Question

Unilateral Z transform of x(n) is equivalent to bilateral Z-transform of

• A. \( \delta(-n) \)
• B. \( x(n) u(-n) \)
• C. \( x(n) u(n) \)
• D. \( x(-n) u(-n) \)

Hint:

Unilateral ZT: \(\sum_{n=0}^{\infty} x(n)z^{-n}\). For causal sequences or initial condition problems.
Bilateral ZT: \(\sum_{n=-\infty}^{\infty} x(n)z^{-n}\).
Multiplying by \(u(n)\) effectively truncates the sequence for \(n<0\).

Answer 1

The Correct Option is C

The unilateral Z-transform of \(x(n)\) is \(X_U(z) = \sum_{n=0}^{\infty} x(n)z^{-n}\). The bilateral Z-transform of a sequence \(g(n)\) is \(G_B(z) = \sum_{n=-\infty}^{\infty} g(n)z^{-n}\). If we let \(g(n) = x(n)u(n)\), where \(u(n)\) is the unit step function (\(u(n)=1\) for \(n \ge 0\), and \(u(n)=0\) for \(n<0\)), then: \(G_B(z) = \sum_{n=-\infty}^{\infty} x(n)u(n)z^{-n} = \sum_{n=0}^{\infty} x(n)(1)z^{-n} = \sum_{n=0}^{\infty} x(n)z^{-n}\). This is identical to \(X_U(z)\). Thus, the unilateral Z-transform of \(x(n)\) is equivalent to the bilateral Z-transform of \(x(n)u(n)\). \[ \boxed{x(n) u(n)} \]