Problem
Compute the energy $ E_\infty $ and the power $ P_\infty $ of this DT signal:
$ x[n] = \left(\frac{5}{6}\right)^n u[n] $
Solution
$ x[n] = \left\{ \begin{array}{ll} \left(\frac{5}{6}\right)^n & \text{ if } n\geq 0,\\ 0 & \text{else}. \end{array} \right. $
Norm of a signal:
$ \begin{align} \left|\left(\frac{5}{6}\right)^n u[n]\right| = \left(\frac{5}{6}\right)^n // \end{align} $
$ \begin{align} E_{\infty}&=\sum_{n=0}^N \left|\left(\frac{5}{6}\right)^n\right|^2 \\ &= \sum_{n=0}^N \left(\frac{5}{6}\right)^{2n} \\ &= \sum_{n=0}^N \left(\frac{25}{36}\right)^{n} \\ &= \frac{1}{1-\frac{25}{36}} \\ &= \frac{36}{11} \\ \end{align} $
$ E_{\infty} = \frac{36}{11} $.
$ \begin{align} P_{\infty} &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^N \left|\left(\frac{5}{6}\right)^n\right|^2 \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} \times \left(\frac{25}{36}\right)^n \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} \times \frac{1}{1-\frac{25}{36}} \\ &= \lim_{N\rightarrow \infty} \left({ \frac{\frac{36}{11}}{2N+1}} \right) \\ &= 0 \\ \end{align} $
$ P_{\infty} = 0 $
Conclusion:
$ E_{\infty} = \frac{36}{11} $, $ P_{\infty} = 0 \\ $
When $ E_{\infty} = \text{finite number} $,
$ P_{\infty} = 0 $
