Question

The ROC of the signal \( x(n) = \delta(n - k), \, k>0 \) is defined as

• A. \( z = 0 \)
• B. \( z = \infty \)
• C. Entire z-plane, except at \( z = 0 \)
• D. Entire z-plane, except at \( z = \infty \)

Hint:

The Z-transform of a delayed delta function \( \delta(n - k) \) is \( z^{-k} \), which is defined everywhere except \( z = 0 \). This is because finite-duration sequences have ROC covering the entire z-plane minus the points where the Z-transform becomes undefined.

Answer 1

The Correct Option is C

The signal \( x(n) = \delta(n - k) \), where \( k>0 \), is a delayed unit impulse.
The Z-transform of \( x(n) \) is: \[ X(z) = \sum_{n=-\infty}^{\infty} \delta(n - k) z^{-n} = z^{-k} \] This is a finite-length signal (single non-zero value), and its Z-transform exists everywhere except where the expression becomes undefined.
Since \( X(z) = z^{-k} \), it is undefined at \( z = 0 \), hence the ROC is the entire z-plane except at \( z = 0 \).