Sum of a finite exponential series

[ElfenKiller]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

Or is the above result based on this more general equation: \sum_{n=0}^{N}ar^n=a\frac{1-r^{N+1}}{1-r}? 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., \frac{1-e^{-j \omega (2N+1)}}{1 - e^{-j\omega}}, but I have no idea why it is multiplied by the term e^{-j\omega N}.

 

[danago]

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.

 

[ElfenKiller]

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 \frac{1−e^{-j\omega(2N+1)}}{1−e^{−jω}}, but the solution states that this fraction is multiplied by e^{−jωN}.

 

[danago]

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.:

\sum^{N}_{n=-N} e^{-j\omega n} = e^{j\omega N} \frac{1-e^{-j\omega(2N+1)}}{1-e^{-j\omega}}

I did it by making the substitution \phi=n+N. 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

 

[ElfenKiller]

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.:

\sum^{N}_{n=-N} e^{-j\omega n} = e^{j\omega N} \frac{1-e^{-j\omega(2N+1)}}{1-e^{-j\omega}}

I did it by making the substitution \phi=n+N. 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

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?

 

[danago]

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 \phi=n+N, then the new limits of the sum will be \phi=0 and \phi=2N. You must then also replace the 'n' in the summand with n=\phi-N. If you do this then you will get the right answer.

EDIT:
The transformed sum will be:

\sum^{2N}_{\phi=0} e^{-j\omega (\phi-N)} = e^{j\omega N} \frac{1-e^{-j\omega(2N+1)}}{1-e^{-j\omega}}

 

[danago]

Maybe it will be easier to understand if we look at why what you did isn't quite correct.

\sum^{N}_{n=-N} e^{n} = e^{-N}+e^{-N+1}+...+1+e^1+...+e^{N-1}+e^N

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

See how they are not the same?

 

[ElfenKiller]

Ah, I see the problem now. Thanks!

 

[danago]

Ah, I see the problem now. Thanks!

No problem!