Change of variables in summations

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SUMMARY

The discussion centers on the change of variables in summations, particularly in the context of digital signal processing and decimation. The user highlights the challenge of applying a substitution (m=2n+5) to a summation that leads to incorrect results due to the nature of decimation, which alters the sequence of terms included in the sum. The correct approach involves recognizing that the substitution must account for the odd values of m, leading to the formulation of the sum as ∑_{n=-\infty}^{+\infty}x_{2n+5} = ∑_{m=-\infty}^{+\infty}x_{m}(1-(-1)^{m+1})/2. This reflects the need for a proper definition of variable changes in summations, particularly when dealing with discrete signals.

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  • Familiarity with summation notation and properties
  • Knowledge of decimation effects in signal processing
  • Basic grasp of variable substitution in mathematical expressions
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This discussion is beneficial for digital signal processing engineers, mathematicians dealing with discrete systems, and anyone interested in the mathematical foundations of summation transformations.

Lajka
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Hi,

although this may sound trivial, I stumbled upon this problem while studying decimation process in digitial signal processing. I can't find anything on the web about some definition for the change of variables in sumations (as there is one for integrations), so maybe someone here could help me.

Consider the sum
1j19K.png

(the summation range is [-inf, +inf])

If I just do the substitution m=2n+5 and get this
BbapZ.png

it wouldn't be right.

Decimation leads to irreversible changes, aka, I should still have the sum of elements x[1], x[3], x[5],..., but, somehow, I now have the sum of all elements of x[n] with this simple substitution.

So, I think the right answer would be
wjmCf.png


I did this ad-hoc, using logic. I was wondering if there is a proper definition for the change of variables in summations, which takes into the account the effects of decimation (which do not exist in continuous case of course)?

Thanks in advance :)
 
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The problem is that, if n goes from minus to plus infinity, then m = 2n + 5 also goes from minus to plus infinity, but taking only odd values. So you could write

\sum_{n=-\infty}^{+\infty}x_{2n+5}=<br /> \sum_{m=-\infty}^{+\infty}x_{m}\frac{1-(-1)^{m+1}}{2}

This because the extra term I introduced is 0 for even m and 1 for odd m.
 
Yeah, that does the same thing as the sequence of deltas, but I kinda like yours more :D

I guess there isn't a rule for this 'substitution' because it's a trivial matter, but I wanted to check it still.

Thank you for your response!
 

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