Z-Transform and Inverse Questions and Answers

Q: Z transform of \(a^{n} x(n)\) is :

The Z transform of \(a^{n} x(n)\) is \(X(z/a)\).This is the scaling in the z-domain property:\[ \begin{aligned} \sum_{n} a^{n} x(n) z^{-n} &= \sum_{n} x(n) \left(\frac{z}{a}\right)^{-n} \\ &= X(z/a) \end{aligned} \]Multiplying the sequence by \(a^{n}\) replaces z by z/a in the transform. X(az) would correspond to multiplying by \(a^{-n}\), and X(a/z) and \(X(z^{n}/a^{n})\) do not follow from the definition.

Q: Consider the signal \(x[n] = 2^{n}\,u[n]\) having Z-transform as X(z) with ROC R. The inverse Z transform for X (2z) will be :

The inverse Z-transform of \(X(2z)\) is \(u[n]\).The transform of \(x[n] = 2^{n} u[n]\) is\[ X(z) = \frac{1}{1 - 2z^{-1}}, \quad |z| \gt 2 \]Replacing \(z\) by \(2z\):\[ X(2z) = \frac{1}{1 - 2(2z)^{-1}} = \frac{1}{1 - z^{-1}} \]which is the transform of the unit step, with ROC \(|z| \gt 1\). The same follows from the scaling property: \(a^{n} x[n] \leftrightarrow X(z/a)\), and \(X(2z) = X(z/\tfrac{1}{2})\), so the sequence is \((\tfrac{1}{2})^{n} 2^{n} u[n] = u[n]\).\(2^{n}u[n]\) is the original signal, and \(2^{2n}u[n] = 4^{n}u[n]\) comes from scaling the wrong way, \(X(z/2)\).

Z transform of \(a^{n} x(n)\) is :

A.\(X(az)\)
B.\(X(a/z)\)
C.\(X(z/a)\)
D.\(X(z^{n}/a^{n})\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The Z transform of \(a^{n} x(n)\) is \(X(z/a)\).

This is the scaling in the z-domain property:

\[ \begin{aligned} \sum_{n} a^{n} x(n) z^{-n} &= \sum_{n} x(n) \left(\frac{z}{a}\right)^{-n} \\ &= X(z/a) \end{aligned} \]

Multiplying the sequence by \(a^{n}\) replaces z by z/a in the transform. X(az) would correspond to multiplying by \(a^{-n}\), and X(a/z) and \(X(z^{n}/a^{n})\) do not follow from the definition.

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Consider the signal \(x[n] = 2^{n}\,u[n]\) having Z-transform as X(z) with ROC R. The inverse Z transform for X (2z) will be :

A.\(u[n]\)
B.\(2^{n}\,u[n]\)
C.\(4^{u}\,u[n]\)
D.\(2^{2n}\,u[n]\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The inverse Z-transform of \(X(2z)\) is \(u[n]\).

The transform of \(x[n] = 2^{n} u[n]\) is

\[ X(z) = \frac{1}{1 - 2z^{-1}}, \quad |z| \gt 2 \]

Replacing \(z\) by \(2z\):

\[ X(2z) = \frac{1}{1 - 2(2z)^{-1}} = \frac{1}{1 - z^{-1}} \]

which is the transform of the unit step, with ROC \(|z| \gt 1\). The same follows from the scaling property: \(a^{n} x[n] \leftrightarrow X(z/a)\), and \(X(2z) = X(z/\tfrac{1}{2})\), so the sequence is \((\tfrac{1}{2})^{n} 2^{n} u[n] = u[n]\).

\(2^{n}u[n]\) is the original signal, and \(2^{2n}u[n] = 4^{n}u[n]\) comes from scaling the wrong way, \(X(z/2)\).

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Z-Transform and Inverse Questions and Answers

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