Difference between revisions of "Signal power energy exercise DT ECE301S18" - Rhea

Revision as of 09:57, 22 January 2018

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 \\ \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}}\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.

Back to ECE301 Spring 2018 Prof. Boutin