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ROC(region of convergence) of Z domain is shown by a circular region while ROC in S domain is shown by a rectangular(approximately looking like rectangle) region

What is the reason of this difference in shapes of ROC regions?

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Because, the region of convergence in the Laplace transform $$ X(s) = \int_{-\infty}^{\infty} x(t) e^{-st} dt $$ is related to the weighting provided by the real part of the complex $s = \sigma + j \omega$; as this will yield the weight $|e^{-st}| = e^{-\sigma t}$ applied on the input signal $x(t)$, and is a function of $\sigma$ alone and is a rectangular (planar) region on the s-plane.

But the region of convergence in the Z-transform $$ X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n} $$ is related to the weighting provided by the magnitude of the complex $z = \sigma + j \omega$ as given by $|z|^{-n} = |\sigma + j\omega|^{-n} = |z|^{-n} |e^{-j n \angle{z}}| = |z|^{-n}$, which is a circular region on th z-plane, due to the magnitude of $z$ being involved.

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