For the signal X(n) =1nu(n– 3) the X(z) will be—5

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For the signal X(n) = 1 ⁿ u(n^{– 3}) the X(z) will be—
5

1. z^{– 3} |z| > 1
  1 – 1 z^{– 1} 5
  5
2. 1 ³ z^{– 3} |z| > 1
  5 1 – 1 z^{– 1} 5
  5
3. 1 ³ z^{– 3} |z| < 1
  5 1 – 1 z^{– 1} 5
  5
4. 1 ³ z^{– 3} |z| < 1
  5 1 + 1 z^{– 1} 5
  5

Correct Option: B

Given that x(n) =			1		ⁿ	u(n– 3) can be written as
			5

x(n) =			1		^{n – 3}	·			1		³	u(n – 3)
			5						5

By using shifting property
If x(n) ← z→ X(z)
x(n – n₀) ←z→ z^{– n₀}.x(z)
Therefore,

			1		³	·			1		^{n – 3}
			5						5

	1		³	z^{– 3}			\|z\| >	1
	5			1 –	1	z^{– 1}	\|z\| >	5
				1 –	5	z^{– 1}

Hence, alternative (B) is correct choice.