For the signal x(n) = b|n|, b > 0 The z-transform is given

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For the signal x(n) = b^|n|, b > 0 The z-transform is given by—

1. x(z) = z – bz , b < |z| < 1
  z – b bz – 1 b
2. x(z) = z – bz , b < |z| < 1
  z – b bz + 1 b
3. x(z) = z – bz , 1 1 b < |z| < b
  z – b bz - 1 b
4. x(z) = z – bz , 1 1 b < |z| < b
  z – b bz + 1 b

Correct Option: A

x(n) = b^|n|, b > 0
or x(n) = b^|n| u(n) + b^{– n} u[– n – 1]
The sequence x(n) = b|n| for |b| < 1,
function is converges while for sequence x(n) = bⁿ for |b| > 1, function is diverges

bⁿ u[n] ←z→	1	, \|z\| > b
	1 – bz^{– 1}

or bⁿ u[n] ←z→	z	, \|z\| > b b
	z – b

b^{– 1} u[– n – 1] ←z→	– 1	\|z\| <	1
	1 – b^{– 1} z^{– 1}		b

or b^{– 1} u[– n – 1] ←z→	– bz	\|z\| <	1
	bz^{– 1}		b

Thus, the z-transform for the composite function is

x(z) =	z	–	bz	, b < \|z\| <	1
	z – b		bz – 1		b

Hence, alternative (A) is the correct choice.