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

To solve this problem, we need to determine the Region of Convergence (ROC) of the Z-transform for the given signal \( x(n) = \delta(n - k) \), where \( k > 0 \).

1. Understanding the Concepts:

- Z-transform: The Z-transform of a discrete-time signal \( x(n) \) is given by:

\[ X(z) = \sum_{n=-\infty}^{\infty} x(n) z^{-n} \]

- Delta function \( \delta(n - k) \): The delta function \( \delta(n - k) \) is defined as:

\[ \delta(n - k) = \begin{cases} 1 & \text{if } n = k \\ 0 & \text{otherwise} \end{cases} \]

For the signal \( x(n) = \delta(n - k) \), the Z-transform is:

\[ X(z) = z^{-k} \]

2. Region of Convergence (ROC):

The ROC of the Z-transform depends on the nature of the signal. For the delta function \( \delta(n - k) \), the Z-transform is a simple power of \( z^{-k} \), and this function is valid for the entire \( z \)-plane except at \( z = 0 \) (where it would lead to an undefined result for \( z^{-k} \)).

Final Answer:

The ROC of the signal \( x(n) = \delta(n - k), \, k > 0 \) is Entire z-plane, except at \( z = 0 \).