Discussion
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 :
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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