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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. – J. M. May 18 '12 at 10:36
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. – J. M. May 18 '12 at 10:40
So the Z-transform of a shifted process diverges then? – lindelof May 18 '12 at 10:42
Not necessarily. You can still take the Z-transform of a shifted unit step function for instance. – J. M. May 18 '12 at 10:46

It seems like you're confusing the unilateral and bilateral versions of the Z transform. The bilateral transform is $\sum_{n=-\infty}^\infty x(n)z^{-n}$; the unilateral transform starts the summation at zero: $\sum_{n=0}^\infty x(n)z^{-n}$.

Clearly, the unilateral transform of $x$ is the bilateral transform of $x(t)u(t)$ where $u$ is the unit step function. In particular, $z/(z-1)$ is the bilateral transform of the unit step function and the unilateral transform of $1$. On the other hand, $1$ does not have a bilateral transform - the sum diverges for all z.

In either case, the linearity of the transform holds, so if you're looking for a unilateral transform, just add $kz/(z-1)$; but if you're looking for a bilateral transform, than for $k\neq 0$, the transforms of $x(t)$ and $x(t)+k$ must have disjoint regions of convergence, since the transform of their difference converges nowhere (I suspect one of them must diverge everywhere, but can't think of a proof right now).

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