(New page: <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→...)
 
Line 2: Line 2:
 
|x[n] |=u[-n] (1/2)^n
 
|x[n] |=u[-n] (1/2)^n
 
E_∞=∑_(n=-∞)^∞▒〖|x[n]|^2〗=∑_(n=-∞)^0▒〖[(1/2)^n ]^2=∑_(n=-∞)^0▒〖(1/4)^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))]=∞
+
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>
 
</math>

Revision as of 18:33, 5 September 2008

$ 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))]=∞ $

Alumni Liaison

To all math majors: "Mathematics is a wonderfully rich subject."

Dr. Paul Garrett