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I know the z-transform of an upsampler is:

$$ y[n] = \begin{cases} x(n/L) &n=0,\pm L, \pm 2L, ...\\ 0&otherwise \end{cases} \longrightarrow Y(z)= X(z^{L}) $$

if $x[n]_L$ is defined to zero for $n <0$ and $n \ge L$ and to be non-zero over interval $[0, L-1]$...then, what's the Z-transform of:

$$ y[n] = \begin{cases} x[n]_L & 0 \le n \le L-1\\ 0&otherwise \end{cases} \longrightarrow Y(z) = ??? $$

its kind of like:

$$ y[n] = x[n]\ u[-n + L]\ u[n] $$

I have this z-transform for LPF that is similar, but doesn't match:

$$ \begin{cases} y[n]=a^{n}&0 \le n \le L-1 \\ \\ 0& otherwise \end{cases} \longrightarrow Y(z) = \frac{1-a^Lz^{-L}}{1-a{\ z}^{-1}} $$

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Ok, its a trick question...its $Y(z) = X(z)$ because $y[n]$ and $x[n]_L$ are identical sequences.

$$ y[n] = \begin{cases} x[n]_L & 0 \le n \le L-1\\ 0&otherwise \end{cases} \longrightarrow Y(z) = X(z) $$

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