
#1
Nov1011, 07:19 AM

P: 9

1. The problem statement, all variables and given/known data
Given is [itex]\sum_{n=N}^{N}e^{j \omega n} = e^{j\omega N} \frac{1e^{j \omega (2N+1)}}{1  e^{j\omega}}[/itex]. I do not see how you can rewrite it like that. 2. Relevant equations Sum of a finite geometric series: [itex]\sum_{n=0}^{N}r^n=\frac{1r^{N+1}}{1r}[/itex] 3. The attempt at a solution Or is the above result based on this more general equation: [itex]\sum_{n=0}^{N}ar^n=a\frac{1r^{N+1}}{1r}[/itex]? Although I think the equation in (2) is just this equation for a=1, right? So, I know how to get to the 2nd term in (1), i.e., [itex]\frac{1e^{j \omega (2N+1)}}{1  e^{j\omega}}[/itex], but I have no idea why it is multiplied by the term [itex]e^{j\omega N}[/itex]. 



#2
Nov1011, 07:42 AM

PF Gold
P: 1,132

Did you notice that the sum you are trying to compute actually starts from n=N and not n=0? I think you can get the answer you want by making a change of variable and then using the geometric series equation you have identified.




#3
Nov1011, 07:52 AM

P: 9





#4
Nov1011, 08:09 AM

PF Gold
P: 1,132

Sum of a finite exponential series
Are you sure that the exponential term in front of the fraction does have a negative sign? I just tried doing the working and ended up with a positive sign, i.e.:
[tex]\sum^{N}_{n=N} e^{j\omega n} = e^{j\omega N} \frac{1e^{j\omega(2N+1)}}{1e^{j\omega}}[/tex] I did it by making the substitution [itex]\phi=n+N[/itex]. I will check my working again. EDIT: I have checked over my working and have convinced myself that the negative should not be there. It is late here so i could easily have made a mistake though 



#5
Nov1011, 08:23 AM

P: 9





#6
Nov1011, 08:29 AM

PF Gold
P: 1,132

You have transformed the upper and lower limits of the sum, however you have not applied the same transformation to the variable n in the summand.
If [itex]\phi=n+N[/itex], then the new limits of the sum will be [itex]\phi=0[/itex] and [itex]\phi=2N[/itex]. You must then also replace the 'n' in the summand with [itex]n=\phiN[/itex]. If you do this then you will get the right answer. EDIT: The transformed sum will be: [tex]\sum^{2N}_{\phi=0} e^{j\omega (\phiN)} = e^{j\omega N} \frac{1e^{j\omega(2N+1)}}{1e^{j\omega}}[/tex] 



#7
Nov1011, 08:38 AM

PF Gold
P: 1,132

Maybe it will be easier to understand if we look at why what you did isn't quite correct.
[tex]\sum^{N}_{n=N} e^{n} = e^{N}+e^{N+1}+...+1+e^1+...+e^{N1}+e^N[/tex] [tex]\sum^{2N}_{n=0} e^{n} = 1+e^{1}+...+e^{2N1}+e^{2N}[/tex] See how they are not the same? 



#8
Nov1011, 08:51 AM

P: 9

Ah, I see the problem now. Thanks!




#9
Nov1011, 09:05 AM

PF Gold
P: 1,132




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