Does ztransform exists for a periodic discrete signal

[Indian]

z-transform of periodic signal

Does ztransform exists for a periodic discrete signal ?

 

[claudiocamera]

discrete z transform of cosine

The Z transform is the discrete counterpart to Laplace transform .

Is there Laplace Transform to periodic signals ? Since as well as Laplace transform the Z transform has a ROC region of convergence. So as Laplace transfor has basic transformation to cos (wt) u(t) or sin(wt) u(t) the Z transform has transformation to cos(nt) u or sin(nt) u.

 

[electronics_kumar]

The Z transform is the discrete counterpart to Laplace transform .

Is there Laplace Transform to periodic signals ? Since as well as Laplace transform the Z transform has a ROC region of convergence. So as Laplace transfor has basic transformation to cos (wt) u(t) or sin(wt) u(t) the Z transform has transformation to cos(nt) u or sin(nt) u.

i/ we could n't understand your answer ...plz be clear sir

 

[claudiocamera]

I can't see a way of being more clear, probably if I improve my english... as it takes time, what I basicaly mean is that there are Z transform for the functions cos(nt) u or sin(nt) u, and probably, but not sure, there are ROC problems for the Z transform of cos(nt) or sin(nt). Thats why the query about Laplace transform of periodic signals.

 

[Indian]

i want to know whether ztransform exists for periodic discrete signal or not. u[n]sin[Ωn] is not a periodic signal . pls tell me whether it can exist for periodic

 

[claudiocamera]

As I quoted before, there might be some problem with the ROC of the Z transform of cos(Ωn) or sin(Ωn). It is just a suspicion, there should be a ROC problem in the interval minus infinite to zero. I try to help giving this opinion, but unfortunately, I am not sure. I wish I had the ready answer.

If I were desperate to know this answer, I would write cos(Ωn) in complex form and compute the z transfor in the above interval, later I would verify if it either diverge or not. As I am lazy to perform this computation I would rather stay with my suspicion and wait to know the answer in case of someone else perform the computation.

Regards.
Claudio