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

Circle the reciprocal of 4

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

 

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