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i try to calculate the $\mathcal z$-transform of $x[a-n]$ (where $n$ is my variable) i can't find any transform. the best suited transform is $x[-n] \longleftrightarrow X(z^{-1})$ i took the sum definition of the $\mathcal z$-transform but i couldnt find.

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There are several ways to derive the result:

  1. Just use the definition of the $\mathcal{Z}$-transform: $$\begin{align}\mathcal{x[a-n]}&=\sum_{n=-\infty}^{\infty}x[a-n]z^{-n}{\Big|}_{k=a-n}\\ &=\sum_{k=-\infty}^{\infty}x[k]z^{k-a}\\ &=z^{-a}\sum_{k=-\infty}^{\infty}x[k]z^{k}=z^{-a}X\left(\frac{1}{z}\right)\end{align}$$

  2. Use the well-known properties of the $\mathcal{Z}$-transform (version 1): $$x[n+a]\longleftrightarrow z^aX(z)\\ [n\rightarrow -n\Rightarrow z\rightarrow 1/z]\\ x[-n+a]\longleftrightarrow z^{-a}X\left(\frac{1}{z}\right)$$

  3. Use the well-known properties of the $\mathcal{Z}$-transform (version 2): $$x[-n]\longleftrightarrow X\left(\frac{1}{z}\right)\\ [n\rightarrow n-a \Rightarrow \times\; z^{-a}] \\ x[-(n-a)]\longleftrightarrow z^{-a}X\left(\frac{1}{z}\right)$$

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  • $\begingroup$ A reference on the $\mathcal{Z}$-transform might be missing on your very first math term. Otherwise, + $\endgroup$ – Laurent Duval Dec 29 '15 at 14:26

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