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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For the Z-transform \(X(z) = 4z^{2} + 2 + 3z^{-1}\), \(0 \lt |z| \lt \infty\) \(x(n)\) is given by

A.\(x(n) = 4\delta[n - 2] + 2\delta[n] + 3\delta[(n + 1)]\)
B.\(x(n) = 4\delta[n + 2] + 2 + 3\delta[n - 1]\)
C.\(x(n) = 4\delta[n + 2] + 2\delta[n] + 3\delta[n + 1]\)
D.\(x(n) = 4\delta[n - 2] + 2\delta[n - 1] + 3\)

Exam Year(s)

UGC NET - 2013,

Answer: B

Each term of the Z-transform is inverted with the shifting property \(z^{k} \leftrightarrow \delta[n + k]\):

\[ 4z^{2} \rightarrow 4\delta[n + 2] \]\[ 3z^{-1} \rightarrow 3\delta[n - 1] \]

The constant 2 corresponds to a unit impulse at n = 0. The keyed option (B) has the terms 4 delta[n + 2] and 3 delta[n - 1] in the right places, and it is the only option with both of them correct. Option B prints the middle term as a bare 2, which is a misprint for 2 delta[n].

Options A and D place the delta of the first term at n - 2, which is the wrong direction, and option C moves the last term to n + 1.

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For finding the inverse of a z-transform :

1. Partial fraction expansion
2. Contour integration
3. Convolution
4. Power series expansion Following is the correct sequence :

A.1, 2, 3, 4
B.1, 2, 4, 3
C.4, 3, 2, 1
D.1, 3, 4, 2

Exam Year(s)

UGC NET - 2013,

Answer: B

The keyed sequence is 1, 2, 4, 3, option (B): partial fraction expansion, contour integration, power series expansion and convolution. The question lists the methods of finding an inverse z-transform and the key takes this order for them.

Options A, C and D put the methods in other orders. The order is a matter of convention in the course text, so it cannot be derived from first principles, and a reviewer may wish to check it against the text.

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Z-Transform and Inverse Questions and Answers

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