I have a question about adding two discrete functions, one of which is an exact copy, but dilated.

Imagine function x[n]. This function exists for all n in the set of natural numbers (1,2,3,etc.). There is also function y[n]. It behaves the same as function x[n], but only exists for all n that are integer multiples of 3 (3,6,9,etc.). It is a dilated function of x[n]. There are values for y[n] for values of n that are not divisible by 3, but these are incorrect and should be ignored.

How do I add these two functions? Do I use two different n variables (n and n') and define them as sets of "all natural numbers" and "all natural numbers divisible by 3" respectively? Or is there a way to put this in the notation of the equation?

Thanks for your help!

  • $\begingroup$ Could the person who gave me a downvote give me an explanation as to what I did wrong in my question? $\endgroup$ – Pep95 Jun 17 '16 at 12:36

I have found a solution to my problem. For future reference, I will post it here. By simply multiplying the function by a summation of delta functions which are positioned at the n values that have usable information, the incorrect values are turned into 0.

My solution

In this picture, n is my variable, while M' is the size of the step between usable outputs. fclk and 2^N are merely multiplications.


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