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.
Discrete sign function and its z-transform
Started by ●September 10, 2011
Reply by ●September 12, 20112011-09-12
On Sep 11, 10:08�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