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If $X(z)$ is the Z-transform of a discrete timeserie $x(t)$, what is the Z-transform of $x(t)+k$ where $k$ is a constant?

From the linearity property of the Z-transform I would expect it to be $X(z) + \mathcal{Z}(k)$, where $\mathcal{Z}(k)$ is the Z-transform of $k$. That, in turn, would be $k$ times the Z-transform of $1$.

In different tables I find the following expression for the Z-transform of $1$: $\frac{z}{z-1}$, but if I do the calculation by hand, (i.e., compute $\sum_{-\infty}^{\infty}1^{-n}$) I obtain $0$. I suspect the former expression to be in fact the Z-transform of the unit step.

Which is correct? Is $\mathcal{Z}(x(t)+k)$ equal to $\mathcal{Z}(x(t)$?

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    $\sum\limits_{-\infty}^{\infty}1^{-n}$ diverges... for that matter, $\sum\limits_{-\infty}^{\infty}z^{-n}$ (what you were supposed to obtain) diverges as well.2012-05-18
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    And yes, if you're taking the doubly-infinite version of the Z-transform, $\dfrac{z}{z-1}$ is the unit step function's transform.2012-05-18
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    So the Z-transform of a shifted process diverges then?2012-05-18
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    Not necessarily. You can still take the Z-transform of a shifted unit step function for instance.2012-05-18

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