<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(j\pi/17)}{2j} \right )^n * u[-n] $


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


$ Pinf = \lim_{N \to \infty}(1/(2N+1))\sum_{n=-N}^\N |x[n]|^2 = \lim_{N \to \infty}(1/(2N+1))* ((1-4^ (N+1))/(1-4)) = \infty $

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010