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I'm looking for the inverse $\mathcal{Z}$-transform of a shifted Dirac delta function in the $z$ domain, i.e.

$$ x[n] = \mathcal{Z}^{-1} \{ \delta(z - z_{0}) \} = \ldots $$

Does an analytic/closed-form expression exist? I know that $ \mathcal{Z} \{ \delta[n] \} = 1 $ so from duality I suppose $\mathcal{Z}^{-1} \{ \delta(z) \} = 2\pi$, and furthermore that time-shifting gives $\mathcal{Z} \{ x[n - n_{0}] \} = z^{-n_{0}} \cdot \mathcal{Z} \{ x[n] \} $, but does an equivalent property exist for "$z$" shifting (or does duality also apply for properties)? I'd be happy to know more, thanks!

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The inverse $\mathcal{Z}$-transform of $\delta(z-z_0)$ doesn't exist. First of all, the Dirac delta impulse isn't even a function (but a distribution), and second, not every function is a valid $\mathcal{Z}$-transform.

The $\mathcal{Z}$-transform of a sequence is a Laurent series and as such it is an analytic function inside its region of convergence. Hence, only functions that are analytic in an annulus of the complex plane can be valid $\mathcal{Z}$-transforms.

Some other examples of (ordinary) functions which don't have an inverse $\mathcal{Z}$-transform are $f(z)=|z|^2$, $f(z)=\textrm{Re}\{z\}$ or $f(z)=z^*$ (where $^*$ means complex conjugation).

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  • $\begingroup$ What's the Z-transform of $x[n]=e^{j \omega n} \qquad \forall n \in \mathbb{Z}$? Seems to me there might be a dirac delta at $e^{j \omega}$. $\endgroup$ Nov 9, 2023 at 22:30
  • $\begingroup$ @robertbristow-johnson: $x[n]=e^{j\omega n}$ doesn't have a Z-transform. It does have a DTFT but that's a different thing. $\endgroup$
    – Matt L.
    Nov 10, 2023 at 7:59
  • $\begingroup$ Thanks! To be sure I understand this correctly (please correct me if this is wrong): every DT signal $x[n]$ has a $\mathcal{Z}$-transform with a RoC equal to a (open) disc or annulus not containing any poles. But $\delta(z - z_{0})$ is only analytic in either $\{ z \in \mathbb{C} : \vert z \vert < z_{0} \}$ or $\{ z \in \mathbb{C} : \vert z \vert > z_{0} \}$ where it is zero everywhere? $\endgroup$ Nov 10, 2023 at 11:23
  • $\begingroup$ @BartWolleswinkel: Not every sequence $x[n]$ has a Z-transform. And the Dirac impulse isn't analytic (in the complex analysis sense) anywhere, because apart from not being an ordinary function, it is not complex-valued (think of the Cauchy-Riemann equations). $\endgroup$
    – Matt L.
    Nov 10, 2023 at 11:40

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