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I have the following signal: $$x[n] = \delta[n+1]+\delta[n]+\delta[n-4]$$ $$y[n] = 2x[2n-1] + x[2n+2]$$ Now, let's say I decompose $y[n]$ into $y_1[n] = x[2n-1]$ and $y_2[n] = x[2n+2]$. In this situation, why is it that $y_1[n]$ will only sample for odd values of $n$ and $y_2[n]$ will only sample for even values of $n$. It doesn't seem intuitive to only sample odd values of $n$. I can see that $(2n-1)$ will be odd for all integer values of $n$. However, I don't see why the constraint is applied just purely to $n$.

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    "I don't see why the constraint is applied just purely to $n$": what do you mean ? – 2017-02-16
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    What I mean is, why is that only for odd values of $n$ is the function sampling instead of odd values that result from the entire entity: $(2n-1)$. Because the former implies it only samples at $..., -1, 1, 3, 5, ...$. The latter interpretation implies that it samples at all values of $n$. – 2017-02-16
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    I believe, that it is said *$x$ in function $y_1$ samples only for odd values*. It doesn't refer to $n$ in $y_1[n]$, but to $2n-1$ in $x[2n-1]$ – 2017-02-16

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There is a confusion about which $n$ is meant.

$y_1$ samples $x$ at indexes $2n-1$, which are indeed odd numbers, so that the $n$ of $x[n]$ is odd. But the $n$ of $y[n]$ is unrestricted.

You can avoid the misinterpretations by using distinct identifiers,

$$x[n] = \delta[n+1]+\delta[n]+\delta[n-4]$$ $$y[m] = 2x[2m-1] + x[2m+2].$$

By the way,

$$y_1[m]=\delta[2m]+\delta[2m-1]+\delta[2m-5],$$ $$y_2[m]=\delta[2m+3]+\delta[2m+2]+\delta[2m-2],$$ and the deltas are taken at both parities.