Revision as of 11:02, 22 January 2018 by Chen2156 (Talk | contribs)

Practice Question on "Signals and Systems"


More Practice Problems


Topic: Signal Energy and Power


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}}2 (1-(\frac{1}{2})^{N+1}) \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} (2-\frac{1}{2^N} \\ &= \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

Alumni Liaison

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

Buyue Zhang