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<math>Einf = \sum_{n=-inf}^\infty |x[n]|^2 = \sum_{k=0}^\infty (1/4)^k = inf</math>
 
<math>Einf = \sum_{n=-inf}^\infty |x[n]|^2 = \sum_{k=0}^\infty (1/4)^k = inf</math>
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<math>Pinf = \lim_{N \to \infty}x_n\sum_{n=-inf}^\infty |x[n]|^2 </math>

Revision as of 18:51, 5 September 2008

<math>x[n]={[e^(jπ/17) ]/2j}u(-n) 
|x[n] |=u[-n] (1/2)^n
E_∞=∑_(n=-∞)^∞▒〖|x[n]|^2〗=∑_(n=-∞)^0▒〖[(1/2)^n ]^2=∑_(n=-∞)^0▒〖(1/4)^n=〗〗 ∞
	
P_∞=lim┬(N→∞)⁡〖1/(2N+1)〗 ∑_(n=-N)^N▒|x[n] |^2=lim┬(N→∞)⁡〖1/(2N+1)〗 ∑_(n=-N)^N▒〖(1/4)^n=〗   lim┬(N→∞)⁡〖1/(2N+1)〗 ∑_(k=0)^∞▒〖4^k=〗   lim┬(N→∞)⁡〖[1/(2N+1)]〗 [(1-4^(N+1))/(1-4))]=∞
</math>

$ x[n]=\left ( \frac{exp(jpi/17)}{2j} \right )^n * u[-n] $


$ Einf = \sum_{n=-inf}^\infty |x[n]|^2 = \sum_{k=0}^\infty (1/4)^k = inf $


$ Pinf = \lim_{N \to \infty}x_n\sum_{n=-inf}^\infty |x[n]|^2 $

Alumni Liaison

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

Buyue Zhang