The z- transform of u{n} is U (z). The region of convergence for which the z-transform of u[n] exists is
A sequence u[n] is defined as
$u\left[ n \right] = \left. {\left\{ \begin{array}{l}1,if{\text{ n}} \ge {\text{0}}\\1,if{\text{ n}} \le {\text{0}}\end{array} \right.{\text{ for n = \{ - }}\infty .... - 1,0,1....\infty \} } \right\}$
The z- transform of u{n} is U (z). The region of convergence for which the z-transform of u[n] exists is
Right Answer is:
|z| > 1
SOLUTION
The z transform of unit step sequence u(n) is given by u(z)
$u(z) = \dfrac{1}{{1 – {Z^{ – 1}}}}$
Because u(z) transform has a zero at |z| = 0 and a pole at |z| = 1, hence the region of convergence is
|z| > 1