Sum of a finite exponential series

  • #1

Homework Statement



Given is [itex]\sum_{n=-N}^{N}e^{-j \omega n} = e^{-j\omega N} \frac{1-e^{-j \omega (2N+1)}}{1 - e^{-j\omega}}[/itex]. I do not see how you can rewrite it like that.

Homework Equations



Sum of a finite geometric series: [itex]\sum_{n=0}^{N}r^n=\frac{1-r^{N+1}}{1-r}[/itex]

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{1-r^{N+1}}{1-r}[/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{1-e^{-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].
 

Answers and Replies

  • #2
danago
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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.
 
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  • #3
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 have identified.
Yes, I've noticed that it starts there. That's why I thought it can be rewritten as [itex]\frac{1−e^{-j\omega(2N+1)}}{1−e^{−jω}}[/itex], but the solution states that this fraction is multiplied by [itex]e^{−jωN}[/itex].
 
  • #4
danago
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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{1-e^{-j\omega(2N+1)}}{1-e^{-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 :tongue:
 
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  • #5
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{1-e^{-j\omega(2N+1)}}{1-e^{-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 :tongue:
Okay, thank you. For me, it is not about the sign in the exponent. I do not see why we have to multiply by the term in front of the fraction. But I think I rewrote the equation in the wrong way. Can you give me your steps?
 
  • #6
danago
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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=\phi-N[/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 (\phi-N)} = e^{j\omega N} \frac{1-e^{-j\omega(2N+1)}}{1-e^{-j\omega}}[/tex]
 
  • #7
danago
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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^{N-1}+e^N[/tex]

[tex]\sum^{2N}_{n=0} e^{n} = 1+e^{1}+...+e^{2N-1}+e^{2N}[/tex]

See how they are not the same?
 
  • #8
Ah, I see the problem now. Thanks!
 
  • #9
danago
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