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Let $x[n]$ be a discrete-time signal, and let

$$y_1[n]=x[2n]$$

You have to show that if:

$x[n]$ is periodic, then $y_1[n]$ is periodic. $y_1[n]$ is periodic, then x[n] is periodic.

So for the first one I determined the period, which differs for $N$ even and $N$ odd. And when I'm solving the second statement I'm getting the period $2N$, but the book says that it isn't periodic? How so?

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For a compressor system $$y[n] = x[2n]$$ given that $y[n]$ is periodic by $N$, in general, you cannot decide whether $x[n]$ was periodic or not; i.e., It's indeterminate.

The reason is that the compressor system is selecting every $M$-th sample ($M$=2 here) of the input signal $x[n]$ and completely discarding the rest. Those values of $x[n]$ which are discarded by the compressor are arbitrary and free. Hence they do not need to obey any structure on them.

A simple nonperiodic example will be enough to prove that $x[n]$ is not necessarily periodic. Let $y[n]$ be periodic in $N=3$ $$y[n] = [1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3]$$

Then the following $x[n]$ is periodic (with $N=6$) $$x_a[n] = [1 , 0 , 2 , 0 , 3 , 0 , 1 , 0 , 2 , 0 , 3 , 0 , 1 , 0 , 2 , 0 , 3 , 0]$$

Whereas the following $x[n]$ is not periodic: $$x_b[n] = [1, 5, 2, 7, 3, 1, 1, 4, 2, 4, 3, 7, 1, 9, 2, 2, 3, 0]$$

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  • $\begingroup$ I don't understand how did you get those values $\endgroup$
    – nor
    Nov 8 '17 at 22:21
  • $\begingroup$ @nor , goldrick answered your question. I hope you can understand it now. $\endgroup$
    – Fat32
    Nov 9 '17 at 8:59
  • $\begingroup$ Yes, thanks to all of you! I have another question that came from this example but you don't have to answer if you don't want to. If the signal was in continuous-time would the statements be different or same as in this one? $\endgroup$
    – nor
    Nov 9 '17 at 9:04
  • $\begingroup$ Unlike the discrete-time case where the compressor is discarding some samples of $x[n]$, in continuous time no input values are discraded, they are just scaled preserving the wave shape hence if $y(t)$ is periodic then will be $x(t)$... $\endgroup$
    – Fat32
    Nov 9 '17 at 9:39
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The values of $y_1$ are those of $x$ taken at it's even indices only: $y_1(0)=x(0)$, $y_1(1)=x(2)$, $y_1(2)=x(4)$... So if $x$ is periodic with period $T$, it is also $2T$ periodic, then $x(2k)$ is periodic hence $y_1$ too. But if $y_1$ is periodic, $x$ is only periodic wrt its subsequence at even indices. For instance, the unknown values at odd samples $2k+1$ might not be periodic. Hence the second assertion is false.

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Just to answer your comment on Fat32's answer (since I dont have enough reputation to post a follow up comment). His $x[n]$ are just examples, we just make these values up. This is fine because a signal can be ANY sequence of values, so it is possible that a signal may look like that. What we're saying is that in the case where the signal does look like that, the $x[n]$ is not periodic, but the $y[n]$ is.

Try working backwards and see; use the $x_b[n]$ that he created (again, just a made up example). Check that it is not periodic. Solve for $y[n] = x_b[2n]$. Clearly, this signal is periodic. Therefore, a periodic $y[n]$ does not always mean $x[n]$ has to be periodic, because we have shown one case where it isn't.

Check:

$x_b[n]=[1,5,2,7,3,1,1,4,2,4,3,7,1,9,2,2,3,0]$ - This IS NOT periodic

$y[n]=x_b[2n]=[1,2,3,1,2,3,1,2,3]$ - This IS periodic

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  • $\begingroup$ I come from statistics and the notation could be different I guess but I'll ask anyway. Should the above be $x_{b}[2n+1]$ because you want to catch the first, third, fifth , seventh , etc indices. thanks. $\endgroup$
    – mark leeds
    Nov 9 '17 at 8:34
  • $\begingroup$ In this particular case I don't believe it matters, since we're just throwing out every other value -- we dont care whether the indices are even or odd. Otherwise, it depends on whether or not you index from 0. In signal processing I think we do typically take the signal to have a value at $n = 0$ (or time $t = 0$), so the first index corresponds to $n = 0$, in which case $x_b[2n]$ is fine here. $\endgroup$
    – goldrik
    Nov 9 '17 at 12:51
  • $\begingroup$ assuming the indices in dsp start at zero, then $x_{b}[2n]$ makes sense. thanks. $\endgroup$
    – mark leeds
    Nov 10 '17 at 6:52
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In the given question, $y_1[n] = x[2n]$ represents a compressor system, as other answers have stated.

To answer the first question, assume $x[n] = x[n + N]$ for some $N \in \mathbb{N}$, so $y_1[n]$ will be periodic with $N_y = N/2$ because the compressor takes every other sample, as to compress the index variable into half its original entries. For $y_1[n]$ to be periodic, $N/2$ must meet the condition $N/2 \in \mathbb{N}$, and $y_1[n]$ will be periodic if and only if $x[n]$'s period is an even number. Thus, I agree with you on the first question.

For the second question, the key is to remember that a compressor system is not invertible. In other words, $y_1[n]$ cannot fully represent all the information that was present in $x[n]$. If the given compressor creates $y_1[n]$ from the even entries of $x[n]$ and that output happens to be periodic, the original sequence $x[n]$ may not have been periodic, since you simply do not know what the odd entries of $x[n]$ might have been.

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