}} \\ |
| + | &= 2 \\ | ||
\end{align}</math> | \end{align}</math> | ||
| − | So <math>E_{\infty} = | + | So <math>E_{\infty} = 2</math> |
| − | + | ||
| − | + | ||
<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= | + | |
| − | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\ | + | \text{Similar to math above, the expression can be derived towards}\\ |
| − | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\ | + | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^N (\frac{1}{2})^n \\ |
| − | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\ | + | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} \frac{1(1-(\frac{1}{2})^{N+1})}{1-\frac{1}{2}} \\ |
| − | &= \lim_{N\rightarrow \infty}{ | + | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}2 (1-(\frac{1}{2})^{N+1}) \\ |
| − | &= \ | + | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} (2-\frac{1}{2^N}) \\ |
| − | &= | + | &= \lim_{N\rightarrow \infty} \left(\frac{2-\frac{1}{2^N}}{2N+1} \right) \\ |
| + | &= \frac{2}{\infty}\\ | ||
| + | &= 0 \\ | ||
\end{align}</math> | \end{align}</math> | ||
| − | So <math>P_{\infty} = | + | So <math>P_{\infty} = 0</math>. |
| − | + | ||
| − | + | ||
=== Answer 2 === | === Answer 2 === | ||
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---- | ---- | ||
| − | [[2011 Spring ECE 301 Boutin|Back to ECE301 Spring | + | [[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2018 Prof. Boutin]] |
[[Category:ECE301Spring2018Boutin]] | [[Category:ECE301Spring2018Boutin]] | ||
Latest revision as of 10:13, 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 |\left(\frac{1}{1+j}\right)^n|^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 \\ \text{Similar to math above, the expression can be derived towards}\\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^N (\frac{1}{2})^n \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} \frac{1(1-(\frac{1}{2})^{N+1})}{1-\frac{1}{2}} \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}2 (1-(\frac{1}{2})^{N+1}) \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} (2-\frac{1}{2^N}) \\ &= \lim_{N\rightarrow \infty} \left(\frac{2-\frac{1}{2^N}}{2N+1} \right) \\ &= \frac{2}{\infty}\\ &= 0 \\ \end{align} $
So $ P_{\infty} = 0 $.
Answer 2
write it here.
Answer 3
write it here.
