The region of convergence of Z - transform of a signal $x (n) = 2^{n} u (n) - 4^{n} u (- n - 1) i s a < | z | < b$ . Then the value of a + b is

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 6
$2^{n} u (n) Z . T ⟷ \frac{z}{z - 2}; R O C | z | > 2$

$- 4^{n} u (- n - 1) Z . T ⟷ \frac{z}{z - 4}; R O C | z | < 4$

$∴ R O C i s 2 < | z | < 4$

$a + b = 2 + 4 = 6$

Suggest Corrections

Similar questions

\frac{1}{A}

x (n) = 2^{n} u (n) - 4^{n} u (- n - 1)

x (n)

X (z)

x (n)

X (z) = \frac{z}{z - 0.1}

\infty \sum n = - \infty n x (n)

x (n) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{(2)^{- n}}{n!}; & n \geq 0 0; & O t h e r w i s e \end{matrix}

The region of convergence of a signal x[n] whose z-transform is represented as x(z),where x (n) = {\begin{matrix} 1; & - 1 \leq n \leq 5 0; & otherwise \end{matrix}

Join BYJU'S Learning Program