# Discrete sign function and its z-transform

Started by September 10, 2011
```Hello.

Let the discrete system whose impulse response is h[n] = u[n] - u[-n-1],
which is really similar to continuous sign function.

I've calculated its DTFT, which is X(Omega) = 2/(1-exp(-j*Omega)).

Using linearity, I've tried to obtain its z-transform. Z-transform of u[n]
is 1/(1-z^(-1)), ROC; |z| > 1, and z-transform of -u[-n-1] is 1/(1-z^(-1)),
ROC: |z| < 1. If I add both results, I obtain 2/(1-z^(-1)), but what about
the ROC? It seems to be null, but that's not possible, because Z-transform
converges (Fourier transform exists).

Thank you.

```
```On Sep 11, 10:08&#2013266080;am, "Bromio" <aLeZX.vb@n_o_s_p_a_m.gmail.com> wrote:
> Hello.
>
> Let the discrete system whose impulse response is h[n] = u[n] - u[-n-1],
> which is really similar to continuous sign function.
>
> I've calculated its DTFT, which is X(Omega) = 2/(1-exp(-j*Omega)).
>
> Using linearity, I've tried to obtain its z-transform. Z-transform of u[n]
> is 1/(1-z^(-1)), ROC; |z| > 1, and z-transform of -u[-n-1] is 1/(1-z^(-1)),
> ROC: |z| < 1. If I add both results, I obtain 2/(1-z^(-1)), but what about
> the ROC? It seems to be null, but that's not possible, because Z-transform
> converges (Fourier transform exists).
>
> So, what about the ROC?
>
> Thank you.

The region of convergence (ROC) is the set of points in the complex
plane for which the Z-transform summation converges. For the
integrator in your example using z^-1 the ROC is the unit circle. |z|
<1 For the other case it's |z|>1. Best to stick to one definition or
another else you get confused. I always use the backwards shift
operator z^-1 (q^-1) which mean all poles of z  lie within the unit
circle. If you use z instead then the opposite holds for the z^-1
plane.

Hardy
```