"Determine the Z-transform for the sequencex[n]=(13)n−2u[n−2]" was answered by our community

Blackettyl2j

Answered question

2022-03-26

Determine the Z-transform for the sequence

x [n] = {(\frac{1}{3})}^{n - 2} u [n - 2]

Answer & Explanation

pautmndu

Beginner2022-03-27Added 10 answers

$x [n] = {(\frac{1}{3})}^{n - 2} u [n - 2]$
$u [n - 2] = {\begin{array}{ll} 1, n \geq 2 \\ 0, n < 2 \end{array}$
Calculate the z-transform of the sequence
$X (z) = \sum_{n = - \infty}^{\infty} ({(\frac{1}{3})}^{n - 2} u (n - 2)) z^{- n}$
$= \sum_{n = - \infty}^{\infty} {(\frac{1}{3})}^{n - 2} z^{- n}$ (Putting the value of $u (n - 2)$ )
Let $n_{1} = n - 2$ in the second term
$X (z) = \sum_{n = - \infty}^{\infty} {(\frac{1}{3})}^{n_{1}} \times z^{- n_{1} - 2}$
$= z^{- 2} \sum_{n = - \infty}^{\infty} {(\frac{1}{3})}^{n_{1}} \times z^{- n_{1}}$
$z^{- 2} \sum_{n = - \infty}^{\infty} {(\frac{1}{3} z^{- 1})}^{n_{1}}$
Apply the sum of an infinite sequence
$= z^{- 2} (\frac{1}{1 - \frac{1}{3} z^{- 1}})$
Since the sequence converges if $| z | > \frac{1}{3}$ , the radius of convergence is $| z | > \frac{1}{3}$

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