$Let X(z) be the Z-transform of a discrete time sequence x [n] = (- 2)^{- n} u [n] . Consider another signal y[n] and its Z-transform is Y(z), given as Y (z) = X (z^{- 2}) . Then the value of y[n] at n=-2 is___.$
-0.5

Open in App

Solution

The correct option is A -0.5
$Given, x [n] = (- 2)^{- n} u [n] = {(- \frac{1}{2})}^{n} u [n]$
$Y (z) = X (z^{- 2}) = \sum_{n = - \infty}^{\infty} x [n] . z^{2 n}$
$Y (z) = \sum_{n = 0}^{\infty} {(- \frac{1}{2})}^{n} z^{2 n}$
$Y (z) = 1 - \frac{1}{2} z^{2} + \frac{1}{4} z^{4} - \frac{1}{8} z^{6} + . . . .$
$At n = -2,we get coefficient of z^{2} term in the expansion of Y(z).$
$∴ y (n) a t n = - 2 is y (- 2) = - \frac{1}{2} = - 0.5$

Suggest Corrections

Similar questions

(7 + 2 i)^{n} = a + i b

53^{n} = a^{2} + b^{2}

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

x = \infty \sum n = 0 a^{n}, y = \infty \sum n = 0 b^{n}, z = \infty \sum n = 0 {(a b)}^{n}

a, b < 1

x z + y z = x y + z

0 < θ, ϕ - \frac{π}{2}

x = \infty \sum n = 0 {cos}^{2 n} θ

y = \infty \sum n = 0 {sin}^{2 n} ϕ, z = \infty \sum n = 0 {cos}^{2 n} θ {sin}^{2 n} ϕ

x z + y z - z = x y

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

n

\frac{(2 i)^{n}}{(1 - i)^{n - 2}}, i = \sqrt{- 1},

Join BYJU'S Learning Program