ROC in Sign function Z-transform

In summary, the Z-transform of the given function x[n] = u[n]-u[-n-1] is X(z) = 2/(1-z^-1), with a region of convergence (ROC) that is the null intersection between |z| > 1 and |z| < 1. This may seem to indicate that the transform is not valid, but it can be evaluated at z = e^(j*Omega) to obtain the correct Fourier transform. However, the DTFT only exists when the unit circle |z| = 1 is in the ROC, so further examination may be needed to determine the correct ROC for this transform. This is a topic that may be covered in a graduate level DSP course in
  • #1
Bromio
62
0

Homework Statement


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

Homework Equations


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

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

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

The Attempt at a Solution


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

Thank you.
 
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  • #2
Isn't it that if there the intersection of the ROC is null then you can't use the z-transform because it diverges?
 
  • #3
SpaceDomain said:
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?
 
  • #4
Bromio said:
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.
 
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  • #5
SpaceDomain said:
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.

SpaceDomain said:
Is this for a graduate level DSP course? Just curious.
Yes. Electrical engineering, more exactly.
 

Related to ROC in Sign function Z-transform

1. What is the ROC in Sign function Z-transform and why is it important?

The ROC, or region of convergence, in Sign function Z-transform is the set of values for which the Z-transform of a discrete-time signal converges. It is important because it determines the range of values for which the Z-transform can be used to analyze a signal, and it also affects the stability and causality of the system.

2. How is the ROC determined in Sign function Z-transform?

The ROC can be determined by examining the Z-transform of a signal and identifying the values for which it converges. This is typically done by factoring the polynomial in the Z-transform and finding the roots, which correspond to the ROC boundaries.

3. What are the possible shapes of the ROC in Sign function Z-transform?

The ROC can take on three possible shapes: a circular region, a half-plane, or an annulus. The shape depends on the poles and zeros of the Z-transform; for example, a circular ROC is associated with a stable and causal system.

4. How does the ROC affect the stability and causality of a system?

The ROC is closely related to the stability and causality of a system. A stable system will have a ROC that includes the unit circle in the complex plane, while an unstable system will have a ROC that does not include the unit circle. A causal system will have a ROC that is either the entire right half-plane or a region that extends to but does not include the unit circle.

5. Can the ROC change for a given signal in Sign function Z-transform?

Yes, the ROC can change for a given signal depending on the values of the poles and zeros. This is because the ROC is determined by the boundaries of the Z-transform, which can change if the poles or zeros are altered. For example, if a zero is added to a signal, the ROC will expand to include that new value.

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