The z-transform of the time function $$\sum\limits_{k = 0}^\infty {\delta \left( {n – K} \right)} $$ is

The correct answer is A.

The z-transform of a sequence $x[n]$ is defined as

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

The z-transform of the unit impulse sequence $\delta[n]$ is

$$X(z) = \sum_{n=-\infty}^{\infty} \delta[n] z^{-n} = 1$$

The z-transform of the shifted unit impulse sequence $\delta[n-k]$ is

$$X(z) = \sum_{n=-\infty}^{\infty} \delta[n-k] z^{-n} = z^{-k}$$

Therefore, the z-transform of the time function

$$\sum_{k=0}^{\infty} \delta[n-k] = \sum_{k=0}^{\infty} z^{-k}$$

is

$$X(z) = \frac{1}{1-z^{-1}} = \frac{z}{z-1}$$

which is option A.

Option B is the z-transform of the time function $e^{-an}$. Option C is the z-transform of the time function $e^{an}$. Option D is the z-transform of the time function $e^{-an} u[n]$.