ROC in Sign function Z-transform

[Bromio]

Homework Statement

Calculate the Z-transform of the function x[n] = u[n]-u[-n-1].

Homework Equations

X(z) = ZT\{x[n]\} = \sum_{n=-\infty}^{\infty}x[n]z^{-n}

ZT\{u[n]\} = \displaystyle\frac{1}{1-z^{-1}}, ROC: |z| > 1.
ZT\{-u[-n-1]\} = \displaystyle\frac{1}{1-z^{-1}}, ROC: |z| < 1.

ZT\{x[n]\} = X(z), ROC: R1
ZT\{y[n]\} = Y(z), ROC: R2
ZT\{ax[n]+by[n]\} = aX(z)+bY(z), ROC: at least R1\cap R2

The Attempt at a Solution

Using formulas in section 2. it is obvious that X(z) = ZT\{x[n]\} = \displaystyle\frac{2}{1-z^{-1}}, but which is the ROC? The intersection between |z| > 1 and |z|< 1 is null.

Thank you.

 

[SpaceDomain]

Isn't it that if there the intersection of the ROC is null then you can't use the z-transform because it diverges?

 

[Bromio]

Isn't it that if there the intersection of the ROC is null then you can't use the z-transform because it diverges?

I thought that, but if you take z = e^(j*Omega), you obtain the correct Fourier transform.

In addition (without rigorousness), this z-transform was asked in an exam and I'm sure it exists. What is happening?

 

[SpaceDomain]

I thought that, but if you take z = e^(j*Omega), you obtain the correct Fourier transform.

So taking the z-transform and evaluating at z = e^(j*Omega) is the definition of the DTFT, right? Isn't the DTFT more prone to instability than the z-transform?

Hmm.. I will think about this more. Maybe someone else is better suited to answer this question as my DSP class begins covering the z-transform this upcoming Monday (good time to start reading up on it though!).

Is this for a graduate level DSP course? Just curious.

 

[Bromio]

So taking the z-transform and evaluating at z = e^(j*Omega) is the definition of the DTFT, right? Isn't the DTFT more prone to instability than the z-transform?

Yes, DTFT only exists if the unit circle |z| = 1 is in the ROC.

So I don't know if I'm making mistakes and this ROC is miscalculated or if there is another explanation to this behavior.

Is this for a graduate level DSP course? Just curious.

Yes. Electrical engineering, more exactly.