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<math>\begin{align}
 
<math>\begin{align}
 
P_{\infty}&=\lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^N |\left(\frac{1}{1+j}\right)^n|^2 \\
 
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=-N}^N {(\sqrt{jj^*})}^2 \\  
+
 
&= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N {(\sqrt{-j^2})}^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=-N}^N 1  \\
 
&= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^{2N} 1  \\
 
&= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^{2N} 1  \\
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Revision as of 10:57, 22 January 2018

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}}\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

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