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Discrete sign function and its z-transform

Started by Bromio 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).

So, what about the ROC?

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