Variable change in a sum has me confused.

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Aedrha2

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Hi there, hope you are having a nice day.
I am studying a course in signal analysis. For what ever reason the engineering program not to give us the relevant math courses before this. So naturally some of the math i find a bit tricky.

I was watching a videolecture today and a varible change was done.
[math]\sum_{L=-\infty}^\infty \sum_{n=LN}^{LN+N-1}x(n) e^{-j2\pi \frac{k}{N}n}=\left[n=n-LN \right]=\sum_{L=-\infty}^\infty \sum_{n=0}^{N-1}x(n) e^{-j2\pi \frac{k}{N}(n-LN)}[/math]
The varible change inside of the sums is trivial. But I can't figure out what how the limits of the second sum to the right changed like that. Maybe it is trival too and I've just stared at it too long.

If anyone can shed some light on this for me I would be grateful!
 
The calculation began as n = LN
Now assume that in second calculation we replace the letter n by h to go from 0 to N -1. You know the inside formula will be executed the same number of times as both upper and lower bounds got subtracted by LN, but if we used h in equation containing e, we will get a wrong answer as h is lower than original n by LN so we LN to h, h is the new definition of n in second equation. Is it clearer now
 
AbdelRahmanShady said:
The calculation began as n = LN
Now assume that in second calculation we replace the letter n by h to go from 0 to N -1. You know the inside formula will be executed the same number of times as both upper and lower bounds got subtracted by LN, but if we used h in equation containing e, we will get a wrong answer as h is lower than original n by LN so we LN to h, h is the new definition of n in second equation. Is it clearer now
Click to expand...
Do you understand why the n in parentheses was not replaced by n-LN? And why n-LN and not n+LN? And what is this strange construction in the middle (in square brackets)?
 
Start with
[math]\sum_{L=-\infty}^\infty \sum_{n=LN}^{n=LN+N-1}x(n) e^{-j2\pi \frac{k}{N}n}[/math]Then, make the substitution n = h + LN , in other words rewrite the expression but everywhere you see "n" in the original just substitute "h + LN". Note that I wrote "n=" in the upper limit of the inner sum to help you to get this correct!

I agree with the comments above, and the suggestion by @AbdelRahmanShady of using "h" after the substitution is very wise. Using the same variable for two different amounts could lead to much confusion.
 
Thank you all!
@AbdelRahmanShady , @lev888 , @Cubist .
What confused me was preciesely that it was n=h-LN and not n=h+LN. But with the new substitution I think understand. So in essence what I am doing here is moving the sum back LN "steps" but the function I am summing I am moving forward LN steps?

This was a part of a much larger proof concerning the DTFT of a function. I checked and it holds up with h=n+LN. The professor must have misspoken.

Thank you all again!
 
lev888 said:
Do you understand why the n in parentheses was not replaced by n-LN? And why n-LN and not n+LN? And what is this strange construction in the middle (in square brackets)?
Click to expand...
Some of the professors at my university use the square brackets to indicate that they made a variable substitution.
 
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