Sum of a finite exponential series

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 $e^{-j\omega N}$.

Related Calculus and Beyond Homework Help News on Phys.org
danago
Gold Member
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.

Last edited:
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
Gold Member
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 :tongue:

Last edited:
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 :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?

danago
Gold Member
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
Gold Member
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?

Ah, I see the problem now. Thanks!

danago
Gold Member
Ah, I see the problem now. Thanks!
No problem!