thanks for taking the time to help with this problem!

I have to prove the time-scaling property:

$$ x_{(m)}[n] = \begin{cases} x[n/m], & n=0,\pm m, \pm 2m,...\\ 0, & otherwise \end{cases} $$

First, for part (a), Show that $$x_{(m)}[n]$$ has a period of $$ mN$$

Then, for part (b), Show that if $$\\x[n]=v[n]+w[n]\\$$ then$$\\x_{(m)}[n]=v_{(m)}[n]+w_{(m)}[n] $$

There's a bit more to the problem after this, but I'm mostly concerned with these first two pieces. I know that for a periodic function $$x[n]=x[n+N]$$

Am I supposed to somehow prove that $$x[n/m+mN]=x[n/m]$$ using the Discrete Fourier Series analysis equation? What is a rigorous way to prove this.

For part (b), I'm unsure of how to prove the linearity in a way that doesn't seem trivial.


  • $\begingroup$ Hint: you are supposed to show that xm[k] = xm[k+mN] for all k $\endgroup$
    – Hilmar
    Commented Nov 3, 2014 at 23:47
  • $\begingroup$ I understand this, but I'm unsure how to formally show that it is true. $\endgroup$
    – jephex
    Commented Nov 4, 2014 at 0:42

1 Answer 1


It looks more like zero padding to create up-sampling resolution to me.

Prove x[n/m+N]=x[n/m], since your argument is now n/m and not n.

  • $\begingroup$ Thanks! I think I worked part (a) out - any thoughts on the second bit, part (b)? $\endgroup$
    – jephex
    Commented Nov 4, 2014 at 9:00
  • $\begingroup$ I think part b just shows that superposition holds for an up or a down sampler, so if I have x(n)=Au(n)+Bv(n) then x(n/m)=Au(n/m)+Bv(n/m). For your case A and B are 1. Just remember samplers are linear systems but time varying, so any kind of a sampler whether it be up or down will be following linearity and superposition but not time invariance. Hence, it is not an LTI system, but an L system. I think the problem given to you has been given with the motive of highlighting that fact. $\endgroup$
    – Ghosal_C
    Commented Nov 4, 2014 at 9:20

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