Line 11: Line 11:
 
\left\{
 
\left\{
 
\begin{array}{ll}
 
\begin{array}{ll}
  \left(\frac{5}{6}\right)^n & n \geq 0,\\
+
  \left(\frac{5}{6}\right)\right^n & n \geq 0,\\
 
  0 & \text{else}.
 
  0 & \text{else}.
 
\end{array}
 
\end{array}

Revision as of 14:56, 1 December 2018

Topic: Energy and Power Computation of a DT geometric signal </center>


Compute the energy $ E_\infty $ and the power $ P_\infty $ of the following DT signal:


$ x[n] = (\frac{5}{6}\right)^n u[n] $


$  x[n] = \left\{ \begin{array}{ll}  \left(\frac{5}{6}\right)\right^n & n \geq 0,\\  0 & \text{else}. \end{array} \right.  $

Norm of a signal: $ \begin{align} |(\frac{5}{6}\right)^n u[n]| = {{je^{3\pi jn}}\times{-je^{-3\pi jn}}} &= {{-j^2}\times{e^{3\pi jn - 3\pi jn}}} &= 1 \end{align} $


$ \begin{align} E_{\infty}&=\lim_{N\rightarrow \infty}\sum_{n=-N}^N |je^{3\pi jn}| \\ &= \lim_{N\rightarrow \infty}\sum_{n=-N}^N 1 \\ &=\infty. \\ \end{align} $


$ E_{\infty} = \infty $.

$ \begin{align} P_{\infty}&=\lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N |je^{3\pi jn}|^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} $


$ P_{\infty} = 1 $

Conclusion:

Therefore, $ E_{\infty} = \infty $, $ P_{\infty} = 1 $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang