Topic: Signals and Systems
The Z-transform of a signal $x[n]=\delta[n-k]$, (where k is a positive integer) is:
- A. $z^{k}$
- B. $z^{-k}$
- C. 1
- D. $\delta[z]$
Answer: B. $z^{-k}$
Explanation: By definition, the Z-transform of a discrete-time signal $x[n]$ is $X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}$.
For $x[n] = \delta[n-k]$, the signal is 1 only when $n = k$ and 0 otherwise.
Therefore, $X(z) = \sum_{n=-\infty}^{\infty} \delta[n-k] z^{-n} = 1 \cdot z^{-k} = z^{-k}$.